JUNIOR CERTIFICATE
EXAMINATION
2012
MARKING SCHEME
MATHEMATICS
ORDINARY LEVEL
PAPER 1
Page 1
GENERAL GUIDELINES FOR EXAMINERS 1.
Penalties of three types are applied to candidates’ work as follows:
 Blunders - mathematical errors/omissions
(-3)
 Slips- numerical errors
(-1)
 Misreadings (provided task is not oversimplified)
(-1).
Frequently occurring errors to which these penalties must be applied are listed in the
scheme. They are labelled: B1, B2, B3,…, S1, S2,…, M1, M2,…etc. These lists are not
exhaustive.
2.
When awarding attempt marks, e.g. Att(3), note that
 any correct, relevant step in a part of a question merits at least the attempt mark for that
part
 if deductions result in a mark which is lower than the attempt mark, then the attempt
mark must be awarded
 a mark between zero and the attempt mark is never awarded.
3.
Worthless work is awarded zero marks. Some examples of such work are listed in the
scheme and they are labelled as W1, W2,…etc.
4.
The phrase “hit or miss” means that partial marks are not awarded – the candidate receives
all of the relevant marks or none.
5.
The phrase “and stops” means that no more work is shown by the candidate.
6.
Special notes relating to the marking of a particular part of a question are indicated by an
asterisk. These notes immediately follow the box containing the relevant solution.
7.
The sample solutions for each question are not intended to be exhaustive lists – there may
be other correct solutions.
8.
Unless otherwise indicated in the scheme, accept the best of two or more attempts – even
when attempts have been cancelled.
The same error in the same section of a question is penalised once only.
9.
10.
Particular cases, verifications and answers derived from diagrams (unless requested)
qualify for attempt marks at most.
11.
A serious blunder, omission or misreading results in the attempt mark at most.
12.
Do not penalise the use of a comma for a decimal point, e.g. €5.50 may be written as €5,50.
Page 2
BONUS MARKS FOR ANSWERING THROUGH IRISH Bonus marks are applied separately to each paper as follows: If the mark achieved is 225 or less, the bonus is 5% of the mark obtained, rounded down. e.g. 198 marks  5% 9.9  bonus 9 marks. If the mark awarded is above 225, the following table applies: Bunmharc
Marc Bónais
Bunmharc
Marc Bónais
(Marks obtained) (Bonus Mark) (Marks obtained) (Bonus Mark)
226
11
261 – 266
5
227 – 233
10
267 – 273
4
234 – 240
9
274 – 280
3
241 – 246
8
281 – 286
2
247 – 253
7
287 – 293
1
254 – 260
6
294 – 300
0
Page 3
QUESTION 1
Part (a)
Part (b)
Part (c)
10 marks
20 (5, 5, 5, 5) marks
20 (10,5,5) marks
(a)
1.
Att (3)
Att (2, 2, 2, 2)
Att (3,2,2)
10 marks
(a)
S = , , , ,
Att 3
P = , ,
Fill the elements of S and P into the following diagram.
S
P
(a)
10 marks
S
Att 3
P
r
p
s
w
u
t
Slips ‐1 S1 Each element incorrectly filled into diagram
S2 Each element omitted from diagram but see W1
S3 Each unlisted element used but see W1 (some relevant element must be present to use S3)
Misreading (-1)
M1 Interchanging S and P totally
Attempts (3 marks)
A1 Totally incorrect filling of the Venn diagram using given elements
A2 Correct number of dots in each set without labels
Worthless (0)
W1 No filling in of the Venn diagram or use of unlisted elements only but see S3
Page 4
(b)
20 marks (5,5,5,5)
Att 2,2,2,2
(b)
A = {1, 2, 3, 6} is the set
of the divisors of 6.
B
A
B = {1, 2, 4, 8} is the set
of the divisors of 8.
.3
.1
.2
.6
C = {1, 2, 4, 5, 10, 20} is the set
of the divisors of 20.
.5
.10
.20
.8
.4
C
List the elements of:
(i)
B∪C
(ii)
A \ B  C 
(iii)
BC
(iv)
the common divisors of 6, 8 and 20.
(i)
5 marks
Att 2
B∪C= {1,2,4,5,8,10,20}
Blunders (-3)
B1 Any incorrect set of the elements of B and C other than the misreading below
Misreading (-1)
M1 B∩C = {1, 2, 4}
Attempts (2 marks)
A1 3or 6 appear in the answer
(ii)
5 marks
A \ B  C   {3,6}
Blunders (-3)
B1 Any incorrect set of the elements of A, B and C
Misreading (-1)
M1 (A\B) ∪C = {3,5, 6,10 ,20,1,2,4 }
Page 5
Att 2
(iii)
5 marks
B  C=
Att 2
{1, 2, 4}
Blunders (-3)
B1 Any incorrect set of the elements of A, B and C other than the misreading below
Misreading (-1)
M1 B  C giving {1, 2, 4, 5, 8, 10, 20}
Attempts (2 marks)
A1 1, 2 or 4 appear in the answer
(iv)
the common divisors of 6, 8 and 20
5 marks
= { 1, 2 }
Att 2
Slips (-1)
S1 Each missing or incorrect element to a max of 3
Attempts ( 2 marks)
A1 Any correct divisors of 6,8 or 20 appears, but see S1
A2 Ans. 120
Worthless (0)
W1 Elements listed that are not divisors of 6, 8 or 20
(c)
20 ( 10,5,5) Marks
Att 3,2,2
(c) In a survey, 60 households were asked if they had a cat (C) or a dog (D).
20 said they had a cat.
25 said they had a dog.
12 said they had both a cat and a dog.
(i)
Represent this information in the Venn diagram below.
C
(ii)
D
How many households had only a cat or a dog?
 (iii) What percentage of households had neither a cat nor a dog?
Page 6
(c)(i)
10 marks
Att 3
C
D
8
12
13
27
Blunders (-3)
B1 Each incorrect or omitted entry (unless consistent error) in Venn diagram subject to S1
below.
Slips (-1)
S1 Numerical errors where work is clearly shown
Misreading (-1)
M1 Interchanges cats and dogs
Attempts (3 marks)
A1 Any one correct/relevant entry
c(ii)
5 marks
Att 2
8 + 13 = 21
*A correct answer written in the space provided takes precedence over an incorrect Venn
diagram.
*Accept candidates work from previous part c (i).
Blunders (-3)
B1 Any incorrect use of the given numbers or the numbers from the candidates incorrect Venn
diagram [Subject to S1].
Slips (-1)
S1 Numerical errors where work is clearly shown
S2 Fails to add their correct relevant 2 figures
Page 7
c(iii)
5 marks

Att 2
27
 100  45%
60
*A correct answer written in the space provided takes precedence over an incorrect Venn
diagram.
*Accept candidates work from previous parts (c) (i), (c) (ii).
Blunders (-3)
B1 No work shown
B2 Mishandles the percentage
B3 Any incorrect use of the given numbers or numbers from the previous work
[Subject to Second *above]
B4 Fails to find the percentage
Misreading (-1)
M1
× 100 or similar and continues
Slips (-1)
S1 Numerical errors where work is clearly shown, to a max of 3
Attempts (2 marks)
A1 Any one correct/relevant step
A2 100 appears
Worthless (0)
W1 Incorrect answer with no work shown
Page 8
QUESTION 2
Part (a)
Part (b)
Part (c)
10 marks
20 (10,5, 5) marks
20 (5, 10, 5) marks
(a)
10 marks
(a)
10 marks
1costs
 Att 3
3 packets of soup cost €3‧51.
What would be the cost of 5 packets of the same soup?
(a)
Att 3
Att (3 ,2, 2)
Att (2, 3,2)
.
= 1.17
Or
3:5
Att 3
Or
.
5 cost 1.17 × 5 =5.85
Ans 5.85
= 1.17
1.17 × 5 = 5.85
3:5 = 3.51 : x
*Correct answer without work
7marks
*Special Case 3.51 2.106 7 marks
.
*Stops at 1.17 or
*Stops at 3.51 × 5 (=17.55)
4marks (no use of 5, B(-3) and B4 or B5)
4 marks ( no use of 3 and possible slips)
Blunders (-3)
B1 Divisor ≠ 3but see above
B2 incorrect multiplier
B3 5:3 = 3.5 : x and continues
B4 Error in decimal point (apply once)
B5 Fails to finish
Slips (-1)
S1 Numerical errors where work is clearly shown, to a max of -3
Attempts (3 marks)
A1 Indicates
or 3:5 or 3.51: x only and stops
or
( only) appear with no work shown
A2
1.17 or 17.55 or
A3
A4
A5
only appears
(3.51×3) or (3.51÷5) and stops
3.51 is multiplied or divided by any wrong number correctly
Worthless (0)
W1 Incorrect answer without work but see A1 and A2
W2 3.51 + 3 = 6.54 or similar, and stops
Page 9
.
=
3x = 3.51 × 5 = 17.55
.
x=
= 5.85
(b)
20 marks (10, 5, 5 )
(b)
(i)
Att (3, 2, 2)
By rounding each of these numbers to the nearest whole number,
24  231
estimate the value of
.
15  6  3  78
.

24  231
is approximately equal to:
15  6  3  78
=
=
–
(ii)
Using a calculator, or otherwise, find the exact value of
24  231
.
15  6  3  78
(iii) Find the difference between the exact value in (ii) and the estimated value
in (i).
b(i)
 10 marks
Att 3
24  231
is approximately equal to:
15  6  3  78
24
24
=
16
–
4
=
2
12
24
24
24
* 16  4 and stops or 16  4 = 12  7 marks
*No penalty if the intermediate step between approximations and final answer not shown i.e.
24
not shown.  10 marks.
12
*Special Case:
24  231
= 2.05
15  6  3  78
in this part → 3 marks.
Page 10
Blunders (-3)
B1
B2
B3
B4
B5
Correct answer without work
Error(s) in rounding off to the nearest whole number (once only if consistent)
Decimal error in calculation of approximate value
An arithmetical operation other than indicated.
– 4 = –2.5 or (24÷4 –16) = –10 (breaking order) or similar and continues
Slips (-1)
S1 Numerical errors to a max of -3.
Attempts (3 marks)
A1 Only one or two approximations made to the given numbers and stops
Worthless (0)
W1 Wrong answer without work but note Special Case above
b(ii)
5 marks
Att 2
24  231
24  231

 2  05
15  6  3  78 11  82
Blunders (-3)
B1 Decimal error or early rounding off
B2 Fails to finish
B3 Treats as (24.231÷15.6) –3.78 = -2.226730769…
B4 Treats as: (24.231÷3.78) –15.6 = -9.18968254…
B5 Treats as: 24.231÷(15.6 + 3.78) = 1.250309598…
B6 Treats as:. 24.231÷(15.6×3.78) = 0.410917785…
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (2 marks)
A1 Any correct relevant step and stops.
A2 Any of the following (see above): -2.226730765…, 9.18968254…, 1.250309598…,
24.231
24.231
0.410917785… or
= 1.553269231 or
= 6.41031746 (minimum 4 decimal
15.6
3.78
places) with or without work
Worthless (0)
W1 Wrong answer without work but see A2
Page 11
b(iii)`
2·05 – 2·00 = 0·05
5 marks
or
-2=
Att 2
*Allow candidate’s figures
Blunders (-3)
B1 Fails to finish
B2 Decimal error (once only if consistent)
B3 Finds the sum of (i) and (ii)
Slips (-1)
S1 Numerical errors to a max of -3
Attempts (2 marks)
A1 Any relevant step i.e. transfers answers from (i) and/or (ii)
Worthless (0)
W1 Incorrect answer without work
(c)`
(c)
20 (5,5,5,5) marks
(i)
Att 2,2,2,2
Using a calculator, or otherwise, multiply 450 000 × 7‧8.
Then express your answer in the form a ×10n, where 1≤ a <10 and n  ℕ.
(ii)
(iii)
a7
in the form a n , where n  ℕ.
a3
117
Hence or otherwise evaluate
.
113
Write
It takes three workers four days to build a wall.
How long would it take two workers to build the same wall?
c(i)

5 marks
450 000 × 7·8 = 3 510 000 = 3·51 ×106
*
3.51 or 3.51 × 106 (without work) → 4 marks
Blunders (-3)
B1 Decimal error
B2 An arithmetic operation other than that indicated e.g. 450000÷7.8 = 57692.30789
Slips (-1)
S1 Numerical errors to a max of -3
S2 Rounds off to 3 ×106
S3 Incorrect format, where a 1 or a 10 and n  Z
S4 Finds 3 510 000 and stops
Attempts (2 marks)
A1 Any relevant step and stops
Page 12
Att 2
c (ii)
5marks
a7
a3
 a73  a 4
or
Att 2
a  a  a  a  a  a  a a7
 3 = a4
aaa
a
* a a a a and stops 4marks
* a7-3 and stops
4marks
Blunders (-3)
B1 Each error in calculation involving indices
B2 Each incorrect number of a’s in the extended form
B3 Each incorrect elimination of the a’s in extended form
Slips (-1)
S1 Numerical errors to a max of -3
Attempts (2 marks)
A1 Some correct manipulation of indices
A2 4 only written down
Worthless (0)
W1 Writes a only or incorrect answer with no work shown other than A2
c(ii)Hence
5 marks
7
11
3
 114  14641
11
*Accept candidate’s answer from above unless it oversimplifies the question
Blunders (-3)
B1 Each error in calculation involving indices
B2 Each incorrect number of 11’s in the extended form
B3 Fails to finish
B4 Each incorrect elimination of the 11’s in extended form
Slips (-1)
S1 Numerical errors to a max of -3
Attempts (2 marks)
A1 Some correct manipulation of indices
A2 112= 121 or similar and stops
A3 Candidate transfers their answers from above
Worthless (0)
W1 Incorrect answer with no work shown
Page 13
Att 2
c(iii)
5 marks
1 man takes
3×4 days = 12days
12
2 men take
 6 days
2
*
Special case:

4 2 8
 →2 marks
3
3
4
( 2)
*
Stops at 2
→2 marks
Blunders (-3)
B1 Incorrect answer without work
B1 Divisor ≠ 2 and continues
B2 Incorrect multiplier (≠3) or fails to multiply, or fails to multiply but see 1st *
Slips (-1)
S1 Numerical errors where work is clearly shown to a max of -3
Attempts (2 marks)
A1 Mentions one man or man days
A2 12 or 2 only appear (no work shown)
4
A3 4×2 or 3 and stops
A4 4 is multiplied or divided by any wrong number, correctly
Worthless(0)
W1 Incorrect answer without work but see A2 above
W2 3 + 4 = 7 or similar
W3 hours only with no mention of 3 or 4 or (96 on its own)
Page 14
Att 2
QUESTION 3
Part (a)
Part (b)
Part (c)
10 marks
20 (10,10) marks
20 ( 10, 10) marks
(a)
3.
10 marks
(a)
Att 3
The cost of a holiday came to €2400.
This was made up of the cost of travel, accommodation and spending money.
3
of the cost was for travel and accommodation.
5
How much spending money was there?
(a)

10 marks
3
× 2400 = 1440
5
2400 –1440 = €960
*
Att 3
Att (3 ,3)
Att (3,3)
3
2
travel + acc => spend.
5
5
2
×2400 = €960
5
Att 3
3
2
=60% => = 40%
5
5
40
2400×
= €960
100
No penalty for omitting € symbol
Blunders (-3)
B1 Correct answer without work
3
B2 2400 ÷
(method 1)
5
2
B3 2400 ÷
(method 2)
5
B4 Calculates the travel and accommodation and stops (method 1 )
B5 Operation other than subtraction in final step or omits final step. (method 1)
B6 Finds 60% 0f 2400 and stops (same as B5)
Slips (-1)
S1
Numerical errors (to max -3)
Attempts (3 marks)
A1
Any attempt at getting
A2
Writes down
3
2
of 2400 or
of 2400
5
5
2
or 40%
5
Page 15
(b)
20 (10,10) Marks
(b) (i)
Amanda borrows €1000.
She agrees to pay it back at €90 per month for a year.
How much interest will she pay?
(ii)
Att (3,3)

A computer is ordered online. It is advertised for €550 plus VAT at 23%.
There’s a delivery charge of €7‧50.
What is the total cost to be paid?
(b) (i)

10 marks
Att 3
 Amanda borrows €1000.
She agrees to pay it back at €90 per month for a year.
How much interest will she pay?
b(i)
10 marks

90×12 = 1080
Int: 1080 – 1000 = €80
* No penalty for omitting € symbol
Blunders (-3)
B1 Correct answer without work
B2 90× 12=1080 and stops
B3
90÷ 12 = 7.5 and continues correctly
B4
Multiplies 90 by some whole number other than 12 and continues
B5
Fails to finish
Slips (-1)
S1 Numerical errors (to max -3)
Attempts (3 marks)
A1
Oversimplification
A2
Multiplies 90 by some number other than 12 and stops
Page 16
Att 3
(b) (ii)

23 % =
VAT =
550 = 126.50
Total Cost
= 550 + 126.50 + 7.50
= €684
10 marks
100% = €550
1% =
123% =
att 3
550
123
1.23
€676.50
Total Cost
= €676.50 + 7.50 = €684
= 5.50 123
= 676.50
Total Cost
= €676.50 + 7 .50 = €684
* No penalty for omitting € symbol
Blunders (-3)
B1 Correct answer without work
B2 Decimal error
and continues (giving answers 2391.30 or 447.51)
B3 Inverts as B4 Mishandles 23 % eg 550 23 550 23 Note: (550 must be used}
B5 550 taken as 123% and finds his 100% and continues
B6 No addition of VAT (as per candidates work) to the bill
B7 No addition of the delivery charge
B8 Subtraction of VAT ( as per candidates work) from the bill
B9 No addition of 550
Slips (-1)
S1
Numerical errors to a max of -3
Misreadings (-1)
M1 Reads as 32% or €500
Attempts (3 marks)
and stops or
and stops
A1
A2 100% = 550 and stops
A3 100
and stops or
and stops
A4 550 23 % and stops
A5 €550 + 7.50 and stops
Worthless (0)
W1 Incorrect answer without work
W2 550 + 23 = €573 and stops or continues
Page 17
Part (c)
20(10,10) marks
(i)
A work of art is priced at €6600. After VAT is added it costs €7491.
Calculate the amount of VAT and the rate of VAT.
(ii)
Att (3,3)

Ronan was given a bicycle which was in need of repair.
For the repairs, he spent €60 on spare parts and €12 on paint.
When it was repaired he sold it for €95.
Calculate the profit he made as a percentage of his costs.
Give the percentage to the nearest whole number.

(c) (i)
10 marks
A work of art is priced at €6600. After VAT is added it costs €7491.
Calculate the amount of VAT and the rate of VAT.
Att3
(c) (i)
Att3

10 marks
7491 – 6600 = 891 = VAT
100 13.5% * No penalty for omitting € symbol
*7991 – 6600 = 891 = 13.5% → 10 marks
*Stops after €891 → 4 marks
(
and stops still only 4 marks)
Blunders (-3)
B1 Correct answer without work.
B2 Decimal error eg 1.35%
B3 Inverts as and continues ( to get 740.74 %)
B4 7461 + ,
by 6600 and continues correctly
B5 Mishandles the finding of the rate of vat
100 to get 11.89% = 12%
B6
B7 Rounds off to 14% without showing 13.5%
B8 Fails to finish
Slips (-1)
S1 Numerical errors (apart from decimal errors) max of -3
Attempts (3marks)
A1 Some use of 100
A2 Some attempt at subtraction
Page 18
(c) (ii)
10 marks
Att3
 Ronan was given a bicycle which was in need of repair.
For the repairs, he spent €60 on spare parts and €12 on paint.
When it was repaired he sold it for €95.
Calculate the profit he made as a percentage of his costs.
Give the percentage to the nearest whole number.
(c) (ii)
60 + 12 = 72
95 – 72 = €23 Profit
100 31.944
= 32 %
10 marks
Att3
60 + 12 = 72
100 131.944% (132% accept)
131.944 – 100 = 31.944
= 32%
* No penalty for omitting € symbol
*Answer €23 →
4 marks
*
100 and stops →
6 marks
Blunders (-3)
B1 Correct answer without work
B2 Adds €95 to €72 and continues
B3 Calculates profit as percentage of selling price. ie.
B4 Divisor not equal to 72
B5 Mishandles the calculation of profit as a percentage
B6 Fails to multiply by 100
Slips (-1)
S1 Numerical errors to a max of -3
S2 Fails to round off to the nearest whole number
Attempts (3 marks)
A1 Some indication of subtraction
A2 Some use of 100
A3 60 +12
(= 72)
Worthless (0 marks)
W1 Incorrect answer without work = 0 marks.
Page 19
100
24.21%
24% QUESTION 4
Part (a)
Part (b)
Part (c)
(a)
(a)
15(10,5) marks
15 (5,5,5) marks
20 ( 5,5, 10) marks
10,5 marks
If a = 4 and b = 5, find the value of:

(i)
2a + b

(ii)
ab – 3
(a)(i)
10 marks
Att 3,2
Att (2,2 ,2)
Att (2,2,3)
Att 3,2
Att 3
(i) 2a + b = 2(4) + 5 = 8 + 5 = 13
*8 +5 (only)
→ 9 marks
*One substitution coupled with an implied substitution leading to correct answer
e.g. = 2a +5 = 13
 10 marks.
Blunders (-3)
B1 Correct answer without work 
B2 Leaves 2(4) in the answer
B3 Breaks order i.e. 2(4+5) = 18
B4 Treats 2(4) as 6 or 24
Slips (-1)
S1 Numerical errors to a max of 3
S2 Values of a and b interchanged.
Misreadings (-1)
M1 Incorrect numerical substitution for either a or b, but not both, and continues (See W1)
or a + 2b calculated out
Attempts (3 marks)
A1 Incomplete substitution and stops e g 2a + 5
Worthless (0)
W1
Incorrect substitution for both a and b
Page 20
(a)(ii)
(ii)
ab – 3 = 4×5 – 3 = 20 – 3 = 17
5 marks
Att 2
*20 – 3 (only)
→ 4 marks
*One substitution coupled with an implied substitution leading to correct answer
e g 4b –3 = 17 or 5a – 3 =17  5 marks
Blunders (-3)
B1
B2
B3
B4
Correct answer without work
Leaves 4(5) in the answer
Breaks order i.e. 4(5-3) = 8
Treats 4(5) as 9 or 45

Slips (-1)
S1 Numerical errors to a max of -3
Misreadings (-1)
M1 Incorrect numerical substitution for either a or b, but not both, and continues (See W1)
Attempts (2 marks)
A1 Incomplete substitution and stops e g 4b -3
Worthless (0)
W1
Incorrect substitution for both a and b.
Page 21
(b)
(b)
15 (5,5,5) Marks
Att 2,2,2
f(x) = 2x – 1.
(i)
Draw a graph of f(x) in the domain –1 ≤ x ≤ 1, xℝ.
(ii)
Use your graph to estimate the value of x when f(x) = 0.
(b)(i)
f(x) = 2x – 1
f(-1) = 2(-1) – 1= -2 -1 = -3 (-1,-3)
f(0) = 2(0 )– 1 = 0 – 1 = -1 (0,-1)
f(1) = 2(1) – 1= 2 – 1 = 1 (1,1)
5 marks
OR
OR
f(x) = 2x – 1
f(-1) = 2(-1) – 1 = -3 (-1,-3)
f(0) = 2(0 )– 1 = -1 (0,-1)
f(1) = 2(1) – 1 = 1 (1,1)
Att 2
x
-1
0
1
 2x
-1
-2
-1
0
-1
+2
-1
f (x)
-3
-1
1
*
Error(s) in each row/column calculation attracts a maximum deduction of 3marks
*
2 points correct (full marks) _ (need not be in domain)
Blunders (-3)
B1 “+2 x ” taken as “2” all the way. [In the row headed “+2 x ” by candidate]
B2 “-1” calculated as “- x ” all the way. [In the row headed “-1” by candidate]
B3 Adds in top row when evaluating f (x) in Box
B4 Omits “-1” row
B5 Omits “+2 x ” row
B6 Takes 2x as 2 + x and applies it in his calculations
B7 Each incorrect image without work i.e. calculation through the function method
Slips (-1)
S1 Numerical errors to a max of -3 in any row / column
Misreadings (-1)
M1 Misreads -1 as +1 and places +1 in the table or function.
M2 Misreads “  2 x ” as “  2 x ” and places “  2 x ” in the table or function
Attempts (2 marks)
A1 Any effort at calculating point(s)
A2 Only one point calculated and stops
Page 22
(b)(ii)
5 marks
Att 2
2
1
-1
1
-1
-2
-3
-4
* Answers need not be written in table. .
*Accept candidate’s value from (i) but see B1 and S4
(see S2)
*Tolerance ± 0.5 ( ± 1Box on grid)
*Correct graph but no table award full marks i.e. (5 + 5)
*Only one correct point graphed correctly but no table  Att 2 + Att 2
*Accept reversed co-ordinates if
(i) if axes not labelled or (ii) if axes are reversed to compensate (see B1 below)
Blunders (-3)
B1 Full domain not covered
B2 Scale error (once)
B3 Reversed co-ordinates plotted against non-reversed axes (once only) {See 6th * above}
Slips (-1)
S1 All points not joined or joined in incorrect order
S2 Each incorrectly plotted point
S3 Each point { 2 points needed } from table not graphed [ See 2nd * above ]
S4 Not a straight line if not already penalised in b(i) or b(ii) but see 2nd *
Attempts (2 marks)
A1 Graduated axes (need not be labelled)
A2 Some effort to plot a point { See 2nd * above}
A3 Random straight line with or without axes
A4 One correct point, with /without work
Page 23
b(iii)
Answer to be written here:
5 marks
x= 0·5
Att 2
when f(x) = 0
*
Allow candidate’s figures
Blunders (-3)
B1 Fails to finish but draws some relevant line
Slips (-1)
S1 Numerical errors to a max of -3
S2 Correct answer indicated and/or written on graph only
Attempts (2 marks)
A1 Some correct indication on graph
A2 Attempts at algebraic evaluation or calculator
A3 Finds answer -1 i.e. find x = 0 (where crosses y-axis)
Worthless (0)
W1 Wrong answer without work
(c)
(c)
(i)
(ii)
20(5,5,10) marks
Conor spent € y on a book.
He then spent € (4y + 6) on a football jersey.
In total, he spent €61.
Write an equation in y to represent this information.
Solve your equation from (i) to find the value of y.
(iii) Solve the equation:
c(i)
x2 – 5x – 14 = 0.
5 marks
Att 2,2,3


Att 2
y + 4y + 6 = 61
5y + 6 =61
Blunders (-3)
B1 Incorrect expression for the cost of a book and football jersey other than misreading below
Slips (-1)
S1 No 61 included in answer
Misreadings (-1)
M1 Answer given as y + 4y – 6 = 61 or similar
Attempts (2 marks)
A1 Any effort at forming an expression (y included)
Worthless (0)
W1 Cost of book given as a constant
Page 24
c(ii)
*
5 marks
Att 2
5y + 6 =61
5y + 6 – 6 = 61 – 6
5y = 55
y = 11
Accept candidates answer from previous work.
Blunders(-3)
B1 Correct answer without work
B2 Error in forming equation
B3 Distribution error
B4 Transposition error
B5 Stops at 5y = 55 or fails to solve equation
B6 Error in collecting like term
Misreadings (-1)
M1 Transfers information in (i) incorrectly if not oversimplied
Slips (-1)
S1 Numerical errors to a max of -3
Attempts (2 marks)
A1 Answer from part c (i) written down and stops.
A2 Any effort at forming an expression
A3 Any effort at solving their equation
A4 Successful Trial and Error
Worthless (0 marks )
W1 Incorrect answer with no work
c(iii)
x2 - 5x – 14 = 0
x2 - 7x + 2x – 14 = 0
x(x -7) + 2(x – 7) = 0
(x + 2)(x – 7) = 0
(x + 2) = 0 or (x – 7) = 0
x = -2
or
10 marks
x - 5x – 14 = 0
Att 3
2
  5 
x
+2
x
7
x=7
(x + 2) = 0 or (x – 7) = 0
x = -2
*
*
or
 52  41 14
21
5  25  56
5 9

2
2
4
14
  2 and
7
2
2
x=7
2 correct solutions by Trial and Error
1 correct solution by Trial and Error
Page 25
10 MARKS
3 MARKS (Attempt)
Blunders (-3)
Factor Method
B1 Correct answers without work 
B2 Incorrect two term linear factors of x2-5x -14 formed from correct (but inapplicable) factors
of x 2 and/or ±14,e.g. (x+14)(x-1)
B3 No roots given, or two incorrect roots (once only)
B4 Incorrect factors of x 2 and/or ±14
B5 Correct cross method but factors not shown and stops [Note: B3 applies also]
B6 x(x–7) + 2(x–7) or similar and stops [Note: B3 applies also].
B7 Error(s) in transposition
Slips (-1)
S1 Numerical errors to a max of -3
S2 One root only from factors
Attempts (3 marks)
A1 Some effort at factorization e.g. (x
A2 States one correct root without work
)(
) or the cross with at least one “x” written in
Worthless (0 marks)
W1 x2 –5x = 14 or similar and stops
W2 Incorrect Trial and error
W3 Oversimplification, resulting in a linear equation
Formula Method
Blunders (-3)
B1 Error in a,b,c substitution (apply once only)
B2 Sign error in substituted formula (apply once only)
B3 Error in square root or square root ignored
B4 Stops at
B5 Incorrect quadratic formula and continues
Slips (-1)
S1 Numerical errors to a max of -3
p
S2 Roots left in the form
q
S3 One root only
Attempts (3 marks)
A1 Correct formula and stops
A2 One correct substitution and stops
Page 26
QUESTION 5
Part (a)
Part (b)
Part (c)
10 marks
20 (5,5.10) marks
20 ( 10, 10) marks
(a)
(a)
10 marks
Simplify fully 2(x + 1) + 5(2x + 3).
(a)
2(x + 1) + 5(2x + 3) = 2x +2 +10x +15
= 12x + 17
Att 3
Att (2,2,3)
Att (3,3)
Att 3

10 marks
Att 3
*
*
Stops after correct removal of brackets
Gathering of terms at most one blunder
7 Marks
Blunders (-3)
B1 Correct answer without work 
B2 Error(s) in distribution (each time)
B3 Combining unlike terms after removal of brackets and continues
B4 Fails to group like terms
B5 Fails to finish
Slips (-1)
S1 Numerical errors to a max of -3
Misreadings (-1)
M1 2(x+2) and continues
Attempts (3 marks)
A1 Any one term correctly multiplied
A2 Combines unlike terms at the start and finishes correctly
Worthless (0)
W1 Combining unlike terms before attempting multiplication and stops e.g. 5(5x) = 25x
Page 27
(b)
20 (5,5,10) marks
(i)
(ii)
(iii)
Att 2,2,3
Factorise 5xy + 3y.
Factorise
ax + 2ay + 3x + 6y.

Solve for x and y: 2x + 5y = 19
3x – y = 3
b((i)

5 marks
Att 2
5xy + 3y = y(5x + 3 )
Blunders (-3)
B1 Removes factor incorrectly
Attempts (2 marks)
A1 Indication of common factor e g underlines y’s and stops
b(ii)
5 marks
ax + 2ay + 3x + 6y = a(x + 2y) + 3(x + 2y) or
x(a+3) + 2y(a + 3)
= (a + 3)(x + 2y)
= (a+3)(x+ 2y)
*Accept also (with or without brackets) for 5 marks any of the following
(a+3) and (x+2y) [The word and is written down.]
(a+3) or (x+2y) [The word or is written down.]
(a+3), (x+2y) [A comma is used]
Blunders (-3)
B1 Correct answer without work 
B2 Stops after first line of correct factorization e.g. a(x + 2y) + 3(x + 2y) or
equivalent i,e. x(a + 3) + 2y(a + 3)
B3 Error(s) in factorising any pair of terms
B4 Correct first line of factorisation but ends as (a+3).2xy or equivalent
Slips (-1)
S1
(a+3)  (x+2y)
Attempts (2 marks)
A1 Pairing off, or indication of common factors and stops
A2 Correctly factorises any pair and stops
Page 28
Att2
b (iii)
10 marks
2x + 5y = 19
3x – y = 3 X5
OR
2x + 5y = 19
15x – 5y = 15
17x
= 34
x= 2
2(2) + 5y = 19
4 + 5y = 19
6x +15y = 57
-6x+2y = -6
Or
17y = 51
y=3
Att 3
3x -3 = y
2x +5(3x-3) = 19
2x +15x – 15 = 19
17x = 19 + 15
2x + 15 = 19
17x = 34
2x = 4
x=2
x=2
4 + 5y = 19
5y = 15
y =3
y=3
*Apply only one blunder deduction (B2 or B3) to any error(s) in establishing the first
equation; in terms of x only or the first equation in terms of y only
*Finding the second variable is subject to a maximum deduction of -3
Blunders (-3)
B1 Correct answers without work (stated or substituted)
B2 Error or errors in establishing the first equation in terms of x only (17x = 34) or the first
equation in terms of y only (17y = 51) through elimination by cancellation (but see S1)
B3 Error or errors in establishing the first equation in terms of x only (17x = 34) or the first
equation in terms of y only (17y = 51) through elimination by substitution (but see S1)
B4 Errors in transposition when finding the first variable
B5 Errors in transposition when finding the second variable
B6 Incorrect substitution when finding second variable
B7 Finds one variable only
Slips (-1)
S1 Numerical errors to a max of -3
Attempt (3 marks)
A1 Attempt at transposition and stops
A2 Multiplies either equation by some number and stops
A3 Incorrect value of x or y substituted correctly to find his correct 2nd variable
A4 One correct answer without work (stated and substituted)
Worthless (0 marks)
W1 Incorrect values for x or y substituted into the equations
Page 29
(c)
20 Marks (10,10)
(i)
Att 3,3
Write as a single fraction

x 3x
.

2 8
(ii)
Solve the equation 3(2x – 7) –5(x – 1) = 0.
Verify your answer.


c (i)
10 marks

x 3x 4 x 3x 7 x




2 8
8
8
8
*
or
or
etc
→ 7 marks
*
+
and stops
→4 Marks
*
x 3x

2 8
=
→0 Marks
Blunders (-3)
B1 Correct answer without work 
B2
Incorrect common denominator and continues
B3 Incorrect numerator from candidate's denominator
B4
Omitted or incorrect denominator
Slips (-1)
S1 Numerical errors to a max of -3
Attempts (3 marks)
A1 Any correct step.
A2 Any correct common denominator found
Worthless (0 marks)
and stops
W1 (
W2 Incorrect answer, with no work
Page 30
Att3
c(ii)
10 marks
Solve
3(2x – 7) –5(x – 1) = 0
6x -21 -5x + 5 = 0
x - 16 = 0
x = 16
Att 3

Verify
3(2x – 7) –5(x – 1)
3(2(16)– 7) –5(16 – 1
3( 32-7) – 5(15)
3(25) – 75 = 0
*Stops after correct removal of brackets
4 Marks
*If changes -5 to +5 at the start: blunder (-3)
*States x=16 (no work) and verifies correctly 7 Marks
*States x=16 (no work) with no verification 4 Marks
*Verifies correctly x=16 (not stated)
Att 3
Blunders (-3)
B1 Correct answer without work 
B2 Error(s) in distribution (each time)
B3 Combining unlike terms (each time) and continues
B4 Fails to group like terms
B5 Error(s) in transposition (each time)
B6 Fails to finish
B7 Fails to verify
Slips (-1)
S1 Numerical errors to a max of -3
S2 Incorrect or no conclusion from their work
Misreadings (-1)
M1 3(2x+7) or similar and continues but see 2nd* above
Attempts (3 marks)
A1 Any one term correctly multiplied
A2 Any correct step
Worthless (0)
W1 combining unlike terms before attempting multiplication and stops e.g. 3(14x) = 42x
W2 Invented answer verified but see * above
W3 Incorrect answer with no work
Page 31
QUESTION 6
Part (a)
Part (b)
Part (c)
10(5,5) marks
30 (15,15 ) marks
10 (5,5) marks
(a)
6.
10(5,5) marks
(a)
Att 2,2
Att (5,5)
Att (2,2)
Att 2,2
P = {(1, a), (2, a), (3, b), (4, c)}.
Write out the domain and range of P.
Domain =
Range =
(a) Domain
5 marks
Att 2
Domain = {1, 2, 3,4}
Slips (-1)
S1 Each incorrect element omitted / included other than the misreading below.
Misreadings (-1)
M1 Correct range { a, b, c } or { a, a, b, c } given.
Worthless (0)
W1 No element of the domain appears.
(a) Range
5 marks
Range = { a, b, c }
*Accept { a, a, b, c } for full marks.
Slips (-1)
S1 Each incorrect element omitted / included other then the misreading below
Misreadings (-1)
M1 Correct domain {1, 2, 3, 4} given
Worthless (0)
W1 No element of the range appears.
Page 32
Att 2
Part (b)
30(15, 15) marks
Att (5, 5)
Draw the graph of the function
f : x  5  2x  x2
in the domain  2  x  4 , where x  R .
Table
15marks
Att 5
f ( x)  5  2 x  x 2
f (2)  5  2(2)  (2) 2
f (1)  5  2(1)  (1) 2
f (0)  5  2(0)  (0) 2
f (1)  5  2(1)  (1) 2
f (2)  5  2(2)  (2) 2
f (3)  5  2(3)  (3) 2
f (4)  5  2(4)  (4) 2
=
=
=
=
=
=
=
 (2,3)
2  (1,2)
5  (0, 5)
6
 (1 , 6)
5
 (2 , 5)
2
 (3 , 2)
 3  (4 , -3).
544 = 3
5  2 1
500
5  2 1
544
569
5  8  16
=
=
=
=
=
=
OR
A
f (2)
=
5
 2(2)
- (2) 2
=
-3
f (1)
=
5
 2(1)
- (1) 2
=
2
f (0)
=
5
 2(0)
- (0) 2
=
5
f (1)
=
5
 2(1)
- (1) 2
=
6
f (2)
=
5
 2(2)
- (2) 2
=
5
f (3)
=
5
 2(3)
- (3)
2
=
2
f (4)
=
5
 2(4)
- (4) 2
=
-3
B
x
5
 2x
x
2
f (x)
-2
5
-1
5
0
5
1
5
2
5
3
5
4
5
-4
-4
-2
-1
0
0
+2
-1
+4
-4
+6
-9
+8
-16
-3
2
5
6
5
2
-3
*Error(s) in each row/column calculation attracts a maximum deduction of 3 marks
Blunders (-3)
B1 Correct answer, without work i.e. 7 correct couples only and no graph
B2 “+2 x ” taken as “2” all the way. [In the row headed “+2 x ” by candidate]
B3 “5” calculated as “5 x ” all the way. [In the row headed “5” by candidate]
B4 Adds in top row when evaluating f (x) in B.
B5 Omits “5” row
B6 Omits “+2 x ” row
B7 Omits a value in the domain (each time).
B8 Each incorrect image without work i.e. calculation through the function method (A)
B9 Misreads “  x 2 ” as “  x 2 ” and places “  x 2 ” in the table or function.
Slips (-1)
S1 Numerical errors to a max of -3 in any row / column
Page 33
Misreadings (-1)
M1 Misreads “  2 x ” as “  2 x ” and places “  2 x ” in the table or function.
M2 Misreads “5” as “-5” and places “-5” in the table or function.
Attempts (5 marks)
A1 Omits “  x 2 ” row from table or treats “  x 2 ” as  x or  2 x .
A2 Any effort at calculating point(s).
A3 Only one point calculated and stops.
Graph
15 marks
Att 5
5
4
3
2
1
-2
-1
1
2
3
4
-1
-2
-3
*
*
*
Only one correct point graphed correctly  Att 5 + Att 5
Correct graph but no table  full marks i.e. (15 + 15) marks.
Accept reversed co-ordinates if
(i) if axes not labelled or (ii) if axes are reversed to compensate (see B1 below)
Blunders ( -3)
B1 Reversed co-ordinates plotted against non-reversed axes (once only) {See 3rd * above}.
B2 Scale error (once only)
B3 Points not joined or joined in incorrect order (once only)
Slips (-1)
S1 Each point of candidate graphed incorrectly {Tolerance  0.25 }
S2 Each point (7 points needed ) from table not graphed [ See 2nd * above ]
Attempts (5 marks)
A1 Graduated axes (need not be labelled)
A2 Some effort to plot a point { See 1st * above}
Page 34
Part (c)
(c) (i)
(ii)
10 (5, 5) marks
Draw the axis of symmetry of the graph you have drawn in 6(b).
Att 2, 2
Use your graph to estimate the value of 5  2 x  x 2 when x  1 5 .
(c) (i)
(c) (i)
5 marks
Draw the axis of symmetry of the graph you have drawn in 6(b).
Att 2
Axis of Symmetry: x = 1
f(1.5) = 5.5
5
4
3
2
1
-2
-1
1
2
3
4
-1
-2
-3
*
Accept any vertical line (parallel to candidate’s y-axis) within tolerance of  0.25 .
Blunders ( -3)
B1 Any vertical line ( parallel to the candidate’s y-axis) outside of the tolerance.
B2 Marks x  1 on the x-axis and stops.
B3 States x  1 but no line is indicated on the graph.
Attempts ( 2 marks)
A1 Any attempt at axial symmetry of f (x) .
A2 y-axis indicated as the axis of symmetry (See B1).
Page 35
(c) (ii)
(c)
5 marks
Att 2
(ii) Use your graph to estimate the value of 5  2 x  x 2 when x  1.5
Work to be shown on the graph and answer to be written here.
5.75
*Correct answer (clearly consistent with candidate’s graph) inside the tolerance without
graphical indication  2 marks.
Blunders (-3)
B1 Correct answer without work
B2 Answer on the diagram but outside of tolerance (  0.25 )
B3 Fails to write down the answer, when indicated correctly on graph
Slips (-1)
S1 Correct answer indicated and/or written on graph only
Attempts (2 marks)
A1 Attempts at algebraic evaluation or calculator
A2 Marks 1.5 in any way on either axis and stops
Worthless (0)
W1 Answer outside of tolerance without graphical indication.
Page 36
Coimisiún na Scrúduithe Stáit
State Examination Commission
Scrúdu
an Teastais Shóisearaigh
JUNIOR CERTIFICATE
EXAMINATION
2011
MARKING SCHEME
MATHEMATICS
ORDINARY LEVEL
PAPER 1
Page 1 GENERAL GUIDELINES FOR EXAMINERS 1.
Penalties of three types are applied to candidates’ work as follows:
 Blunders - mathematical errors/omissions
(-3)
 Slips- numerical errors
(-1)
 Misreadings (provided task is not oversimplified)
(-1).
Frequently occurring errors to which these penalties must be applied are listed in the
scheme. They are labelled: B1, B2, B3,…, S1, S2,…, M1, M2,…etc. These lists are not
exhaustive.
2.
When awarding attempt marks, e.g. Att(3), note that
 any correct, relevant step in a part of a question merits at least the attempt mark for that
part
 if deductions result in a mark which is lower than the attempt mark, then the attempt
mark must be awarded
 a mark between zero and the attempt mark is never awarded.
3.
Worthless work is awarded zero marks. Some examples of such work are listed in the
scheme and they are labelled as W1, W2,…etc.
4.
The phrase “hit or miss” means that partial marks are not awarded – the candidate receives
all of the relevant marks or none.
5.
The phrase “and stops” means that no more work is shown by the candidate.
6.
Special notes relating to the marking of a particular part of a question are indicated by an
asterisk. These notes immediately follow the box containing the relevant solution.
7.
The sample solutions for each question are not intended to be exhaustive lists – there may
be other correct solutions.
8.
Unless otherwise indicated in the scheme, accept the best of two or more attempts – even
when attempts have been cancelled.
9.
The same error in the same section of a question is penalised once only.
10.
Particular cases, verifications and answers derived from diagrams (unless requested)
qualify for attempt marks at most.
11.
A serious blunder, omission or misreading results in the attempt mark at most.
12.
Do not penalise the use of a comma for a decimal point, e.g. €5.50 may be written as €5,50.
Page 2 BONUS MARKS FOR ANSWERING THROUGH IRISH Bonus marks are applied separately to each paper as follows:
If the mark achieved is 225 or less, the bonus is 5% of the mark obtained, rounded down.
(e.g. 198 marks  5% = 9.9  bonus = 9 marks.)
If the mark awarded is above 225, the following table applies:
Bunmharc
Marc Bónais
Bunmharc
Marc Bónais
(Marks obtained) (Bonus Mark) (Marks obtained) (Bonus Mark)
226
11
261 – 266
5
227 – 233
10
267 – 273
4
234 – 240
9
274 – 280
3
241 – 246
8
281 – 286
2
247 – 253
7
287 – 293
1
254 – 260
6
294 – 300
0
Page 3 QUESTION 1
Part (a)
Part (b)
Part (c)
15 (10,5) marks
20 (5,5,5,5) marks
15 (5,5,5) marks
(a)
10,5 marks
(i)
(ii)
(a) S =
Write down a subset of S that has one element.
Write down a subset of S that has three elements.
Att (3,2)
Att (2,2,2,2)
Att (2,2,2)
Att 3,2
(a) (i)
10 marks
{w} or {x} or
{y} or
{z}
*
No penalty for the omission of brackets.
*
No penalty for use of Venn Diagram to show subsets.
Blunders (-3)
B1 Any incorrect set of elements of S other than the misreading as below.
Att 3
Misreadings (-1)
M1 Subset of S with two or three elements. e.g. S = {w, x}.
Attempts (3 marks)
A1 Draws a single bracket & stops.
A2 { } Null set or set itself
Worthless(0)
W1 No relevant element listed without brackets but see A1 above
(a) (ii)
{w,x, y}
*
*
or
{w, x, z }
5 marks
or {w ,y, z}
or
No penalty for omission of brackets.
No penalty for use of Venn Diagram to show subsets.
Blunders (-3)
B1 Any incorrect set of elements of S other than the misreading as below.
Misreadings (-1)
M1 Correct subsets of S with one or two elements e.g. S = {w,x }.etc
Attempts (2)
A1 Draws a single bracket & stops.
A2 { } Null set or set itself
Worthless(0)
W1 No relevant element listed without brackets but see A1 above
Page 4 Att2
{x, y,z}
(b)
5,5,5,5 marks
Att 2,2,2,2
U
U is the universal set.
P
P = {2, 6, 9}
Q = {1, 3, 5, 6}
R = {3, 4, 6, 7, 9}
2
1
9
4
8
6
3
5
7
List the elements of:
(i) R \ Q
(ii) P΄, the complement of set P
(iii) Q  P  R 
(iv) Q  R  \ P
(i)
R \ Q = {4,7,9}
5 marks
Att 2
Blunders (-3)
B1 Any incorrect set of elements of Q and R other than the misreading below.
Misreadings (-1)
M1 Q \ R = {1,5}
Attempts (2 marks)
A1 4 or 7 or 9 appear in the answer.
A2 P  Q / R  ={ }
Worthless(0)
W1 {8}
(ii)
5 marks
Att 2
P΄, the complement of set P = {1,3,4,5,7,8}
Blunders (-3)
B1 Any incorrect set of the elements of P and Q and R other than the misreading below
Misreadings (-1)
M1 P  Q  R giving { 1,2,3,4,5,6,7,9 }
(all needed)
M2 R΄ = {2,1,5,8}
M3 Q΄ giving {2,4,7,8,9 }
M4 {2,6,9}
Attempts (2 marks)
A1 At least one correct entry appears in the answer
Page 5 (iii)
5 marks
Q  P  R   {1,3,5,6,9}
Blunders (-3)
B1 Any incorrect set of elements of Q, P or R other than the misreadings below.
Att 2
Misreadings (-1)
M1 Q \ P  R   1, 5, 3 .
M2 Q  P  R   {6}
M3 Q  P  R   {1,2,3,4,5,6,7,9}
M4 Q  P  R   {3,6}
Attempts (2 marks)
A1 1,3,5,6, or 9 appear in the answer. but see Misreadings above
Worthless(0)
W1 Answer {8}.
(iv)
5 marks
Att2
Q  R  \ P= {3}
Blunders (-3)
B1 Any incorrect set of elements of P and Q and R other than the misreading as below.
B2 (Q ∩ R) = {6,3} and stops
Misreadings (-1)
M1 Q \ R  P   1, 3, 5  .
M2 Q \ ( R  P)  {1,5}
M3 Q  ( R / P)  {1,3,4,5,7}
Attempts (2 marks)
A1 6 or 3 appear in the answer.
Worthless(0)
W1 Answer {8}.
(c)
5,5,10 marks
(i) List all the divisors of 18 and 24.
(ii) Write down the highest common factor of 18 and 24.
(iii) {5, 7, 9, 11, 13, 15} is the set of odd numbers between 4 and 16.
Which of these numbers are not prime numbers?
Give a reason for your answer.
(i)
5 marks
Divisors of 18: = 1, 2 ,3, 6, 9, 18
Divisors of 24: = 1,2, 3, 4, 6, 8, 12, 24
Slips (-1)
S1 Each missing or incorrect element to a max of −3
Attempts (2 marks)
A1 Any one correct element identified
Worthless(0)
W1 Elements listed that are not divisors of 18 or 24
Page 6 Att 2,2,3
Att 2
(ii)
5 marks
Att 2
Highest common factor = 6
*Accept candidate’s answer from c(i)
Blunders (-3)
B1 A common factor that is not the highest
Slips (-1)
S1 Answer written as 2 3
Misreadings (-1)
M1 Writes down LCM = 72
Attempts (2 marks)
A1 Any common factor listed
Worthless(0)
W1 Incorrect answer without work but see M1 or * above
(iii)
5 marks
Not prime numbers: 9 and 15
Reason: “ Each has more than 2 factors”
Blunders (-3)
B1 Each incorrect or omitted entry
Slips (-1)
S1
No or incorrect reason given
Misreadings (-1)
M1 Gives prime numbers only
Attempts (2 marks)
A1 Any one relevant entry between 4 and 16 inclusive
A2 Correct reason as to why numbers are not primes
Worthless(0)
W1 Incorrect answer with no work shown
Page 7 Att 2
QUESTION 2
Part (a)
Part (b)
Part (c)
(a)
10 marks
20(10,5,5) marks
20(5,5,5,5) marks
10 marks
Att 3
Att (3,2,2)
Att (2,2,2,2)
Att 3
€52 is divided between Fiona and Orla in the ratio 9:4.
How much does each receive?
(a)
9+4= 13
52  13  4
9×4 =36
4×4 =16
10 marks
OR 9+4 =13
OR
9x : 4x
=4
x=4
= 36
4x=16
= 16
Orla: 16 or
Fiona: 36
Att 3
13x = 52
52–36 =16
9x = 36
*
Correct answer without work  7 marks
*
Incorrect answer without work  0 marks, except for answers given in A4 below
*
= 13 and = 5.777…/ 5.78 or 5.8 merits 4 marks
Blunders (-3)
B1 Divisor ≠ 13 and continues
B2 Incorrect multiplier or fails to multiply (each time)
B3 Adds instead of subtracts i.e. 36 + 52 = 98
B4 Fails to find second amount
B6 Error in transposition
Slips (-1)
S1 Numerical errors where work is clearly shown to a max of −3
Attempts (3 marks)
A1 Divisor 13 e.g,
A2
A3
A4
A5
A6
and/ or
and stops
Indicates 13 parts or 9 parts or 4 parts or or and stops
Indicates multiplication of 52by 9 and/or 4
and stops
Both answers added together equal 52 (no work shown)
Finds 9% of 52 (4.68) and 4% of 52 (2.08)
One correct answer without work
Worthless(0)
W1 52 +9 = 61 or similar
W2 Incorrect answer without work. (subject to A4)
Page 8 (b)
(i)
10,5,5 marks
Att 3,2,2
By rounding each of these numbers to the nearest whole number, estimate the value of
14  18  4  086
.
1  96
(ii)
Using a calculator, or otherwise, find the exact value of
14  18  4  086
1  96
(iii) Find the difference between the exact value in (ii) and
the estimated value in (i).
(i)
10 marks
Att3
14  18  4  086 14  4 10


5
2
2
1  96
*
*
*
and stops  7 marks.
=
and stops  7 marks.(-3)

No penalty if the intermediate step between approximations and correct final answer is not
shown i.e. not shown
.
.
*
Special Case:
= 5.15 in this part  Attempt 3 marks. (or
or 5 )
.
Blunders (-3)
B1 Error(s) in rounding off to the nearest whole number (once only if consistent)
B2 Decimal error in calculation of final value
B3 An arithmetic operation other than indicated e.g. 14 – (4 ÷ 2) = 7 (breaking order)
B4 Error(s) in the manipulation of the denominator e.g. or
B5 Incorrect cancellation
Slips (-1)
S1 Numerical errors to a max of −3
Attempts (3 marks)
A1 Only one approximation made to the given numbers and stops
A2 Ans. 5 with no preceding rounding off
Worthless (0)
W1 Incorrect answer without work but note Special Case * above
Page 9 (ii)
5 marks
Att2
14  18  4  086 10  094

 5 15
1  96
1  96
Blunders (-3)
B1
B2
Decimal error or early rounding off
.
Leaves as
B3
Treats as 14.18 – 4.086/1.96
.
= 12.09530612
Treats as 14.18 + 4.086 = 9.319387753
1.96
B5 Treats as 14.18 – 4.086 = 3.148603878
1.96
Slips (-1)
S1 Numerical errors to a max of −3
B4
Attempts (2 marks)
A1 Any correct relevant calculation and stops.
A2 Any of the following; (see above)
12.09530612, 9.319387753 or 3.148603878
merits 2 marks (minimum 4 decimal places) (with or without work)
Worthless (0)
W1 Incorrect answer without work but see A2
(iii)
5 marks
5·15-5 = 0·15
*
Allow candidate’s previous answers
Blunders (-3)
B1 Correct answer without work
B2 Decimal error (once only if consistent)
B3 Finds the sum of b(i) and (ii)
Attempts (2 marks)
A1 Any relevant step i.e. transfers answers from b(i) and/or b(ii)
Worthless (0)
W1
Incorrect answer without work
Page 10 Att2
(c)
5,5,5,5 marks
(i)
(ii)
Att 2,2,2,2
Write (a3 )2 in the form an, n  ℕ
Using your answer from (i) or otherwise evaluate (53 )2.
Before going on holidays to the USA Seán changed €500 into dollars.
The exchange rate was €1 = US$1‧22.
(iii) How many dollars did Seán get?
(iv) When Seán came home he changed US$50 back into euro (€).
The exchange rate was the same.
How much, in euro, did Seán receive?
Give your answer to the nearest cent.
(i)
5 marks
3 2
3×2
6
3 2
= a
or (a ) = a3×a3 = a6
(a ) = a
or
a a a a
a = a6
*
a a a a
a and stops
4 marks
*
a3×2 and stops
4 marks
*
6 only written down
2 marks
Blunders (-3)
B1 a3 = a a a and stops
B2 each error in calculation involving indices e.g. (a3 )2 = a5
B3 Each incorrect number of a’s in the extended form
Slips (-1)
S1 Numerical errors to a max of −3
Attempts (2 marks)
A1 (a3 )2 = a3+2 and stops
A2 Some correct manipulation of indices
Worthless (0)
W1 Writes a only
(ii)
5 marks

(53 )2= 5 6 = 15625
or
53 = 125
1252 =15625
*
Accept candidate’s answer fom c(i) unless it oversimplifies the question
Blunders (-3)
B1 Correct answer, without work 
B2 Each error in calculation involving indices
B3 Each incorrect number of 5’s in the extended form
B4 Fails to finish
Slips (-1)
S1 Numerical errors to a max of −3
Attempts (2 marks)
A1 Some correct manipulation of indices
A2 52 = 25 and stops
A3 53 = 125 and stops
A4 Candidate transfers answer from c(i)
Worthless(0)
W1 Incorrect answer with no work shown
Page 11 Att 2
Att 2
(iii)

*
5 marks
Att 2
€500 × 1‧22 = $610
No penalty for omission of € or $ signs
Blunders (-3)
B1 Correct answer, without work 
B2 Incorrect operator i.e. divides by 1.22 correctly : 409.836
B3 Decimal error
B4 Fails to finish i.e. €500 × 1‧ 22 and stops
Slips (-1)
S1 Numerical errors to a max of −3
Attempts (2 marks)
A1 Some correct manipulation of 500 and/ or 1.22
Worthless(0)
W1 Incorrect answer with no work shown
(iv)
5 marks
50
 40  9836  40  98
1  22
Blunders (-3)
B1 Correct answer, without work 
B2 Multiplies by 1.22
`i.e. 50 × 1.22 = 61
.
B3 Incorrect ratio i.e.
or
B4 Decimal error
B5 Fails to finish i.e. leaves answer as
.
Slips (-1)
S1 Numerical errors to a max of −3
S2 Fails to round off or rounds off incorrectly
Attempts (2 marks)
A1 Some manipulation of 50 and/ or 1.22
A2 If answer is 41 or 40.9 with no work shown but see W1
Worthless(0)
W1 Incorrect answer with no work shown but see A2
Page 12 Att 2
QUESTION 3
Part (a)
Part (b)
Part (c)
15 marks
15 (5,5,5) marks
20 (5,5, 5,5) marks
(a)
Att 5
Att (2,2,2)
Att (2,2,2,2)
10 marks
Att 3
Three books were bought. They cost €8·75, €9·50 and €10·55 respectively.
If a €50 note was used to pay for the books, how much change was given?
Part (a)
15 marks
€8.75 + €9.50 + €10.55 = €28.80
€50.00 - €28.80 = €21.20
Change = €21.20
Att 5
€50.00 – (€8.75 + €9.50 + €10.55)
or
50.00 – €8.75 - €9.50 - €10.50 = €21.20
Change = €21.20
*Accept 2120 or 21.2.
*No penalty for the omission of the € sign
*Final subtraction step subject to maximum deduction of −3.
Blunders (-3)
B1 Correct answer without work
12 marks
B2 Fails to find the change.
B3 Operation other than addition when finding the total cost.
B4 Operation other than subtraction when finding the change.
B5 Each missing addition.
B6 Decimal error eg. €2.12 (Note 1st * above).
Slips (−1)
S1 Numerical errors to a max of 3
Attempts (3 marks)
A1 Any attempt at addition or subtraction of the given numbers and stops
Worthless (0)
W1 Incorrect answer without work is 0 marks.
W2 Multiplication or division of the given numbers.
Page 13 (b)
(i)
(i)
5,5,5 marks
Att 2,2,2
A washing machine costs €320 plus VAT at 21·0%.
Calculate the total cost of the washing machine after the VAT is added.
(ii)
A popular breakfast cereal comes in two sizes of packet,
Regular (360 g) and Large (900 g).
A standard portion of cereal is 30 g.
How many portions are there in each size of packet?
(iii)
A Regular box costs €0‧96 and a Large box costs €2‧25.
Using the number of portions per box, or otherwise,
find which size is better value?
5 marks
100%
1% =
121%
Total Bill =
=
320
320
100
=
21% =
× 320
VAT =
× 320
Att 2
320 x 1.21
Total Bill = € 387.20
x 121
= 3.2 x 121
€387.20
= 67.2
Total Bill = 320 + 67.2
Total Bill = €387.20
*
320 + 21% = 387.20
5 marks.
*
320 x 21% = 67.2 and stops
2 marks.
*
320 + 21% and stops or
320 × 21% and stops
2 marks.
*
€67.20 without work and stops
merits
2 marks.
Blunders (-3)
B1 Correct answer without work
B2 Decimal error.
B3 Inverts as 100 or 100 and continues (giving answers 264.46 or 1523.81).
121
21
B4 Mishandles 121% or 21% eg. 320 × 121 or 320 ÷ 121 or similar. (Note: 320 must be used)
B5 320 taken as 121% or 21%.
B6 No addition of VAT (as per candidates work).
B7 Subtraction of VAT (as per candidates work).
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 121 or 21 or 320 and stops.
100
100
100
A2 100% = 320 and stops.
A3 100 × 121 and stops.
320
A4 320 or similar and stops.
121
Worthless (0)
W1 Incorrect answer without work
W2 320 + 21 = 341 and stops or continues.
Page 14 (ii)
5 marks
Att 2
Regular: Number of portions = 360/30 =12

Large:
Number of portions = 900/30 = 30
Blunders (-3)
B1
B2
B3
B4

Correct answers without work
Multiplication instead of division when finding the number of portions (once only)
Finds only one answer
Decimal error
Slips (-1)
S1 Numerical errors to a max of −3
Attempts (2 marks)
A1 Any attempt at division and stops
A2 30 + 30 + …or any correct step
Worthless (0)
W1 Incorrect answer without work
Page 15 b (iii)
5 marks
Att 2
Method 1
Regular: 96 ÷ 12
=
8c per portion
Large:
225 ÷ 30
=
7.5c per portion
Large box is better value.
Method 2
Regular: 360g = 96cent Large: 900g = 225 c
1g = 96
1g = 225
360
900
1g = 0.267cent
1g = 0.25cent
Large box is better value.
Method 3
Regular: 96cent = 360g Large: 225cent = 900g
1 cent = 360
1 cent =
96
1 cent = 3.75g
1 cent = 4g
Large box is better value
Method 4
Regular: 10 boxes = 3600g = 10 x 0.96 = 9.60
Large: 4 boxes = 3600g = 4 x 2.25 = 9.00
Large box is better value.
*
Candidate must indicate in some way that the Large box is better value. See S2.
*
Accept candidate’s previous answer
Blunders (-3)
B1 Operation other than division in unitary methods 1, 2, and 3
B2 Operation other than multiplication in common denominator method 4
B3 Finds unit cost or weight for one size box only
B4 Decimal error
Slips (-1)
S1 Numerical errors to a max of −3
S2 Fails to highlight or indicate Large box as better value
Misreading (-1)
M1 Transposes costs or weight for each box (eg. Regular box costs €2.25 or similar) and
continues.
Attempt (2 marks)
A1 States Larger box without any relevant supporting work.
A2 Some attempt at division or multiplication using either €0.96 or €2.25.
A3 Some attempt at division using 12 or 30 or 360 or 900
A4 12 and 30 or 360 and 900
both multiplied as alternative in method 4
Worthless (0)
W1 Incorrect answer without work
W2 Adds given figures
Page 16 (c)
5,5,5,5 marks
Att 2,2,2,2
Geraldine’s annual wage is €40 000.
She pays income tax at the rate of 20% on the first €33 000 of her wage and
income tax at the rate of 41% on the remainder of her wage.
Geraldine has an annual tax credit of €3500.
(i) Calculate the tax on the first €33 000 of her wage, at the rate of 20%.
(ii) How much of Geraldine’s wage is taxed at the rate of 41%?
(iii) Calculate the amount of tax payable at the rate of 41%.
(iv) Calculate the tax due.
(i)
5 marks
100% = 33000
1% = 330
20% = 6600
Tax = €6600
*
Tax = 33000 × 20
100
Tax = €6600
Att 2
Tax = 33000 × 0.2
20% = 1
5
Tax = €6600
33000 ÷ 5
Tax = €6600
No penalty for omitting € symbol
Blunders (-3)
B1 Correct answer without work.
B2 Mishandles 20% eg. 33000 × 20 = 660000 or 33000 ÷ 20 = 1650
B3 Uses € 40000 instead of €33000
B4 Decimal error.
Slips (-1)
S1 Numerical error to a max of −3.
Attempts (2 marks)
A1 Some use of 100 in attempt to find percentage eg. 20% =
A2
Writes 33000 × 20 and stops
Worthless (0)
W1 Incorrect answer without work
W2 33000 + 20 and stops or continues
Page 17 or 0.2 or and stops
3(c) (ii)
(c) (ii)
How much of Geraldine’s wage is taxed at the rate of 41%?
5 marks
Att 2
€40000 – €33000 = €7000 taxed at 41%
* No penalty for omitting € symbol
Blunders (-3)
B1 Correct answer without work.
B2 Operation other than subtraction used with €40000 or €33000
B3 €6600 or 3500 is used in a subtraction with €40000 or €33000.
.
Slips (-1)
S1 Numerical error to a max of −3.
Attempts (2 marks)
A1 Some subtraction involving €40000 or €33000.
Worthless (0)
W1 Incorrect answer without work.
(c) (iii)
5 marks
100% = 7000
1% = 70
41% = 2870
Tax = €2870
Tax = 7000 × 41
100
Tax = €2870
Att 2
Tax = 7000 × 0.41
Tax = €2870
* No penalty for omitting € symbol
*Accept use of candidate’s answer from (ii) above.
Blunders (-3)
B1 Correct answer without work.
B2
Mishandles 41% eg. 7000 ÷ 41 = 170.73 or similar. Note: (No penalty if already penalised
in (c) (i).... consistent error.)
B3 Does not use €7000 but see 2nd * above.
B4 Decimal error.
Slips (-1)
S1 Numerical error to a max of –3.
Attempts (2 marks)
A1 Some correct use of 100 in attempt to find percentage eg. 41% = or 0.41 and stop
A2 Some correct use of €7000
A3 Uses €40000 or €33000 instead of €7000.
Worthless (0)
W1
Incorrect answer without work
W2
7000 + 41 = 7041 and stops or continues
Page 18 (iv)
5 marks
€6,600 + €2870 = €9470
*
*
*
€9470 - €3500 = €5970
Total Tax
€9470
Tax Credit
€3500
Tax Due
€5970
No penalty for omitting € symbol
Accept use of candidate’s answer from (i) and (iii) above.
If all 3 boxes are correctly filled in award full marks
Blunders (-3)
B1 Correct answer without work.
B2 Subtracts to find gross tax. e.g. 6600 – 2870 = 3730.
B3 Misuse or no use of Tax Credit
B4 Decimal error
B5 Total tax incorrectly calculated
Slips (-1)
S1 Numerical error to a max of –3.
Attempts (2 marks)
A1 Answer from c (i) or (iii) written in this part.
Worthless (0)
W1 Incorrect answer without work.
Page 19 Att 2
QUESTION 4
Part (a)
15(10,5) marks
Part (b)
15(5,10) marks
Part (c)
20(5,5,10) marks
(a)
10,5 marks
If a = 4, find the value of:
(i) 3a + 5
(ii) 3a2 – 20
(i)
10 marks
(i) 3a + 5
3(4) + 5 = 12 + 5 = 17
*
12 +5  9 marks
Blunders (-3)
B1 Correct answer, without work 
B2 Leaves 3(4) in the answer
B3 Incorrect substitution and continues
B4 Breaks order i.e. 3(4+5) = 3(9) = 27
B5 Treats 3(4) as 7 or 34
Slips (-1)
S1 Numerical errors to a max of 3
S2 Treats as 3a – 5
S3 Fails to finish
Misreadings (-1)
M1 Uses 5a + 3
Attempts (2 marks)
A1 Any number substituted for a and stops e.g. 3(6)
A2 Any correct step
A 3 Treats as 15a = 15(4) = 60 or 8a =8(4) = 32
Worthless (0)
W1 Incorrect answer with no work
Page 20 Att (3,2)
Att (2, 3)
Att (2,2,3)
Att 3,2
Att 3
(a)(ii)

*
2
3a – 20
48 -20
5 marks
3(4) – 20 = 3(16) – 20 = 48 – 20 = 28
2
 4 marks
Blunders (-3)
B1 Correct answer without work 
B2 Leaves 42 in the answer
B3 Incorrect substitution and continues
B4 Breaks order e.g. 3(16 – 20) = 3( –4 ) = –12.
B5 Treats 3(16) as 3 + 16
B6 Incorrect squaring eg. 42 = 8
B7 Treats as a2 – 20 i.e omits the 3
Slips (-1)
S1 Numerical errors to a max of −3
S2 Fails to finish but see * above
Misreadings (-1)
2
M1 Treats as 3a +20
Attempts (2 marks)
A1
A3
2
Any substitution for a and stops
Any correct step
Worthless (0)
W1 Incorrect answer, with no work
Page 21 Att2
(b)
5,10 marks
(i)
x 5x
 .
3 6
Multiply (2x – 5) by (3x – 4) and write your answer in its simplest form.
Write as a single fraction
(ii)
(i)
5 marks

=
0 Marks, but allow
*
or
or
5Marks
+
and stops
5 Marks
Blunders (-3)
B1 Correct answer without work 
B2
Incorrect common denominator and continues
B3 Incorrect numerator from candidate's denominator
B4
Omitting denominator
Slips (-1)
S1 Drops denominator
S2 Numerical errors to a max of 3
Attempts (2 marks)
A1 Any correct step.
A2 Any correct common denominator found
Worthless (0)
W1 (
and stops
W2
Att 2
x 5x 2x  5x 7 x



6
3 6
6
*
*
Att 2,3
Incorrect answer, with no work
Page 22 or
etc for full marks
b(ii)
10 marks

(2x – 5) (3x – 4) = 2x(3x – 4) – 5(3x – 4) = 6x2 – 8x – 15x + 20 = 6x2– 23x + 20
*
If 6x2 – 8x – 15x + 20 is correct
(minimum 7 MARKS)
Blunders (-3)
B1 Correct answer without work
B2 Error in distribution each time
B3 Errors in multiplication of powers
B4 Errors in collecting like terms
B5 Mathematical (sign ) errors eg –5 × –4 = – 20
B6 (2x – 5) written as (2x + 5 ) and continues and/or (3x – 4) written as (3x +4) --oversimplifies
Slips (-1)
S1 Numerical errors to a max of −3
Misreadings (-1)
M1 (5x–2)(4x–3) etc and continues
Attempts (3 marks)
A1 One term correctly multiplied and stops e.g. 6x2
A2 2x(3x–4) or –5(3x–4) and stops
A3 2x(3x – 4) –5(3x – 4) and stops
Worthless (0)
W1 Incorrect answer with no work
Page 23 Att 3
(c)
(i)
(ii)
5,10,10 marks
Att 2,3,3
The cost of a DVD is €x. The cost of a CD is €3 less.
What is the cost of a CD in terms of x?
The total cost of 3 DVDs and 2 CDs is €54.
Write an equation in x to represent this information.
Solve your equation to find the cost of a DVD.
c(i)
5 marks
CD : x –3
*
Algebraic work required to earn marks
Blunders (-3)
B1 Incorrect expression for the cost of a CD other than misreading below
Misreadings (-1)
M1 Answer given as
3+x
or
3–x
Attempts (2 marks)
Worthless (0)
W1 Cost of CD given as a constant or x.
Page 24 Att 2
(ii)
5 marks
Equation :

3x + 2(x – 3) =54
3x + 2x – 6 = 54
5x = 60
x = 12
Cost of a DVD = 12
*
B1
B2
B3
B4
B5
Att 2
Accept candidates answer from previous work.
Error in forming equation.
Distribution error
Transposition error
Stops at 5x = 60 or fails to solve equation
Error in collecting like terms
Misreading (-1)
M1 2x + 3(x – 3) = 54 or similar
Slips (-1)
S1 Numerical errors to a max of −3
Attempts (2 marks)
A1 Answer from part c (i) written down and stops.
A2 Any effort at forming an expression.
A3 Writes x = 12
A4 Any effort at solving their equation
A5 Successful Trial and Error
Worthless (0)
W1 Incorrect answer with no work.
Page 25 (iii)
10 marks
Att 3
10 marks
Att 3
Solve for x and y:
x + 3y = 12
3x + 2y = 11
(iii)
5x + 3y = 12 (× –2) OR
5x + 3y = 12 (× 3)
3x + 2y = 11 (×3)
3x+2y =11 (× –5)
OR x =
3(
) +2y =11

–10 x – 6y = –24
9x + 6y = 33
–x
= 9
x
= –9
5(– 9) + 3y = 12
– 45 + 3y = 12
3y = 57
15x +9y =36
−15x −10y=−55
−y = −19
y =19
5x +3(19) =12
5x +57 =12
5x =12 – 57
36 -9y +10y = 55
y =55 − 36
y = 19
x=
x=
x=
y= 19
x =
–9
x = −9
5x = −45
x = −9
*
Apply only one blunder deduction (B2 or B3) to any error(s) in establishing the first
equation; in terms of x only or the first equation in terms of y only.
*
Finding the second variable is subject to a maximum deduction of (3).
Blunders (-3)
B1 Correct answers without work (stated or substituted)
B2 Error or errors in establishing the first equation in terms of x only (–x = 9) or the first
equation in terms of y only (–y = –19) through elimination by cancellation (but see S1)
B3 Error or errors in establishing the first equation in terms of x only (x = –9) or the first
equation in terms of y only (–y = –19) through elimination by substitution (but see S1)
B4 Errors in transposition when finding the first variable
B5 Errors in transposition when finding the second variable
B6 Incorrect substitution when finding second variable
B7 Finds one variable only
Slips (-1)
S1 Numerical errors to a max of –3
Attempt (3 marks)
A1 Attempt at transposition and stops
A2 Multiplies either equation by some number and stops
A3 Incorrect value of x or y substituted correctly to find candidate’s correct 2nd variable
Worthless (0)
W1 Incorrect values for x or y substituted into the equations
Page 26 QUESTION 5
Part (a)
Part (b)
Part (c)
(a)
(a)
10 marks
20(5,5,5,5) marks
20(10,10) marks
10 marks
Att 3
Att (2,2,2,2)
Att (3,3)
Att 3
Write in its simplest form 2( x + 5) + 7( 2x + 3).
(a)
10 marks
Att 3
2( x + 5) + 7( 2x + 3) = 2x +10 +14x +21 = 16x + 31
*Stops after correct removal of brackets
7 Marks
Blunders (-3)
B1
B2
B3
B4
B5

Correct answer without work
Error(s) in distribution (each time)
Combining unlike terms after removal of brackets and continues
Fails to group like terms
Fails to finish
Slips (-1)
S1 Numerical errors to a max of –3
Misreadings (-1)
M1 2(x + 2) and continues.
Attempts (3 marks)
A1 Any one term correctly multiplied
A2 Combines unlike terms at the start and finishes correctly
Worthless (0)
W1 Combining unlike terms before attempting multiplication and stops e.g. 2(5x) = 10x
Page 27 (b)
5,5,5,5 marks
Att 2,2,2,2
Factorise:
(i)
(ii)
(iii)
(iv)

4xy – 8y
xy – xz + 3y – 3z
x² + 7x +12
x² – 64
(i)
5 marks
Att 2
4xy – 8y = 4y(x – 2)
* y(4x–8)
or
2y(2x–4)
or
2(2xy –4y) or
4(xy – 2y) merit 4 Marks
Blunders (-3)
B1 Removes factor incorrectly
Attempts (2 marks)
A1 Indication of common factor e.g. underlines y’s and stops
A2 Lists factors of 4 and factors of 8
(ii)
5 marks
 xy
– xz + 3y – 3z = x(y – z) + 3(y – z) or
= ( y − z )(x + 3)
Att 2
y(x + 3) – z(x + 3)
= ( y − z )(x + 3)
* Accept also (with or without brackets) for 5 marks any of the following
(y−z) and (x+3) [The word and is written down.]
(y−z) or (x+3) [The word or is written down.]
(y−z), (x+3) [A comma is used]
Blunders (-3)
B1 Correct answer without work 
B2 Stops after first line of correct factorisation. e.g. x(y–z) +3 (y–z) or equivalent.
B3 Error(s) in factorising any pair of terms
B4 Correct first line of factorisation but ends as (x+3).–yz or equivalent
Slips (-1)
S1 (y–z)  (x+3)
Attempts (2 marks)
A1 Pairing off, or indication of common factors and stops
A2 Correctly factorises any pair and stops
Page 28 (iii)
x2 + 7x + 12
x2 + 4x + 3x + 12
x(x + 4) +3(x + 2)
(x + 3)(x +4)
5 marks
Att 2
7
x
+3
x
+4
(x + 3)(x +4)
7   4112
21
2
 7  49  48
 7 1

2
2
6
8
  3 and
 4
2
2
(x + 3)(x + 4)
Factor Method
Blunders (-3)
B1 Incorrect two term linear factors of x2+7x+12 formed from correct (but inapplicable) factors
of x 2 and/or ±12. e.g. (x+12)(x−1)
B2 Incorrect factors of x 2 and/or ±12
B3 Correct cross method but factors not shown and stops
B4 x(x+3)+4(x+3)or similar and stops
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (2 marks)
A1 Some effort at factorization e.g. (x
)(
) or the cross with at least one “x” written in
A2 States one correct factor without work
Worthless (0 marks)
W1 x2 + 7x = 12 or similar and stops
W2 Incorrect Trial and error
W3 Oversimplification, resulting in a linear equation
W4 Combines x with numbers and continues or stops
Formula Method
Blunders (-3)
B1 Error in a,b,c substitution (apply once only)
B2 Sign error in substituted formula (apply once only)
B3 Error in square root or square root ignored
 7 1
B4 Stops at
2
B5 Incorrect quadratic formula and continues
B6 No factors from roots or incorrect factors
Slips (-1)
S1 Numerical errors to a max of −3
S3 One factor only
Attempts (2 marks)
A1 Correct formula and stops
Worthless (0 marks)
W1 Combines x with numbers and continues or stops
Page 29 (iv)
5 marks
x² – 64 = x² – (8)2
Att 2
= (x+ 8)(x – 8)
*
Accept also (with or without brackets) for 5 marks any of the following x + 8 and (x–8)
[The word and is written down.]
(x+8) or (x−8) [The word or is written down.]
(x+8) , (x−8) [A comma is used]
*
Quadratic equation formula method is subject to slips and blunders.
*
(x−√64)(x+√64)
merits 5 marks
*
x±8
merits 4 marks
Blunders (-3)
B1 Incorrect two term linear factors of x2–64 formed from correct (but inapplicable) factors of
x 2 and 64 e.g ( x  64)( x  1)
B2 Incorrect factors of −64
B3 Incorrect factors of x 2
B4 (8 − x)(8 + x).
B5 (x − 64)(x + 64)
B6 Answer left as roots. (x = ±8)
Slips (-1)
S1 x−8(x+8)
Attempts (2 marks)
A1 Some effort at factorization e.g. (x
)(
) or the cross with at least one “x” written in
A2
8
appears
2
A3 x – 64 =x.x – 8.8 only
A4 Mention of the difference of two squares .e.g. x2 – 642
A5 Correct quadratic equation formula quoted and stops
A6 √64
Worthless (0)
W1 Combines xs to “numbers” and continues or stops
Page 30 (c)
10,10 marks
5(3x + 1) – 2(5x + 35) = 0.
(i)
Solve the equation
Verify your answer.
(ii)
Solve x2 + 3x – 10 = 0.
(i)
10 marks
Att 3,3
Att 3
5(3x + 1) – 2(5x + 35) = 0
15x + 5 – 10x – 70 = 0
5x – 65 = 0
5x = 65
x = 13
Verify
*
*
*
*
5(3x + 1) – 2(5x + 35)
x = 13
5(3(13) + 1) – 2(5(13) + 35)
5(39 + 1) – 2(65 + 35)
5(40) – 2(100)
200 – 200 = 0
If changes −2 to +2 at the start:
Blunder(−3)
States x =13 (no work) and verifies correctly 7 Marks
States x =13 (no work) with no verification
4 Marks
Verifies correctly x = 13 (not stated)
Att 3
Blunders (-3)
B1 Correct answer without work 
B2 Error(s) in distribution (each time)
B3 Combining unlike terms (each time) and continues
B4 Fails to group like terms
B5 Error(s) in transposition (each time)
B6 Fails to finish
B7 Fails to verify or verifies incorrectly
Slips (-1)
S1 Numerical errors to a max of –3
Misreadings (-1)
M1 5(3x−1) or similar and continues but see * above
Attempts (3 marks)
A1 Any one term correctly multiplied
A2 Any correct step
Worthless (0)
W1 combining unlike terms before attempting multiplication and stops e.g. 5(4x) = 20x
W2 Invented answer verified but see * above
W3 Incorrect answer with no work
Page 31 (ii)
x2 + 3x – 10 = 0
x2 + 5x – 2x – 10 = 0
x(x + 5) – 2x – 10 = 0
x(x + 5) – 2(x + 5) = 0
(x + 5)(x – 2) = 0
(x + 5) = 0 or (x – 2) = 0
10 marks
(x + 5)(x -2) = 0
(x + 5) = 0 or (x – 2) = 0
x = -5
x = -5
or
x=2
x
+5
x
2
or
x=2
Att 3
 3 
3
 41 10 
21
2
 3  9  40
37

2
2
 10
4
  5 and
2
2
2
* 2 correct solutions by Trial and Error
10 Marks
* 1 correct solution by Trial and Error
3 Marks (Attempt)
Factor Method
Blunders (-3)
B1 Correct answers without work 
B2 Incorrect two term linear factors of x2+3x –10 formed from correct (but inapplicable)
factors of x2 and/or ±10,e.g. (x + 10)(x – 1)
B3 No roots given. (once only)
B4 Incorrect factors of x2 and/or ±10
B5 Correct cross method but factors not shown and stops [Note: B3 applies also].
B6 x(x+5)−2(x+5) or similar and stops [Note: B3 applies also].
B7 Error(s) in transposition
Slips (-1)
S1 Numerical errors to a max of –3
S2 One root only from factors
Attempts (3 marks)
A1 Some effort at factorization e.g. (x
)(
) or the cross with at least one “x” written in
A2 States one correct root without work
Worthless (0)
W1 x2 +3x = 10 or similar and stops
W2 Incorrect Trial and error
W3 Oversimplification, resulting in a linear equation
Formula Method
Blunders (-3)
B1 Error in a,b,c substitution (apply once only)
B2 Sign error in substituted formula (apply once only)
B3 Error in square root or square root ignored
B4 Stops at
B5 Incorrect quadratic formula and continues
Slips (-1)
S1 Numerical errors to a max of −3
p
S2 Roots left in the form
q
S3 One root only
Attempts (3 marks)
A1 Correct formula and stops
A2 One correct substitution and stops
Page 32 QUESTION 6
Part (a)
Part (b)
Part (c)
15(5,10) marks
20 (10,10) marks
15 (5,5,5) marks
(a)
Att (2,3)
Att (3,3)
Att (2,2,2)
5, 10 marks
(a)
f (x) = 2x – 7.
(i)
Find:
f(4)
(ii)
f(–3)
(a)
5 marks
(i)
f (x) = 2x – 7
f (4) = 2(4) – 7 = 8 – 7 = 1
Blunders (-3)
B1
B2
Correct answer without work. 
Mathematical error. e.g.  2  4   24,
B3 Leaves 2(4) in the answer.
B4 Combines “ x' s ” to “numbers” and continues e.g. 2 x  7  5 x  5( 4)  20 .
B5 Mathematical error
e.g 8  7  1 .
B6 Breaks order i.e. 2( 4  7 )  2( 3)  6 .
Slips (-1)
S1 Numerical errors to a max of −3.
S2 Leaves x in the answer e.g. 1x
Misreadings (-1)
M1 Correctly substitutes in any number other than 4 and continues.
Attempts (2marks)
A1 Treats as equation and continues or stops. . i.e 2 x  7  4 .
A2 Substitutes for “ x ” and stops. i.e. 2(4).
Worthless (0)
W1 Combines “ x' s ” to “numbers” and stops.
2  7  5 .
W2 Ignores x giving
W3 4[ f ( x)]  8 x  28 .
W4 Replaces coefficient i.e. 2 x  4 x .
W5 Incorrect answer without work.
Page 33 Att 2,3
Att2
(a) (ii)
(a) (ii)
10 marks
Att3
f ( 3)  2( 3)  7 =  6  7 =  13
Blunders (-3)
B1 Correct answer without work.  [Do not penalise if already penalised in part (a) (i) or work
is shown in part (a) (i).]
B2 Mathematical error. i.e. −6 – 7 = 13
B3 Leaves 2(−3) in the answer.
B4 Combines “x’s” to “numbers” and continues e.g. 2 x  7  5 x  5( 3)  15 .
B5 Breaks order i.e. 2(–3 –7) = 2(–10) = –20
Slips (-1)
S1 Numerical errors to a max of –3.
S2 Leaves x in the answer e.g. −13 x
Misreadings (-1)
M1 Substitutes in any negative number other than −3 and continues.
Attempts (3marks)
A1 Treats as equation and continues or stops. i.e. 2 x  7  3 .
A2 Substitutes in any positive number
A3 Substitutes for x and stops. i.e. 2(−3).
Worthless (0)
W1 Ignores x giving 2 7 = −5
W2 −3f(x) = −6x + 21
W3 Combines “x’s” to “numbers” and stops.
W4 Replaces coefficient i.e. 2 x  3x .
W5 Incorrect answer without work.
Page 34 (b)
10 (Table), 10 (Graph) marks
Draw the graph of the function
Att 3,3
g : x  2x² – 4x + 1
in the domain 1  x  3, where x  ℝ.
(b)
10 marks ( table )
Att 3
g : x  2x² – 4x + 1
g(x) =2x² – 4x + 1
g(–1) =2(–1) ² – 4(–1) + 1 = 2 + 4 +1 = 7
g(0) =2(0) ² – 4(0) + 1 =
0 + 0 +1 = 1
g(1) =2(1) ² – 4(1) + 1 =
2 – 4 +1 = – 1
g(2) =2(2) ² – 4(2) + 1 = 8 – 8 +1 = 1
g(3) =2(3) ² – 4(3) + 1 = 18 – 12 +1 = 1
Table
A
10 marks
Att 3
B
f (1) = 2 (1)2 -4(-1) +1 = 7
f (0) = 2 (0)2 -4(0) +1 = 1
f (1)
*
(–1,7)
(0, 1)
(1, – 1)
(2, 1)
(3, 7)
x
2x 2
-1
2
0
0
1
2
2
8
3
18
= 2 (1)2
f ( 2) = 2 (2)2
-4(1)
+1 = -1
-4(2)
+1 = 1
 4x
+1
+4
+1
-0
+1
-4
+1
-8
+1
- 12
+1
f (3)
-4(3)
+1 = 7
f (x )
7
1
-1
1
7
= 2 (3)2
Error(s) in each row/column calculation attracts a maximum deduction of 3marks
Blunders (-3)
B1 Correct answer, without work i.e. 5 correct couples only and no graph
B2 Takes “ 2x 2 ” as “ x 2 ” and places “ x 2 ” in the table or function.
B3 Errors in evaluating “ 2x 2 ”, e.g. 2(1)2  (2)2  4 , once only if consistent.
B4 “–4 x ” taken as “–4” all the way [In the row headed “-4 x ” by candidate]
B5 “+1” calculated as “+1 x ” all the way. [In the row headed “+1” by candidate]
B6 Adds in top row when evaluating f (x ) in table method (B).
B7 Omits “+1” row
B8 Omits “–4 x ” row
B9 Omits a value in the domain (each time).
B10 Each incorrect image, without work, or, calculation through the function method (A).
Page 35 Slips (-1)
S1 Numerical errors to a max of –3 in any row / column
S2 Fails to find a value of Range each time in table to a max of 3
Misreadings (-1)
M1 Misreads “  4 x ” as “  4 x ” and places “  4 x ” in the table or function.
M2 Misreads “+1” as “–1” and places “–1” in the table or function.
Attempts (3marks)
A1 Omits “ 2x 2 ” row or treats “ 2x 2 ” as  2 x or  x , (i.e. evaluates a linear function)
A2 Any effort at calculating point(s) in the Domain
A3 Only one point calculated and stops.
(b)
10 marks (graph)
Att 3
7
6
5
4
3
2
1
-1
1
2
3
-1
*
*
*
*
Accept candidates values from previous work ( 5 co-ordinates needed ) but see S2
Only one correct point graphed correctly  Att 3 + Att 3
Correct graph but no table  full marks i.e. (10 + 10) marks.
Accept reversed co-ordinates if
(i) if axes not labelled or (ii) if axes are reversed to compensate (see B1 below)
Page 36 Blunders ( -3)
B1 Reversed co-ordinates plotted against non-reversed axes (once only) {See 4th * above}.
B2 Scale error (once only)
B3 Points not joined or joined in incorrect order (once only).
Slips (-1)
S1 Each point of candidate graphed incorrectly. {Tolerance  0.25 }
S2 Each point { 5 points needed } from table not graphed [ See 2nd * above ]
Attempts (3 marks)
A1 Graduated axes (need not be labelled)
A2 Some effort at plotting a point { See 2nd * above}
(c)
(c)
5,5,5 marks
Att 2,2,2
Given that y = x – 1, complete the table below.
(i)
x
y
1
2
3
4
On the grid below the graph of the line y = 3 – x is drawn.
Using your answers from (i), draw the graph of y = x – 1 on the same grid.
4
3
y=3–x
2
1
-1
1
2
-1
Page 37 3
4
5
(iii) Use the graphs drawn in 6(c) (ii) to write down the co-ordinates of the point of intersection
of the two lines y = 3 – x and y = x – 1.
Answer to be written here.
(c)
5 marks
(i)
Att 2
Given that y = x – 1, complete the table below.
x
y
1
0
2
1
3
2
4
3
* Accept candidate’s values without work
Slips (-1)
S1 Each ‘ y ’ value omitted or incorrect.
Misreadings (-1)
M1 Treats y  x  1 as y  x  1 . (consistent error)
Attempts (2) marks)
A1 Any one correct ‘ y ’ value.
A2 Any effort at calculating points.
A3 Treats as y = –x and continues
Worthless (0)
W1 Copies x values into y row.
W2 All ‘ y ’ values incorrect with no work shown but (See M1 and A3 above)
Page 38 (ii)
5 marks
Att 2
On the grid below the graph of the line y = 3 – x is drawn.
Using your answers from (i), draw the graph of y = x – 1 on the same grid.
4
3
y=3–x
2
1
-1
1
2
-1
*
Accept candidates values from previous work
Blunders ( -3)
B1 Reversed co-ordinates plotted.
B2 Points not joined or joined in incorrect order.
Slips (-1)
S1 Each point of candidate graphed incorrectly. {See B1}.
S2 Each point from table not graphed.
Attempts (2 marks)
A1 Any one correct point plotted.
A2 Any incorrect straight line drawn
Worthless (0)
W1 No correct point plotted. {See B1 above}.
Page 39 3
4
5
(iii)
5 marks
(2,1)
* Accept correct answer based on candidate’s graph from c(ii),
otherwise, attempt marks at most.
Blunders (-3)
B1 Answer beyond tolerance (  0.25 ).
B2 Answer given with co-ordinates reversed, i.e. ( y , x ) .
Slips (-1)
S1 Correct answer written on graph but not presented in the answer box.
Attempts (2 marks)
A1 Algebraic evaluation. (fully correct)
A2 Point of intersection clearly indicated correctly on graph, but not written down.
Worthless (0)
W1 Answer outside of tolerance without graphical indication.
W2 Incorrect answer from candidate’s graph.
Page 40 Att 2
JUNIOR CERTIFICATE
EXAMINATION
2010
MARKING SCHEME
MATHEMATICS
ORDINARY LEVEL
PAPER 1
GENERAL GUIDELINES FOR EXAMINERS
1.
Penalties of three types are applied to candidates’ work as follows:
• Blunders - mathematical errors/omissions
(-3)
• Slips- numerical errors
(-1)
• Misreadings (provided task is not oversimplified)
(-1).
Frequently occurring errors to which these penalties must be applied are listed in the
scheme. They are labelled: B1, B2, B3,…, S1, S2,…, M1, M2,…etc. These lists are not
exhaustive.
2.
When awarding attempt marks, e.g. Att(3), note that
• any correct, relevant step in a part of a question merits at least the attempt mark for that
part
• if deductions result in a mark which is lower than the attempt mark, then the attempt
mark must be awarded
• a mark between zero and the attempt mark is never awarded.
3.
Worthless work is awarded zero marks. Some examples of such work are listed in the
scheme and they are labelled as W1, W2,…etc.
4.
The phrase “hit or miss” means that partial marks are not awarded – the candidate receives
all of the relevant marks or none.
5.
The phrase “and stops” means that no more work is shown by the candidate.
6.
Special notes relating to the marking of a particular part of a question are indicated by an
asterisk. These notes immediately follow the box containing the relevant solution.
7.
The sample solutions for each question are not intended to be exhaustive lists – there may
be other correct solutions.
8.
Unless otherwise indicated in the scheme, accept the best of two or more attempts – even
when attempts have been cancelled.
9.
The same error in the same section of a question is penalised once only.
10.
Particular cases, verifications and answers derived from diagrams (unless requested)
qualify for attempt marks at most.
11.
A serious blunder, omission or misreading results in the attempt mark at most.
12.
Do not penalise the use of a comma for a decimal point, e.g. €5.50 may be written as €5,50.
Page 1
QUESTION 1
Part (a)
Part (b)
Part (c)
Part (a)
(i)
(ii)
10(5, 5) marks
25(10, 5, 5, 5) marks
15(5, 5, 5) marks
Att(2, 2)
Att(3, 2, 2, 2)
Att(2, 2, 2)
10(5, 5) marks
Att (2, 2)
Using the Venn diagram below, shade in the region that represents A∩B.
Using the Venn diagram below, shade in the region that represents A \ B
A
(a)(i)
B
5 marks
A B
Att 2
A
B
Blunders (-3)
B1 Any incorrect indication other than the misreading below
Misreadings (-1)
M1 A ∪ B indicated.
Worthless(0)
W1 No filling in of the Venn Diagram
(a)(ii)
A/B
5 marks
A
Att 2
B
Blunders (-3)
B1 Any incorrect indication other than misreading below
Misreadings (-1)
M1 B/A indicated
Worthless(0)
W1 No filling in of the Venn diagram
Page 2
Part (b)
U is the universal set
25(10, 5, 5, 5)
P
P = {2, 4, 6, 8, 10, 12}
10
2
Q = {3, 6, 9, 11, 12}
4
1
R = {4, 8, 11, 12}
Att(3, 2, 2, 2)
3
6
8
5
12
9
Q
11
R
(b)(i)
(i) List the elements of P ∩ Q ∩ R .
10 marks
Att 3
(b)(i)
10 marks
Att 3
P ∩ Q ∩ R = {12}
Blunders (-3)
B1 Any incorrect set of the elements of P and Q and R other than the misreading below
Misreadings (-1)
M1 P ∪ Q ∪ R giving {2,3,4,6,8,9,10,11,12} (all needed)
Attempts (3 marks)
A1 1 or 5 appear in the answer
(b)(ii)
5 marks
(ii) List the elements of R', the complement of the set R.
(ii)
R' = {1, 2, 3, 5, 6, 9, 10}
Blunders (-3)
B1 Any incorrect set of elements of R' other than the misreadings below.
Misreadings (-1)
M1 R\Q giving {4, 8}, R\P giving {11}, R\(P ∪ Q) giving { }
M2 P' = {3, 9, 11, 5} or Q' = {1, 2, 4, 5, 8, 10}
Attempts (2 marks)
A1 1,2,3,5,6,9, or 10 appear in the answer.
A2 R or any proper subset of R.
Page 3
Att 2
.
(b) (iii)
(iii)
5 marks
Att 2
List the elements of P\ ( Q ∩ R ) .
(iii)
P\ ( Q ∩ R ) = {2, 4, 6, 8, 10}
Blunders (-3)
B1 Any incorrect set of elements of P and Q and R other than the misreadings below
Misreadings (-1)
M1 P\(Q ∪ R) giving {2,10}
or (Q ∩ R)\ P giving {11}
Attempts (2 marks)
A1 1 or 5 appear in the answer
(b) (iv)
5 marks
Att 2
(iv) Write down # (Q ∪R).
(iv)
#(Q ∪R) = 7
Blunders (-3)
B1 Any incorrect cardinal number (Q ∪R) ≤11 other than the misreadings as below
Misreadings (-1)
M1 Q ∪ R giving {3,4,6,8,9,11,12}
M2 #(Q ∪ R)' = 4
M3 #(Q∩R) = 2
Attempts (2 marks)
A1 Some understanding of notation e.g. Cardinal numbers or number of elements
Worthless (0)
W1 Any number greater than 11
Page 4
(c)
15(5, 5, 5) marks
Att(2, 2, 2)
In a survey, a group of 72 students were asked if they played basketball or tennis
37 of these students said they played basketball (B)
30 of these students said they played tennis (T)
28 of these students said they played basketball but not tennis
(c)(i)
5 marks
(i) Represent this information in the Venn diagram below.
(c)(i)
5 marks
B
Att2
Att2
T
28
9
21
14
Blunders (-3)
B1 Each incorrect or omitted entry but see S1 and M1 below
Slips (-1)
S1
Numerical errors, where work is clearly shown, to a max of 3
Misreadings (-1)
M1 Interchanges Basketball and Tennis
Attempts (2 marks)
A1 Any one relevant entry
A2 #B = 37 or #T = 30 or #U = 72
(c)(ii)
5 marks
Att2
(ii) How many students played neither basketball nor tennis?
(c)(ii)
5 marks
Att2
c(ii)
28+21+9 = 58 72-58 =14
*
Any correct answer written here in the space provided takes precedence over an
incorrect Venn diagram
(Subject to S1)
*
Accept candidate’s work from previous part c(i)
Blunders (-3)
B1 Any incorrect use of the given numbers or the numbers from an incorrect Venn diagram
(Subject to S1)
Slips (-1)
S1
Numerical errors where work is clearly shown, to a max of 3
Attempts (2 marks)
A1 Any one relevant sum where work is clearly shown
Worthless (0)
W1 Incorrect answer with no work shown
Page 5
(c)(iii)
5 marks
(iii) What percentage of the students surveyed played both basketball and tennis?
Att2
(c)(iii)
Att2

*
*
5 marks
9
× 100 = 12.5%
72
or
100
= 12.5%
8
Any correct answer written here in the space provided takes precedence over an
incorrect Venn diagram
(Subject to S1)
Accept candidate’s work from previous part c(i)
Blunders (-3)
B1 Any incorrect use of the given numbers or the numbers from an incorrect Venn diagram
(Subject to S1)
B2 Correct answer without work 
Slips (-1)
S1
Numerical errors, where work is clearly shown, to a max of 3
Attempts (2 marks)
A1 Mention of 9 or candidate’s work from c(i)
A2 Some use of 100
Worthless (0)
W1 Incorrect answer with no work shown
Page 6
QUESTION 2
Part (a)
Part (b)
Part (c)
Part (a)
10 marks
25(10, 5, 10)marks
15(5, 5, 5)marks
10 marks
There is €1200 in a prize fund. The first prize is
Find the value of the first prize.
(a)
Method (1)
10
=1200
10
1 1200
=
=120
10 10
7
=120 × 7=840
10
= €840
Att 3
Att(3, 2, 3)
Att(2, 2,2)
Att 3
7
of the fund.
10
10 marks
Att 3
Method (2)
Method (3)
Method (4)
10 parts=1200
1200
×7
10
10 parts=1200
120 × 7
3 parts=120 × 3=360
1 part=
1200
=120
10
7 parts=120 × 7
€840
= €840
7 parts=1200-360
= € 840
Correct answer without work ⇒ 7 marks
1200
× 10 = 1714.28 ⇒ 7 marks
*
Special Case
7
1200
*
Stops at
or 120 ⇒ 4 marks
10
*
Stops at 1200 ×7[= 8400] ⇒ 4 marks
*
Incorrect answer without work ⇒ 0 marks, except for answers given in A2 below
Blunders (-3)
B1 Divisor ≠ 10 and continues but see 2nd *
B2 Incorrect multiplier i.e. ≠ 7 and continues but see 2nd *
3
B3 Gets
of 1200 only
10
B4 Decimal error (once only)
B5 Fails to finish
Slips (-1)
S1 Numerical errors where work is clearly shown to a max of 3
Attempts (3 marks)
10
1
3
A1 Indicates 0.7, or 7: 10, or
, or
, or
, or .3 only, and stops
10
10
10
A2 120 or 8400 or 360 or 1714.29 only, appears (no work shown)
1200
A3
or 1200 × 10
and stops
7
A4 1200 is multiplied or divided by any wrong number correctly
*
Worthless(0)
W1 1200 +7 = 1207 or similar
Page 7
Part (b)
25(10, 5, 10) marks
Att (3, 2, 3)
(i) By rounding each of these numbers to the nearest whole number,
9 ⋅15 × 2 ⋅196
5 ⋅ 5815
estimate the value of
9 ⋅15 × 2 ⋅196
5 ⋅ 5815
(ii) Using a calculator, or otherwise, find the exact value of
3
9
(iii) Using a calculator, or otherwise, write
and
as decimals.
8
25
Hence, or otherwise, put the following numbers in order, starting with the smallest
And finishing with the largest
3
9
,
, 0.37
8
25
Part (b)(i)
10 marks
Att 3

9 ⋅15 × 2 ⋅196
is approximately equal to:
5 ⋅ 5815
9
×
6
2
18
=
=
3
6
9
× 2 and stops ⇒ 7 marks.
6
*
No penalty if the intermediate step between approximations and correct final answer is not
18
shown i.e.
not shown
6
9 ⋅15 × 2 ⋅196
 18   3 
*
Special Case:
= 3.6 or    3  in this part ⇒ Attempt 3 marks.
5 ⋅ 5815
 5   5
18
*
and stops ⇒ 7 marks.(-3)
6
Blunders (-3)
B1 Error(s) in rounding off to the nearest whole number (once only if consistent)
B2 Decimal error in calculation of final value
B3 An arithmetic operation other than indicated
9 2
18
B4 Error(s) in the manipulation of the denominator e.g. × =
6 6
36
B5 Incorrect cancellation
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (3 marks)
A1 Only one approximation made to the given numbers and stops
A2 Ans. 3 with no preceding rounding off
Worthless (0)
*
Page 8
W1 Incorrect answer without work
(b)(ii)
9.15 × 2.196 = 20.0934 ÷ 5.5815 = 3.6 or
5 marks
18
or
5
or
9.15 × 2.196
5.5815
20.0934
⇒
=3.6
5.5815
Att 2
Blunders (-3)
B1 Decimal error or early rounding off
9.15
2.196
B2 Treats as
×
= 1.639344262 × 0.393492623=0.644987906
5.5815 5.5815
9.15 − 2.196 6,954
B3 Reads as
=
=1.245901639
5.5815
5.5815
9.15 + 2.195
B4 Reads as
=2.032786885
5.5815
5,5815
5.5815
B5 Treats as
=
=0.27777777…
9.15 × 2.196 20.0934
5.5815
5.5815
=
=0.491935483
B6 Treats as
9.15 + 2.196 11.346
5.5815
5.5815
B7 Treats as
=
=0.802631578
9.15 − 2.195 6.954
20.0934
B8 Leaves as
5,5815
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (2 marks)
A1 Any correct relevant calculation and stops.
9.15
2.196
=1.639344262 or
=0.393442623
e.g. 9.15 × 2.196=20.0934 or
5.5815
5.5815
A2 Any of the following; (see above)
0.644987906 , 1.245901639, 2.032786885
0.27777777…, 0.491935483 or 0.802631578 merit 2 marks (minimum 4 decimal
(with or without work)
places)
Worthless (0)
W1 Incorrect answer without work but see A2
Page 9
(b)(iii)
10marks
3
= 0.375
8
9
25
*
Accept:
*
Note:
Att 3
9
= 0.36
25
3
8
0.37
0.36, 0.37, 0.375 for 10 marks.
3
9
= 0,375 or
= 0.36 merits 4 marks
25
8
Blunders (-3)
B1 Fails to write a fraction as a decimal (each time)
B2 Writes fraction as incorrect decimal (each time)
B3 Decimal error (once only if consistent)
B4 Inverts fraction and continues. (each time)
B5 Incorrect order or fails to order
Attempts (3 marks)
37
A1
and stops
0.37 =
100
A2 Attempt at ordering using all 3 given numbers
A3
Any 2 of the given numbers in the correct order i.e.(
Worthless (0)
W1 Nothing correct
0.37
0.37
W2
or
or similar
8
25
Page 10
9
9 3
3
, 0.37 ), ( 0.37, ), (
, )
25
25 8
8
Part(c)
15(5, 5, 5)
(i) Using a calculator, or otherwise, divide 1120 by 0∙035.
Express your answer in the form a × 10 n , where 1 ≤ a < 10 and n ∈ N
Att(2, 2, 2)
a5 × a 2
, Give your answer in the form a n , where n ∈N
3
a×a
65 × 6 2
(iii) Using your answer to part (ii), or otherwise, find the value of
6 × 63
(ii) Simplify
(c)(i)

*
5 marks
Att2
1120
= 32000
0.035
= 3.2 ×
3.2 × ( without work) 4 marks
Blunders (-3)
B1 Decimal error
B2
B3
0.035
=0.00003125
1120
Multiplies fraction i.e. 1120 × 0.035=39.2 (3.92 × 10)
Inverts fraction
Slips (-1)
S1 Numerical errors to a max of 3
S2 Incorrect format, where a < 1 or a ≥ 10
S3 32000
and stops
S4 Error in index when forming Scientific Notation
Attempts (2 marks)
A1
A2
A3
Any relevant step. e.g. Partial long division e.g.
1
=
1000
0.035
35
1120000
35
=
200
7
Page 11
1120
=3
0.0355
(c)(ii)

5 marks
.
a × a 2 a.a.a.a.a.a.a
=
= a.a.a = a3
3
a×a
a.a.a.a
5
2
a ×a
a7
=
= a 7-4 = a 3
or
3
4
a×a
a
Att2
5
or
a5 × a 2
1
= a 4 × = a3
3
a×a
a
a × a × a (as answer) ⇒ 4 marks
a7
*
and stops ⇒ 2 marks
a4
a 7 and stops ⇒ 2 marks
*
a 3 × a and stops ⇒ 2marks
*
Blunders (-3)
*
B1 Correct answer, without work 
B2 Each error in calculation involving indices
B3 Each incorrect number of a’s in the extended form
B4 Each incorrect elimination of a’s in the extended form
Slips (-1)
a7
1
S1
= 3 or −3 as final answer
4
a
a
Attempts (2 marks)
A1
Some correct manipulation of indices
Worthless(0)
W1 Incorrect answer with no work shown
(c)(iii)
5 marks

5
or
or
2
6 ×6
= 63 =216
or
3
6× 6
7776 × 36
279936
=
= 216
6 × 216
1296
6.6.6.6.6.6.6
= 6.6.6 = 216
6.6.6.6
5
2
6 ×6
1
=64 × =63=216
3
6× 6
6
or
63=216
*Accept candidate’s answer from c(ii) unless it oversimplifies the question.
Blunders (-3)
B1 Correct answer, without work
B2 Each error in calculation involving indices
B3 Each incorrect number of 6’s in the extended form
B4 Each incorrect elimination of 6’s in the extended form
B5 Fails to finish
Attempts (2 marks)
A1 Some correct manipulation of indices
A2 Writes answer from c(ii) in c(iii)
Worthless(0)
W1 Incorrect answer with no work shown
Page 12
Att2
QUESTION 3
Part (a)
Part (b)
Part (c)
15 marks
20(10, 5, 5) marks
15(10, 5) marks
Att 5
Att(3, 2, 2)
Att(3, 2)
Part (a)
15 marks
Carol buys a magazine which costs €2·83.
In her purse she only has the following
Three 50 cent coins
Four 20 cent coins
Seven 10 cent coins
How much money will she have left after paying for the magazine?
Att 5
(a)
Att 5

15 marks
50 × 3 = 150
20 × 4 = 80
10 × 7=70
Total amount (150+80+70) = 300
⇒Change⇒ € 3.00 – €2.83
= €0.17
*
*
*
Accept 17c, (€0.17)
Final subtraction step subject to maximum deduction of 3 marks.
No penalty for the omission of € symbol.
Blunders (-3)
B1
B2
B3
B4
B5
B6
B7

Correct answer without work
Fails to find the change.
Fails to find total cost i.e. no addition
Operation other than subtraction when finding the change
Operation other than addition when finding total cost
Decimal error (Note: * above)
Each missing multiplication or addition
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (5 marks)
A1 Any attempt at addition / subtraction /multiplication of the given numbers
Worthless (0)
W1 Incorrect answer without work
Page 13
Part (b)
20(10, 5, 5) marks
Att(3, 2, 2)
(i) A bicycle costs €305. There is a 15% discount on the cost during a sale
What is the sale price of the bicycle?
(ii) David wishes to get some bars for a party
A packet of 12 bars cost €4.08 in Shop A.
A packet of 7 bars costs €2·17 in Shop B.
Find the unit cost (cost of one bar) in each shop
(iii) If David buys 84 bars, how much will he save by buying the bars in the shop offering the
better value?
(b)(i)

10 marks
305 ×
15
= €45.75
100
100%-15% = 85%
305 − 45.75 = €259.25
305 ×
Att 3
305 × .85 = €259.25
85
= €259.25
100
*
305 – 15% = 259.25  10 marks
*
305 × 15% = 45.75 and stops  7 marks
*
305 – 15% and stops  4 marks or
305 × 15% and stops  4 marks
*
€45.75 without work and stops merits 4 marks
Blunders (-3)
B1
B2

B4
B5
Correct answer without work
Decimal error
100 100
Inverts as
or
and continues (giving answers 358.82 or 2033.33 )
85
15
Mishandles 85 % or 15% e.g. 305 × 85 or 305 ÷ 85 or similar Note: {305 must be used}
305 taken as 85%or15%
B6
No subtraction (as per candidates work)
B3
B7 Addition of discount ( as per candidates work)
B8 305×1.15=€350.75
Slips (-1)
S1 Numerical errors to a max of 3
Misreadings (-1)
M1 Reads as €350 instead of €305
Attempts (3 marks)
15
85
305
A1
or
or
and stops
100
100
100
A2 100% = 305 and stops
85
A3 100×
and stops
305
305
A4
or similar and stops
85
Worthless (0)
W1 Incorrect answer without work
W2 305-15 =290 and stops or continues
Page 14
(b)(ii)

5 marks
Shop A.
unit cost 
408
= €0.34
12
Shop B
Unit cost 
2.17
= €0.31
7
*
Accept 34c and/or 31c
Blunders (-3)

B1 Correct answers without work.
B2 Operation other than division when finding the unit cost (once only)
B3 Finds only one unit cost
B4 Decimal error
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (2 marks)
A1 Any attempt at division
Worthless (0)
W1 Incorrect answer without work
W2 Addition or subtraction of the given numbers
(b)(iii)
5 marks
34−31 = 3
84×34 = 2856

84 × 3=252
= €2.52
84×31 = 2604
Savings = 2856 − 2604
= 252
= €2.52
*
Accept 2.52 or 252 or 252 c
Blunders (-3)
B1
B2
B3
B4
B5
Att 2

Correct answer without work
Operation other than subtraction or multiplication where appropriate
Finds only one unit saving
Decimal error
Fails to subtract
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (2 marks)
A1 Any attempt at subtraction or multiplication
A2 Any one correct step
Worthless (0)
W1 Incorrect answer without work
Page 15
Att 2
Part (c)
20(10, 10) marks
Att(3, 3)
(i) €12000 is invested at 2% per annum
What is the amount of the investment at the end of the first year?
(ii) Using central heating oil for 6 hours a day, a tank full of oil will last for 90 days
If the oil were used for only 5 hours a day, how much longer would it last?
(c)(i)
10 marks
Att 3
€12000 is invested at 2%per annum.
What is the amount of the investment at the end of the first year?
(c)(i)
10 marks

12000 × 1.02 = €12240 or
12000 ×
2
= €240
100
Att 3
12000 ×
12000+240 = €12240
*
*
*
*
*
*
P×R
100
12000
=
×2
100
= €240
12000+240 = €12240
102
= €12240
100
Finds interest only, €240 and stops ⇒ 7 marks
€240 without work and stops ⇒4 marks
€12000 + 2% = €12240 ⇒ 10 marks
€12000 × 2% = €240 and stops ⇒ 7 marks
€12000 + 2% and stops ⇒ 4 marks
€12000 × 2% and stops ⇒ 4 marks
Blunders (-3)
B1 Correct answer, without work 
B2 Mishandles 2% of 12,000. {Must use 12,000}
B3 Decimal error
B4 Fails to finish
Slips (-1)
S1
Numerical errors to a max of 3
Attempts (3 marks)
A1
Some use of 100 in attempt to find percentage e.g. 2% =
A2
Correct formula (with or without substitution) and stops
Worthless (0)
W1 Incorrect answer without work
W2 12000+2 (=12002) and stops
Page 16
2
and stops
100
I =
(c)(ii)
5 marks
Att2
Using central heating oil for 6 hours a day, a tank full of oil will last for 90 days
If the oil were used for only 5 hours a day, how much longer would it last?
(c)(ii)
5 marks

6 × 90 = 540
540
= 108
5
108-90 = 18
*
*
6 : 5= x : 90
6-5 = 1
5x = 6(90) = 540 x =108
1 × 90 = 90
108-90=18
90 ÷ 5 = 18
90
= 18 att. only ⇒2 marks
5
Answer 108 (work shown) ⇒ 4 marks
Answer
Att2
i.e.(no mention of 6)
Blunders (-3)
B2 Adds instead of subtracts
B3 Incorrect ratio
B4 Incorrect division or similar
B5 Fails to finish (method 3)
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (2 marks)
A1 Some indication of subtraction or ratio
A2 Some correct use of 6 or 5 but see W2
A3 Answer 108 or 18 (no work shown)
A4 6-5 [=1]
Worthless (0)
W1 Incorrect answer without work
90
W2
[=15] and stops
6
Page 17
QUESTION 4
Part (a)
Part (b)
Part (c)
10 marks
20(10, 10) marks
20(5, 5, 10) marks
Part (a)
10(5, 5)marks
Att(2, 2)
Att(3, 3)
Att(2, 2, 3)
Att(2, 2)
If a = 3 and b = 5, find the value of

(i)
(ii)
a +2b
ab − 6
(a)(i)

*
*
5 marks
a + 2b
=3+2(5)
=3+10
=13
3+10
 4 marks
5+2(3) = 5+6=11  4 marks
Blunders (-3)
B1 Correct answer, without work
B2 Leaves 2(5), in the answer
B3 Incorrect substitution and continues
B4 Breaks order i.e. 3+2(5) = (5)(5) = 25
B5 Treats 2(5) as 7 or 25
Slips (-1)
S1 Numerical errors to a max of 3
S2 Treats as a – 2b
Misreadings (-1)
M1 a and b interchanged see * above
Attempts (2 marks)
A1 Any number substituted for a or b and stops e.g. 2(8)
A2 Writes 5 or 3 in this part
A3 Any correct step
Worthless (0)
W1 Incorrect answer with no work
Page 18
Att2
(a)(ii)
5 marks

*
ab − 6
3(5) − 6
= 15 − 6
=9
15 − 6  4 marks
Blunders (-3)

B1 Correct answer without work
B2 Leaves 3(5) in the answer
B3 Incorrect substitution and continues
B4 Breaks order e.g. 3(5 − 6)
B5 Treats 3(5) as 8 or 35
Slips (-1)
S1 Numerical errors to a max of 3
S2 Treats as ab + 6
Misreadings (-1)
M1 a and b interchanged but no penalty here if already penalised in a(i)
Attempts (2 marks)
A1 Any substitution for either a or b and stops
A2 writes 5 or 3 in this part
A3 Any correct step
Worthless (0 marks)
W1 Incorrect answer, with no work
Page 19
Att2
Part (b)
(i)
(ii)
20(10, 10) marks
Write in it’s simplest form (3x + 2y) −2(x + 3y −4)
Solve 3x − 2≤7, x ∈ N
(b)(i)
10 marks

(i)
*
3x + 2y −2x − 6y + 8
Att 3
(3x + 2y) −2(x + 3y −4)
= 3x + 2y −2x − 6y + 8
= x − 4y + 8
Blunders (-3)
(stops or continues)
 7 marks (at least )

B1 Correct answer without work
B2 Error in distributive law and continues (each time)
B3 Fails to finish
Slip (-1)
S1 Numerical errors to a max of 3
Attempts (3 marks)
A1 Any correct step.
A2 Combines “x’s” to numbers and continues with any correct step
Worthless (0 marks)
W1 Combines “x’s” to numbers and stops
W2 Incorrect answer, with no work
Page 20
Att(3, 3)
(b)(ii)

10 marks
3x −2≤7
⇒ 3x≤7 + 2
⇒ 3x≤9
⇒ x≤3
{1, 2, 3} (not necessary)
-5
*
Att3
-4
-3
-2
-1
0
1
2
3
4
5
Do not penalize for inclusion of 0 in answer
Blunders (-3)
B1 Correct answer without work
B2 Error in transposition
B3 No plotting on number line
B4 Mishandles the direction of inequality e.g. 3 x ≥ 9
B5 Treats inequality as equality and continues
Slips (-1)
S1 Numerical errors to a max of 3
S2 ≤ taken as <
S3 Missing elements or incorrect elements on the number line (each time), but see S4
S4 Correct range but shaded in
Misreadings (-1)
M1 3 x + 2 ≤ 7 , and continues
M2 x∈Z or x∈R correctly mapped
Attempts (3 marks)
A1 Attempt at transposition and stops
A2 0 or 1 or 2 or 3 substituted for x
A3 Number line drawn with one of the correct elements only clearly indicated
A4 Combines “x's” to “numbers” e.g. x≤7 and continues
Worthless (0)
W1 Incorrect answer with no work e.g.{1,2,3,4,5,6,7,8,…..}.
Page 21
Part (c)
(i)
(ii)
20(5, 5, 10) marks
Eoin is t years of age
Katie is 4 years older than Eoin
Laura is twice as old as Eoin
Write Katie’s age and Laura’s age in terms of t
From part(i), the sum of Eoin’s age, Katie’s age and Laura’s age is 52
Write down an equation in t to represent this information
Solve your equation to find Eoin’s age in years
(iii) Solve for x and for y:
(c)(i)
Att(2, 2, 3)
Katie’s age = t + 4
7x + 2y = 11
4x + y = 7
5 marks
Att2
Laura’s age = 2t
Blunders (-3)
B1 Each incorrect expression
Misreading (-1)
M1 Substitutes x (or similar) for t
M2 Treats Laura’s age as twice Katie’s age i.e. 2(t + 4)
Attempts (2 marks)
A1 Any attempt at forming an expression but numbers written on their own are worthless
(c)(ii)
5marks
t + t + 4 + 2t = 52
4t + 4=52
4t = 52−4
4t = 48
t = 12
i.e.( Eoin’s age = 12 ) (not necessary)
*
Accept candidates’ expression from previous work.
Blunders (-3)
B1 Correct answer without work (t = 12 stated or substituted)
B2 Errors in transposition
B3 Stops at 4t = 48
B4 error in forming equation
B5 Fails to solve equation
Slip (-1)
S1 Numerical errors to a max of 3
S2 Leaves as 48 or similar
4
Attempts (2 marks)
A1 Answer from part c(i) written down and stops
A2 Any correct step
Worthless (0 marks)
W1 Incorrect answer, with no work
Page 22
Att2


(c)(iii)
Solve for x and y:
*
*
Att3
10 marks
Att3
7x + 2y = 11
4x + y = 7
(c)(iii)

I
7x+
7x +2y
2y ==1111
4x
4x++yy==77
7x +2y = 11
-8x
7x+
– 2y
2y==-14
11
-x
–8x –2y==-3
–14
x ==3–3
–x
4(3)+y
x = −=3 7= 3
12 + y−=1 7
y = 7-12
y y= =-5–5
10 marks
I
II
7x7x+
+ 2y2y==1111
4x4x
++
y =y =
77
28x+8y = 44
-28x-7y
28x +8y==-49
44
–28x – 7y = – 49
y =y–5
= -5
4x-5 = 7
4x =7+5
4x = 12
 x = x3 = 3
III
II
y =y 7–
= 7-4x
4x
7x+2(7-4x) = 11
7x + 14-8x = 11
7x + 2(7– 4x) = 11
7x +-x14=–11-14
8x = 11
–x = –3
x = − 3-x==3-3
−1
x=3
 y y==–5-5
Apply only one blunder deduction (B2 or B3) to any error(s) in establishing the first
equation; in terms of x only or the first equation in terms of y only.
Finding the second variable is subject to a maximum deduction of (−3).
Blunders (-3)
B1 Correct answers without work (stated or substituted)
B2 Error or errors in establishing the first equation in terms of x only (-x = -3) or the first
equation in terms of y only (y = –5) through elimination by cancellation (but see S1)
B3 Error or errors in establishing the first equation in terms of x only (-x = -3) or the first
equation in terms of y only (y = –5) through elimination by substitution (but see S1)
B4 Errors in transposition when finding the first variable
B5 Errors in transposition when finding the second variable
B6 Incorrect substitution when finding second variable
B7 Finds one variable only
Slip (-1)
S1 Numerical errors to a max of 3
Attempt (3 marks)
A1 Attempt at transposition and stops
A2 Multiplies either equation by some number and stops
A3 Incorrect value of x or y substituted correctly to find his correct variable
Worthless (0 marks)
W1 Incorrect values for x or y substituted into the equations
Page 23
QUESTION 5
Part (a)
Part (b)
Part (c)
10 marks
20(5, 5, 10) marks
20(5, 5, 10) marks
Att 3
Att(2, 2, 3)
Att(2, 2, 3)
Part (a)
10 marks
(a) Solve the equation 3(x −2) =2x + 5.
Att 3
(a)
Att 3

10 marks
3x − 6 = 2x + 5
3x −2x = 6 + 5
x =11
Blunders (-3)
B1 Correct answer without work 
B2 Error(s) in distribution (each time)
B3 Combining unlike terms (each time) and continues
B4 Fails to group like terms
B5 Error(s) in transposition (each time)
B6 Fails to finish
Slips (-1)
S1 Numerical errors to a max of 3
Misreadings (-1)
M1 3(x + 2) and continues.
Attempts (3 marks)
A1 Any correct multiplication.
A2 Any correct step
Worthless (0)
W1 combining unlike terms before attempting multiplication and stops e.g. 3(-2x) = 2x + 5
Page 24
Part (b)
20(5, 5, 10) marks
(i)
Factorise −25
(ii)
Factorise ab − 2ax + mb − 2mx
(iii)
Factorise + 4x − 12.
Hence solve the equation + 4x −12 = 0.
(b) (i)
5 marks
Att(2, 2, 3)
Att 2
2
x − 25
*
*
*
x 2 − 52
(x − 5)(x + 5)
Accept also (with or without brackets) for 5 marks any of the following
(x − 5) and (x + 5) [The word and is written down.]
(x − 5) or (x + 5) [The word or is written down.]
(x − 5) , (x + 5) [A comma is used]
Quadratic equation formula method is subject to slips and blunders.
x − 25 x + 25 merits 5 marks
(
)(
)
Blunders (-3)
B1 Incorrect two term linear factors of − 25 formed from correct (but inapplicable) factors of
x 2 and ± 25 .e.g ( x − 25)( x + 1)
B2 Incorrect factors of 25
B3 Incorrect factors of x 2
B4 (5 − x )(5 + x ) .
B5 ( x − 25)( x + 25) .
B6 Answer left as roots. (x = ±5 )
Slips (-1)
S1 ( x − 5) ± ( x + 5)
Attempts (2 marks)
A1 Correct factors of x 2 only
A2 Correct factors of ± 25 only
A3
± x or ± 5 appears.
A4 x 2 − 25 = x × x − 5 × 5
A5 Mention of the difference of two squares .e.g. x 2 − 25 2
A6 Correct quadratic equation formula quoted and stops
A7
25
Worthless (0 marks)
W1 Combines xs to “numbers” and continues or stops
Page 25
(b) (ii)

*
5 marks
ab −2ax + mb − 2mx
b(a+ m) −2x(a+m)
(a + m)(b −2x)
ab −2ax + mb − 2mx
a(b−2x)+ m(b−2x)
(a + m)(b −2x)
or
Accept also (with or without brackets) for 5 marks any of the following
(a+ m) and (b−2x) [The word and is written down.]
(a+ m) or (b−2x) [The word or is written down.]
(a+ m), (b−2x) [A comma is used]
Blunders (-3)
B1
B2
B3
B4
Att 2

Correct answer without work
Stops after first line of correct factorisation. e.g. a(b − 2x) + m(b − 2x)or equivalent.
Error(s) in factorising any pair of terms
Correct first line of factorisation but ends as (a+ m).-2bx or equivalent
Slips (-1)
S1 (a + m) ± (b−2x)
Attempts (2 marks)
A1 Pairing off, or indication of common factors and stops
A2 Correctly factorises any pair and stops
(b) (iii)
10 marks
Att 3
x 2 + 4 x − 12 = 0
x
x 2 + 6 x − 2 x − 12 = 0
x(x + 6 ) − 2(x + 6 ) = 0
(x + 6)(x − 2) = 0
⇒ x = −6 and x = 2
+6
x
−2

⇒ (x + 6 )(x − 2 ) = 0
⇒ x = −6 and x = 2
* 2 correct roots without work or by substitution
Page 26
− (4 ) ±
(4)2 − 4(1)(− 12)
2(1)
− 4 ±8
− 4 ± 16 + 48
=
2
2
4
− 12
= − 6 and
=2
2
2
⇒ x = −6 and x = 2
⇒ (x + 6 )(x − 2 ) = 0
4 MARKS
Blunders (-3)
B1
B2
B3
B4
B5
B6
B7
Factor Method

Correct answers without work
Incorrect two term linear factors of + 4x−12 formed from correct (but inapplicable) factors
of x 2 and/or ±12. e.g. (x + 12)(x − 1)
No roots given.
Incorrect factors of x 2 and/or ±12
Correct cross method but factors not shown and stops [Note: B3 applies also].
x(x+6)−2(x+6) or similar and stops [Note: B3 applies also].
Error(s) in transposition
Slips (-1)
S1 Numerical errors to a max of 3
S2 One root only
Attempts (3 marks)
A1 Some effort at factorisation
A2 States one correct root without work
Worthless (0 marks)
W1 x 2 + 4 x = 12 or similar and stops
W2 Incorrect Trial and error
W3 Oversimplification, resulting in a linear equation
Formula Method
Blunders (-3)
B1 Error in a,b,c substitution (apply once only)
B2 Sign error in substituted formula (apply once only)
B3 Error in square root or square root ignored
−4±8
B4 Stops at
2
B5 Incorrect quadratic formula and continues
B6 No factors from roots
Slips (-1)
S1 Numerical errors to a max of 3
p
S2 Roots left in the form
q
S3 One root only
Attempts (3 marks)
A1 Correct formula and stops
A2 One correct substitution and stops
Page 27
Part (c)
(i)
(ii)
20(5, 5, 10) marks
5x − 1
4x − 9
+
as a single fraction.
2
3
Give your answer in its simplest form.
Express
Verify your answer to part (i) by substituting x = 3 into
and into your answer to part (i).
(iii)
(c)(i)

(i)
Att(2, 2, 3)
5x − 1 4 x − 9
+
2
3
Multiply ( x − 2) by ( −3x +11)
Write your answer in its simplest form
5 marks
3(5 x − 1) + 2(4 x − 9)
6
23 x − 21
15 x − 3 + 8 x − 18
⇒
=
6
6
Att2
5x − 1
4x − 9
9 x − 10
Zero marks
+
=
2
3
5
Blunders (-3)
*

B1
B2
B3
B4
Correct answer without work
Error(s) in distribution e.g. 3(5 x − 1) =15 x −1
Mathematical error e.g. −3 −18= +21
Incorrect common denominator and continues
B5
Incorrect numerator from candidate's denominator e.g.
B6
B7
No simplification of numerator
Omitting denominator
Slips (-1)
S1
S2
Drops denominator
Numerical errors to a max of 3
S3
Answer not in simplest form. e.g.
46 x − 42
12
Attempts (2 marks)
A1
6 only or a multiple of 6 only appears
A2
Any correct step
Worthless (0)
 5 x − 1  4 x − 9 
W1


 and stops
 2  3 
Page 28
2(5 x − 1) + 3(4 x − 9 )
6
(c) (ii)
5 marks
Att2
5(3) − 1 4(3) − 9
+
2
3
15 − 1 12 − 9
=
+
2
3
14 3
=
+
2 3
= 7 +1
=8
23(3) − 21
6
69 − 21
=
and
6
48
=
6
=8
*
Accept candidates answer from previous section [May result in inequality].
*
Accept usage of a value other than 3 for verification.
Blunders (-3)

B1
B2
B3

Correct answer without work
Substitutes into one expression only
Manipulation to force equality
Slips (-1)
S1 Numerical errors to a max of 3
S2 Conclusion missing if unequal
Attempts (2 marks)
A1 Writes answer from previous part in this section
A2 Substitutes a value into one expression and stops
(c) (iii)

10 marks
( x − 2) ( −3x +11)
x( −3x +11) −2( −3x +11)
− +11x − +6x −22
− +17x −22
*
If
− +11x − +6x −22
is correct (minimum 7 MARKS)
Blunders (-3)
B1 Errors in distribution each time
B2 Errors in multiplication of powers
B3 Errors in grouping of terms (apply once)
B4 Mathematical errors eg -2.−3x = -6x

B5 Correct answer without work
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (3 marks)
A1 One correct multiplication
Page 29
Att 3
QUESTION 6
Part(a)
Part (b)
Part (c)
Part(a)
10(5, 5) marks
30(15, 15) marks
10(5, 5) marks
(5, 5)marks
P = {(1, 5), (2, 8), (2,9), (3, 10)}
Write out the domain and range of P
(a)(i)
5marks
Att(2, 2)
Att(5, 5)
Att(2, 2)
att(2,2)
att2
Domain = {1, 2, 3}
*
Accept {1, 2, 2, 3} for full marks.
Slips (-1)
S1 Each incorrect element omitted / included other than the misreading below
Misreadings (-1)
M1 Correct range {5, 8, 9, 10} given
Attempts (2 marks)
A1 Any one correct element of the Domain
Worthless (0)
W1 No element of the domain appears, but note M1.
(a)(ii)
5marks
Range = {5, 8, 9, 10}
Slips (-1)
S1 Each incorrect element omitted / included other then the misreading below
Misreadings (-1)
M1 Correct domain {1, 2, 3} or {1, 2, 2, 3} given
Attempts (2 marks)
A1 Any one correct element of the Range
A2 Range 5  10
Worthless (0)
W1 No element of the range appears but note M1
Page 30
att2
Part(b)
30(15, 15)marks
att (5, 5)
Draw the graph of the function
f : x → 3 + 2x − x2
in the domain − 1 ≤ x ≤ 3 , where x ∈ R .
Table
A

15 marks
f (−1)
=
3
+ 2(−1)
-(-1)2 =
0
f (0)
=
3
+ 2(0)
- (0) 2
=
3
f (1)
=
3
+ 2(1)
- (1) 2
=
4
f (2)
=
3
+ 2(2)
- (2) 2
=
3
f (3)
=
3
+ 2(3)
- (3) 2
= 0
Att 5
B
x
3
+ 2x
−x
2
f (x)
-1
3
0
3
1
3
2
3
3
3
-2
-1
0
0
+2
-1
+4
-4
+6
-9
0
3
4
3
0
*
Error(s) in each row/column calculation attracts a maximum deduction of 3marks
Blunders (-3)
B1
B2
B3
B4
B5
B6
B7
B8
B9

Correct answer, without work i.e. 5 correct couples only and no graph
- taken as + all the way but see M1
“+2 x ” taken as “2” all the way. [In the row headed “+2 x ” by candidate]
“3” calculated as “3 x ” all the way. [In the row headed “3” by candidate]
Adds in top row when evaluating f (x) in B
Omits “3” row
Omits “+2 x ” row
Omits a value in the domain (each time).
Each incorrect image without work i.e. calculation through the function method (A)
Slips (-1)
S1 Numerical errors to a max of 3 in any row / column
Misreadings (-1)
M1 Misreads “ − x 2 ” as “ + x 2 ” and places “ + x 2 ” in the table or function
M2 Misreads “ + 2 x ” as “ − 2 x ” and places “ − 2 x ” in the table or function
M3 Misreads “3” as “-3” and places “-3” in the table or function
Attempts (5 marks)
A1 Omits “ − x 2 ” row from table or treats “ − x 2 ” as ± x or ± 2 x
A2 Any effort at calculating point(s)
A3 Only one point calculated and stops
Page 31
Graph

15 marks
Att 5
(i)
4
3
2
(ii)
1
-1
*
*
*
1
2
3
accept candidates values from previous work ( 5 co-ordinates needed ) but see S2
Only one correct point graphed correctly ⇒ Att 5 + Att 5
Correct graph but no table ⇒ full marks i.e. (15 + 15) marks.
Blunders ( -3)
B1 Reversed co-ordinates plotted against non-reversed axes (once only) {See S3 below}.
B2 Scale error (once only)
B3 Points not joined or joined in incorrect order (once only).
Slips (-1)
S1 Each point of candidate graphed incorrectly. {Tolerance ± 0.25 }
S2 Each point { 5 points needed } from table not graphed [ See * above ]
S3 reversed co-ordinates if
(i) axes not labelled or (ii) axes are reversed to compensate (see B1 above)
Attempts (5 marks)
A1 Graduated axes (need not be labelled)
A2 Some effort to plot a point { See * above}
Page 32
Part (c)
(c)
10(5, 5) marks

(i)
Att(2, 2)
Draw the axis of symmetry of the graph you have drawn in part (b) above.
Work to be shown on the graph.
(ii) Use the graph you have drawn in part (b) to estimate the value of
3 + 2 x − x 2 when x = 2.5
(c) (i)
5 marks
Att 2
(i)
4
3
2
1
-1
(ii)
1
2
3
*
Accept any vertical line (parallel to candidate’s y-axis) within tolerance of ± 0.25 .
Blunders ( -3)
B1 Any vertical line (parallel to the candidate’s y-axis) outside of the tolerance.
B2 Marks x = 1 on the x-axis and stops.
B3 States x = 1 but no line is indicated on the graph
Attempts ( 2 marks)
A1 Any attempt at axial symmetry of f (x) .
A2 y-axis indicated as the axis of symmetry (See B1).
(c) (ii)
5 marks
Att 2
Work to be shown on the graph and answer to be written here.
1.8
*
Correct answer (clearly consistent with student’s graph) inside the tolerance without
graphical indication ⇒ 2 marks.
Blunders (-3)
B1 Correct answer without work 
B2 Answer on the diagram but outside of tolerance ( ± 0.25 )
B3 Fails to write down the answer, when indicated correctly on graph
Slips (-1)
S1 Answer not written in box when written on graph
Attempts (2 marks)
A1 Attempt at algebraic evaluation or calculator.
A2 Marks 2.5 in any way on either axis and stops.
Worthless (0)
W1 Answer outside of tolerance without graphical indication.
Page 33
JUNIOR CERTIFICATE
EXAMINATION
2009
MARKING SCHEME
MATHEMATICS
ORDINARY LEVEL
PAPER 1
Page 1
GENERAL GUIDELINES FOR EXAMINERS
1.
Penalties of three types are applied to candidates’ work as follows:
 Blunders - mathematical errors/omissions
(-3)
 Slips- numerical errors
(-1)
 Misreadings (provided task is not oversimplified)
(-1).
Frequently occurring errors to which these penalties must be applied are listed in the
scheme. They are labelled: B1, B2, B3,…, S1, S2,…, M1, M2,…etc. These lists are not
exhaustive.
2.
When awarding attempt marks, e.g. Att(3), note that
 any correct, relevant step in a part of a question merits at least the attempt mark for that
part
 if deductions result in a mark which is lower than the attempt mark, then the attempt
mark must be awarded
 a mark between zero and the attempt mark is never awarded.
3.
Worthless work is awarded zero marks. Some examples of such work are listed in the
scheme and they are labelled as W1, W2,…etc.
4.
The phrase “hit or miss” means that partial marks are not awarded – the candidate receives
all of the relevant marks or none.
5.
The phrase “and stops” means that no more work is shown by the candidate.
6.
Special notes relating to the marking of a particular part of a question are indicated by an
asterisk. These notes immediately follow the box containing the relevant solution.
7.
The sample solutions for each question are not intended to be exhaustive lists – there may
be other correct solutions.
8.
Unless otherwise indicated in the scheme, accept the best of two or more attempts – even
when attempts have been cancelled.
9.
The same error in the same section of a question is penalised once only.
10.
Particular cases, verifications and answers derived from diagrams (unless requested)
qualify for attempt marks at most.
11.
A serious blunder, omission or misreading results in the attempt mark at most.
12.
Do not penalise the use of a comma for a decimal point, e.g. €5.50 may be written as €5,50.
Page 2
QUESTION 1
Part (a)
Part (b)
Part (c)
10 marks
20(5,5,5,5) marks
20(10,5,5) marks
Part (a)
P = {w, x, y, z}
Att 3
Att 2,2,2,2
Att 3,2,2
10 marks
Att 3
Q = {v, w, x}
Fill the elements of P and Q into the following diagram.
P
(a)
Q
10 marks
Att 3
Q
P
y
z
w x
v
*
• Not necessary
Slips (-1)
S1 Each element incorrectly filled into the diagram
S2 Each element omitted from the diagram but see W1
S3 Each unlisted element used
Misreadings (-1)
M1 Interchanging P and Q totally
Attempts (3 marks)
A1 Totally incorrect filling of the Venn diagram using given elements
Worthless
W1 No filling in of the Venn diagram or use of unlisted elements only
Page 3
(b)
(b)
20(5,5,5,5) marks
Att 2,2,2,2
U
U is the universal set.
A = {1, 5, 6, 9, 10}
B = {1, 3, 5, 8}
C = {4, 5, 8, 10}
A
9
6
1
5
10
B
3
●8
2
4
C
7
(i) List the elements of B  C .
(ii) List the elements of A' , the complement of the set A.
(iii) List the elements of ( B  C ) \ A .
(iv) Write down #B.
(b)(i)
5 marks
B  C = {1, 3, 4, 5, 8, 10}
Blunders (-3)
B1 Any incorrect set of the elements of B and C other than the misreading as below
Misreadings (-1)
M1 B∩C giving {5, 8}
Attempts (2 marks)
A1 2, 6, 9 or 7 appear in the answer
Page 4
Att 2
(b) (ii)
5 marks
A' = {2, 3, 4, 7, 8}
Att 2
Blunders (-3)
B1 Any incorrect set of elements of A' other than the misreadings below.
Misreadings (-1)
M1 A\B giving {6,9,10}.A\C giving {6,9,1} or A\( B  C ) giving{6,9}.
Attempts (2 marks)
A1 2, 4, 7, 8 or 3 appear in the answer.
A2 A or any proper subset of A
.
(b) (iii)
5 marks
(B  C) \ A = 8
Att 2
Blunders (-3)
B1 Any incorrect set of elements of A and B and C other than the misreading as below.
Misreadings (-1)
M1 ( B  C )/A giving {3, 4, 8,}, A\(B∩C) giving {1,6,9,10}
Attempts (2 marks)
A1 2 or 7 appear in the answer.
(b) (iv)
5 marks
#B.= 4
Blunders (-3)
B1 Any incorrect cardinal number of B ≤ 10 other than the misreading as below.
Misreadings (-1)
M1 Set B giving {1, 3, 5, 8}.
M2 #B = 6 i.e, # B'
Attempts (2 marks)
A1 Some understanding of notation e.g. Cardinal numbers or number of elements
A2 #B = 17 or 120
Worthless
W1 Any number greater than 10, but see A2
Page 5
Att 2
Part(c)
20(10,5,5) marks
Att3,2,2
1(c)
In a survey, a group of students were asked if they were studying French or German
at school.
80 of these students said they were studying French (F).
24 of these students said they were studying German (G).
15 of these students said they were studying both French and German.
11 of these students said they were studying neither of the two languages.
(i) Represent this information in the Venn diagram below.
(ii) How many students were in the group?
(iii) How many students did not study German?
(c)(i)
(c)(i)
10 marks
Att 3
F
G
65
15
9
11
*
.Failing to subtract 15 from 80 and/or 24 is one blunder only(-3)
Blunders (-3)
B1 Each incorrect or omitted entry but see S1 and M1 below and * above
Slips (-1)
S1 Numerical errors, where work is clearly shown to a max of 3
Misreadings (-1)
M1 Interchanges French and German
Attempts (3 marks)
A1 Any one correct relevant entry
Page 6
(c)(ii)
c(ii)
*
*
5 marks
65 + 15 + 9 + 11 = 100
Att 2
Any correct answer written here in the space provided takes precedence over an incorrect
Venn diagram (Subject to S1)
Accept candidate’s work from previous part c(i)
Blunders (-3)
B1 Any incorrect use of the given numbers or the numbers from an incorrect Venn diagram
(Subject to S1)
B2 Number of students = 11+15+24+80 = 130
Slips (-1)
S1 Numerical errors where work is clearly shown, to a max of 3
S2 Fails to finish
Attempts (2 marks)
A1 Any one correct relevant sum where work is clearly shown
Worthless
W1 Incorrect answer with no work shown
(c)(iii)
5 marks
c(iii)
65 + 11 = 76 or 100-24
*
*
Att 2
Any correct answer written here in the space provided takes precedence over an incorrect
Venn diagram (Subject to M1)
Accept candidate’s work from previous part c(i)
Blunders (-3)
B1 Any incorrect use of the given numbers or the numbers from an incorrect Venn diagram
(Subject to S1)
Slips (-1)
S1 Numerical errors, where work is clearly shown to a max of 3
S2 Fails to finish.
Misreadings (-1)
M1 German read as French (Ans. = 20).
Attempts (2 marks)
A1 Mention of 65 or 11 or candidate’s work from c(i)
Worthless
W1 Incorrect answer with no work shown
Page 7
QUESTION 2
Part (a)
Part (b)
Part (c)
10 marks
20(5,10,5) marks
20(10,5,5) marks
Att 3
Att 2,3,2
Att 3,2,2
Part (a)
10 marks
(a) 9 metres of cloth cost €13·95. Find the cost of 20 metres of the same cloth.
(a)
Method (1)
Method (2)
9m = 13.95
9:20
9:20 = 13.95:x
13.95
 1.55
9
20m=1.55  20=31
13.95
 1.55
9
1.55  20=31
9
13.95
=
20
x
9x = 13.95  20=279
279
x=
= 31
9
1m =
*
*
*
*
10 marks
Method (3)
Att 3
Method (4)
13.95
 20
9
1.55  20
31
Correct answer without work  7 marks
9
Special Case
×13.95 = 6.2775  7 marks
20
13.95
Stops at 1.55 or
[=1.55] 4 marks (no use of 20(-3) and B4 or B5
9
Stops at 13.95 ×20[= 279]  4 marks (no use of 9 and possible slips)
 31 
Incorrect answer without work  0 marks except 279,155 or equivalent  
 20 
Blunders (-3)
B1 Divisor ≠ 9 and continues but see 2nd *
B2 Incorrect multiplier i.e. ≠ 20 and continues but see 2nd *
B3 20 : 9 =13.95 : x and continues
B4 Error in decimal point (once only)
B5 Fails to finish
Slips (-1)
S1 Numerical errors where work is clearly shown to a max of 3
Attempts (3 marks)
20
A1 Indicates
or 9 : 20 or 13.95 : x ,only, and stops
9
 21 
A2 279 or 1.55 or   , only, appears
 20 
1
A3
only appears
9
A4 13.95  9 or 13.95  20 and stops or continues
A5 13.95 is multiplied or divided by any wrong number correctly
Worthless
W1 13.95 +9 = 22.95 or similar
*
Page 8
Att 3
Part (b)
(i)
(ii)
20(5,10,5) marks
Att 2,3,2
a9  a3
6
2
Simplify a  a , giving your answer in the form a n , where n  N.
By rounding each of these numbers to the nearest whole number, estimate the value of
18  207
3  7  2  08 .
(iii) Using a calculator, or otherwise, find the exact value of
(b)(i)
5 marks
a a
a
=
 a4
8
6
2
a a
a
9
(i)
or
*
*
*
*
*
18  207
.
3  7  2  08
3
12
Att 2
a a
= a3  a = a 4
6
2
a a
9
or
3
a9  a3
a a a a a a a a a a a a
=
= a4
6
2
a a a a a a a a
a a
a12
and stops  2 marks
a8
a 12 and stops  2 marks
Correct answer without work  2 marks
a 3 ×a and stops  2marks
a × a ×a × a as answer  2 marks
Blunders (-3)
B1 Correct answer, without work
B2 Each error in calculation involving indices
B3 Each incorrect number of a’s in the extended form
B4 Each incorrect elimination of a’s in the extended form
Slips (-1)
1
a 12
S1
= 4 or  4 as final answer
8
a
a
Attempts (2 marks)
A1
Some correct manipulation of indices
Worthless
W1 Incorrect answer with no work shown
Page 9
(b)(ii)
10 marks
18
18
4
+
Att 3
2
=
6
=
3
18
and stops  4 marks.
42
*
No penalty if the intermediate step between approximations and correct final answer is not
18
shown i.e.
not shown
6
18.207
 63 
*
Special Case:
= 3.15 or  
presented in this part  Attempt 3 marks.
3.7  2.08
 20 
18
and stops  7 marks.
*
6
Blunders (-3)
B1 Error(s) in rounding off to the nearest whole number (once only)
B2 Decimal error in calculation of final value
B3 An arithmetic operation other than indicated
B4 Error(s) in the manipulation of the denominator
B5 Incorrect cancellation
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (3 marks)
A1 Only one or two approximations made to the given numbers and stops.
A2 Ans. 3 with no preceding rounding off
Worthless (0)
W1 Incorrect answer without work
*
(b)(iii)
5 marks
Att 2
18.207
 63 
 3.15 or  
5.78
 20 
*
Any of the following; 7.00081081
13.6741762
2.365774428
11.23888889
merit 2 marks (with or without work)
10.23528649
or 12.45336538
Blunders (-3)
B1 Decimal error
B2 Fails to finish
Slips (-1)
S1 Numerical errors to a max of 3
S2 Any rounding off.
Attempts (2 marks)
A1 Any correct relevant calculation and stops.
18.207
e.g.
.  4.9208 or similar
3.7
Worthless (0)
W1 Incorrect answer without work but see *
Page 10
Part (c)
2(c)
20(10,5,5) marks
(i)
(ii)
(iii)
*
Note:
Att 3,2,2
1
13
and
as decimals.
8
80
Hence or otherwise, put the following numbers in order,
starting with the smallest and finishing with the largest:
1 13
,
, 0·1525.
8 80
Using a calculator, or otherwise, write
1
2
Using a calculator, or otherwise, find the exact value of (3  61) .
Using a calculator, or otherwise, evaluate
1
94  09  (2  75) 2 −
.
0  3125
Give your answer correct to two decimal places.
1
13
 0 125 . or
 0.1625 . merits 4 marks.
8
80
(c)(i)
10marks
1
= 0.125
8
1
8
*
Accept:
*
Note:
Att 3
13
= 0.1625
80
0.1525
0.125, 0.1525, 0.1625, merits 10 marks.
1
13
 0  125
or
 0  1625
merits 4 marks
8
80
Blunders (-3)
B1 Fails to write a fraction as a decimal (each time)
B2 Writes fraction as incorrect decimal (each time)
B3 Decimal error (once only if consistent)
B4 Inverts fraction and continues. (each time)
B5 Incorrect order or fails to order.
Attempts (3 marks)
1525
A1
and stops
0.1525 
10000
A2 Attempt at ordering
Worthless(0)
W1 Nothing correct
Page 11
13
80
(c)(ii)
5 marks
 19 
1.9 or  
 10 
Att 2
Blunders (-3)
B1 Squares
B2 Decimal error
Attempts (2 marks)
A1 mentions square root or power
Worthless(0)
W1 Dividing by 2 or multiplying by 2
(c)(iii)
*
*
*
*
5 marks
9.7 × 7.5625 –3.2 = 70.15625 = 70.16
answer 70.15625  2 marks
answer 70.15625 = 70.16  5 marks
2245
[
] as final answer  0 marks but = 70.15625  4 marks
32
Ans 70.15 (no work shown)  2 marks
Att 2
Blunders (-3)
B1
B2
B3
B4
Correct answer, without work
Decimal error
Inverts fraction
Incorrect operator

Slips (-1)
S1 Numerical errors to a max of 3
S2 Fails to give answer to 2 dec. places
S3 Each premature rounding off, that effects final answer,( to a maximum of 3marks)
Attempts (2 marks)
A1 Any relevant step. e.g. Partial long division or similar
Page 12
QUESTION 3
Part (a)
Part (b)
Part (c)
10 marks
20(10,10) marks
20(10,10) marks
Att 3
Att 3,3
Att 3,3
Part (a)
10 marks
(a) Aideen owns 6000 shares in a certain company.
She sells two-thirds of her shares.
How many shares does she now own in the company?
Att 3
(a)
Att3
10 marks

6000 ÷ 3 = 2000
or
Number of shares sold:
Shares now owned:
6000 x ⅔ = 4000
6000 - 4000 = 2000
Blunders (-3)
B1 Correct answer without work
B2 6000  ⅔
B3 Calculates the number of shares sold and stops
B4 Operation other than subtraction in final step
Slips (-1)
S1 Numerical errors (to max -3)
S2 Early rounding off
Attempts (3 marks)
A1 Any attempt at getting ⅔ of 6000
1
6000
A2 Writes down or
 3000
3
2
Page 13
Part (b)
(i)
(ii)
20(10,10) marks
Att 3,3
Brian’s gross annual pay is €26 000. His annual tax credit is €2800. He pays income
tax at the rate of 20%. What is his annual take-home pay?
A dealer buys a car for €17 500. He sells the car for €23 800.
Calculate his profit as a percentage of the cost price.
(b)(i)
10 marks
Att 3
(i) Brian’s gross annual pay is €26 000. His annual tax credit is €2800. He pays income tax at
the rate of 20%. What is his annual take-home pay?
(b)(i)
10 marks
€26 000
Gross Pay

*
Att 3
Tax @ 20%
5200
Tax Credit
€2800
Tax Due
2400
Take-home Pay
23600
26000  20
 5200 5200 – 2800 = 2400
100
Finds Tax Due 2400 and stops  7 marks
26000 – 2400 = 23600
(at least 2 out 3 boxes filled in)
Blunders (-3)
B1 Correct answer, without work.
B2 Mishandles 20% of 26,000. {Must use 26,000}
B3 Decimal error
B4 Misuse of Tax Credit
B5 Incorrect use of Tax Amount e.g. 26000 + 5200
B6 Fails to finish. {B4 may apply}
Slips (-1)
S1
Numerical errors to a max of 3
Attempts (3 marks)
A1 Some use of 100 in attempt to find percentage e.g. 20% = 20/100 and stops.
Worthless (0)
W1 Incorrect answer without work
Page 14
(b) (ii)
10 marks
Att 3
(b) (ii)
A dealer buys a car for €17 500. He sells the car for €23 800.
Calculate his profit as a percentage of the cost price.
(b)(ii)

10 marks
23800 – 17500 = 6300
Att 3
6300
 100  36%
17500
23800
 100  136  136 – 100 = 36%
17500
*
Answer 6300  4 marks
6300
*
 17500  1102500  7 marks
100
Blunders (-3)
B1 Correct answer without work
B2 Adds €17 500 to €23 800.
B3 Calculates profit as percentage of selling price.
B4 Divisor not equal to 17500
or
Method 2:
B5
Mishandles the calculation of profit as a percentage e.g.
B6
B7
B8
Incorrect cancellation(s)
Fails to multiply by 100
Fails to finish
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (3 marks)
A1 Some indication of subtraction
A2 Some use of 100
Page 15
6300
 17500
100
Part (c)
20(10,10) marks
3(c) (i) €20 000 is invested at 5  2% per annum.
What is the amount of the investment at the end of one year?
Att 3,3
(ii) €5000 is withdrawn from this amount at the beginning of the second year.
The interest rate for the second year is 6  25% per annum.
What is the amount of the investment at the end of that year?
(c)(i)

10 marks
20000  5.2
 1040 20000 + 1040 = €21040
100
or
or
20000
1%=
100
20000
5.2% =
 5.2
100
Interest = 1040
Amount = 20000  1040
Amount = 21040
*
*
*
*
Att3
or
PR
I 
100
20000
I =
 5.2
100
Interest = 1040
Amount = 20000  1040
Amount = €21040
20000 × 1.025 = 21040
or
Amount = 20000 1  052
Amount = €21040
€ 1040 (without work) and stops  4 marks.
Writes down 20000 +5.2% = 21040  10 marks
Writes down 20000 ×5.2% = 1040 and stops  7 marks.
Writes down 20000 ×5.2% and stops, or 20000 +5.2% and stops  4 marks.
Blunders (-3)
B1
B2
B3
B4
B5
B6
Correct answer without work 
20000
Mishandles 5.2 %. e.g.
 100 Note: {20000 must be used}.
5.2
Decimal error (once only)
Stops at interest i.e. fails to calculate amount.
Subtracts to calculate amount.
1  052 treated as 1 52 .
Slips (-1)
S1 Numerical errors to a max of 3
Misreadings (-1)
M1 Reads as €2000
Page 16
Attempts (3 marks)
A1 Correct formula with or without substitution and stops
A2
Some use of 100 in attempt to find percentage e.g. 5.2 % 
5.2
or 1 052 and stops.
100
Worthless (0)
W1 Incorrect answer without work
W2 20000 + 5.2 = 20005.2 and stops or continues.
(c)(ii)
*
*
*
*
*
*
10 marks
Att3
16040  6.25
21040 – 5000 = 16040
 1002.5

100
16040 + 1002.5 = €17042.5 [ or 16040 × 1.0625 = 17042.5 ]
Accept candidates answer from (i)
€ 16040 (without work) and stops  4 marks.
10 marks
Writes down 16040 + 6.25% = 17042.5 
Writes down 16040 ×6.25% = 1002.5 and stops  7 marks.
Writes down 16040 ×6.25% and stops, or 16040 + 6.25% and stops  4 marks.
Uses 5000 (-3)(-3). Uses 20000 (-3)
Blunders (-3)
B1 Correct answer without work 
B2 Fails to subtract 5000
B3 Mishandles 6.25%
B4 Decimal error (once only).
B5 Stops at interest i.e. fails to calculate amount.
B6 Subtracts to calculate amount.
B7 Incorrect Principal
Slips (-1)
S1 Numerical errors to a max of 3
Misreadings (-1)
M1 Reads as €500 or similar.
Attempts (3 marks)
A1 Correct formula with or without substitution and stops
A2 Some use of 100 in attempt to find percentage and stops.
A3 21040-5000 = 16040 and stops
Worthless (0)
W1 Incorrect answer without work
W2 21040 + 6.25 and stops or continues
Page 17
QUESTION 4
Part (a)
Part (b)
Part (c)
10 marks
20(10,10) marks
20(5,5,10) marks
Part (a)
(a) If a = 5, find the value of


(a)(i)
(i)
*
20 + 1 => 4 marks
10 (5,5)marks
Att 2,2
Att 3,3
Att 2,2,3
Att 2,2
(i) 4a + 1
(ii)
a2 – 3a + 6
5 marks
4(5) + 1 = 21
Blunders (-3)
B1 Correct answer, without work
B2 Leaves 4(5), in the answer
B3 Incorrect substitution and continues
B4 Breaks order i.e. 4(5 + 1) = 4.6 = 24
B5 Treats 4(5) as 9 or 45
Slips (-1)
S1 Numerical errors to a max of 3
S2 Treats as 4a - 1
Attempts (2 marks)
A1 Any number substituted for a and stops e.g. 4(8).
A2 Writes 5 in this part
A3 Any correct step.
Worthless (0)
W1 Incorrect answer with no work.
Page 18
Att2
(a)(ii)
(5)2 – 3(5) + 6 = 16
(ii)
*
31 - 15 or 10 + 6 => 4 marks
5 marks
or
25-15 + 6 = 10 + 6 = 16
Blunders (-3)
B1 Correct answer without work
B2 Leaves 52 or -3(5) in the answer
B3 Incorrect substitution and continues.
B4 Breaks order e.g. -3(5+6).
B5 Treats -3(5) as 2 or -35.
B6 Fails to finish but see * above
Slips (-1)
S1 Numerical errors to a max of 3
S2 Treats as a2 – 3a – 6
Attempts (2 marks)
A1 Any substitution for either a2 or -3a and stops e.g. (8) etc.
A2 writes 5 in this part.
A2 Any correct step.
Worthless (0 marks)
W1 Incorrect answer, with no work.
Page 19
Att2
Part (b)
4(b) (i)
Solve the equation
20(10,10) marks
5 x  10  3( x  2) .
Att 3,3
Multiply ( x  3) by (2 x  1) .
Write your answer in its simplest form.
(ii)
(b)(i)
10 marks
5x – 10 = 3x + 6 => 5x – 3x = 6 + 10 => 2x = 16 => x = 8
(i)
Blunders (-3)
B1 Correct answer without work (x = 8 stated or substituted).
B2 Error in distributive law and continues, e.g. 5x – 10 = 3x + 2.
B3 Errors in transposition (each time)
B4 Stops at 2x = 16 or similar.
Att 3
Slips (-1)
S1 Numerical errors to a max of 3
S2 Leaves as 16 or similar.
2
Attempts (3 marks)
A1 Any substitution for values of x other than x = 8.
A2 Any correct step.
A3 Combines “x’s” to numbers and continues with any correct step e.g. 5x – 10 = -5x.
Worthless (0 marks)
W1 Combines “x’s” to numbers and stops.
W2 Incorrect answer, with no work
Page 20
(b)(ii)
(ii) 2x(x – 3) + 1(x – 3)
=> 2x2 – 6x + x – 3
=> 2x2 – 5x – 3
*
10 marks
or
Att3
x(2x + 1) – 3(2x + 1)
=> 2x2 + x – 6x – 3
=> 2x2 – 5x – 3
2x2 + x – 6x –3 => 7 marks
Blunders (-3)
B1 Correct answer without work
B2 Error(s) in distribution.(each time)
B3 Fails to group or groups incorrectly
Slip (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 Any correct multiplication e.g. 2x2 etc.
A2 Any correct grouping of terms.
A3 Any correct step.
A4 Substitutes a value of “x” and continues correctly.
A5 Treats as (x – 3)±(2x + 1) to give 3x – 2 or –x – 4
A6 Combines “x’s” to numbers and continues with correct step e.g. x – 3= – 3x or 2x + 1 = 3.x
Worthless (0 marks)
W1 Combines “x’s” to numbers and stops.
W2 No distribution but A2 or A5 may apply to subsequent work e.g. gathering of
terms.
Page 21
Part (c)
20(5,5,10) marks
Att 2,2,3
The cost of a cinema ticket is € t for an adult and €5 for a child.
The cost of tickets for 2 adults and 3 children is €33.
(i)
Write down an equation in t to represent this information.
(ii)
Solve the equation you formed in part (i) above, for t.
(iii) Solve for x and for y:
5x – 4y = 16
2x + 3y = 11
2t + 3(5) = 33
or
(c)(i)
5 marks
Att2
2t + 15 = 33
Blunders (-3)
B1 Each incorrect term in equation
Misreading (-1)
M1 Substitutes x (or similar) for t
Attempt (2 marks)
A1 Any attempt at forming an equation but numbers written on their own (except 15 or 33) are
worthless
(c)(ii)
2t + 15 = 33
*
5marks
=> 2t = 18 => t =9
Accept candidates’ equation from previous work.
Blunders (-3)
B1 Correct answer without work (t = 9 stated or substituted).
B2 Errors in transposition
B3 Stops at 2t = 18 or similar
Slip (-1)
S1 Numerical errors to a max of 3
S2 Leaves as 18 or similar.
2
Attempts (2 marks)
A1 Answer from part c(i) written down and stops
A2 Any correct step e.g. 3.5 = 15
Worthless (0 marks)
W1 Incorrect answer, with no work
Page 22
Att 2
(c)(iii)
10 marks
Att 3
I
5x – 4y = 16
2x + 3y = 11
15x – 12y = 48
8x + 12y = 44
23x = 92
II
5x – 4y = 16
2x + 3y = 11
4y = 5x – 16
y = 5 x  16
10x – 8y = 32
–10x – 15y = –55
–23y = –23
x = 92 = 4
23
=> y = 1
y =  23 = 1
 23
4
x
5
2x + 3(  16 ) = 11
4
8x + 15x – 48 = 44
23x = 92
x=4
=> y = 1
=> x = 4
Blunders (-3)
*
Apply only one blunder deduction (B2 or B3) to any error(s) in establishing the first
equation; in terms of x only or the first equation in terms of y only.
*
Finding the second variable is subject to a maximum deduction of (3).
Blunders (-3)
B1 Correct answers without work(stated or substituted)
B2 Error or errors in establishing the first equation in terms of x only (23x = 92) or the first
equation in terms of y only (–23y = –23) through elimination by cancellation (but see S1)
B3 Errors in transposition when finding the first variable.
B4 Errors in transposition when finding the second variable
B5 Incorrect substitution when finding second variable
B6 Finds one variable only
Slips (-1)
S1 Numerical errors to a max of 3
Attempt (3 marks)
A1 Attempt at transposition and stops
A2 Multiplies either equation by some number and stops
A3 Incorrect value of x or y substituted correctly to find his correct 2nd variable
Worthless (0 marks)
W1 Incorrect values for x or y substituted into the equations
Page 23
QUESTION 5
Part (a)
Part (b)
Part (c)
10 marks
15(5,5,5) marks
25(5,10,10) marks
Att 3
Att 2,2,2
Att 2,3,3
Part (a)
10 marks
(a) Write in its simplest form 3(x + 2) + 4(3x + 1).
Att 3
(a)
Att 3
10marks

*
*
3x + 6 + 12x + 4 = 15x + 10
Stops after correct removal of brackets  7marks
Ignore excess work 53x  2 
Blunders (-3)
B1
B2
B3

Correct answer without work
Error(s) in distribution (each time)
Combining unlike terms
Attempts (3 marks)
A1 Any correct multiplication
B4 Fails to group like terms
Slips (-1)
S1 Numerical errors to a max of 3
Misreadings (-1)
M1 3 x  2   43 x  1 and continues
Worthless (0)
W1 combining unlike terms, before attempting multiplication and stops
Page 24
Part (b)
5(b)
15(5,5,5) marks
Att 2,2,2
Factorise
(i)

5cd  7 d
(ii)
ax  3ay  4 x  12 y
(iii)
x 2  49
(b)(i)
5 marks
Att 2
d 5c  7 
Blunders (-3)
B1 Removes factor incorrectly.
Attempts (2 marks)
A1 Indication of common factor e.g. underline ds and stops.
(b) (ii)

5marks
ax  3ay  4 x  12 y
a x  3 y   4 x  3 y 
ax  4 x  3ay  12 y
xa  4  3 y a  4 
or
a  4x  3 y 
Att 2
x  3 y a  4
*
Accept also (with or without brackets) for 5 marks any of the following
a  4 and x  3 y  [The word and is written down.]
a  4 or x  3 y  [The word or is written down.]
a  4 , x  3 y  [A comma is used]
Blunders (-3)
B1
B2
B3
B4
B5

Correct answer without work
Stops after first line of correct factorisation e.g. a  x  3 y   4x  3 y  or equivalent.
Error(s) in factorising any pair of terms (each time)
(B2 will apply)
Incorrect common factor and continues. e.g. xa  4   y 3a  12 
Correct first line of factorisation but ends as x  3 y 4 a .
Slips (-1)
S1 x  3 y   a  4 
Attempts (2 marks)
A1 Pairing off, or indication of common factors and stops.
A2 Correctly factorises any pair and stops.
Page 25
(b) (iii)
5 marks
Att 2
x  49
2
*
*
*
x2  72
x  7 x  7 
Accept also (with or without brackets) for 5 marks any of the following
x  7  and x  7  [The word and is written down.]
x  7  or x  7  [The word or is written down.]
x  7  , x  7  [A comma is used]
Quadratic equation formula method is subject to slips and blunders.
x  49 x  49 merits 5 marks



Blunders (-3)
B1 Incorrect two term linear factors of x 2  49 formed from correct (but inapplicable) factors
of x 2 and  49 .e.g. x  49 x  1
B2 Incorrect factors of 49
B3 Incorrect factors of x 2
B4 7  x 7  x  .
B5 x  49  x  49  .
B6 Answer left as roots. ( x   7 )
Slips (-1)
S1  x  7  ±  x  7 
Attempts (2 marks)
A1 Correct factors of x 2 only
A2 Correct factors of  49 only
A3
 x or  7 appears.
A4 x 2  49  x  x  7  7
A5 Mention of the difference of two squares .e.g. x 2  49 2
A6 Correct quadratic equation formula quoted and stops.
49
A7
Worthless (0 marks)
W1 Combines xs to “numbers” and continues or stops.
Page 26
Part (c)
(i)
(ii)
25(5,10,10) marks
Att 2,3,3
5x  1
x6

as a single fraction.
3
5
Give your answer in its simplest form.
Express
Verify your answer to part (i) by substituting x = 4 into
5x  1 x  6

3
5
and into your answer to part (i).
(iii)
Solve the equation x 2  4 x  21  0.
(c)(i)
(i)
5 marks
5(5 x  1)  3( x  6)
25 x  5  3x  18
22 x  13
=
=
15
15
15
5x  1
4x  7
x  6


3
5
2
Blunders (-3)
*
Zero marks

B1
B2
B3
B4
Correct answer, without work
Error(s) in distribution e.g. 55 x  1  5 x  1 .
Mathematical error e.g. 5-18=13 , -3(6) =18
Incorrect common denominator and continues
B5
Incorrect numerator, from candidate's denominator e.g.
B6
B7
No simplification of numerator
Omitting denominator
Slips (-1)
S1 Drops denominator
S2 Numerical errors to a max of 3
S3
Att2
Answer not in simplest form. e.g.
44 x  2 6
.
30
Attempts (2 marks)
A1 15 only or a multiple of 15 only appears.
A2 Any correct step.
Worthless (0)
 5 x  1  x  6 
W1 

 and stops.
 3  5 
Page 27
35 x  1  5x  6
.
15
Part(c) (ii)

*
*
10 marks
5(4)  1 4  6

3
5
20  1 10


5
3
21 10


3 5
72
5
and
Att 3
22 x  14
15
22(4)  13

15
88  13

15
75

15
5
Accept candidates answer from previous section [May result in inequality].
Accept usage of a value other than 4 for verification.
Blunders (-3)
B1
B2
B3
B4

Correct answer, without work
Substitutes into one expression only (B4 will also apply)
Manipulation to force equality
Conclusion missing
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (3 marks)
A1 Writes answer from previous part in this section
A2 Substitutes a value into one expression and stops
Page 28
(c) (iii)
10 marks
Att 3
x 2  4 x  21  0
x
x 2  7 x  3 x  21  0
x x  7   3 x  7   0
+3
x
7
x  3x  7   0
 x  3 and x  7
  x  3 x  7   0
 x  7 and x  3
  4  
 42  41 21
21
4  16  84
2
14
 7 and
2
 x  7 and

4  10
2
6
3
2
x  3
Factor Method
Blunders (-3)
B1
B2
B3
B4
B5
B6
B7

Correct answers without work
Incorrect two term linear factors of x 2  4 x  21 formed from correct (but inapplicable)
factors of x 2 and/or ±21. e.g. (x+21)(x-1)
No roots given.(each time)
Incorrect factors of x 2 and/or ±21.
Correct cross method but factors not shown and stops [Note: B3 applies also].
x(x-7) +3(x-7) or similar and stops [Note: B3 applies also].
Error in transposition (each time)
Slips (-1)
S1 Numerical errors, to a max of 3
Attempts (3 marks)
A1 Some effort at factorisation
A2 One correct answer without work
Worthless (0 marks)
W1 x 2  4 x  21 , or similar, and stops.
W2 Trial and error
W3 Oversimplification, resulting in a linear equation
Formula Method
Blunders (-3)
B1
B2
B3
B4

Correct answers without work.
Error in a,b,c substitution (apply once only)
Sign error in substituted formula (apply once only)
Error in square root or square root ignored.
Page 29
B5
B6
4  10
2
Incorrect quadratic formula and continues.
Stops at
Slips (-1)
S1 Numerical errors to a max of 3
p
S2 Roots left in the form
q
Attempts (3 marks)
A1 Correct formulas and stops
A2 One correct substitution and stops
Page 30
QUESTION 6
Part (a)
Part (b)
Part (c)
10(5,5) marks
25(10,15) marks
15(10,5) marks
Att 2,2
Att 3,5
Att 3,2
Part (a)
(a)
10 (5,5)marks
Att 2,2
f (x) = 4x – 5.
Find:
(i) f(3)
(ii) f(-2)
(a)(i)
5 marks

f (3)  4(3)  5  12  5  7
*
Answer
12-5  4 marks
Blunders (-3)
B1 Correct answer no work.
B2 Leaves 4(3) in the answer
B3 Mathematical error e.g. treats 4(3) as 43.
B4 Breaks order i.e. [ 4 3  5 4(2)  8 ].
Slips (-1)
S1 Numerical errors to a max of 3
S2 Leaves x in the answer e.g. 7x
Misreadings (-1)
M1 Correct substitution of any number other than 3 and continues.
Attempts (2 marks)
A1 Substitutes for x and stops e.g. 4(3)
A2 Treats as an equation and continues or stops 4 x  5  3
A3 Combines "x"s to “numbers” and continues. e.g. 4 x  5  x  (3)
Worthless (0)
W1
Combines "x"s to “numbers” and stops.
W2
Ignores x giving 4  5  1
3 f ( x )  12 x 15
W3
W4
Replaces coefficient i.e. 4 x  3x
W5
Incorrect answer, without work
Page 31
Att 2
(a) (ii)
5 marks

f (2)  4(2)  5  8  5  13
*
Answer
-8-5  4 marks (stops or continues)
*
-8x -5x = 13x  4marks
but -8x -5x = 13  5marks (rectified error)
Blunders (-3)
B1 Correct answer no work
B2 Leaves 4(-2) in the answer
B3 Mathematical error e.g. treats 4(-2) as 42 .
B4 Breaks order i.e. [ 4 2  5 4(7)  28 ].
Slips (-1)
S1 Numerical errors to a max of 3
S2 Leaves x in the answer e.g. -13x
A3 Combines "x"s to “numbers” and continues. e.g. 4 x  5  x  (2)  2
A4 Substitutes positive value for x and continues correctly
Misreadings (-1)
M1 Correct substitution of any negative number other than -2 and continues
Attempts (2 marks)
A1 Substitutes for x and stops e.g. 4(-2)
A2 Treats as an equation and continues or stops 4 x  5  2
Worthless (0)
W1
Combines "x"s to “numbers” and stops
W2
Ignores x giving 4  5  1
W3
2 f ( x )  8 x  10
W4
Replaces coefficient i.e. 4 x  2 x
W5
Incorrect answer, without work
Page 32
Att 2
Part (b)
25(10,15) marks
Draw the graph of the function
f : x  x 2  2x  1
(b)
in the domain 1  x  3,
where x  R.
Table
*
Att 3,5
2
10marks
2
x
f(-1)
=
(-1)
-2(-1)
-1 =
f(0)
=
(0)2
-2(0)
-1 = -1
f(1)
=
(1)2
-2(1)
-1 = -2
f(2)
=
(2)2
-2(2)
-1 = -1
f(3)
=
(3)2
-2(3)
-1 =
2
Att 3
-1
0
1
2
3
x2
1
0
1
4
9
-2 x
2
0
-2
-4
-6
-1
-1
-1
-1
-1
-1
f x 
2
-1
-2
-1
2
Error(s) in each row /column calculation attracts a maximum deduction of 3 marks
Blunders (-3)
B1 Correct answer, without work i.e. 5 correct couples only and no graph
B2 “ 2 x ” taken as “2” all the way. [In row headed ” 2 x ” by candidate]
B3 “-1” calculated as “–x” all the way. [In row headed “-1” by candidate]
B4 Adds in top row when evaluating f (x) .
B5 Omits “-1” row
B6 Omits “-2 x ” row
B7 Omits a value in the domain (each time).
B8 Each incorrect image without work i.e. calculation through the function method
Slips (-1)
S1 Numerical errors to a max of 3 in any row / column
Misreadings (-1)
M1 Misreads “ x 2 ”as “  x 2 ”and places ''- x 2 '' in the table or function.
M2 Misreads “-2x”as “2x” and places “2x” in the table or function.
M3 Misreads “-1”as “1” and places “1” in the table or function
Attempts (3 marks)
A1 Omits “ x 2 ”row from table or treats “ x 2 ” as  xor  2 x .
A2 Any effort at calculating point(s).
A3 Only one point calculated and stops.
Page 33
Graph
15 Marks
Att 5
y-axis
3
2
1
x-axis
-1
1
2
3
-1
-2
*
*
*
*
Accept candidate's values from previous work.( 5 co-ordinates needed ) but see S2
Only one correct point graphed correctly  Att 3 + Att 5
Correct graph but no table  full marks i.e. (10+15) marks.
Accept reversed co-ordinates if
(i) if axes not labelled or (ii) if axes are reversed to compensate (see B1 below)
Blunders (-3)
B1 Reversed co-ordinates plotted against non-reversed axes (once only) {See 4th * above}.
B2 Scale error (once only).
B3 Points not joined or joined in incorrect order (once only).
Slips (-1)
S1 Each point of candidate graphed incorrectly. {Tolerance  0.25 }
S2 Each point ( 5 points needed ) from table not graphed [See 2nd * above].
Attempts (5 marks)
A1 Graduated axes (need not be labelled)
A2 Some effort to plot a point {See 2nd * above}
Page 34
Part (c)
15(10,5) marks
Use the graph drawn in 6(b) to estimate:
(i)
Att 3,2
the values of x for which x 2  2 x  1  0
(ii) the value of f(x) when x 1 5 .
(c) (i)

10 marks
x  2.4 and x  0.4
work to be shown on graph for correct answer
*
*
Accept candidate's values from previous work.
2 indications on graph and 2 values written down (blunder each time)
Blunders(-3)
B1 Answers beyond tolerance. {Tolerance  0.25 }
Misreading (-1)
M1 Answers not presented in designated box (but elsewhere)
.
Attempts (3marks)
A1 One point of intersection indicated only or one value of x written down
A2 Algebraic evaluation ( x = 1  √2)
Worthless (0)
W1 Answers outside of tolerance without graphical indication
W2 f(0) gives -1 as answer.
Page 35
Att 3
(c) (ii)

*
5 marks
f(x) = -1.75
work to be shown on graph for correct answer
Accept candidate's values from previous work.
Blunders (-3)
B1 Answer beyond tolerance. {Tolerance  0.25 }.
B2 Correct answers no work
B3 Sign error
Misreading (-1)
M1 Answers not presented in designated box (but elsewhere)
Attempts (2 marks)
A1 Point indicated only.
A2 Algebraic evaluation or correct calculator calculation.
A3 Testing x value for y = 1.5
Worthless(0)
W1 Answers outside of tolerance without graphical indication.
Page 36
Att 2
JUNIOR CERTIFICATE 2008
MARKING SCHEME
MATHEMATICS
ORDINARY LEVEL
PAPER 1
GENERAL GUIDELINES FOR EXAMINERS
1.
Penalties of three types are applied to candidates’ work as follows:
• Blunders - mathematical errors/omissions
(-3)
• Slips- numerical errors
(-1)
• Misreadings (provided task is not oversimplified)
(-1).
Frequently occurring errors to which these penalties must be applied are listed in the
scheme. They are labelled: B1, B2, B3,…, S1, S2,…, M1, M2,…etc. These lists are not
exhaustive.
2.
When awarding attempt marks, e.g. Att(3), note that
• any correct, relevant step in a part of a question merits at least the attempt mark for
that part
• if deductions result in a mark which is lower than the attempt mark, then the attempt
mark must be awarded
• a mark between zero and the attempt mark is never awarded.
3.
Worthless work is awarded zero marks. Some examples of such work are listed in the
scheme and they are labelled as W1, W2,…etc.
4.
The phrase “hit or miss” means that partial marks are not awarded – the candidate
receives all of the relevant marks or none.
5.
The phrase “and stops” means that no more work is shown by the candidate.
6.
Special notes relating to the marking of a particular part of a question are indicated by
an asterisk. These notes immediately follow the box containing the relevant solution.
7.
The sample solutions for each question are not intended to be exhaustive lists – there
may be other correct solutions.
8.
Unless otherwise indicated in the scheme, accept the best of two or more attempts –
even when attempts have been cancelled.
9.
The same error in the same section of a question is penalised once only.
10.
Particular cases, verifications and answers derived from diagrams (unless requested)
qualify for attempt marks at most.
11.
A serious blunder, omission or misreading results in the attempt mark at most.
12.
Do not penalise the use of a comma for a decimal point, e.g. €5·50 may be written as
€5,50.
Page 1
QUESTION 1
Part (a)
Part (b)
Part (c)
10(5, 5) marks
20 (5, 5, 5, 5) marks
20 (5, 5, 5, 5) marks
Att 4(2, 2)
Att 8 (2, 2, 2, 2)
Att 8 (2, 2, 2, 2)
Part (a) (i)
5 marks
S = {a, b, c}
1(a) (i) Write down a subset of S that has one element
Att 2
Part (a) (i)
Att 2
*
*
5 marks
{a} or {b} or {c}.
No penalty for the omission of brackets.
No penalty for use of Venn Diagram to show subsets.
Blunders (-3)
B1 Any incorrect set of elements of S other than the misreading below.
Misreadings (-1)
M1 Subset of S with two elements. e.g. S = {a, b}.
Attempts (2 marks)
A1 Draws a single bracket & stops.
A2 { } Null set.
Part (a) (ii)
1(a) (ii)
5 marks
S = {a, b, c}
Write down a subset of S that has two elements
Part (a) (ii)
*
*
Att 2
5 marks
{a,b} or {a,c} or {b,c}
No penalty for omission of brackets.
No penalty for use of Venn Diagram to show subsets.
Blunders (-3)
B1 Any incorrect set of elements of S other than the misreading below.
Misreadings (-1)
M1 Subset of S with one element. e.g. S = {a}.
Attempts (2)
A1 Draws a single bracket & stops.
A2 { } Null set.
Page 2
Att 2
Part (b)
1(b)
U is the universal set.
20 (5, 5, 5, 5)
Att 8 (2,2,2,2)
P = {2, 3, 6, 7, 8}
P
Q = {1, 6, 8, 10}
3
7
R = {2, 4, 6}
1
8
10
6
2
Q
9
4
R
Part (b) (i)
1 (b) (i)
5 marks
5
Att 2
List the elements of: P ∩ Q
Part (b) (i)
5 marks
P ∩ Q = {6, 8}
Att 2
Blunders (-3)
B1 Any incorrect set of elements of P and Q other than the misreading below.
Misreadings (-1)
M1 P ∪ Q = {1,2 , 3, 6, 7, 8,10}.
Attempts (2 marks)
A1 4 or 5 or 9 appear in the answer.
Part (b) (ii)
1 (b) (ii)
5 marks
Att 2
List the elements of: Q \ R
Part (b) (ii)
5 marks
Q \ R = {1, 8,10}
Blunders (-3)
B1 Any incorrect set of elements of Q and R other than the misreading below.
Misreadings (-1)
M1 R \ Q = { 2, 4 }.
Attempts (2 marks)
A1 3 or 5 or 7 or 9 appear in the answer.
Page 3
Att 2
Part (b) (iii)
1 (b) (iii)
5 marks
List the elements of: (Q ∪ R )
Part (b) (iii)
Att 2
/
5 marks
(Q ∪ R ) = {3, 5, 7, 9}
Att 2
/
Blunders (-3)
B1 Any incorrect set of elements of Q and R other than the misreadings below.
B2 (Q ∪ R ) = {1, 2, 4, 6, 8,10}
Misreadings (-1)
/
M1 (Q ∩ R ) = {1,2 , 3, 4, 5, 7,8,9,10}
M2 Q / ∪ R / = {1, 2, 3, 4, 5, 7, 8, 9,10}
Attempts (2 marks)
A1 Any incorrect listing of element(s) other than the misreadings above.
Part (b) (iv)
1 (b) (iv)
5 marks
Att 2
List the elements of: (P ∩ R ) \ Q
Part (b) (iv)
5 marks
(P ∩ R ) \ Q = {2}
Att 2
Blunders (-3)
B1 Any incorrect set of elements of P and Q and R other than the misreadings below.
B2 P ∩ R = {2, 6 }
Misreadings (-1)
M1 Q \ (P ∩ R ) = {1, 8,10 }.
M2 (P ∪ R ) \ Q = {2,3,4,7}
Attempts (2 marks)
A1 5 or 9 appear in the answer.
Page 4
Part(c)
20(5, 5, 5, 5)
Att 8 (2, 2, 2, 2)
M is the set of natural numbers from 1 to 36, inclusive
Part (c) (i)
5 marks
1(c) (i) List the elements of M that are multiples of 6.
Att 2
Part (c) (i)
Att 2
5 marks
6, 12, 18, 24, 30, 36.
Slips (-1)
S1 Each missing or incorrect element subject to a max of 3.
Attempts (2 marks)
A1 Any one correct element identified.
A2 Any correct factor(s) of 6.
Worthless (0)
W1 Elements listed are not multiples or factors of 6.
Part (c) (ii)
1(c) (ii)
Part (c) (ii)
5 marks
List the elements of M that are multiples of 9.
5 marks
9, 18, 27, 36.
Slips (-1)
S1 Each missing or incorrect element subject to a max of 3.
Attempts (2 marks)
A1 Any one correct element identified.
A2 Any correct factor(s) of 9.
Worthless (0)
W1 Elements listed are not multiples or factors of 9.
Page 5
Att 2
Att 2
Part (c) (iii)
5 marks
Att 2
1(c) (iii)
Write down the lowest common multiple of 6 and 9.
*
Accept candidate's L.C.M. from incorrect answers in part (i) and part (ii) for full marks.
*
Accept indication of candidate's L.C.M.
Part (c) (iii)
5 marks
L.C.M. = 18
Att 2
Blunders (-3)
B1 An incorrect common multiple that is not the lowest.e.g. {36}.
Slips (-1)
S1 Answer written as 2 × 3 × 3 and stops.
Misreadings (-1)
M1 Writes down H.C.F. = 3.
Attempts (2 marks)
A1 Any multiple of either 6, or 9 written in this part.
Worthless (0)
W1 Elements listed are not multiples of 6 or 9.
Part (c) (iv)
1(c) (iv)
5 marks
Express 30 as the product of three prime numbers,
Part (c) (iv)
*
5 marks
30 = 2 × 3 × 5
Att 2
Att 2
2, 3, 5, listed merits full marks.
Blunders (-3)
B1 Each correct prime constituent omitted and/or each incorrect prime constituent included.
Misreadings (-1)
M1 Write as sum of 3 primes (30 = 2 + 5 + 23) or (30 = 2 + 11 + 17).
Attempts (2 marks)
A1 Writes a prime number ∈ M.
Worthless (0)
W1 No prime number appears.
Page 6
QUESTION 2
Part (a)
Part (b)
Part (c)
Part (a)
2(a)
10 marks
20 (5, 10, 5)
20 (10, 5, 5)
Att 3
Att 7 (2, 3, 2)
Att7 (3, 2, 2)
10 marks
€260 is shared between Mark and Una in the ratio 6:7.
How much does each receive?
Part (a)
10 marks
Att 3
Att 3
"
6 parts : 7parts
13 Parts = 260
260
⇒1 Part =
= 20
13
Mark = 20 × 6 = 120
Una = 20 × 7 = 140
or 260 − 120 =140
6x : 7x
⇒13 x = 260
⇒ x = 20
⇒ 6 x = 120
⇒ 7 x = 140
6 + 7 = 13
1
= 20
13
6
⇒ =120 (Mark)
13
⇒ 260 − 120 = 140 (Una)
7
= 140
13
⇒ 260 −140 = 120
or
260 −140 =120
Blunders (-3)
B1
B2
B3
B4
B5
B6
B7
"
Correct answers without work.
Divisor ≠ 13 only and continues.
Incorrect multiplier or fails to multiply. (each time).
Error in transposition.
Fails to find second amount.
Adds instead of subtracts.e.g.260 + 120 = 380 or 260 +140 = 400.
Finds 6% of 260 (15 ⋅ 6) and 7% of 260 (18 ⋅ 2)
Slips (-1)
S1 Numerical errors to a max of 3.
Misreadings (-1)
M1 Interchanges Mark and Una.
Attempts (3 marks)
A1
A2
A3
A4
260
260
and/or
and stops.
6
7
6
7
Indicates 6 parts or 7 parts or 13 parts or
or
or 6 + 7=13 and stops.
13 13
Indicates multiplication of 260 by 6 and/or 7 and stops.
Both answers added together equal €260. (No work shown).
Divisor ≠ 13 e.g.
Worthless (0)
W1 Incorrect answer without work. {Subject to A4}.
Page 7
Part (b) (i)
2(b)
(i)
5 marks
Att 2
On a day when €1 = £ 0 ⋅ 68 , find the value in euro of £816.
Part (b) (i)
5 marks
Att2
"
€1 = £ 0⋅ 68
€? = £816
?=
£ 0⋅ 68 = €1
1
⎛ 25 ⎞
⇒ £1 = €
⎜ ⎟
0 ⋅ 68
⎝ 17 ⎠
816
= € 1200
0 ⋅ 68
⇒ £816 = 816 ×
1
= €1200
0 ⋅ 68
*
No penalty for the omission of € or £ symbols.
⎛ 25 ⎞
⎛ 13872 ⎞
⎛ 1 ⎞
*
Note: Natural Display calculator gives ⎜
⎟.
⎟ = ⎜ ⎟ , 816 × 0 ⋅ 68 = ⎜
⎝ 17 ⎠
⎝ 25 ⎠
⎝ 0.68 ⎠
Blunders (-3)
"
B1
B2
Correct answer without work.
Incorrect operation. i.e. 816 × 0 ⋅ 68 = 554 ⋅ 88
0 ⋅ 68
68
or
B3 Incorrect ratio
.
816
81600
B4 Decimal error.
816
and stops.
B5 Fails to finish, leaves as
0 ⋅ 68
Slips (-1)
S1 Numerical errors to a max of 3.
1
=1 ⋅ 470588235 = 1 ⋅ 47 (€ 1199 ⋅ 52 ) or 1⋅ 5 (€1224 )
S2 Rounds off too early. e.g.
0 ⋅ 68
Attempts (2 marks)
1
A1 £1 = €
0 ⋅ 68
and stops.
Worthless (0)
W1 Adds or subtracts 816 to 0 ⋅ 68 and stops.
W2 Incorrect answer without work.
Page 8
Part (b) (ii)
10 marks
2(b) (ii) By rounding each of these numbers to the nearest whole number,
5 ⋅ 8 × 8 ⋅ 148
estimate the value of
.
11 ⋅ 64
Att 3
Part (b) (ii)
Att 3
"
10 marks
5 ⋅ 8 × 8 ⋅ 148
is approximately equal to
11 ⋅ 64
6
×
8
48
=
*
*
4
12
12
*
=
6×8
and stops ⇒ 4 marks.
12
No penalty if the intermediate step between approximations and final answer is not
48
shown. i.e
not shown.
12
5 ⋅ 8 × 8 ⋅ 148
⎛ 203 ⎞
Special Case:
= 4 ⋅ 06 . ⎜
⎟ presented in this part ⇒ Attempt 3 marks.
11 ⋅ 64
⎝ 50 ⎠
Blunders (-3)
B1 Correct answer without work.
B2 Error(s) in rounding off to the nearest whole number. (once only)
B3 Decimal error in calculation of final value.
B4 An arithmetic operation other than indicated.
6 8
48
B5 Error(s) in the manipulation of the denominator. e.g.
.
× =
12 12 144
6 × 8 3 × 4 12
=
= =2
B6 Incorrect cancellation e.g.
12
6
6
"
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 Only one or two approximations made to the given numbers and stops.
Worthless (0)
W1 Incorrect answer without work.
Page 9
Part (b) (iii)
2(b) (iii)
5 marks
Using a calculator, or otherwise, find the exact value of
Part (b) (iii)
*
Att 2
Note:
⎛ 5.8 ⎞
⎜
⎟
⎝ 11.64 ⎠
5 ⋅ 8 × 8 ⋅ 148
.
11 ⋅ 64
5 marks
5 ⋅ 8 × 8 ⋅ 148 47 ⋅ 2584
⎛ 203 ⎞ ⎛ 3 ⎞
=
= 4 ⋅ 06 ⎜
⎟ ⎜4 ⎟
11 ⋅ 64
11 ⋅ 64
⎝ 50 ⎠ ⎝ 50 ⎠
145
⎛ 8.148 ⎞ 7
,
=
⎜
⎟= . .
291
⎝ 11.64 ⎠ 10
Att 2
Blunders (-3)
B1 Decimal error.
B2
B3
B4
B5
B6
B7
B8
47 ⋅ 2584
.
11 ⋅ 64
5 ⋅ 8 8 ⋅ 148
⎛ 203 ⎞
×
= 0 ⋅ 4982817869 × 0 ⋅ 7 = 0 ⋅ 3487972509 . ⎜
Treats as
⎟.
11 ⋅ 64 11 ⋅ 64
⎝ 582 ⎠
5 ⋅ 8 + 8 ⋅ 148 13 ⋅ 948
⎛ 3487 ⎞
Reads as
=
= 1 ⋅ 198281787 . ⎜
⎟.
11 ⋅ 64
11 ⋅ 64
⎝ 2910 ⎠
5 ⋅ 8 − 8 ⋅ 148 − 2 ⋅ 348
⎛ − 587 ⎞
=
Reads as
= − 0 ⋅ 2017182131 . ⎜
⎟.
11 ⋅ 64
11 ⋅ 64
⎝ 2910 ⎠
11 ⋅ 64
11 ⋅ 64
⎛ 50 ⎞
=
Treats as
= 0 ⋅ 2463054187 . ⎜
⎟.
5 ⋅ 8 × 8 ⋅ 148 47 ⋅ 2584
⎝ 203 ⎠
11 ⋅ 64
11 ⋅ 64
⎛ 2910 ⎞
Treats as
=
= 0 ⋅ 8345282478. ⎜
⎟.
5 ⋅ 8 + 8 ⋅ 148 13 ⋅ 948
⎝ 3487 ⎠
11 ⋅ 64
11 ⋅ 64
⎛ − 2910 ⎞
Treats as
=
= − 4 ⋅ 957410562 . ⎜
⎟.
5 ⋅ 8 − 8 ⋅ 148 − 2 ⋅ 348
⎝ 587 ⎠
Leaves answer as
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Any rounding off.
Attempts (2 marks)
A1 Any correct relevant calculation and stops.
⎛ 5 ⋅ 8 ⎞ 145
= 0 ⋅ 4982817869 ,
e.g. 5 ⋅ 8 × 8 ⋅ 148 = 47.2584, ⎜
⎟=
⎝ 11 ⋅ 64 ⎠ 291
Worthless (0)
W1 Incorrect answer without work.
Page 10
⎛ 8 ⋅ 148 ⎞ 7
⎜
⎟ = =0⋅7.
⎝ 11 ⋅ 64 ⎠ 10
Part(c) (i)
10 marks
Att 3
1
11
and
as decimals.
5
50
Hence, or otherwise, put the following numbers in order,
starting with the smallest and finishing with the largest:
1 11
0 ⋅ 25,
,
.
5 50
2(c) (i) Using a calculator, or otherwise, write
Part(c) (i)
1
=0⋅ 2
5
10 marks
1
,
5
*
Accept:
*
Accept:
*
Note:
11
,
50
0·2, 0·22, 0·25, merits 10 marks.
1
No penalty for writing 0·25 as .
4
1
11
or
=0⋅ 2
= 0 ⋅ 22
5
50
Blunders (-3)
B1 Fails to write a fraction as a decimal.
B2 Writes fraction as incorrect decimal.
B3 Decimal error.
B4 Inverts fraction and continues.
B5 Incorrect order or fails to order.
Attempts (3 marks)
1
0 ⋅ 25 = and stops.
A1
4
Worthless(0)
W1
Att 3
11
= 0 ⋅ 22
50
0 ⋅ 25
0 ⋅ 25
, or .
or similar.
5
50
Page 11
0·25,
merits 4 marks
Part(c) (ii)
2(c) (ii)
5 marks
Att 2
Using a calculator, or otherwise, divide 1170 by 0⋅ 45 and express
your answer in the form a ×10 n , where 1 ≤ a < 10 and n∈ N
Part(c) (ii)
"
5 marks
Att 2
1170 ÷ 0 ⋅ 45 = 2600 = 2 ⋅ 6 × 10 3
Blunders (-3)
"
B1
B2
Correct answer without work.
Decimal error.
B3
⎛ 1 ⎞
Inverts 0 ⋅ 45 ÷1170 = 0 ⋅ 00038 ⎜
⎟.
⎝ 2600 ⎠
B4
⎛ 1053 ⎞
Multiplies 1170 × 0 ⋅ 45 = 526 ⋅ 5 = 5 ⋅ 256 × 10 2 ⎜
⎟.
⎝ 2 ⎠
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Incorrect format, where a < 1 or a ≥ 10 and n ∉ N.
S3 2600 and stops.
Attempts (2 marks)
A1
Any relevant step. e.g. Partial long division.e.g.
A2
1
20
=
0 ⋅ 45 9
Page 12
1170
=2
0 ⋅ 45
Part(c) (iii)
2(c) (iii)
5 marks
Using a calculator, or otherwise, evaluate
5 ⋅ 32
2
= (3 ⋅ 9) −
× 1 ⋅ 81
0.64
Give your answer correct to two decimal places
Part(c) (iii)
5 marks
5 ⋅ 32
= 15 ⋅ 21 −
× 1 ⋅ 81
0⋅8
= 15 ⋅ 21 − 6 ⋅ 65 × 1 ⋅ 81
= 15 ⋅ 21 − 12 ⋅ 0365
⎛ 6347 ⎞
= 3 ⋅ 1735 ⎜
⎟
⎝ 2000 ⎠
= 3 ⋅ 17
*
Correct answer (without work) incorrectly rounded off ⇒ 2 marks.
e.g. = 3 ⋅ 1735 , = 3 ⋅ 174
5.32
⎛ 1521 ⎞
⎛ 133 ⎞
2
*
Note: Natural Display calculator gives (3 ⋅ 9) = ⎜
=⎜
⎟,
⎟,
0 ⋅ 64 ⎝ 20 ⎠
⎝ 100 ⎠
Blunders (-3)
Att 2
Att 2
"
"
B1
B2
Correct answer without work.
Mishandles (3 ⋅ 9) 2 .
B3
B4
B5
B6
B7
Mishandles 0 ⋅ 64 .
Error in 5 ⋅ 32 ÷ 0 ⋅ 8 or 9 ⋅ 6292 ÷ 0 ⋅ 8 .
Error in 5 ⋅ 32 × 1 ⋅ 81 or 6 ⋅ 65 × 1 ⋅ 81 .
Decimal error.
Subtracts before Division (15 ⋅ 21 − 5 ⋅ 32 ) ÷ 0 ⋅ 8 ×1 ⋅ 81 = 9 ⋅ 89 ÷ 0 ⋅ 8 × 1 ⋅ 81 = 22 ⋅ 376125 .
B8
Subtracts before Multiplication
(15 ⋅ 21 − 6 ⋅ 65) ×1 ⋅ 81 = 8 ⋅ 56 × 1 ⋅ 81 =15 ⋅ 4936 .
B9
Use of mathematical operator other than that which is indicated.
5 ⋅ 32
= 27 ⋅ 5301 − 6 ⋅ 65 = 20 ⋅ 8801 = 20 ⋅ 88 .
B10 Works as 15 ⋅ 21 × 1 ⋅ 81 −
0⋅8
⎛ 5 ⋅ 32
⎞
B11 Works as ⎜
×1 ⋅ 81⎟ −15 ⋅ 21 =12 ⋅ 0365 −15 ⋅ 21 = − 3 ⋅ 1735 = − 3 ⋅ 17 .
⎝ 0⋅8
⎠
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Each premature rounding off that effects the final answer to a max of 3.
S3 Fails to round off or rounds off incorrectly when giving final answer.
6347
S4 Leaves as
.
2000
Attempts (2 marks.)
A1 Any correct relevant step e.g. (3 ⋅ 9) 2 = 15 ⋅ 21 , 0 ⋅ 64 = 0 ⋅ 8 .
Worthless (0)
W1 Incorrect answer without work. (Note 1st *).
Page 13
QUESTION 3
Part (a)
Part (b)
Part (c)
10 marks
20 (10, 10) marks
20 (5, 5, 5, 5,) marks
Att 3
Att 6(3, 3)
Att 8(2, 2, 2, 2)
Part (a)
10 marks
Att 3
3(a)
Kate went to the cinema. She bought a ticket at €8·50 and then bought popcorn
costing €4·40. How much change did Kate get from a €20 note?
Part (a)
10 marks
Att 3
"
*
€8 ⋅ 50 + €4 ⋅ 40 = €12 ⋅ 90
€20 ⋅ 00 − €12 ⋅ 90 = €7 ⋅ 10
*
*
*
**
€20 ⋅ 00 − €8 ⋅ 50 = €11 ⋅ 50
€11 ⋅ 50 − €4 ⋅ 40 = €7 ⋅ 10
***
€20 ⋅ 00 − €4 ⋅ 40 = €15 ⋅ 60
€15 ⋅ 60 − €8 ⋅ 50 = €7 ⋅ 10
Accept 710c, ( 7 ⋅ 10 )
Final subtraction step subject to maximum deduction of 3 marks.
No penalty for the omission of € symbol.
Blunders (-3)
B1
B2
B3
B4
B5
B6
"
Correct answer without work.
Calculates the cost of the ticket and the popcorn but fails to find the change.
Fails to find total cost i.e. no addition.
Operation other than subtraction when finding the change.
Operation other than addition when finding total cost.
Decimal error e.g. €1 ⋅ 29 (Note: 1st* above).
Slips (-1)
S1
Numerical errors to a max of 3.
Attempts (3 marks)
A1 Any attempt at addition /subtracation.
Worthless (0)
W1 Incorrect answer without work.
W2 Multiplication or division of the given numbers.
Page 14
Part (b) (i)
3(b) (i)
10 marks
VAT at 13 ⋅ 5 % is added to a bill of €860
Calculate the total bill.
Part (b) (i)
Att3
10 marks
Att3
"
860
100
860
113 ⋅ 5 % =
× 113 ⋅ 5
100
= 8 ⋅ 60 × 113 ⋅ 5
13 ⋅ 5
100
13 ⋅ 5
VAT =
× 860
100
⎛ 1161 ⎞
= 116 ⋅ 10 ⎜
⎟
⎝ 10 ⎠
Total Bill = 860 + 116 ⋅ 10
Total Bill = € 976 ⋅ 10
Total Bill = € 976 ⋅ 10
100 % = 860
1%
*
=
13 ⋅ 5% =
860 ×1 ⋅ 135
Total Bill = € 976 ⋅ 10
€ 116 ⋅ 10 without work and stops merits 4 marks.
Blunders (-3)
B1
B2
"
B4
B5
Correct answer without work.
Decimal error.
100
100
or
and continues (giving answers 757 ⋅ 71 or 6370 ⋅ 37 ).
Inverts as
113 ⋅ 5 13 ⋅ 5
Mishandles 13 ⋅ 5% e.g. 860 ×13 ⋅ 5 or 860 ÷13 ⋅ 5 . Note: {860 must be used}.
860 taken as 113 ⋅ 5 %.
B6
No addition of VAT (as per candidates work) to the bill.
B7
Subtraction of VAT ( as per candidates work) from the bill.
B3
Slips (-1)
S1 Numerical errors to a max of 3.
Misreadings (-1)
M1 Reads as 15·3% or €680.
Attempts (3 marks)
13 ⋅ 5
860
and stops.
A1
and stops. or
100
100
A2 100% = 860 and stops.
13 ⋅ 5
860
A3 100 ×
and stops. or
and stops.
860
13 ⋅ 5
A4
860 + 13 ⋅ 5% and stops.
Worthless (0)
W1 Incorrect answer without work.
W2 860 + 13 ⋅ 5 = 873 ⋅ 5 and stops or continues.
Page 15
Part (b) (ii)
3(b) (ii)
10 marks
€4750 is invested at 3 ⋅ 7 % per annum
What is the amount of the investment at the end of one year?
Part (b) (ii)
10 marks
Att 3
Att 3
"
4750
100
4750
3⋅ 7 % =
× 3⋅7
100
1%=
⎛ 703 ⎞
Interest = 175 ⋅ 75 . ⎜
⎟
⎝ 4 ⎠
Amount = 4750 + 175 ⋅ 75
P×R
100
4750 × 3 ⋅ 7
I =
100
Interest = 175 ⋅ 75
I =
Amount = 4750 ×1 ⋅ 037
⎛ 19703 ⎞
Amount= € 4925 ⋅ 75 ⎜
⎟
⎝ 4 ⎠
Amount = € 4925 ⋅ 75
Amount = € 4925 ⋅ 75
*
€ 175 ⋅ 75 (without work) and stops ⇒ 4 marks.
Blunders (-3)
B1
B2
B3
B4
B5
B6
B7
Correct answer without work. "
4750 × 100
{No penalty if already penalised in b (i)}
Mishandles 3 ⋅ 7 %. e.g.
3⋅7
Decimal error (once only).
Stops at interest i.e. fails to calculate amount.
Subtracts to calculate amount.
4750
Mathematical error(s) working with
×3 ⋅ 7 .
100
1⋅ 037 treated as 1⋅ 37 .
Slips (-1)
S1 Numerical errors to a max of 3.
Misreadings (-1)
M1 Reads as €4570.
M2 3 ⋅ 7% written as 7 ⋅ 3%.
Attempts (3 marks)
A1 Correct formula with or without substitution and stops.
A2
Some use of 100 in attempt to find percentage e.g. 3 ⋅ 7 % =
A3
4750 + 3 ⋅ 7 % and stops.
Worthless (0)
W1 Incorrect answer without work.
W2 4750 + 3 ⋅ 7 = 4753⋅ 7 and stops or continues.
Page 16
3⋅7
or 1⋅ 03 7 and stops.
100
Part (c)
20 (5, 5, 5, 5) marks
Att 8 (2, 2, 2, 2)
3(c) Darragh’s annual wage is €48 000.
He pays income tax at the rate of 20% on the first €34000 of his wage
and income tax at the rate of 41% on the remainder of his wage.
Darragh has an annual tax credit of €3600.
Part (c) (i)
5 marks
Find the tax on the first €34000 of his wage, calculated at the rate of 20%.
3(c) (i)
Att 2
Part (c) (i)
Att 2
5 marks
"
100% = 34000
1% = 340
20% = 6800
Tax = €6800
*
34000
× 20
100
Tax = €6800
Tax =
Tax = 34000 × 0 ⋅ 2
Tax = €6800
1
5
34000 ÷ 5
Tax = €6800
20 % =
No penalty for missing € symbol.
Blunders (-3)
B1
B2
B3
B4
Correct answer without work. "
Mishandles 20%, e.g. 34000 × 20 = 680000 or 34000 ÷ 20 =1700
Uses €48000 instead of €34000.
Decimal error.
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (2 marks)
A1
A2
A3
Some use of 100 in attempt to find percentage e.g. 20% =
1
5
Writes 34000 + 20% and stops.
20 % =
Worthless (0)
W1 Incorrect answer without work
Page 17
20
or 0 ⋅ 2 and stops.
100
Part (c) (ii)
3(c) (ii)
5 marks
Att 2
Find the tax on the remainder of his wage, calculated at the rate of 41%.
Part (c) (ii)
"
(ii)
100% = 14 000
5 marks
Remainder of wage = €48 000 − €34 000= €14,000
Tax = 14 000 × 0 ⋅ 41
14 000
Tax =
× 41 = 5 740
100
1% = 140
41% = 5740
Tax = 5740
*
Att 2
Tax = 5 740
No penalty for missing € symbol.
Blunders (-3)
B1
B2
B3
B4
"
Correct answer without work.
Mishandles 41%, e.g. 14 000 ÷ 41 [No penalty if already penalised in (c) (i)].
Does not use €14 000.
Decimal error.
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (2 marks)
A1
Some use of 100 in attempt to find percentage e.g. 41% =
A2
48 000 - 34 000 or 14 000 and stops.
Worthless (0)
W1 Incorrect answer without work.
Page 18
41
= 0 ⋅ 41 and stops.
100
Part (c) (iii)
3(c) (iii)
5 marks
Hence calculate Darragh’s gross tax.
Att 2
Part (c) (iii)
5 marks
Att 2
"
*
*
(iii)
Darragh’s gross tax = €6800 + €5740 = €12 540
Allow candidates incorrect answers from parts (i) and (ii).
No penalty for missing € symbol.
Blunders (-3)
B1
B2
B3
"
Correct answer without work.
Subtracts to find gross tax e.g. €6800 − €5740 = €1060
Misuse of tax credit.
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (2)
A1 Answer from c (i) and /or c (ii) written in this part.
Worthless (0)
W1 Incorrect answer without work
Part (c) (iv)
3(c) (iv)
5 marks
Calculate Darragh’s take home pay.
Att 2
Part (c) (iv)
5 marks
Att 2
"
*
*
Tax due = €12 540 - €3600
= €8940
Take home pay = €48 000 - €8940
Take home pay = €39 060
Allow candidate’s incorrect gross tax figure from (c) (iii).
No penalty for missing € symbol.
Blunders (-3)
B1
B2
B3
Correct answer without work. "
Misuse or no use of tax credit e.g. €12 540 + €3600
Decimal error.
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (2)
A1 Answer from c (iii) written in this part.
Worthless (0)
W1 Incorrect answer without work.
Page 19
QUESTION 4
Part (a)
Part (b)
Part (c)
10 (5, 5) marks
20 (10, 10) marks
20 (5, 5, 5, 5) marks
Part (a)
4(a)
10(5, 5) marks
If a = 5 and b = 7, find the value of:
"
(i)
"
(ii)
Part (a) (i)
4(a) (i)
*
ab + 13
5 marks
If a = 5 and b = 7, find the value of:
(i)
9a + b
(i)
45 + 7 ⇒ 4 marks.
5 marks
9a + b = 9(5) + 7 = 45 + 7 = 52
Blunders (-3)
B1
B2
B3
B4
B5
B6
Att 4(2,2)
9a + b
Part (a) (i)
"
Att 4(2, 2)
Att 6(3, 3)
Att 8(2, 2, 2, 2)
"
Correct answer without work.
Leaves 9(5) in the answer.
Incorrect substitution and continues
Breaks order i.e. [ 9(5 + 7 ) = 9(12 ) = 108 ].
Treats 9(5) as 14 or 95.
Combines 9a + b to give 9ab and continues.
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Treats as 9a − b .
S3 Treats as 9b + a .
Attempts (2 marks)
A1 Substitutes for either a or b and stops e.g. 9(5).
A2 Writes 7 or 5 in this part.
A3 Any correct step.
Worthless (0)
W1 Incorrect answer with no work.
W2 Writes as 9ab and stops.
Page 20
Att 2
Att 2
Part (a) (ii)
4(a) (ii)
5 marks
If a = 5 and b = 7, find the value of:
(ii)
ab + 13
Part (a) (ii)
5 marks
"
ab + 13 = 5(7 )+13 = 35 + 13 = 48
*
35 + 13 ⇒ 4 marks.
Blunders (-3)
B1
B2
B3
B4
B5
B6
Correct answer without work. "
Leaves 5(7) in the answer.
Incorrect substitution and continues
Breaks order i.e. [ 5(7 + 13) = 5(20 ) = 100 ].
Treats 5(7) as 12 or 57.
Combines ab + 13 to give 13ab and continues.
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Treats as ab −13 .
Attempts (2 marks)
A1 Substitutes for either a or b and stops e.g. a (7).
A2 Writes 7 or 5 in this part.
A3 Any correct step.
Worthless (0)
W1 Incorrect answer with no work.
W2 Writes as 13ab and stops.
Page 21
Att 2
Att 2
Part (b) (i)
4(b) (i)
10 marks
Att 3
Solve the equation 3(2 x −1) = 4 x + 9
Part (b) (i)
10 marks
"
Att 3
3(2 x − 1) = 4 x + 9
6x − 3 = 4x + 9
6x − 4x = 3 + 9
2 x = 12
12
2
x=6
x =
Blunders (-3)
B1
B2
B3
B4
B5
"
Correct answer without work.
( x = 6 stated or substituted)
Error in distributive law and continues. e.g. 6 x − 1 = 4 x + 9
Error in transposition. (each time).
Combines “x's”to “numbers” and continues. e.g. 6 x − 3 = 3x or 4 x + 9 = 13 or 2 x −1 = x
Stops at 2 x = 12 or similar.
Slips (-1)
S1 Numerical errors to a max of 3.
12
or similar.
S2 Leaves as x =
2
Attempts (3 marks)
A1 Any substitution for values of x other than x = 6 .
A2
Any correct step. e.g.
(2 x −1) = 4 x + 9
3
Worthless (0)
W1 Combines “x's” to “numbers” and stops
W2 Incorrect answer with no work.
Page 22
Part (b) (ii)
4(b) (ii)
10 marks
Multiply (5 x − 2) by (3x + 4) .
Write your answer in its simplest form.
Part (b) (ii)
10 marks
Att 3
Att 3
"
(5 x − 2)(3x + 4)
(3x + 4)(5 x − 2)
15 x + 20 x − 6 x − 8
15 x − 6 x + 20 x − 8
(5 x − 2)(3x + 4)
5 x(3 x + 4 ) − 2(3 x + 4 )
15 x 2 + 14 x − 8
15 x 2 + 14 x − 8
15 x 2 + 20 x − 6 x − 8
2
2
15 x 2 + 14 x − 8
*
15 x 2 + 20 x − 6 x − 8 merits 7 marks.
Blunders (-3)
B1
B2
B3
B4
"
Correct answer without work.
Error in distribution.
Combines “x's”to “numbers” and continues. e.g. 5 x − 2 = 3 or 3x + 4 = 7 x .
Fails to group or groups incorrectly.
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 Any correct multiplication. e.g. 15 x .
A2 Any correct grouping of terms.
A3 A correct step.
A4 Substitutes a value of “ x ” and continues correctly.
A5 Treats as (5 x − 2 ) ± (3x + 4 ) to give 8 x + 2 or 2 x − 6.
Worthless (0)
W1 Combining unlike terms and stops.
W2 No distribution but A2 or A5 may apply to subsequent work e.g. gathering of terms.
Page 23
Part (c)
20 (5, 5, 5, 5) marks
4(c) Shane is x years old. Eileen is three years younger than Shane.
Att 8 (2, 2, 2, 2)
Part (c) (i)
5 marks
4(c) (i)
Find Eileen's age in terms of x .
Att 2
Part (c) (i)
Att 2
5 marks
x −3
*
Algebraic work required to earn marks.
Blunders (-3)
B1 Incorrect expression of Eileen's age other than mireadings below.
Misreadings (-1)
M1 Answer given as x + 3 .
M2 Answer given as 3 x .
Attempts (2 marks)
A1 Any effort at forming an expression.e.g. Eileen’s age = y .
Worthless (0 marks)
W1 Eileen's age given as a constant or x .
W2 Inequality appears.
Part (c) (ii)
5 marks
If the sum of Shane’s age and Eileen’s age is 47, write down an
4(c) (ii)
equation in x to represent this information.
Att 2
Part (c) (ii)
Att 2
*
5 marks
x + x − 3 = 47
Accept candidates answer from previous work.
Blunders (-3)
B1 Error in forming equation.
Attempts (2 marks)
A1 Answer from part c (i) written down and stops.
A2 Any effort at forming an expression.
A3 Writes Shane's age and stops. i.e. x
Page 24
Part (c) (iii)
4(c) (iii)
5 marks
Solve the equation that you formed in part (ii) above, for x.
Part (c) (iii)
5 marks
x + x − 3 = 47
2 x = 47 + 3
2 x = 50
"
50
2
x = 25
x=
*
Accept candidates answer from previous work.
Blunders (-3)
B1
B2
B3
B4
B5
"
( x = 25 stated or substituted)
Correct answer without work.
Error in transposition.
Error in grouping terms or fails to group.
Combines “x's”to “numbers” and continues. e.g. x + x − 3 = − x
Stops at 2 x = 50 or similar.
Slips (-1)
S1 Numerical errors to a max of 3.
50
or similar.
S2 Leaves as x =
2
Attempts (2 marks)
A1 Any substitution for values of x other than x = 25 .
A2 Answer from part c(ii) written down and stops.
A3 Any effort at solving equation.
A4 Any correct step.
Worthless (0)
W1 Combines “x's” to “numbers” and stops. ( Note A3 above)
W2 Incorrect answer with no work.
Page 25
Att 2
Att 2
Part (c) (iv)
4(c) (iv)
5 marks
Att 2
When Eileen is 2 x + 5 years old, find the sum of Shane’s age and Eileen’s age
Part (c) (iv)
5 marks
"
Eileen's age
→ 2 x + 5 = 2(25) + 5 = 50 + 5 = 55
Shane's age → 55 + 3= 58
Sum of Shane's age and Eileen's age
= 55 + 58 = 113
*
Att 2
Allow
Eileen's age = 2 x + 5
Shane's age = 2 x + 8
Sum of Shane's age and Eileen's age
= 2 x + 5 + 2 x + 8 = 4 x + 13
Accept candidates answer from previous work.
Blunders (-3)
B1
B2
B3
B4
"
Correct answer without work.
Error in grouping terms or fails to group.
Combines “x's”to “numbers” and continues. e.g. 2 x + 5 + 2 x + 8 = 7 x + 10 .
Only one age calculated.
Misreadings (-1)
M1 Takes Shane's age as x and continues 2 x + 5 + x = 3x + 5 .
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Leaves as 55 + 58 or similar.
Attempts (2 marks)
A1 Any correct step.
A2 Writes answer from c (iii) and stops.
Worthless (0)
W1 Incorrect answer with no work.
W2 Works with 2 x + 5 = 47
Page 26
QUESTION 5
Part (a)
Part (b)
Part (c)
5 marks
20(5, 5, 5, 5) marks
25(5, 5, 15) marks
Part (a)
5 marks
Att 2
Find the values of x for which 3x + 2 ≤ 8, x ∈ N
5 (a)
Part (a)
"
Att 2
Att 8(2, 2, 2, 2)
Att 9(2, 2, 5)
5 marks
Att 2
3x + 2 ≤ 8
3x ≤ 8 − 2
or
3x ≤ 6
x≤2
{0 ,1, 2}
0
1
2
Blunders (-3)
B1
B2
B3
B4
B5
"
Correct answer without work.
Error in transposition.
Combines “x's” to “numbers”. e.g.. 5 x ≤ 8 and continues.
Mishandles the direction of inequality e.g. 3x ≥ 6
Treats inequality as equality and continues.
Slips (-1)
S1 Numerical errors to a max of 3.
S2 ≤ taken as <
S3 Each missing or incorrect value of x subject to a max of 3.
Misreadings (-1)
M1 2 x + 3 ≤ 8 , and continues.
Attempts (2 marks)
A1 Attempt at transposition and stops.
A2 0, 1 or 2 substituted for x .
A3 Number line with one or more of the correct elements clearly indicated. (Ignore arrows)
Worthless (0)
W1 Incorrect answer with no work.e.g.{1, 2, 3, 4, 5, 6, 7, 8,…..}.
Page 27
Part (b) (i)
5(b) (i)
5 marks
Factorise:
Part (b) (i)
Att 2
4 a + ab
5 marks
a (4 + b )
Att 2
Blunders (-3)
B1 Removes factor incorrectly.
Attempts (2 marks)
A1 Indication of common factor. e.g. underline a 's and stops.
Part (b) (ii)
5(b) (ii)
Factorise:
Part (b) (ii)
"
*
5 marks
2 x − 2 y + cx − cy
Att 2
5 marks
Att 2
2 x − 2 y + cx − cy
2 x − 2 y + cx − cy
or
x(2 + c ) − y (2 + c )
2( x − y ) + c ( x − y )
(2 + c )(x − y )
(x − y )(2 + c )
Accept also (with or without brackets) for 5 marks any of the following
(2 + c ) and (x − y ) [The word and is written down.]
(2 + c ) or (x − y ) [The word or is written down.]
(2 + c ) , (x − y ) [A comma is used]
Blunders (-3)
B1
B2
B3
B4
"
Correct answer without work.
Stops after first line of correct factorisation. e.g. 2(x − y ) + c(x − y ) or equivalent.
Error(s) in factorising any pair of terms.
Incorrect common factor and continues. e.g. y (− 2 − c ) + x(2 + c )
Slips (-1)
S1 (2 + c ) ± ( x − y )
S2 Correct first line of factorisation but ends as 2c( x − y ) or xy(2 + c ) .
Attempts (2 marks)
A1 Pairing off, or indication of common factors and stops.
A2 Correctly factorises any pair and stops.
Page 28
Part (b) (iii)
5(b)
5 marks
Att 2
x − 2 x − 24
2
Factorise:
(iii)
Part (b) (iii)
5 marks
Att 2
x 2 − 2 x − 24
x
+4
x 2 − 6 x + 4 x − 24
x( x − 6 ) + 4( x − 6 )
(x + 4)(x − 6)
x
−6
*
− (− 2 ) ±
(− 2)2 − 4(1)(− 24)
2(1)
2 ± 10
2 ± 4 + 96
=
2
2
12
−8
= 6 and
=− 4
2
2
⇒ ( x − 6 )( x + 4 )
⇒ ( x + 4 )( x − 6 )
Accept also (with or without brackets) for 5 marks any of the following
(x − 6) and (x + 4) [The word and is written down.]
(x − 6) or (x + 4) [The word or is written down.]
(x − 6) , (x + 4) [A comma is used]
Blunders (-3)
B1 Incorrect two term linear factors of. x 2 − 2 x − 24 formed from correct (but
inapplicable) factors of x 2 and − 24 .e.g ( x − 12 )( x + 2 ) .
B2 Incorrect factors of x 2 .
B3 Incorrect factors of − 24 .
B4 Correct cross method but factors not shown and stops.
B5 x( x − 6 ) + 4( x − 6 ) or similar and stops.
B6 Incorrect common factor and continues.
B7 Incorrect quadratic formula and continues.
B8 Error in quadratic formula.
B9 Answer left as roots.
B10 Sign error in substituted formula.
B11 Error in square root or square root ignored.
Slips (-1)
S1 Numerical errors to a max of 3.
S2 (x + 4 ) ± ( x − 6 ) .
Attempts (2 marks)
A1 Correct quadratic equation formula quoted and stops
A2 Correct factors of either x 2 or ± 24.
A3 Any correct step.
Worthless (0 marks)
W1 x 2 − 2 x = 24 or similar and stops.
W2 Combines “x's”to “numbers” and continues or stops.
Page 29
Part (b) (iv)
5(b)
5 marks
144 − y 2
Factorise:
(iv)
Att 2
Part (b) (iv)
*
*
*
5 marks
(12 − y )(12 + y )
Accept also (with or without brackets) for 5 marks any of the following
(12 − y ) and (12 + y ) [The word and is written down.]
(12 − y ) or (12 + y ) [The word or is written down.]
(12 − y ) , (12 + y ) [A comma is used]
Quadratic equation formula method is subject to slips and blunders.
144 − y 144 + y merits 5 marks.
(
)(
Att 2
)
Blunders (-3)
B1 Incorrect two term linear factors of 144 − y 2 formed from correct (but inapplicable)
factors of 144 and − y 2 .e.g (144 − y )(1 + y ) .
B2 Incorrect factors of 144 .
B3 Incorrect factors of − y 2 .
B4 ( y − 12 )( y + 12 ) .
B5 (144 + y )(144 − y ) .
B6 Answer left as roots. ( y = ± 12 )
Slips (-1)
S1 (12 − y ) ± (12 + y )
Attempts (2 marks)
A1 Correct factors of 144 only.
A2 Correct factors of ± y 2 only.
A3
±12 or ± y appears.
A4 144 − y 2 = (12)(12) − ( y ) y and stops.
A5
A6
A7
Mention of the difference of two squares .e.g. { (144) − y 2 }
Correct quadratic equation formula quoted and stops.
144
2
Worthless (0 marks)
W1 Combines “y's” to “numbers” and continues or stops.
Page 30
Part(c) (i)
5(c)(i)
5 marks
x −1 x − 2
–
as a single fraction
5
7
and give your answer in its simplest form.
Express
Part(c) (i)
5 marks
"
Att2
x −1
x−2
−
5
7
7( x − 1) − 5( x − 2 )
35
7 x − 7 − 5 x + 10
35
2x + 3
35
Blunders (-3)
B1 Correct answer without work. "
B2 Error(s) in distribution. e.g. 7( x − 1) = 7 x − 1
B3 Mathematical error e.g. -7 +10 = -3. 2(-1) = 2.
B4 Incorrect common denominator and continues.
B5
Incorrect numerator from candidate's denominator e.g.
B6
B7
No simplification of numerator.
Omitting denominator.
Slips (-1)
S1 Drops denominator.
S2 Numerical error to a max of 3.
S3
Att 2
Answer not in simplest form. e.g.
4x + 6
.
70
Attempts (2 marks)
A1 35 only or a multiple of 35 only appears.
A2 Any correct step.
Worthless (0)
⎛ x − 1 ⎞⎛ x −
W1 ⎜
⎟⎜
⎝ 5 ⎠⎝ 7
x − 1
x
−
W2
5
2⎞
⎟ and stops.
⎠
− 2
2x − 3
=
7
−2
Page 31
5( x −1) − 7( x − 2)
.
35
Part(c) (ii)
5(c) (ii)
5 marks
Hence, or otherwise, solve the equation
Part(c) (ii)
"
Att2
x −1 x − 2
−
=1
5
7
5 marks
2x + 3
35
2x + 3
2x
2x
= 1
= 35
= 35 − 3
= 32
32
2
x = 16
x=
*
Accept candidates answer from previous work.
Blunders (-3)
B1
B2
"
Correct answer without work.
Error in transposition. (each time)
Slips (-1)
S1 Numerical error to a max of 3.
32
S2 Leaves as
.
2
Attempts (2 marks)
A1 Answer from (c) (i) written in this part or worked again in this part.
A2 Any correct step and stops.
Page 32
Att2
Part(c) (iii)
5(c) (iii)
15 marks
Solve for x and for y:
Att 5
3x + 2 y = 73
4 x + y = 59
.
Part(c) (ii)
15 marks
Att 5
"
3 x + 2 y = 73
4 x + y = 59
I
II
2 y = 73 − 3 x
73 − 3 x
4 x + y = 59
y =
2
3 x + 2 y = 73
12 x + 8 y = 292
⎛ 73 − 3 x ⎞
− 8 x − 2 y = − 118
− 12 x − 3 y = − 177
4x + ⎜
⎟ = 59
⎝ 2 ⎠
− 5 x = − 45
5 y = 115
8 x + 73 − 3 x = 118
5 x = 45
115
y =
= 23
5 x = 118 − 73
5
45
x=
=9
5 x = 45
⇒ x=9
5
45
⇒ y = 23
x=
=9
5
⇒ y = 23
*
Apply only one blunder deduction (B2 or B3) to any error(s) in establishing the first
equation in terms of x only or the first equation in terms of y only.
*
Finding the second variable is subject to a maximum deduction of (3).
Blunders (-3)
B1
B2
B3
B4
B5
B6
B7
3x + 2 y = 73
"
Correct answers without work.
Error(s) in establishing the first equation in terms of x only [ 5 x = 45 ]or the first
equation in terms of y only [ 5 y =115 ] through elimination by cancellation.
Error(s) in establishing the first equation in terms of x only [ 5 x = 45 ]or the first
equation in terms of y only [ 5 y =115 ] through elimination by substitution.
Errors in transposition in solving the first one variable equation.
Errors in transposition when finding the second variable.
Incorrect substitution when finding second variable.
Finds one variable only.
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (5 marks)
A1 Attempt at transposition and stops.
A2 Multiplies either equation by some number and stops .
Worthless (0 marks)
W1 Incorrect values for x or y substituted into the equations.
Page 33
QUESTION 6
Part (a)
Part (b)
Part (c)
10 (5, 5) marks
25 (15, 10) marks
15 (5, 5, 5) marks
Part (a)
6 (a)
10(5, 5) marks
f (x) = 3x – 1.
(i)
f (5)
(ii)
f (−4)
Part (a) (i)
6(a) (i) f (x) = 3x – 1.
Part (a) (i)
(i)
Att 4(2, 2,)
Att 8(5, 3)
Att 6(2, 2, 2)
Att 4(2, 2)
Find:
5 marks
Att 2
5 marks
f (5) = 3(5) −1 = 15 − 1 = 14
Att2
Find:
f (5)
Blunders (-3)
B1 Leaves 3(5) in the answer
B2 Combines “x's”to “numbers” and continues e.g. 3x − 1 = 2 x = 2(5) = 10
B3 Mathematical error. e.g. 15 −1 = − 14
B4 Breaks order i.e. 3(5 − 1) = 3(4 ) = 12
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Leaves x in the answer e.g. 14 x
Misreadings (-1)
M1 Correct substitution of any number other than 5 and continues.
Attempts (2marks)
A1 Substitutes for x and stops. i.e. 3(5)
A2 Treats as equation and continues or stops. i.e. 3x −1 = 5
Worthless (0)
W1 Combines “x's”to “numbers” and stops.
W2 Ignores x giving 3 −1 = 2.
W3 5[ f ( x )] = 15 x − 5
W4 Replaces coefficient i.e. 3 x → 5 x .
W5 Incorrect answer without work.
Page 34
Part (a) (ii)
6(a) (ii)
f (x) = 3x – 1.
Part (a) (ii)
(ii)
Find:
5 marks
f (−4)
5 marks
f (− 4) = 3(− 4) −1 = − 12 − 1 = − 13
Blunders (-3)
B1 Leaves 3(− 4 ) in the answer
B2 Combines “x's”to “numbers” and continues e.g. 3x − 1 = 2 x = 2(− 4) = − 8
B3 Mathematical error. e.g. − 12 − 1 =13
B4 Breaks order i.e. 3(− 4 − 1) = 3(− 5) = − 15
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Leaves x in the answer e.g. − 13x .
Misreadings (-1)
M1 Correct substitution of any negative number other than − 4 and continues.
Attempts (2marks)
A1 Substitutes for x and stops. i.e. 3(− 4 )
A2 Treats as equation and continues or stops i.e. 3x − 1 = − 4
Worthless (0)
W1 Combines “x's”to “numbers” and stops.
W2 Ignores x giving 3 −1 = 2.
W3 − 4[ f ( x )] = − 12 x − 4 .
W4 Replaces coefficient .i.e. 3 x → − 4 x .
W5 Incorrect answer without work.
Page 35
Att 2
Att2
Part (b)
6(b)
25 (15, 10) marks
Draw the graph of the function
f : x → x 2 − 3x − 1
in the domain −1 ≤ x ≤ 4,
where x ∈ R
Part (b) Table
f (x )
=
f (− 1) =
*
f (0 )
=
f (1)
=
f (2 )
=
f (3)
=
f (4 )
=
Att 8 (5, 3)
15 marks
x2
(− 1)2
(0)2
(1)2
(2)2
(3)2
(4)2
− 3x
-1
− 3(− 1)
-1
=
3
− 3(0 )
-1
=
-1
− 3(1)
-1
=
-3
− 3(2 )
-1
=
-3
− 3(3)
-1
=
-1
− 3(4 )
-1
=
3
x
x2
− 3x
−1
f (x )
Att 5
-1
1
3
-1
3
0
0
0
-1
-1
1
1
-3
-1
-3
2
4
-6
-1
-3
Error(s) in each row /column attract a maximum deduction of 3.
Blunders (-3)
B1 “ − 3x ” taken as “ − 3 ”all the way. [In row headed − 3x by candidate]
B2 “ − 1 ” calculated as “ − x ” all the way.[In row headed “ − 1 ” by candidate]
B3 Adds in top row when evaluating f (x) .
B4 Omits “ − 1 ” row.
B5 Omits“ − 3 x ”row.
B6 Omits a value in the domain (each time).
B7 Each incorrect image without work.
Slips (-1)
S1 Numerical errors to a max of 3 in any row / column.
Misreadings (-1)
M1 Misreads “ x 2 ” as “ − x 2 ” and places “ − x 2 ” in the table or function
M2 Misreads “ − 3 x ”as “ 3 x ”and places “ 3 x ”in the table or function.
M3 Misreads “ − 1 ” as “ 1 ” and places “ 1 ” in the table or function
Attempts (5 marks)
A1 Omits “ x 2 ” row from table or treats “ x 2 ” as ± 2 x .
A2 Any effort at calculating point(s).
A3 Only one point calculated and stops.
Page 36
3
9
-9
-1
-1
4
16
-12
-1
3
Part (b) Graph
10 marks
Att 3
"
4
3
2
1
0
-2
-1
0
1
2
3
4
5
-1
-2
-3
-4
*
*
*
*
Accept candidate's values from previous work.
Only one correct point graphed correctly ⇒ Att 5 + Att 3
Correct graph but no table ⇒ full marks i.e. (15+10) marks.
Accept reversed co-ordinates if
(i) if axes not labelled or (ii) if axes are reversed to compensate (see B1 below)
Blunders (-3)
B1 Reversed co-ordinates plotted against non-reversed axes (once only) {See 4th * above}.
B2 Scale error (once only).
B3 Points not joined or joined in incorrect order (once only).
Slips (-1)
S1 Each point of candidate graphed incorrectly. {Tolerance ± 0.25 }
S2 Each point from table not graphed [See 2nd * above].
Attempts (3 marks)
A1 Axes drawn (need not be labelled).
A2 Some effort to plot a point{See 2nd * above}.
Page 37
Part(c) (i)
6(c) (i)
5 marks
Given that y = x + 2, complete the table below
Part(c) (i)
*
Att 2
5 marks
Att 2
x
-1
0
1
2
y
1
2
3
4
Accept candidate's values without work.
Slips(-1)
S1 Each “ y ” value omitted or incorrect.
Misreadings (-1)
M1 Misreads y = x + 2 as y = x − 2 .
Attempts(2marks)
A1 Any one correct “ y ” value.
A2 Any effort at calculating points.
Worthless (0)
W1 Copies x values into y row.
W2 All y values incorrect with no work shown. (See M1 above).
Page 38
Part(c) (ii)
6(c) (ii)
5 marks
Att 2
On the grid below, the graph of the line y = 4 − x is drawn.
Using your answers from (i), draw the graph of y = x + 2 on the same grid.
Part(c) (ii) Graph
5 marks
Att 2
5
4
3
2
y=x+2
1
y=4-x
0
-2
-1
0
1
2
3
4
-1
*
Accept candidate's values from previous work.
Blunders (-3)
B1 Reversed co-ordinates ( y , x ) plotted.
B2 Points not joined or joined in incorrect order.
Slips (-1)
S1 Each point of candidate graphed incorrectly. {See B1}
S2 Each point from table not graphed or not contained on the candidate’s graph.
Attempts (2 marks)
A1 Any one correct point plotted.
Worthless (0)
W1 No correct point plotted. (See B1 above).
Page 39
5
Part(c) (iii) Intersection
5 marks
Att2
6(c) (iii)
Use the graphs drawn in 6 (c) (ii) to write down the coordinates
of the point of intersection of the two lines y = 4 − x and y = x + 2 .
Part(c) (iii) Intersection
"
*
5 marks
Point of intersection = (1, 3)
Accept correct answer based on candidate's graph fully plotted.
i.e. 4 points correctly plotted from c(i) otherwise attempt marks at most.
Blunders(-3)
B1 Answer not presented in designated box.
B2 Answer beyond tolerance. {Tolerance ± 0.25 }.
B3 Reversed co-ordinates ( y , x ) plotted.
Attempts(2marks)
A1 Point of intersection highlighted on graph.
Worthless(0)
W1 Answers outside of tolerance without graphical indication.
W2 Incorrect answer from candidate's graph.
Page 40
Att2
MARKING SCHEME
JUNIOR CERTIFICATE EXAMINATION 2007
MATHEMATICS - ORDINARY LEVEL - PAPER 1
GENERAL GUIDELINES FOR EXAMINERS
1.
Penalties of three types are applied to candidates’ work as follows:
• Blunders - mathematical errors/omissions
(-3)
• Slips- numerical errors
(-1)
• Misreadings (provided task is not oversimplified)
(-1).
Frequently occurring errors to which these penalties must be applied are listed in the
scheme. They are labelled: B1, B2, B3,…, S1, S2,…, M1, M2,…etc. These lists are not
exhaustive.
2.
When awarding attempt marks, e.g. Att(3), note that
• any correct, relevant step in a part of a question merits at least the attempt mark for
that part
• if deductions result in a mark which is lower than the attempt mark, then the attempt
mark must be awarded
• a mark between zero and the attempt mark is never awarded.
3.
Worthless work is awarded zero marks. Some examples of such work are listed in the
scheme and they are labelled as W1, W2,…etc.
4.
The phrase “hit or miss” means that partial marks are not awarded – the candidate
receives all of the relevant marks or none.
5.
The phrase “and stops” means that no more work is shown by the candidate.
6.
Special notes relating to the marking of a particular part of a question are indicated by
an asterisk. These notes immediately follow the box containing the relevant solution.
7.
The sample solutions for each question are not intended to be exhaustive lists – there
may be other correct solutions.
8.
Unless otherwise indicated in the scheme, accept the best of two or more attempts –
even when attempts have been cancelled.
9.
The same error in the same section of a question is penalised once only.
10.
Particular cases, verifications and answers derived from diagrams (unless requested)
qualify for attempt marks at most.
11.
A serious blunder, omission or misreading results in the attempt mark at most.
12.
Do not penalise the use of a comma for a decimal point, e.g. €5.50 may be written as
€5,50.
Page 1
QUESTION 1
Part (a)
Part (b)
Part (c)
Part (a) (i)
1(a) (i)
10(5, 5) marks
20(5, 5, 5, 5) marks
20(5, 5, 5, 5) marks
Att 4(2, 2)
Att 8(2, 2, 2, 2)
Att 8(2, 2, 2, 2)
5 marks
Att 2
Using the Venn diagram below, shade in the region that represents A ∪ B.
A
B
Part (a) (i)
5 marks
A
Att 2
B
Blunders (-3)
B1 Any incorrect indication other than the misreading below.
Misreadings (-1)
M1 A ∩ B indicated.
Page 2
Part (a) (ii)
5 marks
Att 2
1(a) (ii)
Using the Venn diagram below, shade in the region that represents A ∩ B.
A
B
Part (a) (ii)
5 marks
A
Att 2
B
Blunders (-3)
B1 Any incorrect indication other than the misreading below.
Misreading (-1)
M1 A∪ B indicated.
Page 3
Part (b)
1(b)
U is the universal set
20(5, 5, 5, 5) marks
Att(2,2,2,2)
P
P = {4, 7, 8}
4•
•7
R = {1, 2, 3, 6, 7}
•5
•8
Q = {1, 2, 5, 7, 8}
•3
1•
•2
6•
R
Part (b) (i)
1(b) (i)
List the elements of: P ∪ Q.
Part (b) (i)
Q
•9
5 marks
5 marks
P ∪ Q = {1, 2, 4, 5, 7, 8.}
Att 2
Att 2
Blunders (-3)
B1 Any incorrect set of elements of P and Q other than the misreading as below.
Misreadings (-1)
M1 P ∩ Q giving {7, 8}.
Attempts ( 2 marks)
A1 3 or 6 or 9 appear in the answer.
Part (b) (ii)
1(b) (ii) List the elements of: P \ R.
Part (b) (ii)
5 marks
Att 2
5 marks
P \ R = { 4, 8}
Att 2
Blunders (-3)
B1 Any incorrect set of elements of P and R other than the misreading as below.
e.g. { P \ (Q ∪ R )} = {4} .
Misreadings (-1)
M1 R \ P giving {1, 2, 3, 6}.
Attempts (2 marks)
A1 5 or 9 appear in the answer.
Page 4
Part (b) (iii)
1(b) (iii)
5 marks
Att 2
List the elements of: (P ∪ R ) ∩ Q
Part (b) (iii)
5 marks
(P ∪ R ) ∩ Q = {1, 2, 7, 8}
Att 2
Blunders (-3)
B1 Any incorrect set of elements of P and Q and R other than the misreading as below.
B2 P ∪ R = {1,2,3,4,6,7,8} .
Misreadings (-1)
M1 (P ∩ R ) ∪ Q giving {1, 2, 5, 7, 8}.
M2 (P ∪ R ) ∪ Q giving {1, 2, 3, 4, 5, 6, 7, 8}.
M3 (P ∩ R ) ∩ Q giving {7}.
Attempts (2 marks)
A1 9 appears in the answer.
Part (b) (iv)
1(b) (iv)
List the elements of: (P ∪ Q )
Part (b) (iv)
5 marks
Att 2
5 marks
Att 2
/
(P ∪ Q )/ = {3, 6, 9}
Blunders (-3)
B1 Any incorrect set of elements of P and Q other than the misreading as below.
B2 (P ∪ Q ) = {1, 2, 4, 5, 7, 8} in this part.
Misreadings (-1)
/
M1 (P ∩ Q ) giving {1, 2, 3, 4, 5, 6, 9,}.
M2 P / ∪ Q / giving {1, 2, 3, 4, 5, 6, 9}.
Attempts (2 marks)
A1 Any incorrect listing of elements other than the misreadings above.
Page 5
Part (c)
20(5,5,5,5)marks
1(c) In a class, all the students study Science (S) or Technical Graphics (T).
A number of the students study both of these subjects.
22 students study Science. 12 students study Technical Graphics
8 study both subjects.
Part (c) (i)
1(c) (i)
5 marks
Att(2,2,2,2)
Att 2
Represent this information in the Venn diagram below.
S
T
Part (c) (i)
5 marks
Att 2
S
T
14
4
8
Blunders (-3)
B1 Incorrect Venn diagram subject to S1 below.
Slips (-1)
S1 Numerical errors where work is clearly shown to a max of 3.
Misreadings (-1)
M1 Interchanges Technical Graphics and Science.
Attempts (2 marks)
A1 Any one correct relevant entry.
A2 Incorrect work with numbers 8, 12, and 22. (work shown)
Page 6
Part(c) (ii)
1(c) (ii)
5 marks
How many students study Science only?
Att 2
Part(c) (ii)
*
*
*
5 marks
Att 2
14
A correct answer written here in the space provided takes precedence over an incorrect
Venn diagram.
Accepts candidates work from previous part c (i).
If no work appears here, award 2 marks if the correct answer appears in the Venn
diagram.
Blunders (-3)
B1 Any incorrect use of the given numbers or the numbers from the candidates incorrect
Venn diagram. {Subject to S1}.
Slips (-1)
S1 Numerical errors where work is clearly shown to a max of 3.
Misreadings (-1)
M1 Science read as Technical Graphics.
Attempts (2 marks)
A1 Incorrect work with numbers 8, and/or 22. ( work shown)
Part(c) (iii)
1(c) (iii)
5 marks
How many students are there in the class?
Att 2
Part(c) (iii)
*
*
5 marks
Att 2
26
A correct answer written here in the space provided takes precedence over an incorrect
Venn diagram.
Accepts candidates work from previous part c (i), c (ii).
Note: Answer c (ii) + 12 added correctly merits full marks.
Blunders (-3)
B1 Any incorrect use of the given numbers or numbers from the candidates incorrect Venn
diagram. {Subject to 2nd * above}.
Slips (-1)
S1 Numerical errors where work is clearly shown to a max of 3.
S2 Written as 14 + 8 + 4.
Attempts (2 marks)
A1 Incorrect work with numbers 14, 8,4,12 or 22.
Page 7
Part(c) (iv)
1(c) (iv)
5 marks
How many students study only one of the two subjects?
Part(c) (iv)
Att 2
5 marks
Att 2
18
*
A correct answer written here in the space provided takes precedence over an incorrect
Venn diagram.
*
Accepts candidates work from previous part c (i), c (ii) and c (iii).
Note: Answer c (iii) - 8 merits full marks.
Blunders (-3)
B1 Any incorrect use of the given numbers or numbers from the candidates incorrect Venn
diagram. {Subject to 2nd * above}.
Slips (-1)
S1 Numerical errors where work is clearly shown to a max of 3.
S2 Written as 14 + 4.
Attempts (2 marks)
A1 Incorrect work with numbers 14, 8, 4, 12, or 22.
Page 8
QUESTION 2
Part (a)
Part (b)
Part (c)
10 marks
20(5, 10, 5) marks
20(5, 5, 10) marks
Att 3
Att 7(2, 3, 2)
Att 7(2, 2, 3)
Part (a)
10 marks
2(a) €6650 was shared between Ciarán and Sheila in the ratio 2:5.
How much did each receive?
Att 3
Part (a)
Att 3
10 marks
"
2 parts : 5 parts
6650
⇒
= 950
7
Ciaran = 950 × 2 = €1900
Sheila = 950 × 5 = €4750
2+5=7
1
= 950
7
2
⇒ = €1900 (C)
7
⇒ 6650-1900 = €4750 (S)
2x :5x
⇒ 7 x = 6650
⇒ x = 950
⇒ 2 x = €1900 (C)
⇒ 5 x = €4750 (S)
Blunders (-3)
B1
B2
B3
B4
B5
B6
"
Correct answer without work.
Divisor ≠ 7 only and continues.
Incorrect multiplier or fails to multiply. (each time).
Error in transposition.
Fails to find second amount.
Adds instead of subtracts. e.g. 6650 + 1900 = 8550.
Slips (-1)
S1 Numerical errors to a max of 3.
Misreadings (-1)
M1 Interchanges Ciaran and Sheila.
Attempts (3 marks)
A1
A2
A3
A4
6650
6650
and/or
and stops.
2
5
2
5
Indicates 7 parts or 2 parts or 5 parts or or or 2+5=7 and stops.
7
7
Indicates multiplication of 6500 by 2 and/or 5 and stops.
Both answers added together equal €6650. (No work shown).
Divisor ≠ 7 e.g.
Worthless (0)
W1 Incorrect answer without work. {Subject to A4}.
Page 9
Part (b) (i)
5 marks
2(b) (i)
Simplify
a 8 × a 10
, giving your answer in the form, a n where n ∈ N.
5
7
a ×a
Part (b) (i)
5 marks
"
a ×a
a
= 12 = a 6
5
7
a ×a
a
8
10
Att 2
18
⎛ a8
⎜⎜ 5
⎝a
⎞ ⎛ a 10
⎟⎟ × ⎜⎜ 7
⎠ ⎝a
Att 2
⎞ 3
⎟⎟ = a × a 3 = a 6
⎠
a
a a
× ……. ……. = a 6
a a
a
Blunders (-3)
B1 Correct answer without work.
B2 Error in calculation involving indices.
B3 Error in number of a’s in the extended form.
B4 Error in elimination in the extended form.
B5 Fails to finish.
"
Slips (-1)
a 18
S1
=6
a 12
S2 Answer left a × a × a × a × a × a .
Attempts (2 marks)
A1
Some correct manipulation of indices. e,g, 8+10,
Worthless (0)
W1 Incorrect answer without work.
Page 10
18 3 5
, a , a , or a and stops.
12
Part (b) (ii)
2(b) (ii)
10 marks
Att 3
By rounding each of these numbers to the nearest whole number,
24 ⋅ 092
estimate the value of
.
6 ⋅ 1 − 2 ⋅ 93
Part (b) (ii)
10 marks
" 6 24⋅1 −⋅ 092
2 ⋅ 93
Att 3
is approximately equal to
24
6
_
24
=
3
=
8
3
24
and stops ⇒ 4 marks.
6−3
*
No penalty if the intermediate step between approximations and final answer is not
shown.
24
not shown.
i.e
3
24 ⋅ 092
*
Special Case:
= 7 ⋅ 6 presented in this part ⇒ Attempt 3 marks.
6 ⋅ 1 − 2 ⋅ 93
Blunders (-3)
B1 Correct answer without work.
B2 Error(s) in rounding off to the nearest whole number.
B3 Decimal error in calculation of final value.
B4 An arithmetic operation other than indicated.
24 24
−
or similar.
B5 Error(s) in the manipulation of the denominator. e.g.
6 3
*
"
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 Only one or two approximations made to the given numbers and stops.
A2 No rounding off applied to given numbers.
Worthless (0)
W1 Incorrect answer without work.
Page 11
Part (b) (iii)
2(b)(iii)
Part (b) (iii)
5 marks
Using a calculator, or otherwise, find the exact value of
Att 2
24 ⋅ 092
.
6 ⋅ 1 − 2 ⋅ 93
5 marks
24 ⋅ 092
24 ⋅ 092
=
=7 ⋅6
6 ⋅ 1 − 2 ⋅ 93
3 ⋅ 17
Blunders (-3)
B1 Decimal error.
24 ⋅ 092
B2 Treats as
− 2.93 = 3 ⋅ 949508197 − 2 ⋅ 93 =1 ⋅ 019 ….{B1 may occur}.
6 ⋅1
24 ⋅ 092
B3 Treats as 24 ⋅ 092 −
= 24 ⋅ 092 − 8 ⋅ 2225 = 15 ⋅ 8695. . {B1 may occur}.
2 ⋅ 93
24 ⋅ 092
24 ⋅ 092
B4 Treats as
=
= 2 ⋅ 66799 . {B1 may occur}.
6 ⋅ 1 + 2 ⋅ 93
9 ⋅ 03
24 ⋅ 092
24 ⋅ 092
B5 Treats as
=
= 1 ⋅ 347955016 . {B1 may occur}.
6 ⋅ 1 × 2 ⋅ 93 17 ⋅ 873
24 ⋅ 092
B6 Treats as 24 ⋅ 092 −
= 24 ⋅ 092 − 7 ⋅ 6 = 16 ⋅ 492. {B1 may occur}.
3 ⋅ 17
24 ⋅ 092 24 ⋅ 092
B7 Treats as
−
= 3 ⋅ 9495 − 8 ⋅ 2225.... = − 4 ⋅ 2755 . {B1 may occur}.
6 ⋅1
2 ⋅ 93
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Any rounding off.
Attempts (2 marks)
A1 Any correct relevant calculation and stops.
Worthless (0)
W1 Incorrect answer without work.
Page 12
Att 2
Part(c) (i)
2(c) (i)
5 marks
Att 2
1
2
Using a calculator, or otherwise, find the exact value of (2 ⋅ 25) .
Part(c) (i)
5 marks
1
2
(2 ⋅ 25) =
Att 2
3
=1·5
2
Blunders (-3)
1
B1
B2
Mishandles (2 ⋅ 25) 2 e.g. (2 ⋅ 25) = 5 ⋅ 0625 .
Decimal error.
2
Attempts (2 marks)
A1
2 ⋅ 25 and stops.
1
A2 2 ⋅ 25 × = 1 ⋅ 125 .
2
Worthless(0)
W1
2 ⋅ 25 × 2 or 2 ⋅ 75 .
Part(c) (ii)
2(c) (ii)
5 marks
Att 2
Using a calculator, or otherwise, multiply 54·5 by 60
and express your answer in the form a × 10 n , where 1 ≤ a < 10 and n ∈ N
Part(c) (ii)
"
5 marks
54 ⋅ 5 × 60 = 3270 = 3 ⋅ 27 ×10 3
Blunders (-3)
B1
B2
Correct answer without work.
Decimal error.
"
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Rounds off to, 3 ⋅ 3×10 3 or 3 ⋅ 0 ×10 3
S3 Incorrectly rounds off. e.g. 3 ⋅ 2 ×10 3 also attracts S2.
S4 Incorrect format, where a < 1 or a ≥ 10 and n ∉ N.
Attempts (2 marks)
A1 Any relevant step. e.g. Partial multiplication.
Page 13
Att 2
Part(c) (iii)
2(c) (iii)
10 marks
Using a calculator, or otherwise, evaluate
(6 ⋅ 9) 2 − 139 ⋅ 8 ÷ 3 ⋅ 55.
Give your answer correct to two decimal places
Part(c) (iii)
"
*
10 marks
Att 3
Att 3
= 47 ⋅ 61 − 11 ⋅ 823705 ÷ 3 ⋅ 55
= 47 ⋅ 61 − 3 ⋅ 330621127
= 44 ⋅ 27937887
= 44 ⋅ 28
Correct answer (without work) incorrectly rounded off ⇒ 6 marks
Blunders (-3)
"
B1
B2
Correct answer without work.
Mishandles (6 ⋅ 9) 2 .
B3
B4
B5
B6
B7
B8
B9
Mishandles 139 ⋅ 8 .
Error in 11 ⋅ 823705 ÷ 3 ⋅ 55 or candidate's equivalent from previous work.
Error in 47 ⋅ 61 − 3 ⋅ 330621127 or candidate's equivalent from previous work.
Decimal error.
Subtracts before Division 35 ⋅ 786295 ÷ 3 ⋅ 55 = 10 ⋅ 08064648 = 10 ⋅ 08 {Note S2,S3 }
Use of mathematical operator other than that which is indicated.
Works as 47 ⋅ 61 ÷ 3 ⋅ 55 − 11 ⋅ 823705 = 1 ⋅ 587562606 = 1 ⋅ 59 .{Note S2,S3 }
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Each premature rounding off that effects the final answer to a max of 3.
S3 Fails to round off or rounds off incorrectly when giving final answer.
Attempts (3 marks.)
A1 Any correct relevant step e.g. (6 ⋅ 9) 2 = 47 ⋅ 61 , 139 ⋅ 8 = 11 ⋅ 823705 .
Page 14
QUESTION 3
Part (a)
Part (b)
Part (c)
10 marks
20(10, 10) marks
20(10, 10) marks
Att 3
Att 6(3, 3)
Att 6(3, 3)
Part(a)
10 marks
3(a) In one week Bríd sent 26 text messages on her mobile phone
11 of these messages cost 8c each
The rest of the text messages cost 12c each.
Find the total cost of Bríd’s texting.
Att 3
Part (a)
Att 3
10 marks
"
26 – 11 = 15
11 × 8 = 88
15 × 12 = 180
Total Cost = 268 c (€ 2 ⋅ 68 )
*
*
*
26 – 11 = 15
8 + 8 …..11 Times = 88
12 + 12 ….15 Times = 180.
Total Cost = 268 c (€ 2 ⋅ 68 )
No penalty for omission of € symbol.
Accept 268c, (€ 2 ⋅ 68 )
Adds 8 + 12 = 20 and stops merits 3 marks (Oversimplification).
Blunders (-3)
B1
B2
B3
B4
B5
B6
B7
Correct answer without work. "
Fails to subtract 11 from 26.
Each missing product when finding each cost e.g. 11 not multiplied by 8.
Each missing item when finding total cost e.g. Expensive texts omitted.
Fails to find total cost i.e. no addition.
Operation other than addition when finding total cost.
Decimal error e.g. € 26 ⋅ 8 (Note: 1st* above).
Slips (-1)
S1 Numerical errors to a max of 3.
Misreadings (-1)
M1 15 texts @ 8c and 11 texts @ 12c.
Attempts (3 marks)
A1 Any attempt at addition /multiplication.
Worthless (0)
W1 Incorrect answer without work.
Page 15
Part (b) (i)
3(b) (i)
10 marks
John’s gross pay is €23 000. His tax credit is €3400
He pays income tax at the rate of 20%
Find John’s take-home pay.
Part (b) (i)
10 marks
"
Gross Pay
Tax @ 20%
Tax Credit
Tax-Due
Take-home Pay
€23 000
€4600
€3400
€1200
€21,800
20
= 4600
100
or
23,000 × 0.2 = 4600
Tax Due = 4600 – 3400
= 1200
Take-home Pay = 23000 - 1200
Tax = 23,000 ×
= €21,800
Blunders (-3)
B1
B2
B3
B4
B5
B6
Correct answer without work. "
Mishandles 20% of 23,000. {Must use 23 000}
Decimal error.
Misuse of Tax Credit
Incorrect use of Tax Amount.
Fails to finish. {B4 may apply}
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (3 marks)
20
A1 Some use of 100 in attempt to find percentage e.g. 20% = 100
and stops.
Worthless (0)
W1 Incorrect answer without work
Page 16
Att 3
Att 3
Part (b) (ii)
3(b) (ii)
10 marks
VAT at 21% is added to a bill of €255
Calculate the total bill.
Part (b) (ii)
Att3
10 marks
Att3
"
100 % = 255
255
100
255
121 % =
× 121
100
= 2 ⋅ 55 121
1%
=
Total Bill = € 308 ⋅ 55
*
21
100
21
VAT =
× 255
100
= 53.55
21 % =
255 ×1 ⋅ 21
Total Bill = € 308 ⋅ 55
Total Bill = 255 + 53 ⋅ 55
Total Bill = € 308 ⋅ 55
€ 53⋅ 55 without work and stops merits 4 marks.
Blunders (-3)
B1
B2
B3
B4
B5
B6
B7
Correct answer without work. "
Decimal error
121
21
or
and continues (giving answers 210 ⋅ 74 or 1214 ⋅ 29 ).
Inverts
100
100
Mishandles 21%. e.g.255 × 21 or 255 ÷ 21. Note: {255 must be used}.
255 taken as 121%.
No addition of VAT (as per candidates work) to the bill.
Subtraction of VAT ( as per candidates work) from the bill.
Slips (-1)
S1 Numerical errors to a max of 3.
Misreadings (-1)
M1 Reads as €225.
Attempts (3 marks)
21
and stops.
A1
100
A2 100% = 255 and stops.
255
A3
and stops.
100
21
A4 100 ×
and stops.
255
255
and stops.
A5
21
A6 Use of any other %.
A7 255 + 21% and stops.
Page 17
Part(c) (i)
3(c) (i)
10 marks
€15 000 is invested at 3% per annum
What is the amount of the investment at the end of the first year?
Part(c) (i)
10 marks
Att 3
Att 3
"
15000
100
15000
3%=
×3
100
Interest = 450
1%=
P×R
100
15000 × 3
I =
100
I = 450
I =
Amount = 15000 + 450
Amount = 15000 + 450
Amount = €15 450
Amount = €15 450
*
Amount =15000 × 1 ⋅ 03
Amount = €15 450
€450 (without work) and stops ⇒ 4 marks.
Blunders (-3)
B1
B2
B3
B4
B5
B6
B7
Correct answer without work. "
15000 × 100
Mishandles 3 %. e.g.
Note:{15000 must be used}.
3
Decimal error (once only).
Stops at interest i.e. fails to calculate amount.
Subtracts to calculate amount.
15000 × 3
Mathematical error(s) working with
.
100
1 ⋅ 03 treated as 1 ⋅ 3 .
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 Correct formula with or without substitution and stops.
A2
Some use of 100 in attempt to find percentage e.g. 3 % =
A3
15000 + 3% and stops.
Worthless (0)
W1 Incorrect answer without work.
Page 18
3
or 1 ⋅ 03 and stops.
100
Part(c) (ii)
3(c) (ii)
10 marks
Att 3
€1450 is withdrawn from this amount at the beginning of the second year.
The interest rate for the second year is 3·5%.
What is the amount of the investment at the end of that year?
Part(c) (ii)
10 marks
Att 3
"
Principal for second year = 15450 - 1450 = 14000
Amount =14000 × 1 ⋅ 035
P×R
14000
1% =
I =
100
100
Amount = €14 490
14000 × 3 ⋅ 5
14000
3.5 % =
× 3⋅5
I =
100
100
Interest = 490
I = 490
Amount = 14000 + 490
Amount = 14000 + 490
Amount = €14 490
Amount = €14 490
*
*
*
*
No penalty for consistent error(s) already penalised in (c) (i).
Accept candidates work from previous part (c) (i).
€490 (without work) and stops ⇒ 4 marks.
€14000 (without work) and stops ⇒ 3 marks.
Blunders (-3)
B1
B2
B3
B4
B5
B6
B7
B8
B9
Correct answer without work. "
Incorrect principal for second year.
Incorrect interest rate for second year.
14000 × 100
Mishandles 3.5 %. e.g.
see (1st * above)
3.5
Decimal error (once only).
Stops at interest i.e. fails to calculate amount.
Subtracts to calculate amount.
14000 × 3 ⋅ 5
Mathematical error(s) working with
100
1 ⋅ 035 treated as 1 ⋅ 35
Note:{14000 must be used}.
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 Correct formula with or without substitution and stops.
A2
Some use of 100 in attempt to find percentage e.g. 3 ⋅ 5 % =
A3
14000 + 3·5% and stops.
Worthless (0)
W1 Incorrect answer without work.
Page 19
3⋅5
or 1 ⋅ 035 and stops.
100
QUESTION 4
Part (a)
Part (b)
Part (c)
Part(a)(i)
4(a)(i)
15(10, 5) marks
15(10, 5) marks
20(10, 10) marks
10 marks
If x = 3 , find the value of :
(i)
Part(a)(i)
"
*
Att 3
4x + 5
10 marks
4x + 5 = 4(3) + 5 = 12 + 5 = 17
12 + 5 ⇒ 9 marks.
Blunders (-3)
B1
B2
B3
B4
B5
B6
Correct answer without work. "
Leaves 4(3) in the answer.
Incorrect substitution and continues
Combines “x's”to “numbers” and continues. e.g. 4 x + 5 = 9 x = 9(3) = 27 .
Breaks order i.e. [ 4(3 + 5) = 32 ].
Treats 4(3) as 7 or 43 or similar.
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 Substitutes for x and stops e.g. 4(3)
A2 Any correct step.
Worthless (0)
W1
Combines “x's”to “numbers” and stops.
Page 20
Att 5(3, 2)
Att 5(3, 2)
Att 6(3, 3)
Att 3
Part (a) (ii)
4(a) (ii)
5 marks
If x = 3 , find the value of : (ii)
Part (a) (ii)
2 x 2 − 11
5 marks
"
*
Att 2
2 x 2 − 11 = 2(3) − 11 = 2(9) - 11 = 18 - 11 = 7
2
18 - 11 ⇒ 4 marks.
Blunders (-3)
B1
B2
B3
B4
B5
B6
B7
B8
B9
Correct answer without work.
Leaves 2(9) in the answer.
2
2
Mishandles (3) e.g. (3) = 6.
"
Mishandles 2(3) e.g 2(3) = (6) .
Mathematical error. e.g. 18 - 11 = - 7.
Incorrect substitution and continues.
Combines “x's ”to “numbers” and continues. e.g. 2 x 2 − 11x = − 9 x 2
Breaks order i.e. [ 2(9 − 11) = − 4 ].
Treats 2(9) as 11 or 29 or similar.
2
2
2
Slips (-1)
S1 Numerical errors to a max of 3.
.
Attempts (2 marks)
2
A1 Substitutes for x and stops e.g. 2(3)
A2 Any correct step.
Worthless (0)
W1
Combines “ x 2 ” to “numbers” and stops.
Page 21
Att 2
Part (b) (i)
4(b) (i)
Part (b) (i)
"
10 marks
Solve the equation 4(5 x + 6) = 84 .
4(5 x + 6 ) = 84
20 x + 24 = 84
20 x
20 x
x
10 marks
= 84 − 24
= 60
=3
Att 3
4(5 x + 6 ) = 84
84
5x + 6
=
4
5x + 6
= 21
5x
= 21 − 6
5x
= 15
=3
x
Blunders (-3)
B1
B2
B3
B4
B5
Correct answer without work. " e.g. x = 3 stated or substituted.
Error in distributive law and continues,e.g. 20 x + 6 = 84 (once only).
Error in transposition. (each time).
Combines “x's”to “numbers” and continues. e.g.. 20 x + 24 = 44 x
Stops at 20 x = 60 or similar.
Slips (-1)
S1 Numerical errors to a max of 3.
60
or similar.
S2 Leaves as
20
Attempts (3 marks)
A1 Any correct step.
A2 Particular case verified for any value of x other than 3.
Worthless (0)
W1 Combines “x's” to “numbers” and stops.
Page 22
Att 3
Part (b) (ii)
4(b) (ii)
5 marks
Write in its simplest form
3 x 2 − 2 x + 6 − x(2 x − 3)
Part (b) (ii)
5 marks
3x − 2 x + 6 − x(2 x − 3)
2
"
3x 2 − 2 x + 6 − 2 x 2 + 3x
x2 + x + 6
Blunders (-3)
B1
B2
B3
B4
B5
Correct answer without work. "
Error(s) in distribution.
Combining unlike terms.
Fails to group or groups incorrectly.
Treats as 3x 2 − 2 x + 6 − x (2 x − 3) and continues.
(
)
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (2 marks)
A1 Any correct multiplication.e.g. 3 x
A2 Any correct grouping of terms.
A3 A correct step.
A4 Substitutes a value of “ x ” and continues.
Worthless (0)
W1 Combining unlike terms and stops.
W2 No attempt at distribution but A2 may apply to subsequent work.
Page 23
Att 2
Att 2
Part(c) (i)
10 marks
4(c) (i)
Liam drove from Town A to Town B, a distance of x km.
He then drove from Town B to Town C, a distance of (2x + 1) km.
The total distance that he drove was 56 km.
Find the value of x, correct to the nearest kilometre.
Att3
Part(c) (i)
Att3
10 marks
x + 2 x + 1 = 56
3x + 1 = 56
= 56 − 1
3x
= 55
3x
"
x
x
1
55
or 18 ⋅ 33333 or 18
3
3
= 18
=
Blunders (-3)
B1
B2
B3
B4
B5
B6
Correct answer without work. "
Error(s) in forming equation for distance travelled.
Error in grouping terms. e.g. 2 x + 1 = 56 and continues.(once only).
Error in transposition.(each time).
Combines “x's”to “numbers”. e.g. 4 x = 56 and continues.
Stops at 3x = 55 or candidate's equivalent. {S2 also applies}
Slips (-1)
S1 Numerical errors to a max of 3.
1
55
S2 Leaves as
or 18 ⋅ 333 or 18 or candidate's equivalent.
3
3
Attempts (3 marks)
A1 Any correct step.
A2 Illustrates information on a diagram and stops.
Worthless (0)
W1 Combines “x's” to “numbers” and stops.
W2 Incorrect answer no work e.g. x = 56 .
Page 24
Part(c) (ii)
4(c) (ii)
Solve for x and for y:
10 marks
Att3
3x + 5 y = 13
x + 2y = 5
.
Part(c) (ii)
"
10 marks
I
II
3 x + 5 y = 13
x + 2y = 5
3 x + 5 y = 13
x + 2y = 5
6 x + 10 y = 26
− 5 x − 10 y = − 25
x = 1
⇒ y=2
*
*
Att3
or
3 x + 5 y = 13
− 3x − 6 y = − 15
or
− y = −2
y = 2
⇒ x =1
x = 5 − 2y
3(5 − 2 y ) + 5 y = 13
15 − 6 y + 5 y = 13
−y = −2
y =2
⇒ x =1
Apply only one blunder deduction (B2 or B3) to any error(s) in establishing the first
equation in terms of x only or the first equation in terms of y only.
Finding the second variable is subject to a maximum deduction of (3).
Blunders (-3)
B1
B2
Correct answers without work. " e.g. x =1, y = 2. stated or substituted.
Error(s) in establishing the first equation in terms of x only [ x = 1 ] or the first
equation in terms of y only [ − y = − 2 ] through elimination by cancellation.
B3
Error(s) in establishing the first equation in terms of x only [ x = 5 − 2 y ] or the first
equation in terms of y only [ 5 y = 13 − 3x ] through elimination by substitution.
B4
B5
B6
B7
Errors in transposition in solving the first one variable equation.
Errors in transposition when finding the second variable.
Incorrect substitution when finding second variable.
Finds one variable only.
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (3 marks)
A1 Attempt at transposition and stops.
A2 Multiplies either equation by some number and stops.
Page 25
QUESTION 5
Part (a)
Part (b)
Part (c)
10 marks
20(5, 5, 5, 5) marks
20(5, 5, 10) marks
Part (a)
10 marks
5(a)
Att 3
Find the values of x for which 3x + 2 < 11, x ∈ N
Part (a)
"
10 marks
3x + 2 < 11
3x < 11 − 2
3x < 9
x <3
{ 0, 1, 2}
Blunders (-3)
B1
B2
B3
B4
B5
B6
B7
Att 3
Att 8(2, 2, 2, 2)
Att 7(2, 2, 3)
Correct answer without work. "
Error in transposition. (each time).
Combining unlike terms.
Mishandles the direction of inequality e.g. 3 x > 9
Treats inequality as equality and continues. {S3 may apply}
Combines “x's” to “numbers”. e.g.. 5 x < 11 and continues.
x < 3 and stops.
Slips (-1)
S1 Numerical errors to a max of 3.
S2 < taken as ≤ .
S3 No listing or incorrect listing of values. {Subject to max penalty of 3}.
Misreadings (-1)
M1 3x + 2 < 1 , and continues.
Attempts (3 marks)
A1 Attempt at transposition and stops.
A2 Particular case verified.
Page 26
Att 3
Part (b) (i)
5(b) (i)
Factorise:
Part (b) (i)
5 marks
16 xy + 11 y
Att 2
5 marks
y (16 x + 11)
Att 2
Blunders (-3)
B1 An incorrect factor
B2 Removes factor incorrectly.
Attempts (2 marks)
A1 Indication of common factor. e.g. underline y 's and stops.
Part (b) (ii)
5(b) (ii)
Factorise:
Part (b) (ii)
"
*
5 marks
5 x + 10 y + ax + 2ay
Att 2
5 marks
5 x + 10 y + ax + 2ay
Att 2
5 x + 10 y + ax + 2ay
or
x(5 + a ) + 2 y (5 + a )
5( x + 2 y ) + a ( x + 2 y )
(5 + a )(x + 2 y )
(x + 2 y )(5 + a )
Accept also (with or without brackets) for 5 marks any of the following
(5 + a ) and (x + 2 y ) {The word and is written down.}
(5 + a ) or (x + 2 y ) {The word or is written down.}
(5 + a ) , (x + 2 y ) {A comma is used}
Blunders (-3)
B1
B2
B3
B4
Correct answer without work. "
Stops after first line of correct factorisation. e.g. . 5( x + 2 y ) + a (x + 2 y ) or equivalent.
Error(s) in factorising any pair of terms.
Incorrect common factor and continues. e.g. 2(ay + 5 y ) + x(a + 5)
Slips (-1)
S1 (5 + a ) ± ( x + 2 y )
S2 Correct first line of factorisation but ends as 5a ( x + 2 y ) .
Attempts (2 marks)
A1 Pairing off, or indication of common factors and stops.
A2 Correctly factorises any pair and stops.
Page 27
Part (b) (iii)
5(b) (iii)
5 marks
Att 2
x − x − 90
2
Factorise:
Part (b) (iii)
5 marks
Att 2
x 2 − x − 90
x
x 2 + 9 x − 10 x − 90
x( x + 9 ) − 10( x + 9 )
(x − 10)(x + 9)
+9
x
−10
⇒ ( x − 10 )( x + 9 )
*
− (− 1) ±
(− 1)2 − 4(1)(− 90)
2(1)
1 ± 19
1 ± 1 + 360
=
2
2
− 18
20
= 10
=−9
2
2
⇒ ( x − 10 )( x + 9)
Accept also (with or without brackets) for 5 marks any of the following
(x − 10) and (x + 9) {The word and is written down.}
(x − 10) or (x + 9) {The word or is written down.}
(x − 10) , (x + 9) {A comma is used}
Incorrect two term linear factors of. x 2 − x − 90 formed from correct (but inapplicable)
factors of x 2 and − 90 .e.g (x − 45)( x + 2 ) .
B2 Incorrect factors of x 2 .
B3 Incorrect factors of − 90 .
B4 Correct cross method but factors not shown and stops.
B5 x( x + 9) − 10( x + 9 ) or similar and stops.
B6 Incorrect common factor and continues.
B7 Incorrect quadratic formula and continues.
B8 Error in quadratic formula. (each time).
B9 Answer left as roots.
B10 Sign error(s) in substituted formula.
B11 Error in square root or square root ignored.
B1
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Uses quadratic equation formula, but has wrong sign in factors.
Attempts (2 marks)
A1 Correct quadratic equation formula quoted and stops
A2 Correct factors of either x 2 or ± 90.
A3 Any correct step.
Worthless (0 marks)
W1 x 2 − x = 90 or similar and stops.
W2 Combines “x's”to “numbers” and continues or stops.
Page 28
Part (b) (iv)
5 marks
5(b) (iv)
Factorise:
Att 2
x 2 − 121
Part (b) (iv)
*
*
*
5 marks
(x − 11)(x + 11)
Accept also (with or without brackets) for 5 marks any of the following
(x − 11) and (x + 11) {The word and is written down.}
(x − 11) or (x + 11) {The word or is written down.}
(x − 11) , (x + 11) {A comma is used}
Quadratic equation formula method is subject to slips and blunders.
x − 121 x + 121 merits 5 marks.
(
)(
Att 2
)
Blunders (-3)
B1 Incorrect two term linear factors of x 2 − 121 formed from correct (but inapplicable)
factors of x 2 and − 121 .e.g (x − 121)(x + 1) .
B2 Incorrect factors of x 2 .
B3 Incorrect factors of −121 .
B4 (11 − x )(11 + x ) .
B5 (x − 121)(x + 121) .
B6 Answer left as roots.
Slips (-1)
S1 (x − 11) ± (x + 11)
Attempts (2 marks)
A1 Correct factors of x 2 only.
A2 Correct factors of ± 121 only.
A3 x or ± 11 appears.
A4 x 2 − 121 = x.x − 11.11 and stops.
A5
A6
A7
Mention of the difference of two squares .e.g. { x 2 − (121) }
Correct quadratic equation formula quoted and stops.
121
2
Worthless (0 marks)
W1 Combines “x's” to “numbers” and continues or stops.
Page 29
Part(c) (i)
5(c)(i)
5 marks
Att2
2x − 1
x+7
+
as a single fraction.
5
2
Give your answer in its simplest form.
Express
Part(c) (i)
"
5 marks
2x − 1
x+7
+
5
2
2(2 x − 1) + 5( x + 7 )
10
4 x − 2 + 5 x + 35
10
9 x + 33
10
Att2
2x − 1
3x + 6
x + 7
Zero marks.
+
=
5
2
7
Blunders (-3)
*
B1
B2
B3
B4
Correct answer without work. "
Error(s) in distribution. e.g 2(2 x − 1) = 4 x − 1 .
Mathematical error e.g. -2 +35 = -33. 2(-1) = 2.
Incorrect common denominator and continues.
B5
Incorrect numerator from candidate's denominator e.g.
B6
B7
No simplification of numerator.
Omitting denominator.
Slips (-1)
S1 Drops denominator.
S2 Numerical error to a max of 3.
S3
Answer not in simplest form. e.g.
18 x + 66
.
20
Attempts (2 marks)
A1 10 only or a multiple of 10 only appears.
A2 Any correct step.
Worthless (0)
x 8x
W1
+
, or
5
2
⎛ 2 x − 1 ⎞⎛ x + 7 ⎞
⎟ and stops.
⎟⎜
⎜
⎝ 5 ⎠⎝ 2 ⎠
Page 30
5(2 x − 1) + 2( x + 7 )
.
10
Part(c) (ii)
5(c) (ii)
5 marks
Hence, or otherwise, solve the equation
Part(c) (ii)
*
2x − 1 x + 7
+
= 6.
5
2
5 marks
9 x + 33
10
9 x + 33
9x
9x
"
Att2
= 6
= 60
= 60 − 33
= 27
= 3
x
Accept candidates answer from previous work.
Blunders (-3)
B1
B2
Correct answer without work. "
Error in transposition. (each time)
Slips (-1)
S1 Numerical error to a max of 3.
27
.
S2 Leaves as
9
Attempts (2 marks)
A1 Answer from (c) (i) written in this part or worked again in this part.
A2 Any correct step and stops.
A3 Particular case verified.
Page 31
Att2
Part(c) (iii)
10 marks
5(c) (iii)
Solve the equation:
Att3
x 2 + 5 x − 36 = 0.
Part(c) (iii)
10 marks
Att3
"
− (5) ±
x 2 + 5 x − 36 = 0
x 2 + 9 x − 4 x − 36 = 0
x( x + 9 ) − 4( x + 9 ) = 0
x
+9
⇒ x=−9
x
−4
(x + 9)(x − 4) = 0
x =4
(5)2 − 4(1)(− 36)
2(1)
− 5 ± 25 + 144 − 5 ± 169 − 5 ± 13
=
=
2
2
2
8
− 18
=4
=−9
2
2
(x + 9)(x − 4)
⇒ x =−9
x =4
Blunders (-3)
B1
B2
Correct answers without work. " e.g. x = 4, x = − 9 stated or substituted.
Incorrect two term linear factors of x 2 + 5 x − 36 formed from correct (but
inapplicable) factors of x 2 and − 36 . e.g. (x − 12 )( x + 3)
B3
B4
B5
B6
B7
B8
B9
B10
Incorrect factors of x 2 .
Incorrect factors of − 36 .
Correct cross method and factors not shown and stops.{B8 also applies}
x( x + 9) − 4( x + 9 ) or similar and stops. {Note: B8 also applies }.
Incorrect root(s) from factors.
No roots given.
One root only
Error in quadratic formula. (each time).
Slips (-1)
S1 Numerical errors to a max of 3.
p
S2 Leaves as .
q
Attempts (3 marks)
A1 Correct factors of x 2 only
A2 Correct factors of ± 36 only.
A3 Some effort at factorisation.
A4 Correct quadratic equation formula quoted and stops
A5 Any correct step.
Worthless (0)
W1 Combines unlike terms and continues or stops.
Page 32
QUESTION 6
Part (a)
Part (b)
Part (c)
Part(a)
6(a)
10(5, 5) marks
25(15, 10) marks
15(5, 5, 5) marks
10(5,5) marks
P = {(1,3) (4,6) (5,8) (7,9)}
Write out the domain and range of P.
Part(a) Domain
Att 4(2, 2)
Att 8(5, 3)
Att 6(2, 2, 2)
Att 4(2,2)
5 marks
Domain
=
Att 2
{1, 4, 5, 7}
Slips (-1)
S1 Each correct element omitted and/or each incorrect element included. {See M1}
Misreadings (-1)
M1 Correct range. i.e. { 3,6,8,9}given.
Attempts (2 marks)
A1 One element of domain.
A2 Domain { 1 → 7 }
Worthless (0)
W1 No element of the domain appears. {See M1}
Part(a) Range
5 marks
Range
=
{3, 6, 8, 9}
Slips (-1)
S1 Each correct element omitted and/or each incorrect element included. {See M1}
Misreadings (-1)
M1 Correct domain. i.e. {1, 4, 5, 7} given.
Attempts (2 marks)
A1 One element of range.
A2 Range{ 3 → 9 }
Worthless (0)
W1 No element of the range appears. {See M1}
Page 33
Att 2
Part (b)
6(b)
25(15, 10) marks
Draw the graph of the function
f : x → 2 + 3x − x 2
in the domain −1 ≤ x ≤ 4, where x ∈ R
Part (b) Table
Att 8(5, 3)
15 marks
Att 5
"
f ( −1) = 2 + 3 ( −1) − ( −1) =
-2
x
-1
0
1
2
3
4
f (0)
= 2 + 3 (0) − (0)
=
2
2
2
2
2
2
2
2
f (1)
= 2 + 3 (1) − (1)
2
=
4
+3x
-3
0
3
6
9
12
f ( 2)
= 2 + 3 ( 2) − ( 2)
=
4
−x 2
-1
0
-1
-4
-9
-16
f ( 3)
= 2 + 3 ( 3) − ( 3)
=
2
f (x )
-2
2
4
4
2
-2
f ( 4)
= 2 + 3 ( 4) − ( 4)
2
*
2
2
2
2
= -2
Error(s) in each row /column attract a maximum deduction of 3.
Blunders (-3)
B1 Treats − x 2 taken as x 2 and places '' x 2 '' in the table or function..
B2 − x 2 taken as −2x all the way. [In row headed − x 2 by candidate]
B3 + 3x taken as + 3 all the way. [In row headed + 3x by candidate]
B4 2 calculated as 2 x all the way.[In row headed 2 by candidate]
B5 Adds in top row when evaluating f (x) .
B6 Omits “2” row or omits “3 x ” row.
B7 Omits a value in the domain (each time).
B8 Each incorrect image without work.
Slips (-1)
S1 Numerical errors to a max of 3 in any row / column.
Misreadings (-1)
M1 Misreads “ + 3x ” as “-3 x ” and places “-3 x ” in the table or function
M2 Misreads “2”as “-2” and places “-2” in the table or function.
Attempts (5 marks)
A1 Omits − x 2 row from table or treats − x 2 as ±x.
A2 Any effort at calculating point(s).
A3 Only one point calculated and stops.
Page 34
Part (b) Graph
10 marks
Att 3
"
5
4
3
2
1
0
-2
-1
0
1
2
3
4
5
-1
-2
-3
*
*
*
*
Accept candidate's values from previous work.
Only one correct point graphed correctly ⇒ Att 5 + Att 3
Correct graph but no table ⇒ full marks i.e. (15+10) marks.
Accept reversed co-ordinates if
(i) if axes not labelled or (ii) if axes are reversed to compensate (see B1 below)
Blunders (-3)
B1 Reversed co-ordinates plotted against non-reversed axes (once only) {See 4th * above}.
B2 Scale error (once only).
B3 Points not joined or joined in incorrect order (once only).
Slips (-1)
S1 Each point of candidate graphed incorrectly. {Tolerance ± 0.25 }
S2 Each point from table not graphed [See 2nd * above].
Attempts (3 marks)
A1 Graduated axes (need not be labelled).
Page 35
Part(c) (i)
6(c) (i)
5 marks
Given that y = x + 1, complete the table below
Part(c) (i)
*
5 marks
Att 2
x
0
1
2
3
y
1
2
3
4
Accept candidate's values without work.
Slips(-1)
S1 Each y value omitted or incorrect.
Attempts(2marks)
A1 Any one correct value of y .
A2 Any effort at calculating point where work is shown.
Page 36
Att 2
Part(c) (ii)
6(c) (ii)
5 marks
Att 2
On the grid below, the graph of the line y = 3 − x is drawn.
Using your answers from (i), draw the graph of y = x + 1 on the same grid
Part(c) (ii) Graph
5 marks
Att 2
4
y =3−x
3
2
1
0
-2
-1
0
1
2
3
4
-1
-2
*
*
Accept candidate's values from previous work.
Only one point listed and graphed correctly ⇒ Att 2 + Att 2
Blunders (-3)
B1 Reversed co-ordinates ( y , x ) plotted.
B2 Points not joined or joined in incorrect order.
Slips (-1)
S1 Each point of candidate graphed incorrectly. {See B1}
S2 Each point from table not graphed or not contained on the candidate’s graph.
Attempts (2 marks)
A1 Any straight line drawn.
Page 37
5
Part(c)(iii) Intersection
5 marks
6(c)(iii)
Use the graphs drawn in 6 (c) (ii) to write down the coordinates
of the point of intersection of the two lines y = 3 − x and y = x + 1 .
Att2
Part(c)(iii) Intersection
Att2
"
5 marks
Point of intersection = (1,2)
*
Accept previous graph from c (ii).
Blunders(-3)
B1 Answer not presented in designated box.
B2 Answer beyond tolerance. {Tolerance ± 0.25 }.
Attempts(2marks)
A1 Indicates correctly either x or y co-ordinate of point of intersection.
A2 Point of intersection indicated.
A3 Algebraic evaluation.
Worthless(0)
W1 Answers outside of tolerance without graphical indication.
Page 38
Coimisiún na Scrúduithe Stáit
State Examinations Commission
JUNIOR CERTIFICATE EXAMINATION 2006
MATHEMATICS - ORDINARY LEVEL - PAPER 1
1.
GENERAL GUIDELINES FOR EXAMINERS
Penalties of three types are applied to candidates’ work as follows:
• Blunders - mathematical errors/omissions
(-3)
• Slips- numerical errors
(-1)
• Misreadings (provided task is not oversimplified)
(-1).
Frequently occurring errors to which these penalties must be applied are listed in the
scheme. They are labelled: B1, B2, B3,…, S1, S2,…, M1, M2,…etc. These lists are not
exhaustive.
2.
When awarding attempt marks, e.g. Att(3), note that
• any correct, relevant step in a part of a question merits at least the attempt mark for
that part
• if deductions result in a mark which is lower than the attempt mark, then the attempt
mark must be awarded
• a mark between zero and the attempt mark is never awarded.
3.
Worthless work is awarded zero marks. Some examples of such work are listed in the
scheme and they are labelled as W1, W2,…etc.
4.
The phrase “hit or miss” means that partial marks are not awarded – the candidate
receives all of the relevant marks or none.
5.
The phrase “and stops” means that no more work is shown by the candidate.
6.
Special notes relating to the marking of a particular part of a question are indicated by
an asterisk. These notes immediately follow the box containing the relevant solution.
7.
The sample solutions for each question are not intended to be exhaustive lists – there
may be other correct solutions.
8.
Unless otherwise indicated in the scheme, accept the best of two or more attempts –
even when attempts have been cancelled.
9.
The same error in the same section of a question is penalised once only.
10.
Particular cases, verifications and answers derived from diagrams (unless requested)
qualify for attempt marks at most.
11.
A serious blunder, omission or misreading results in the attempt mark at most.
12.
Do not penalise the use of a comma for a decimal point, e.g. €5.50 may be written as
€5,50.
Page 1
QUESTION 1
Part (a)
Part (b)
Part (c)
10 marks
20(5, 5, 5, 5) marks
20(5, 5, 5, 5) marks
Part (a)
(a) A = {a, b, c, d, e}
Att 3
Att (2, 2, 2, 2)
Att (2, 2, 2, 2)
10 marks
Att 3
B = {c, d, f, g}
Fill the elements of A and B into the following Venn diagram:
A
B
Part (a)
10 marks
Att 3
A
B
a b e
• • •
*
*
dc
• •
f g
• •
Only one correct element correctly placed in the Venn diagram merits 4 marks.
• Not necessary
Slips (-1)
S1 Each element incorrectly filled into the diagram.
S2 Each element omitted from the diagram.
Attempts (3 marks)
A1 Totally incorrect filling of the Venn diagram
Worthless (0)
W1 No filling of the Venn diagram.
Page 2
Part (b)
20(5, 5, 5, 5) marks
1(b) U is the universal set.
P = {1, 4, 5, 7}
Att (2, 2, 2, 2)
U
P
Q
•5
Q = {4, 6, 7, 9, 10}
•4
R = {1, 7, 8, 10}
•6
7
•9
•10
•1
•8
•2
•3
R
Part(b)(i)
(i) List the elements of Q ∪ R.
5 marks
Att 2
Part(b)(i)
Q ∪ R = {1, 4, 6, 7, 8, 9, 10}
5 marks
Att 2
Blunders (-3)
B1 Any incorrect set of the elements of Q and R other than the misreading as below.
Misreadings (-1)
M1 Q ∩ R giving {7, 10}.
Attempts (2 marks)
A1 2 or 5 or 3 appear in the answer.
Part (b) (ii)
(ii) List the elements of Q \ ( P ∪ R ).
5 marks
Att 2
Part (b) (ii)
Q \ ( P ∪ R ). = {6, 9}
5 marks
Att 2
Blunders (-3)
B1 Any incorrect set of elements of P and Q and R other than the misreading as below.
Misreadings (-1)
M1 ( P ∪ R ) \ Q giving {1, 5, 8}. Q \ ( P ∩ R ) giving {4,6,9,10} or ( P ∩ R ) \ Q giving{1}.
Attempts (2 marks)
A1 2 or 3 appear in the answer.
Page 3
Part (b) (iii)
5 marks
Att 2
(iii) List the elements of P′, the complement of the set P.
Part (b) (iii)
P′ = {2,3,6,8,9,10}
5 marks
Att 2
Slips (-1)
S1 Each correct element omitted and/or each incorrect element included. (Max -3)
Attempts (2 marks)
A1 P or any proper subset of P.
Part (b) (iv)
(iv) Write down # R.
5 marks
Att 2
Part (b) (iv)
5 marks
Att 2
4
Blunders (-3)
B1 Incorrect #R ≤ 10. (See M2)
Misreadings (-1)
M1 R= {1, 7, 8, 10}
M2 #R' = 6.
Attempts (2 marks)
A1 Uses phrase “number of elements” or “cardinal number”.
A2 #R = 26 or 560.
Page 4
Part (c)
1(c)
Part(c) (i)
20(5, 5, 5, 5) marks
Att (2, 2, 2, 2)
There are 30 students in a class.
21 own a mobile phone (M) and 12 own a computer (C).
7 own both a mobile phone and a computer.
5 marks
(i) Represent this information in the Venn diagram below.
Part(c) (i)
Att 2
5 marks
Att 2
M
C
[14]
[7]
[4]
Blunders (-3)
B1 Incorrect Venn diagram subject to S1 below.
Slips (-1)
S1 Numerical errors where work is clearly shown.
Attempts (2 marks)
A1 Any one correct/relevant entry.
Page 5
[5]
Part(c) (ii)
5 marks
(ii) How many students own a mobile phone but not a computer?
Att 2
Part(c) (ii)
*
*
*
5 marks
Att 2
14
A correct answer written in the space provided takes precedence over an incorrect Venn
diagram.
Accept candidates work from previous part c (i).
If no work appears here, award 2 marks if correct answer appears in Venn Diagram.
Blunders (-3)
B1 Any incorrect use of the given numbers or the numbers from the candidates incorrect
Venn diagram. [ Subject to S1].
Slips (-1)
S1 Numerical errors where work is clearly shown.
Misreadings (-1)
M1 C\M
Part(c) (iii)
5 marks
(iii) How many students own neither a mobile phone nor a computer?
Att 2
Part(c) (iii)
Att 2
*
*
*
5 marks
4
A correct answer written in the space provided takes precedence over an incorrect Venn
diagram.
Accept candidates work from previous parts (c) (i), (c) (ii).
If no work appears here, award 2marks if correct answer appears in Venn Diagram.
Blunders (-3)
B1 Incorrect Venn diagram.[ Subject to Second *above].
B2 Any incorrect use of the given numbers or numbers from the previous work.
[Subject to Second *above].
Slips (-1)
S1 Numerical errors where work is clearly shown.
Page 6
Part(c) (iv)
Part(c) (iv)
*
*
5 marks
(iv) How many students do not own a mobile phone?
5 marks
9
Att2
Att2
A correct answer written in the space provided takes precedence over an incorrect Venn
diagram.
Accept candidates work from previous parts (c) (i), (c) (ii), and (c) (iii).
Blunders (-3)
B1 Incorrect Venn diagram. [ Subject to Second *above].
B2 Any incorrect use of the given numbers or numbers from the previous work.
[Subject to Second * above].
Slips (-1)
S1 Numerical errors where work is clearly shown
Page 7
QUESTION 2
Part (a)
Part (b)
Part (c)
10 marks
20(5, 10, 5) marks
20(5, 5, 10) marks
Att 3
Att (2, 3, 2)
Att (2, 2, 3)
Part (a)
10 marks
(a) In a school of 646 pupils the ratio of girls to boys is 9:8.
Find the number of girls and the number of boys in the school.
Att 3
Part (a)
Att 3
"
10 marks
9 + 8 = 17
9 parts : 8 parts
9 x :8 x
646
⇒
= 38
17
Girls = 38 × 9 = 342
Boys = 38 × 8 = 304
⇒17 x = 646
⇒ x = 38
⇒ 9 x = 342
⇒ 8 x = 304
1
= 38
17
9
⇒
= 342(Girls)
17
⇒ 646 − 342 = 304( Boys).
Blunders (-3)
B1 Correct answers without work. "
B2 Divisor = 8 or 9 only and continues.
B3 Incorrect multiplier or fails to multiply. (each time)
B4 Error in transposition (x method).
B5 Fails to find second number. (Number of boys or girls only).
B6 Adds instead of subtracting e.g. 646 + 342 = 988.
Slips (-1)
S1 Numerical errors to a max of 3
.
Attempts (3 marks)
646
646
and/or
and st ops.
A1 Divisor ≠ 17 e.g.
9
8
A2 Indicates 17 parts or 9 parts or 8 parts or 9/17 or 8/17 or 9+8=17 only and stops.
A3 5814:5168 only. i.e. multiplies 646 by 9 and by 8.
A4 Divide by 2 and stops or continues. (Oversimplification).
A5 Both answers added to equal 646. (If no work shown).
Worthless (0)
W1 Incorrect answer without work.
Page 8
Part (b) (i)
5 marks
2(b) (i) On a day when €1 = $ 1⋅ 21 , find the value in euro of $6655.
Att 2
Part (b) (i)
Att2
"
*
5 marks
$1.21 = €1
1
⇒ $1 = €
1.21
€1= $1.21
€ ? = $6655
6655
= €5500
?=
1.21
⇒ $6655 = 6655 ×
No penalty for the omission of € or $ symbols.
Blunders (-3)
B1 Correct answer without work. "
B2 Incorrect multiplier i.e.6655 × 1.21=8052.55
1.21
121
B3 Incorrect ratio
or
.
6655
665500
B4 Decimal error.
6655
B5 Fails to finish, leaves as
and stops.
1.21
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Rounds off too early. i.e. (0.83).
Attempts (2 marks)
1
A1 $1= €
and stops.
1.21
Worthless (0)
W1 Adds or subtracts 6655 and 1.21.
W2 Incorrect answer without work.
Page 9
1
= €5500
1.21
Part (b) (ii)
10 marks
2 (b) (ii) By rounding each of these numbers to the nearest whole number,
4 ⋅ 368 + 10 ⋅ 92
estimate the value of
.
3 ⋅ 12
Att3
Part (b) (ii)
Att3
"
10 marks
4 ⋅ 368 + 10 ⋅ 92
is approximately equal to:
3 ⋅ 12
4
+
11
15
=
3
*
*
*
=
5
3
4 + 11
and stops ⇒ 4 marks.
3
No penalty if the intermediate step between approximations and final answer not
15
not shown.
shown.i.e.
3
4.368 + 10.92
Special Case:
= 4.9 ⇒ 3 marks.
3.12
Blunders (-3)
B1 Correct answer without work. "
B2 Error(s) in rounding off to the nearest whole number.
B3 Decimal error in calculation of approximate value.
B4 An arithmetical operation other than indicated.
4
B5
+ 11 or similar and continues.
3
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 Only one or two approximations made to the given numbers & stops.
Page 10
Part (b) (iii)
Part(b)
5marks
(iii) Using a calculator, or otherwise, find the exact value of
Part (b) (iii)
Att2
4 ⋅ 368 + 10 ⋅ 92
.
3 ⋅ 12
5marks
15.288
4 ⋅ 368 + 10 ⋅ 92
=
= 4.9
3.12
3 ⋅ 12
Blunders (-3)
B1 Decimal error.
4.368
B2 Treats as:
+ 10.92 = 1.4 + 10.92 = 12.32 .[ B1 may occur].
3.12
10.92
B3 Treats as: 4.368 +
= 4.368 + 3.5 = 7.868 .[ B1 may occur].
3.12
4.368 −10.92
B4 Treats as:
= − 2.1 [B1 may occur].
3.12
4.368 ×10.92
B5 Treats as:
= 15.288 .[B1 may occur].
3.12
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (2 marks)
A1 Some correct calculation done.
Page 11
Att2
Part (c) (i)
5 marks
Att 2
( )
3
Using a calculator, or otherwise, find the exact value of 4 2 .
2(c) (i)
Part(c) (i)
5marks
(4 ) = 4096
Att2
2 3
*
46 and stops. = 4 marks.
Blunders (-3)
B1
Mishandles ( 42 ) .e.g. 45 = 1024,
3
( 4 ) = 8, ( 4 ) = 2.5198421.
3
3
2
Attempts (2 marks)
3
A1 ( 4 ) = 64.
A2
( 4)
A3
4 × 3× 2 = 24 .
2
= 16.
Part (c) (ii)
5 marks
Using a calculator, or otherwise, multiply 65 ⋅ 5 by 40 and express your
answer in the form a × 10 n , where 1 ≤ a < 10 and n ∈ Z.
Att 2
Part (c) (ii)
Att 2
"
5 marks
65.5 × 40 = 2620 = 2.62 ×103
Blunders (-3)
B1 Correct answer without work. "
B2 Decimal error.
B3 Incorrect format, where a ≤1 or a ≥10 and n ∉ Z .
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Rounds off to 3 × 103 , 2.6 × 103 .
S3 Incorrectly rounds off. e.g. 2.7 × 103 also attracts S2.
Attempts (2 marks)
A1 2620 and stops.
A2 Any relevant step.e.g. Partial multiplication.
Page 12
Part (c) (iii)
10 marks
(iii)
Using a calculator, or otherwise, evaluate
86 ⋅ 49
1
× 7 ⋅ 48 .
+
0 ⋅ 0125
15 ⋅ 5
Give your answer correct to two decimal places.
Att 3
Part (c) (iii)
Att3
"
10 marks
9.3
= 80 +
× 7.48
15.5
= 80 + 0.6 × 7.48
= 80 + 4.488
= 84.488
= 84.49
*
Correct answer (without work) incorrectly rounded off ⇒ 6 marks
Blunders (-3)
B1 Correct answer without work. "
1
B2 Mishandles
.
0 ⋅ 0125
B3 Mishandles 86.49
9.3
or candidate's equivalent from previous work.
B4 Error in
15.5
B5 Error in multiplication of 0.6 × 7.48 or candidate's equivalent from previous work.
B6 Decimal error.
1
86 ⋅ 49
+
× 7 ⋅ 48 .= 602.888.
B7 Adds before Multiplication:
0 ⋅ 0125
15 ⋅ 5
B8 Incorrect denominator.
B9 Incorrect numerator.
B10 Works as 80 × 7.48 + 0.6 = 599 .
B11 Multiplies instead of adds.
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Each premature rounding off to a max of 3.
S3 Fails to round off or rounds off incorrectly when giving final answer.
Attempts (3 marks.)
A1
Any relevant step e.g.
1
= 80,
0 ⋅ 0125
86.49 = 9.3
Page 13
QUESTION 3
Part (a)
Part (b)
Part (c)
Part (a)
3.
(a)
10 marks
20(10, 10) marks
20(5, 5, 5, 5) marks
Att 3
Att (3, 3)
Att (2, 2, 2, 2)
10 marks
Find the total cost of the following bill:
Att 3
"
6 litres of milk at €1.05 a litre
3 loaves of bread at €1 .20 a loaf
5 apples at 65c each
Part (a)
10 marks
1.05 × 6 = 6.3
1.20 × 3 = 3.6
OR
0.65 × 5 = 3.25
Total Cost = €13.15
1.05 + 1.05...6 Times = 6.30
+ 1.20 + 1.20....3Times = 3.60
+ 0.65 + 0.65....5 Times = 3.25
Total Cost = €13.15
*
Accept 1315, 13.15.
*
No penalty for missing € symbol.
*
Adds 1.05 + 1.20 + 0.65 = 2.90 and stops ⇒ 3 marks. (Oversimplification).
Blunders (-3)
B1 Correct answer without work. "
B2 Each missing product when finding items cost e.g. 1.05 not multiplied by 6.
B3 Each missing item when finding total cost e.g. cost of bread omitted.
B4 Fails to find total cost i.e. no addition.
B5 Operation other than addition of items to find total cost.
B6 Decimal error e.g. 131.5 (Note: First *).
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 Any attempt at addition /multiplication.
Worthless (0)
W1 Incorrect answer without work.
Page 14
Att 3
Part (b) (i)
10marks
3(b) (i)
V.A.T .at 21% is added to a bill of €750.
Calculate the total bill.
Part (b) (i)
10marks
Method 1
100% = 750
Method 2
750
1% =
100% = 750
100
750
750
121% =
× 121
1% =
100
100
= 7.5 ×121
750
× 21 = 157.50
21% =
100
Total bill = €907.5
Total Bill =157.5 + 750 = €907.50
Method 3
"
21
Method 4
100
750 ×1.21 = 907.5
21
V . AT
. .=
× 750.
Total bill = €907.50
100
Total bill =157.5 + 750 = €907.5
*
€157.50 (without work) and stops ⇒ 4 marks.
*
No penalty for missing € symbol.
Blunders (-3)
B1 Correct answer without work. "
121
21
B2 Inverts
or
and continues (giving answers 619.83 or 3571.43).
100
100
B3 Mishandles 21%. e.g.750× 21 or 750 ÷ 21 (750 must be used).
B4 750 taken as 121%
B5 No addition of V.A.T. (as per candidates work) to the bill.
B6 Subtraction of V.A.T. ( as per candidates work) from the bill.
21% =
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
21
and stops.
A1
100
A2 100% = 750 and stops.
750
A3
and stops.
100
21
A4 100 ×
and stops.
750
750
A5
and stops.
21
A6 Use of any other %
Page 15
Att 3
Att 3
Part (b) (ii)
(ii)
10marks
€7450 is invested at 2 ⋅ 6 % per annum.
What is the amount of the investment at the end of one year?
Att 3
Part (b) (ii)
10marks
Att 3
"(ii)
7450
100
2.6% = 74.50 × 2.6
1% =
P × R 7450 × 2.6
=
= 193.7
100
100
Amount = €7643.70
I=
7450 × 1 ⋅ 026
= 7643.7
Amount = €7643.7
Interest = €193.70
Amount = €7643.70
*
*
€193.70 (without work) and stops ⇒ 4 marks.
No penalty for missing € symbol.
Blunders (-3)
B1 Correct answer without work. "
B2 Mishandles 2.6%. e.g. 7450 × 2.6 or 7450 ÷ 2.6 (7450 must be used).
B3 Decimal error (once only).
B4 Stops at interest i.e. fails to calculate amount.
B5 Subtracts to calculate amount.
7450 × 2.6
.
B6 Illegal cancellation(s) in
100
B7 1 ⋅ 026 = 1 ⋅ 26.
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 Correct formula with or without substitution and stops.
A2
Some use of 100 in attempt to find percentage e.g. 2.6% =
Worthless (0)
W1 Incorrect answer without work.
Page 16
2.6
and stops.
100
Part (c)
20(5, 5, 5, 5) marks
3(c) John’s weekly wage is €730.
He pays income tax at the rate of 20% on the first €440 of his wage
and income tax at the rate of 42% on the remainder of his wage.
John has a weekly tax credit of €65.
Att (2, 2, 2, 2)
Part (c) (i)
5 marks
(i) Find the tax on the first €440 of his wage, calculated at the rate of 20%.
Att 2
Part (c) (i)
Att 2
5 marks
"(i)
1% = 4.4
20% = 88
Tax = €88
*
Tax =
440
× 20 = €88
100
440 × 0 ⋅ 2 = €88
.
No penalty for missing € symbol.
Blunders (-3)
B1 Correct answer without work. "
B2 Mishandles 20%, e.g. 440 × 20 = 8800 or 440 ÷ 20 = 22.
B3 Uses €730 instead of €440.
B4 Decimal error.
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (2 marks)
20
A1 Some use of 100 in attempt to find percentage e.g. 20% = 100
and stops
Worthless (0)
W1 Incorrect answer without work
Page 17
Part (c) (ii)
5 marks
(ii) Find the tax on the remainder of his wage, calculated at the rate of 42%.
Att 2
Part (c) (ii)
Att 2
" (ii)
1% = 2 ⋅ 9
5 marks
Remainder of wage = €730 − €440 = €290
Tax =
290
× 42 = €121.8
100
or
290 × 0 ⋅ 42 = €121.8
42% = 121.8
Tax = €121.8.
*
No penalty for missing € symbol.
Blunders (-3)
B1 Correct answer without work. "
B2 Mishandles 42%, e.g. 290 × 42 or 290 ÷ 42 . [No penalty if already penalised in (c) (i)].
B3 Uses €730 or €440 instead of €290.
B4 Decimal error.
B5 730 - 440 = 290 and stops.
or
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (2 marks)
42
and stops.
A1 Some use of 100 in attempt to find percentage e.g. 42% = 100
Worthless (0)
W1 Incorrect answer without work.
Part (c) (iii)
(iii) Hence calculate John’s gross tax.
Part (c) (iii)
5 marks
Att 2
5 marks
Att 2
" (iii)
John’s gross tax = €88 + €121.80= €209.80
*
Allow candidates incorrect answers from parts (i) and (ii).
*
No penalty for missing € symbol.
Blunders (-3)
B1 Correct answer without work. "
B2 €88 − €121.80 = - €33.80
B3 Misuse of tax credit.
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (2)
A1 Answer from c (i) and /or c (ii) written in this part.
Worthless (0)
W1 Incorrect answer without work
Page 18
Part (c) (iv)
(iv)
5 marks
Calculate John’s take home pay.
Att 2
Part (c) (iv)
5 marks
Att 2
"
*
*
Tax payable = €209.80 - €65
Take home pay = €730 - €144.80
Take home pay = €585.20
Allow candidate’s incorrect gross tax figure from (c) (iii).
No penalty for missing € symbol.
Blunders (-3)
B1 Correct answer without work. "
B2 Misuse of tax credit e.g. 209.80 + 65 = 274.80.
B3 Decimal error.
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (2)
A1 Answer from c (iii) written in this part.
Worthless (0)
W1 Incorrect answer without work.
Page 19
QUESTION 4
Part (a)
Part (b)
Part (c)
Part (a)
(i)
10(5, 5) marks
20(10, 10) marks
20(5, 5, 5, 5) marks
10(5, 5) marks
If a = 2 and b = 5 , find the value of 3a + b
Part (a) (i)
"
*
*
Att (2, 2)
Att (3, 3)
Att (2, 2, 2, 2)
5 marks
(i)
Att 2,2
Att 2
3a + b = 3 ( 2 ) + 5 = 6 + 5 = 11
6 + 5 ⇒ 4 marks.
One substitution coupled with an implied substitution leading to correct answer
⇒ 5 marks.e.g. = 3a +5 = 11
Blunders (-3)
B1 Correct answer without work. "
B2 Leaves 3(2) in the answer.
B3 Breaks order i.e. [ 3 ( 2 + 5 ) = 21 ].
B4
Treats 3(2) as 5 or 32.
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Values of a and b interchanged.
Misreadings (-1)
M1 Incorrect numerical substitution for either a or b, but not both, and continues. (See W1)
Attempts (2 marks)
A1 Incomplete substitution and stops e.g. 3a + 5,
Worthless (0)
W1
Incorrect substitution for both a and b.
Page 20
Part (a)
(ii)
10(5, 5) marks
If a = 2 and b = 5 , find the value of ab − 3
Part (a) (ii)
"
*
5 marks
(ii)
Att 2, 2
Att 2
ab − 3 = 2 ( 5 ) − 3 = 10 − 3 = 7
10 − 3 ⇒ 4 marks.
Blunders (-3)
B1 Correct answer without work.
"
B2
B3
Leaves 2(5) in the answer.
Breaks order i.e. [ 2 ( 5 − 3) = 4 ].
B4
Treats 2(5) as 25, 7, or 52.
[Do not penalise if already penalised in part (a) (i) or
work is shown in part (a) (i).]
Slips (-1)
S1 Numerical errors to a max of 3.
Misreadings (-1)
M1 Incorrect numerical substitution for either a or b but not both, and continues. (See W1)
Attempts (2 marks)
A1 Incomplete substitution and continues or stops e.g. 2b − 3, 5a − 3
Worthless (0)
W1 Incorrect substitution for both a and b.
Page 21
Part (b) (i)
4(b) (i)
Solve the equation
Part (b) (i)
10marks
2 ( x − 3) = x +1.
Att 3
10marks
Att 3
2 ( x − 3) = x + 1
"
2 x − 6 = x +1
2 x − x = 1+ 6
x =7
Blunders (-3)
B1 Correct answer without work. "
B2 Error in distributive law and continues,e.g. 2 x − 3 = x + 1, 2 x − 6 = 2 x + 2 (once only).
B3 Error(s) in transposition.
B4 Combines “ x ”to “numbers” and continues. e.g. 2 x − 6 = − 4 x .
B5 Fails to finish.
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 Any correct step.
Worthless (0)
W1 Combines “ x ”to “numbers” and stops.
Page 22
Part (b) (ii)
10marks
Att 3
Multiply ( x − 5) by (2 x + 3) .
Write your answer in its simplest form.
Part (b) (ii)
"
10marks
( x − 5)( 2 x + 3) = x ( 2 x + 3) − 5 ( 2 x + 3)
= 2 x 2 + 3 x − 10 x − 15
= 2 x 2 − 7 x −15
*
First line = x ( 2 x + 3) − 5 ( 2 x + 3) or 2 x ( x − 5 ) + 3 ( x − 5 ) = 4 marks.
Blunders (-3)
B1 Correct answer without work. "
B2 Error(s) in distribution.
B3 Combining unlike terms.
B4 Fails to group or groups incorrectly.
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 Any correct multiplication.
A2 Oversimplification of question.
A3 A correct step.
Worthless (0)
W1 ( x − 5 ) ± ( 2 x + 3) stops or continues.
W2 Combining unlike terms before attempting multiplication and stops.
Page 23
Att 3
Part(c) (i)
10(5, 5) marks
Att (2, 2)
The cost of 2 jumpers and 3 shirts is €84.
The cost of 4 jumpers and 1 shirt is €78.
Let €x be the cost of a jumper and let €y be the cost of a shirt.
(i)
Write down two equations, each in x and y
to represent the above information.
Write down two equations
"
10(5, 5) marks
First equation:
2 x + 3 y = 84
Second equation: 4 x + y = 78
*
Special Case: 2 + 3 = 84, 4 + 1 = 78. Award 7 marks.
Blunders (-3)
B1
Correct answer without work.
"
Apply to both equations
Slips (-1)
S1 Incorrect coefficient of x (other than zero).
S2 Incorrect coefficient of y (other than zero).
S3 Incorrect constant.
Attempts (2 marks)
A1 Any effort at a linear equation in x only or a linear equation in y only.
A2 2 x only or 4 x only or 3 y only appear.
Page 24
Att(2,2)
Part(c) (ii)
5 marks
ii) Solve these equations to find the cost of a jumper and the cost of a shirt.
Att 2
Part(c) (ii)
Att 2
5 marks
2 x + 3 y = 84
4 x + y = 78
2 x + 3 y = 84
4 x + y = 78
"
2 x + 3 y = 84
−12 x − 3 y = − 234
or
− 4 x − 6 y = − 168
4 x + y = 78
−10 x = − 150
x = €15
y = €18
*
*
*
*
− 5 y = − 90
y = 78 − 4 x
2 x + 3 ( 78 − 4 x ) = 84
or
5 y = 90
y = €18
2 x + 234 − 12 x = 84
− 10 x = − 150
x = €15
y = €18
x = €15
Apply only one blunder deduction (B2 or B3) to any error(s) in establishing the first
equation in terms of x only or the first equation in terms of y only.
Finding the second variable is subject to a maximum deduction of (-3).
If the candidates equations in (c)(i) are such that they lead to an over simplification of
the work in (c)(ii) then Attempt marks apply at most.
No penalty for missing € symbol.
Blunders (-3)
B1 Correct answers without work. "
B2 Error(s) in establishing the first equation in terms of x only [ −10 x = − 150 ] or the first
equation in terms of y only [ −5 y = − 90 ] through elimination by cancellation.
B3 Error(s) in establishing the first equation in terms of x only [ 4 x = 60 ] or the first
equation in terms of y only [ 3 y = 54 ] through elimination by substitution.
B4 Errors in transposition in solving the first one variable equation.
B5 Errors in transposition when finding the second variable.
B6 Incorrect substitution when finding second variable.
B7 Finds one variable only.
Slips (-1)
S1 Numerical errors to a max of 3
Attempts (2 marks)
A1 Attempt at transposition and stops.
A2 Multiplies either equation by some number and stops.
Page 25
Part(c) (iii)
(iii) Verify your result.
5marks
Att2
Part(c) (iii)
5marks
Att2
"
2(15) + 3(18) = 84
4(15) + 18 = 78
*
Accept candidates answers from previous work in this part.
Blunders (-3)
B1 Correct answers without work. "
B2 Verifies only one equation.
B3 Error in substitution to either equation.
B4 Forces equality
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Conclusion missing.
Attempts (2 marks)
A1 Substitutes into one equation and stops.
A2 Writes the equations in this section.
A3 Answers from (c) (ii) written in this part.
Page 26
QUESTION 5
Part (a)
Part (b)
Part (c)
10 marks
20 (5, 5, 5, 5) marks
20 (10, 5, 5) marks
Att 3
Att (2, 2, 2, 2)
Att (3, 2, 2)
Part (a)
10 marks
Write in its simplest form
4( x + 3) + 2(5 x + 4) .
Att 3
Part (a)
Att3
"
*
*
10 marks
4 x + 12 + 10 x + 8
14 x + 20
Stops after correct removal of brackets ⇒ 7 marks.
Ignore excess work 2 ( 7 x + 10 )
Blunders (-3)
B1 Correct answer without work.
B2 Error(s) in distribution.
B3 Combining unlike terms.
B4 Fails to group like terms.
"
Slips (-1)
S1 Numerical errors to a max of 3.
Misreadings (-1)
4( x + 3) × 2(5 x + 4) and continues.
M1
Attempts (3 marks)
A1 Any correct multiplication.
Worthless (0)
W1 Combining unlike terms before attempting multiplication and stops.
Page 27
Part (b) (i)
(i)
Factorise:
5 marks
xy + wy
Att 2
Part (b) (i)
(i)
5 marks
y(x + w)
Att 2
Blunders (-3)
B1 An incorrect factor.
B2 Removes factor incorrectly.
Attempts (2 marks)
A1 Indication of common factor. e.g. underline y 's and stops.
Part (b) (ii)
(ii)
Factorise:
5 marks
ax − ay + bx − by
Part (b) (ii)
" (ii)
5 marks
ax − ay + bx − by
a ( x − y) + b( x − y)
Att 2
ax + bx − ay − by
x (a + b) − y (a + b)
or
( x − y )( a + b )
*
Att 2
( a + b )( x − y )
Accept (with or without brackets) for 5 marks any of the following
( x − y ) and ( a + b ) . [The word and is written down.]
( x − y ) or ( a + b ) .
( x − y) , (a + b) .
[The word or is written down.]
[A comma is used]
Blunders (-3)
B1 Correct answer without work. "
B2 Stops after first line of correct factorisation. e.g. a ( x − y ) + b ( x − y ) or equivalent.
B3
B4
Error(s) in factorising any pair of terms.
Incorrect common factor and continues. e.g.
a ( x − y ) + b ( x + y ) = ( a + b )( x − y )
Slips (-1)
S1 ( a + b ) ± ( x − y )
S2
Correct first line of factorisation but ends as ab ( x − y ) .
Attempts (2 marks)
A1 Pairing off, or indication of pairing off, and stops.
A2 Correctly factorises any pair and stops.
Page 28
Part (b) (iii)
Factorise:
5 marks
Att 2
5 marks
p − 36
Att 2
p 2 − 36
Part (b) (iii)
2
P 2 − 62
(iii)
( p − 6 )( p + 6 )
*
Accept (with or without brackets) for 5 marks any of the following
( p + 6 ) and ( p − 6 ) . [The word and is written down.]
( p + 6 ) or ( p − 6 ) .
[The word or is written down.]
( p + 6), ( p − 6) [A comma is used]
*
Quadratic equation formula is subject to slips and blunders.[See 5(c)(i)]
Blunders (-3)
B1 Incorrect two term linear factors of p 2 − 36 formed from correct (but not applicable)
factors of p 2 and ±36 e.g. ( p − 9 )( p + 4 ) .
B3
( 6 + p )( 6 − p ) .
( p − 36 )( p + 36 ) .
B4
Incorrect factors of p 2 and/or 36.
B2
Slips (-1)
S1 Solves p 2 = 36 to give p = 6 and p = −6 and stops.
S2 ( p + 6) ± ( p − 6)
Attempts (2 marks)
A1 Correct factors of p 2 only.
A2 Correct factors of 36 or −36 only.
A3 p or ± 6 appears.
A4 p 2 − 36 = p. p − 6.6 and stops.
A5 Mention of the difference of two squares.
Page 29
Part (b) (iv)
5 marks
Factorise: 4a + 8a
(iv)
Part (b) (iv)
5 marks
4 a 2 + 8a
(iv)
*
Att2
2
Att2
4a ( a + 2 )
Accept
4a 2 +8a
4 ( a 2 + 2a )
or
4a 2 + 8a
or
2 ( 2a 2 + 4a )
4a 2 + 8a
a ( 4a + 8 )
or
2a ( 2a + 4 )
Blunders (-3)
B1 An incorrect factor.
B2 Stops after some correct effort at factorisation. e.g. 4.a.a + 4.2 a
B3 Mathematical blunder 4a 2 =16a 2 & continues.
Attempts (2 marks)
A1 4a ( a ) and / or 8 ( a ) or effort at brackets.
A2
Common factor identified or indicated and stops. e.g. 4 a a + 4 2 a or similar.
Page 30
Part (c) (i)
(i)
Solve the equation:
10 marks
x2 - 5x - 14 = 0
Att 3
10 marks
Att 3
Part (c) (i)
" (iii)
x
x
x 2 − 5 x − 14 =
= x 2 − 7 x + 2 x − 14 = 0
= x ( x − 7) + 2 ( x − 7) = 0
+2
−7
⇒ ( x − 7)( x + 2) = 0
⇒ x = 7 and x = − 2
or
= ( x − 7 )( x + 2 ) = 0
⇒ x = 7 and x = − 2
x=
− ( −5 ) ±
( −5 ) − 4 (1)( −14 )
2 (1)
2
5 ± 25 + 56 5 ± 9 14
−4
=
=
and
2
2
2
2
⇒ x = 7 and x = − 2
⇒
or
Factor Method:
B1 Correct answers without work. "
B2 Incorrect two term linear factors of. x 2 − 5 x − 14 formed from correct (but inapplicable
factors of x 2 or ± 14 .
B3 No roots given.
B4 Incorrect factors of x 2 and/or ±14.
B5 Correct cross method but factors not shown and stops [Note: B3 applies also].
B6 x ( x − 7 ) + 2 ( x − 7 ) or similar and stops. [Note: B3 applies also].
B7
B8
Error(s) in transposition.
One root only.
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 Some effort at factorisation.
A2 Oversimplification resulting in a linear equation & continues.
Worthless (0 marks)
W1 x 2 − 5 x = 14 or similar and stops.
W2 Trial and error.
Page 31
Formula Method
Blunders (-3)
B1 Correct answers without work."
B2 Error in a,b,c, substitution (apply once only).
B3 Sign error in substituted formula (apply once only).
B4 Error in square root or square root ignored.
5±9
B5 Stops at
.
2
B6 Incorrect quadratic formula and continues.
B7 One root only.
p
B8 Roots left in the form
q
Slips (-1)
S1 Numerical errors to a max of 3.
Attempts (3 marks)
A1 Correct formula and stops.
A2 One correct substitution and stops.
A3 Oversimplification of formula.
Page 32
Part (c) (ii)
(ii)
Att2
3x + 2 x + 4
as a single fraction.
−
4
5
Give your answer in its simplest form.
Express
Part (c) (ii)
"
5 marks
5 marks
3x + 2 x + 4
−
4
5
5 ( 3x + 2 ) − 4 ( x + 4 )
=
20
15 x + 10 − 4 x − 16
=
20
11x − 6
=
( 5 marks )
20
3x + 2 x + 4 2 x + 6
Zero marks.
−
=
4
5
9
Blunders (-3)
B1 Correct answer without work. "
B2 Error(s) in distribution. e.g 5 ( 3 x + 2 ) = 15 x + 2.
B3 Mathematical error e.g. 10 -16 =6, -4(4) = 16.
B4 Incorrect common denominator and continues.
Att2
*
B5
Incorrect numerator from candidate's denominator e.g.
B6
No simplification of numerator.
Slips (-1)
S1 Correct common denominator implied.
S2 Numerical error to a max of 3.
Attempts (2 marks)
A1 20 only or a multiple of 20 only appears.
Worthless (0)
5x 4 x
⎛ 3x + 2 ⎞ ⎛ x + 4 ⎞
− , or ⎜
W1
⎟ and stops.
⎟⎜
4
5
⎝ 4 ⎠⎝ 5 ⎠
Page 33
4 ( 3x + 2 ) − 5 ( x + 4 )
20
.
Part (c) (iii)
(iii)
5 marks
Verify your answer to part (ii) by letting x = 6 .
Part (c) (iii)
5 marks
=
"
*
*
Att2
11x − 6
20
11( 6 ) − 6
20
66 − 6
=
20
60
=
20
=3
Att2
3(6) + 2
and
−
( 6) + 4
4
5
18 + 2
10
=
−
4
5
20
10
=
−
4
5
=
5 − 2
=
3
Accept candidates answer from previous section. [May result in inequality].
Accept usage of a value other than 6 for verification.
Blunders (-3)
B1 Correct answer without work. "
B2 Substitutes into one expression only.
B3 Manipulation to force equality.
Slips (-1)
S1 Numerical errors to a max of 3.
S2 Conclusion missing if unequal.
Attempts (2marks)
A1 Writes answer from previous part in this section.
A2 Substitutes a value into one expression and stops.
Page 34
QUESTION 6
Part (a)
Part (b)
Part (c)
10 (5, 5) marks
30 (20, 10) marks
10 (5, 5) marks
Part (a) (i)
(i)
f ( x) = 2 x − 1.
Part (a) (i)
"
5 marks
Find: f (4)
Att 2
5 marks
f (4) = 2(4) − 1
Att2
(i)
= 8 −1
=7
Blunders (-3)
B1
B2
Att (2, 2,)
Att (7, 3)
Att (2, 2)
Correct answer without work. "
Mathematical error. e.g. ( 2 )( 4 ) = 24,
Slips (-1)
S1 Numerical errors to a max of 3.
Misreadings (-1)
M1 Correctly substitutes in any number other than 4 and continues.
Attempts (2marks)
A1 Treats as equation and continues or stops.
Worthless (0)
W1 Ignores x giving 2 −1 = 1.
W2 4 [ f ( x) ] = 8 x − 4
Page 35
Part (a) (ii)
(ii)
Part (a) (ii)
"
5 marks
Att2
5 marks
f (−5) = 2(−5) − 1
Att2
Find: f ( −5)
(ii)
Blunders (-3)
B1 Correct answer without work.
work is shown in part (a) (i).]
B2 Mathematical error.
= −10 − 1
= −11
"[Do not penalise if already penalised in part (a) (i) or
Slips (-1)
S1 Numerical errors to a max of 3.
Misreadings (-1)
M1 Substitutes in any negative number other than -5 and continues.
Attempts (2marks)
A1 Treats as equation and continues or stops.
A2 Substitutes in any positive number other than 4.
Worthless (0)
W1 Ignores x giving 2 −1 = 1.
W2 −5 [ f ( x) ] = −10 x + 5
Page 36
Part (b) Table
Draw the graph of the function
20 marks
Att 7
f : x → 1 + 4 x − x2
in the domain −1≤ x ≤ 5
where x ∈ R.
Part (b) Table
20 marks
"
f (−1)
= 1 +
4 ( −1)
_
f (0)
= 1 +
4 ( 0)
_
f (1)
= 1 +
4 (1)
_
f (2)
= 1 +
4 ( 2)
_
f (3)
= 1 +
4 ( 3)
_
f (4)
= 1 +
4 ( 4)
_
f (5)
= 1 +
4 ( 5)
_
Att7
( −1)
2
( 0)
2
(1)
2
( 2)
2
( 3)
2
( 4)
2
( 5)
2
= -4
= 1
= 4
= 5
= 4
= 1
= -4
or
x
1
4x
− x2
f(x)
*
-1
1
-4
-1
-4
0
1
0
0
1
1
1
4
-1
4
2
1
8
-4
5
3
1
12
-9
4
4
1
16
-16
1
Error in each row or column attracts a maximum deduction of 3 marks.
Blunders (-3)
B1
B2
B3
B4
B5
B6
B7
B8
5
1
20
-25
-4
− x 2 taken as x 2 .
− x 2 taken as −2x all the way. [In row headed − x 2 by candidate]
+ 4 x taken as + 4 all the way. [In row headed + 4 x by candidate]
1 calculated as x all the way.[In row headed 1 by candidate]
Adds in top row when evaluating f ( x) .
Omits ‘1’ row or omits ‘4 x ’ row.
Omits a value in the domain.
Each incorrect image without work.
Page 37
Slips (-1)
S1 Numerical errors to a max of 3
Misreadings (-1)
M1 Misreads ‘4 x ’ as ‘-4 x ’ and places ‘-4 x ’ in the table
M2 Misreads ‘ + 1 ’ as ‘ − 1 ’ and places ‘ − 1 ’ in the table.
Attempts (7marks)
A1 Omits − x 2 row from table or treats − x 2 as ±x.
A2 Table with only f (x) = ± x 2
A3 Any effort at calculating point(s).
A4 Only one point calculated and stops.
Page 38
Part (b) Graph
*
*
*
*
10 marks
Att3
Accept candidate's values from previous work.
Only one point graphed correctly ⇒ Att 7 + Att 3
Correct graph but no table ⇒ full marks i.e. 30 marks.
Accept reversed co-ordinates if
(i) if axes not labelled or (ii) if axes are reversed to compensate (see B1 below)
Blunders (-3)
B1 Reversed co-ordinates plotted against non-reversed axes (once only) [See 4th * above].
B2 Scale error (once only).
B3 Points not joined or joined in incorrect order (once only).
Slips (-1)
S1 Each point of candidate graphed incorrectly.
S2 Each point from table not graphed [See 2nd * above].
Attempts (3 marks)
A1 Graduated axes (need not be labelled).
Page 39
Part (c) (i)
5 marks
(i) Draw the axis of symmetry of the graph drawn in 6 (b) above.
"
(c) (i)
*
*
Work to be shown on the graph.
5 marks
Att 2
Accept any vertical line (parallel to candidate's y-axis) within a tolerance of (± 0.25).
A candidate's incorrect graph may merit full marks for this section subject to the same
tolerance)
Blunders (-3)
B1 Any vertical line (parallel to the candidate's y-axis) outside of tolerance.
B2 Marks x = 2 on the x -axis and stops.
B3 States x = 2 but no line is indicated on the graph.
Attempts (2marks)
A1 Any attempt at axial symmetry of f ( x ) .
A2
Att 2
y -axis as the axis of symmetry (See B1)
Page 40
Part (c) (ii)
5 marks
Use the graph drawn in 6 (b) to estimate the value of f (x) when x = 3 ⋅ 5.
(ii)
"
Att 2
Work to be shown on the graph and answer to be written here.
2.75
Part (c) (ii)
*
*
5 marks
Att2
Correct answer (clearly consistent with graph) inside tolerance without graphical
indication ⇒ 2 marks.
A candidates incorrect graph can earn up to full marks for this section (see tolerance)
Page 41
Blunders (-3)
B1 Correct answer without work. "
B2 Answer on diagram but outside of tolerance ( ± 0 ⋅ 25 ).
B3 Fails to write down the answer.
Attempts (2 marks)
A1 Algebraic evaluation or calculator.
A2 Marks 3 ⋅ 5 in any way on either axis and stops.
Worthless (0)
W1 Answer outside of tolerance without graphical indication.
W2 f ( 0 ) =1 as answer.
Page 42
MARKING SCHEME
JUNIOR CERTIFICATE EXAMINATION 2005
MATHEMATICS – ORDINARY LEVEL – PAPER 1
GENERAL GUIDELINES FOR EXAMINERS
1.
Penalties of three types are applied to candidates’ work as follows:
• Blunders
- mathematical errors/omissions
(-3)
• Slips
- numerical errors
(-1)
• Misreadings (provided task is not oversimplified)
(-1).
Frequently occurring errors to which these penalties must be applied are listed in the scheme.
They are labelled: B1, B2, B3,…, S1, S2,…, M1, M2,…etc. These lists are not exhaustive.
2.
When awarding attempt marks, e.g. Att(3), note that
• any correct, relevant step in a part of a question merits at least the attempt mark for that
part
• if deductions result in a mark which is lower than the attempt mark, then the attempt mark
must be awarded
• a mark between zero and the attempt mark is never awarded.
3.
Worthless work is awarded zero marks. Some examples of such work are listed in the scheme
and they are labelled as W1, W2,…etc.
4.
The phrase “hit or miss” means that partial marks are not awarded – the candidate receives all
of the relevant marks or none.
5.
The phrase “and stops” means that no more work is shown by the candidate.
6.
Special notes relating to the marking of a particular part of a question are indicated by an
asterisk. These notes immediately follow the box containing the relevant solution.
7.
The sample solutions for each question are not intended to be exhaustive lists – there may be
other correct solutions.
8.
Unless otherwise indicated in the scheme, accept the best of two or more Attempts – even when
Attempts have been cancelled.
9.
The same error in the same section of a question is penalised once only.
10.
Particular cases, verifications and answers derived from diagrams (unless requested) qualify for
attempt marks at most.
11.
A serious blunder, omission or misreading results in the attempt mark at most.
12.
Do not penalise the use of a comma for a decimal point, e.g. €5.50 may be written as €5,50.
QUESTION 1
Part (a)
Part (b)
Part (c)
10 (5, 5) marks
20 (5, 5, 5, 5) marks
20 (5, 5, 5, 5) marks
Part (a)(i)
5 marks
(i)
Att (2, 2)
Att (2, 2, 2, 2)
Att (2, 2, 2, 2)
Att 2
P = {x, y, w}
Write down a subset of P that has one element
(i)
{x} or {y} or {w}
Blunders (−3)
B1
Any inapplicable subset of P. [proper or improper]
Misreadings (−1)
M1
Subsets of P with two elements.
Part (a)(ii)
(ii)
5 marks
P = {x, y, w}
Write down a subset of P that has two elements.
(ii)
{x, y} or {x, w} or {y, w}
Blunders (−3)
B1
Any inapplicable subset of P. (proper or improper)
Misreadings (−1)
M1
Subsets of P with one element.
Page 2 of 37
Att 2
Part (b)
20 (5, 5, 5, 5) marks
Att( 2, 2, 2, 2)
U
U is the universal set.
A
8•
A = {1, 2, 4, 8},
the set of divisors of 8.
•6
•1
•2
•4
B = {1, 2, 3, 4, 6, 12},
the set of divisors of 12.
C = {1, 2, 4, 5, 10, 20},
the set of divisors of 20.
•5
•12
•7
•10
•20
Part (b) (i)
(i)
B
•3
•9
C
5 marks
Att 2
A ∩ C ={1, 2, 4}
Blunders(−3)
B1 Any incorrect set of the elements of U other than the misreading as below.
Misreadings(−1)
M1 A ∪ C giving {1, 2, 4, 5, 8, 10, 20}.
Part (b) (ii)
5 marks
Att 2
B′ = {5, 7, 8, 9, 10, 20}
(ii)
Slips(−1)
S1 Each correct element omitted and/or each incorrect element included.
Attempts (2 marks)
A1
B or any proper subset of B.
Part (b) (iii)
(iii)
5 marks
C \ ( A ∩ B ) = {5, 10, 20}
Blunders (−3)
B1 Any incorrect set of the elements of U other than the misreading as below.
Misreadings (−1)
M1 ( A ∩ B ) \ C giving the Null Set.(Ø)
Page 3 of 37
Att 2
Part (b) (iv)
(iv)
5 marks
Using the Venn diagram above, or otherwise,
find the highest common factor of 8, 12 and 20.
Att 2
(iv)
H.C.F. = 4
4 is the highest factor in A ∩ B ∩ C. Common factors are 1, 2, 4 ⇒ H.C.F. = 4.
Or 8 = 2 × 2 × 2 : 12 = 2 × 2 × 3 : 20 = 2 × 2 × 5 : ⇒ H. C. F = 2 × 2 = 4
Blunders (−3)
B1 An inapplicable element of U. [But see S1].
B2 A listing of elements of U with 4 included.
B3 Correct factors of 8 or12 or 20 but no conclusion drawn re H.C.F.
Slips (−1)
S1 1 or 2 as H.C.F.
Misreadings (−1)
M1
L.C.M. given i.e. 120.
Attempts (2 marks)
A1 Incorrect factors of 8 and/or 10 and/or 20 e.g. 8 = 2 × 3.
Worthless (0)
W1 Any number ∉ U except 120. [See M1].
W2 A listing of elements of U with 4 not included.
Part (c) (i)
5 marks
M is the set of natural numbers from 1 to 20, inclusive.
(i)
List the elements of M that are multiples of 3.
(i)
3, 6, 9, 12, 15, 18.
Slips (−1)
S1
Each correct element omitted and/or each incorrect element included.
Worthless (0)
W1 No applicable multiple of 3 appears.
Page 4 of 37
Att 2
Part (c) (ii)
5 marks
M is the set of natural numbers from 1 to 20, inclusive.
(ii)
List the elements of M that are multiples of 5.
Att 2
(ii)
5, 10, 15, 20.
Slips (−1)
S1
Each correct element omitted and/or each incorrect element included.
Worthless (0)
W1
No applicable multiple of 5 appears.
Part (c) (iii)
(iii)
5 marks
Att 2
Write down the lowest common multiple of 3 and 5.
15
(iii)
3, 6, 9, 12, 15, 18.
*
*
5, 10, 15, 20.
Or 3 × 5 = 15
Accept candidate’s least common number from their incorrect answers in parts (i) and (ii) for
full marks.
Accept an indication of candidate’s L.C.M. for full marks.
Blunders (−3)
B1 An inapplicable multiple of 3 only and/or an inapplicable multiple of 5 only.
Misreadings (−1)
M1 H.C.F = 1.
Worthless (0)
W1
Numbers other than multiples of 3 or 5 (But see 1st *)
Page 5 of 37
Part (c) (iv)
(iv)
(iv)
*
5 marks
Att 2
Express 10 as the sum of three prime numbers
10 = 2 + 3 + 5.
Accept a listing of 2, 3, 5 for full marks.
Blunders (−3)
B1
Each correct prime constituent omitted and/or each incorrect constituent included.
Attempts (2 marks)
A1
Some attempt at product e.g. 1 × 2 × 5. (But see W1)
Worthless (0)
W1
No applicable prime number appears i.e. 2 or 3 or 5 do not appear.
Page 6 of 37
QUESTION 2
Part (a)
Part (b)
Part (c)
10 marks
20 (5, 10, 5) marks
20 (5, 5, 10) marks
Att 3
Att (2,3,2)
Att (2,2,3)
Part (a)
10 marks
Att 3
If 12 m2 of carpet cost €504, find the cost of 15 m2 of the same carpet.
(a)
12 m 2 ≡ 504
504
= 42
12
⇒ 15m 2 ≡ 42 × 15 = 630
⇒ 1m 2 ≡
*
Correct answer without work ⇒ 7 marks.
12
*
15 × 504 = 403 ⋅ 2 ⇒ 7 marks (B1).
4
*
5 × 504 = 403 ⋅ 2 ⇒ 7 marks (B1).
5
*
Indicates 15
12 or 4 only and stops. ⇒ 4 marks. (No use of 504 (−3) and possible Slips (−3)).
504
504
*
12 or 12 = 42 and stops ⇒ 4 marks (No use of 15 (−3) and possible Slips (−3)).
*
504 × 15 or 504 × 15 = 7560 ⇒ 4 marks.
*
Incorrect answer without work ⇒ 0 marks.
Blunders (−3)
12
4
B1
× 504 or × 504 and continues. [ 403 ⋅ 20 as answer.]
15
5
B2
Divisor ≠ 12 and continues. [But see B1]
B3
Incorrect multiplier i.e. ≠ 15 and continues. [But see B1]
B4
Divisor ≠ 4 and continues. [But see B1]
B5
Incorrect multiplier i.e. ≠ 5 and continues. [But see B1]
B6
15 : 12 = 504 : x and continues.
B7
Error in decimal point.
Slips (−1)
S1
Numerical errors (max –3).
Attempts (3 marks)
504
A1
Divisor ≠ 12 or ≠ 4 e.g. 504
15 or 5 and stops.
4
A2
Indicates 12
15 or 5 or12: 15 or 504: x only and stops.
A3
6048 only.i.e. multiplies 504 by 12.
1
A4
12 only appears
Worthless (0)
W1
504 + 15 = 519.
Page 7 of 37
Part (b) (i)
5 marks
Att 2
a9 × a5
(i) Simplify 6
, giving your answer in the form a n , where n ∈ N.
2
a ×a
(i)
a 9 × a 5 a 14
= 8 = a6 .
a6 × a2
a
a×a×a×a×a×a×a×a×a×a×a×a×a×a
= a × a × a × a × a × a = a6.
a×a×a×a×a×a×a×a
*
*
*
*
a 14
and stops ⇒ 2 marks.
a8
a 14 and stops ⇒ 2 marks.
a 3 × a 3 and stops ⇒ 2 marks.
Correct answer without work ⇒ 2 marks.
Blunders (−3)
B1 Each error in calculation involving indices.
B2 Each incorrect number of a’s in the extended form.
B3 Each incorrect elimination in the extended form.
B4 a 3 × a 3 as an answer.
Slips (−1)
a 14
a 14
1
S1
=
6
or
= −6 as final answers.
8
8
a
a
a
S2 a × a × a × a × a × a as answer.
Attempts (2 marks)
A1 Some manipulation of indices e.g. a 9 × a 5 = a 45 only.
Page 8 of 37
Part (b) (ii) Estimate
10 marks
Att 3
By rounding each of these numbers to the nearest whole number, estimate
56 ⋅ 214
the value of
.
2 ⋅ 31 + 5 ⋅ 79
(ii)
(ii)
56 ⋅ 214
is approximately equal to:
2 ⋅ 31 + 5 ⋅ 79
56
56
=
=
2
*
*
+
7
8
6
56
and stops ⇒ 4 marks.
2+6
No penalty if the intermediate step between approximations and final answer not shown
56
e.g.
not shown.
8
Blunders (−3)
B1 Error in rounding off to the nearest whole number (each time).
B2 Decimal point error in calculation of approximate value.
B3 An arithmetical operation other than indicated.
56 56
+
and continues.
B4
2
6
56
B5
+ 6 = 28 + 6 = 34 .
2
Slips (−1)
S1 Numerical errors in arithmetical operations.
Attempts (3 marks)
A1 Only one or two approximations made to the given numbers.
Worthless (0)
W1 No approximations made to given numbers.
Page 9 of 37
Part (b) (iii) Calculator
(iii)
5 marks
Using a calculator, or otherwise, find the exact value of
(iii)
56 ⋅ 214
= 6 ⋅ 94
8 ⋅1
Blunders (−3)
B1 Otherwise: Error(s) in decimal point.
56 ⋅ 214 56 ⋅ 214
B2 Otherwise:
+
= 24.335 + 9.709 = 34 ⋅ 044 .
2 ⋅ 31
5 ⋅ 79
56 ⋅ 214
+ 5 ⋅ 79 = 30 ⋅ 12506494.
B3 Otherwise:
2 ⋅ 31
B4 Calculator: Incorrect Answer.
Slips (−1)
S1 Otherwise: numerical errors in addition or division. (max −3).
Attempts (2 marks)
A1 Some correct calculation done.
Page 10 of 37
Att 2
56 ⋅ 214
.
2 ⋅ 31 + 5 ⋅ 79
Part (c) (i)
(i)
5 marks
Att 2
Using a calculator, or otherwise, find the exact value of: 49
1
2
(i)
7
Blunders (−3)
1
2
1
2
1
2
1
2
1
2
1
2
B1 Mishandles 49 e.g. 49 = 2401 , 49 = 24 ⋅ 5 , 49 = 49 ⋅ 5 , 49 = 98 , 49 =
B2 Calculator: Incorrect Answer.
B3 Otherwise: error in use of Maths. Tables e.g. 2 ⋅ 214 (wrong page).
Misreadings (−1)
1
1
1
M1 49 2 = 2 =
= 0 ⋅ 0004164 .
2401
49
99
2
.
1
2
M2 49 = 3 49 = 3 ⋅ 6593.
Attempts (2 marks)
A1
is mentioned.
Part (c) (ii)
(ii)
(ii)
5 marks
Using a calculator, or otherwise, find the exact value of
Att 2
1
.
6⋅4
1
= 0 ⋅ 15625
6⋅4
Blunders (−3)
1
B1
= 6 ⋅ 4 = 2 ⋅ 529 or (6 ⋅ 4) 2 = 40 ⋅ 96 .
6⋅4
1
B2 6 ⋅ 4 × 1 = 6 ⋅ 4 or
.
24
B3 Calculator: Incorrect Answer.
B4
Otherwise: Decimal point error in division or in use of Maths. Tables e.g.
Slips (−1)
S1 Otherwise: numerical errors (max. of −3).
S2 Maths. Tables: 0 ⋅ 1563 .
S3 Rounded off to 0 ⋅ 2, 0 ⋅ 16, 0 ⋅ 156, 0 ⋅ 1563.
S4 Incorrectly rounded off e.g. 0 ⋅ 1562, also attracts S3.
Attempts (2 marks)
A1 Some correct calculation done.
1
10
5
A2
=
or =
and stops.
6 ⋅ 4 64
32
Page 11 of 37
1
= 1563 .
6⋅4
Part (c) (iii)
(iii)
(iii)
*
*
B4
B5
B6
Att 3
Using a calculator, or otherwise, evaluate
3 ⋅ 14
65 ⋅ 61 ×
− (2 ⋅ 42) 2 .
0 ⋅ 47
Give your answer correct to two decimal places.
8 ⋅ 1 × 6 ⋅ 6808511 − 5.8564 (3 marks)
= 54 ⋅ 114894 − 5 ⋅ 8564
(6 marks)
= 48 ⋅ 258494
(9 marks)
= 48 ⋅ 26.
(10 marks)
Correct answer without work ⇒ 7 marks.
Correct answer (without work) incorrectly rounded off ⇒ 6 marks.(See 1st * and S3).
Blunders (−3)
B1
Mishandles
B2
B3
10 marks
65 ⋅ 61 e.g. (65 ⋅ 61) 2 = 4304 ⋅ 6721.
Mishandles (2 ⋅ 42) 2 e.g. 2 ⋅ 42 × 2 = 4 ⋅ 84.
An arithmetical operation other than a given one e.g. + for × .
65 ⋅ 61 × 3 ⋅ 14 − (2 ⋅ 42) 2
= 41 ⋅ 65 (breaking order). [Check candidate’s calculations]
0 ⋅ 47
3 ⋅ 14
65 ⋅ 61 − (2 ⋅ 42) 2 ×
= 14 ⋅ 99 (breaking order). [Check candidate’s calculations]
0 ⋅ 47
Error in decimal point.
Slips (−1)
S1
Numerical errors in arithmetical operations (to max−3).
S2
Each rounding off which would affect the final rounded off answer (max −3). [Check
candidate’s calculations]
S3
Fails to round off or incorrectly rounds off when giving final answer.
Attempts (3 marks)
A1
Calculator: incorrect answer without work.
A2
65 ⋅ 61 = 8 ⋅ 1 and stops.
3 ⋅ 14
A3
= 6 ⋅ 680851064 and stops.
0 ⋅ 47
A4
65 ⋅ 61 × 3 ⋅ 14 = 25 ⋅ 434 and stops.
A5
(2 ⋅ 42) 2 = 5 ⋅ 8564 and stops.
Page 12 of 37
QUESTION 3
Part (a)
Part (b)
Part (c)
10 marks
20 (10, 10) marks
20 (10, 10) marks
Part (a)
10 marks
(a)
Att 3
Aoife bought 3 compact discs at € 16 ⋅ 50 each and 2 magazines at
€ 4 ⋅ 20 each. How much did she pay altogether?
(a)
16 ⋅ 50 × 3 = 49 ⋅ 50
or
4 ⋅ 20 × 2 = 8 ⋅ 40
⇒ total cost : 49 ⋅ 50 + 8 ⋅ 40 = 57 ⋅ 90.
*
*
*
*
*
Att 3
Att (3, 3)
Att (3, 3)
16 ⋅ 50 + 16 ⋅ 50 + 16 ⋅ 50 = 49 ⋅ 50.
4 ⋅ 20 + 4 ⋅ 20 = 8 ⋅ 40
⇒ total cost : 49 ⋅ 50 + 8 ⋅ 40 = 57 ⋅ 90.
Accept 5790, 57 ⋅ 90 or 57 ⋅ 9 regardless of subsequent labelling or work.
Final addition step subject to maximum deduction of −3.
Adds 16 ⋅ 50 to 4 ⋅ 20 = 20 ⋅ 70 and stops ⇒ 3 marks. [Oversimplification].
Correct answer without work ⇒ 7 marks.
Incorrect answer without work ⇒ 0 marks.
Blunders (−3)
B1
Each missing product when finding items cost e.g. 16 ⋅ 50 not multiplied by 3.
B2
Incorrect number of additions when finding items cost e.g. 16 ⋅ 50 + 16 ⋅ 50 only.
B3
Fails to find total cost i.e. no addition.
B4
49 ⋅ 50 − 8 ⋅ 40 = 41 ⋅ 10 .
B5
Error in decimal point.
Slips (−1)
S1
Numerical errors (to max –3).
Page 13 of 37
Part (b) (i)
(i)
(i)
10(5, 5) marks
Att (2,2)
Patrick bought a car for €14 080 and sold it for €16 000.
Calculate his profit as a percentage of the selling price.
Profit: €16 000− €14 080 = €1920
Percentage of the selling price:
1920
× 100 = 12%
16000
Profit: (5 marks)
*
Correct answer without work ⇒ 2 marks.
*
Incorrect answer without work ⇒ 0 marks.
16000
+ 100 = 8 ⋅ 3 + 100 = 108 ⋅ 3 ⇒ 2 marks.
*
1920
Blunders (−3)
B1 Adds €14 080 to €16 000.
Slips (−1)
S1 Numerical errors in arithmetical operations.
Attempts (2 marks)
A1 Some indication of subtraction.
Percentage of selling price: (5 marks)
Blunders (−3)
B1 As percentage of cost price.
16000
× 100 = 833 ⋅ 3 .
1920
B2
Mishandles the calculation of profit as a percentage e.g.
B3
B4
Error in decimal point.
Illegal cancellation(s) in correct method of calculation of profit as a percentage.
Slips (−1)
S1 Numerical errors in arithmetical operations. (to a max −3).
Attempts (2 marks)
A1
Some use of 100 or of the given data or the calculated profit.
A2
" Profit" " Selling Price " or " Profit" " Selling Price "×100 and stops i.e. no substitution of
values.
Page 14 of 37
Part (b) (ii)
10 marks
Att 3
€6000 is invested at 5% per annum.
What is the amount of the investment at the end of one year?
(ii)
1% = 60
5% = 300
Amount = €6300
P × T × R 6000 × 1 × 5
=
= 300
100
100
Amount = €6300
I=
6000 × 1 ⋅ 05
= 6300
Amount = €6300
6000 @ or + 5% = 300 (use of % button, calculator)⇒ €6300 as total.
*
*
*
*
*
€300 (with work shown) and stops ⇒ 7 marks.
6000 × 5 = 30000 and stops ⇒ 4 marks (B1 + B3).
6000 × 5 = 30000 + 6000 = 36000 ⇒ 7 marks (B1).
Correct answer without work ⇒ 7 marks.
Incorrect answer without work ⇒ 0 marks.
Blunders (−3)
B1
Mishandles 5%, e.g. 6000 × 5 or 6000 ÷ 5 (6000 must be used).
B2
Error in decimal point (once only).
B3
Stops at interest i.e. fails to calculate amount.
B4
Subtracts to calculate amount.
B5
Incorrect substitution into formula and continues. [say T = 2: but 6000 must be used ].
6000 × 1 × 5
B6
Illegal cancellation(s) in
.
100
B7
6000 × ⋅05 = 300 and stops.
B8
1 ⋅ 05 = 1 ⋅ 5.
Slips (−1)
S1 Numerical errors in arithmetical operations. (to a max −3)
Attempts (3 marks)
A1
correct formula with or without substitution and stops.
A2
some use of 100 in attempt to find percentage e.g. 5% =
Page 15 of 37
5
and stops.
100
3(c)
20(5, 5, 5, 5) marks
Att (2, 2, 2, 2)
Helen’s weekly wage is €850.
She pays income tax at the rate of 20% on the first €600 of her wage
and income tax at the rate of 42% on the remainder of her wage.
Helen has a weekly tax credit of €54.
Part (c) (i)
5 marks
Att 2
Calculate the tax payable at the rate of 20% on the first €600 of her wage.
(i)
1% = 6
20% = 120
Tax = €120.
Tax =
600
× 20 = €120
100
600 × 0 ⋅ 2 = €120.
.
Correct answer without work ⇒ 2 marks.
Incorrect answer without work ⇒ 0 marks.
*
*
Blunders (−3)
B1
Mishandles 20%, e.g. 600 × 20 = 12 000 or 600 ÷ 20 = 30.
B2
Uses €850 instead of €600.
B3
Error in decimal point.
B4
Illegal cancellation(s) in correct method of calculation of tax.
Slips (−1)
S1 Numerical errors in arithmetical operations. (to a max −3)
Attempts (2 marks)
20
and stops.
A1 Some use of 100 in attempt to find percentage e.g. 20% = 100
Part (c) (ii)
(ii)
5 marks
Att 2
Calculate the tax payable at the rate of 42% on the remainder of her wage.
(ii)
1% = 2 ⋅ 5
42% = 105
Tax = €105
Remainder of wage = €850 − €600 = €250
Tax =
250
× 42 = €105
100
250 × 0 ⋅ 42 = €105
Blunders (−3)
B1
Mishandles 42%, e.g. 250 × 42 or 250 ÷ 42 . (But no penalty if the error is as in Part (c) (i)).
B2
Uses €850 or €600 instead of €250.
B3
Error in decimal point.
B4
Illegal cancellation(s) in correct method of calculation of tax.
Slips (−1)
S1 Numerical errors in arithmetical operations. (to a max −3).
Attempts (2 marks)
42
A1 Some use of 100 in attempt to find percentage e.g. 42% = 100
and stops.
Page 16 of 37
Part (c) (iii)
5 marks
Hence calculate Helen’s gross tax.
(iii)
Helen’s gross tax = €120 + €105 = €225
(iii)
*
*
Att 2
Incorrect answer without work ⇒ 0 marks.
Allow candidate’s incorrect answers from parts (i) and (ii).
Blunders (−3)
B1
€120 − €105 = €15.
B2
Misuse of tax credit.
Slips (−1)
S1
Numerical errors in arithmetical operations. (to a max −3).
Part (c) (iv)
(iv)
(iv)
*
*
*
*
5 marks
Calculate the tax payable by Helen.
Tax payable = €225 − €54 = €171
No use of tax credit ⇒ 0 marks.
Incorrect answer without work ⇒ 0 marks.
Allow candidate’s incorrect gross tax figure from Part (iii).
171 only ⇒ 2 marks.
Blunders (−3)
B1
Misuse of tax credit e.g. 225 + 54 = 279.
Slips (−1)
S1
Numerical errors in arithmetical operations. (to a max −3)
Page 17 of 37
Att 2
QUESTION 4
Part (a)
Part (b)
Part (c)
10(5, 5) marks
20(10, 10) marks
20(5, 5, 5, 5) marks
Part (a) (i)
*
*
*
5 marks
(i)
If x = 4, find the value of: 5 x + 3
(i)
5 x + 3 = 5(4) + 3 = 20 + 3 = 23
20 + 3 ⇒ 4 marks.
Correct answer without work ⇒ 2 marks.
Incorrect answer without work ⇒ 0 marks.
Blunders (−3)
B1
Incorrect numerical substitution for x and continues.
B2
Leaves 5(4) in the answer.
B3
Breaks order i.e. [ 5(4 + 3) = 35 ].
B4
5(4) taken as 54 .
Slips (−1)
S1
Numerical errors.( to a max −3).
Attempts (2 marks)
A1
Substitution and stops e.g. 5(4) only.
Worthless (0)
W1
Incorrect substitution for x and stops.
Page 18 of 37
Att (2, 2)
Att (3, 3)
Att (2, 2, 2, 2)
Att 2
Part (a) (ii)
5 marks
(ii)
If x = 4, find the value of: x 2 − x + 7
x 2 − x + 7 = ( 4) 2 − 4 + 7
(ii)
= 16 − 4 + 7
= 19
*
*
*
*
*
16 – 4 + 7 ⇒ 3 marks.
16 – 4 + 7 = 16 − 11 = 5 ⇒ 4 marks.
12 + 7 ⇒ 4 marks.
Correct answer without work ⇒ 2 marks.
Incorrect answer without work ⇒ 0 marks.
Blunders (−3)
B1
Incorrect numerical substitution for x and continues.
B2
Mishandles (4) 2 i.e. (4) 2 = 8 or leaves (4) 2 in answer.
B3
Mishandles − (4) i.e. = 4 .
B4
Breaks order i.e. [ 16 − 1(4) = 15(4) = 60 ].
B5
− 1(4) taken as − 1 + 4 .
B6
−1(4) clearly taken as −14.
Slips (−1)
S1
Numerical errors (to max –3).
Attempts (2 marks)
A1
Substitution and stops i.e. (4) 2 − (4) + 7 only.
A2
Incomplete substitution and continues or stops.
A3
4x substituted for x in both terms with x and continues or stops.
Page 19 of 37
Att 2
Part (b) (i)
(i)
(i)
10 marks
Att 3
Multiply (3 x − 2) by (4 x + 5) and write your answer in its simplest form.
(3 x − 2)(4 x + 5) = 3 x(4 x + 5) − 2(4 x + 5)
= 12 x 2 + 15 x − 8 x − 10
= 12 x 2 + 7 x − 10.
*
*
Correct answer without work ⇒ 7 marks.
Incorrect answer without work ⇒ 0 marks.
Blunders (−3)
B1
Each incorrect term or each term omitted on multiplication.
Slips (−1)
S1
Each incorrect term or each term omitted in final simplification. (to a max of −3)
Attempts (3 marks)
A1
Any correct multiplication.
A2
3 x(4 x + 5) − 2(4 x + 5) and stops.
A3
4 x(3 x − 2) + 5(3 x − 2) and stops.
Worthless (0)
W1
(3 x − 2) ± (4 x + 5) stops or continues.
W2
Adding unlike terms before attempt at multiplication.
Page 20 of 37
Part (b) (ii)
10 marks
(ii)
Att 3
Write in its simplest form
(4 x 2 − 3 x + 7) + ( x 2 − 2 x − 8)
(ii)
( 4 x 2 − 3 x + 7 ) + ( x 2 − 2 x − 8)
= 4 x 2 − 3x + 7 + x 2 − 2 x − 8
= 5x 2 − 5x − 1
*
*
Stops after correct removal of brackets ⇒ 7 marks.
Incorrect answer without work ⇒ 0 marks.
Blunders (−3)
B1
Each incorrect term or each term omitted on bracket removal.
Slips (−1)
S1
Each incorrect term or each term omitted in final simplification.(max −3)
Misreadings (-1)
M1
(4 x 2 − 3x + 7) × ( x 2 − 2 x − 8) . Apply B1 as each incorrect term or each term omitted on
multiplication.
Attempts (3 marks)
A1
Any correct addition of a pair of like terms.
A2
(4 x 2 − 3x + 7) = 4 x 2 − 3 x + 7 only or ( x 2 − 2 x − 8) = x 2 − 2 x − 8 only.
A3
Treats as equation e.g. 4 x 2 − x 2 − 3 x + 2 x + 7 + 8 .
Worthless (0)
W1
Adding unlike terms before removal of brackets.
Page 21 of 37
Part (c)
20 (5, 5, 5, 5) marks
A rectangle has a length ( x + 6) cm and width x cm,
as in the diagram
Part (c) (i)
x+6
x
5 marks
Att 2
Find the perimeter of this rectangle in terms of x.
(i)
Att (2, 2, 2, 2)
(i)
x + x + ( x + 6) + ( x + 6)
2( x + x + 6)
4 x + 12
*
*
*
*
*
*
Accept either x + x + ( x + 6) + ( x + 6) or 2( x + x + 6) for full marks.
If x + x + ( x + 6) + ( x + 6) present give full marks for this section, irrespective of any
subsequent errors within the section.
Brackets as above not required, accept x + x + x + 6 + x + 6.
Incorrect answer without work ⇒ 0 marks.
4 x + 12 only ⇒ 2 marks.
4 x + 12 and diagram as in Att 2 ⇒ 5 marks.
Blunders (−3)
B1
Adding only any two of the four sides required e.g. x + ( x + 6) .
B2
x × ( x + 6) or x × x or ( x + 6) × ( x + 6) .
B3
x + x + ( x + 6) .
B4
x + ( x + 6) + ( x + 6) .
Slips (−1)
S1
Numerical errors. (max −3)
Attempts (2 marks)
A1
P = 2( L + B ) or P = ( L + B ) .
A2
Diagram as over:
Worthless (0)
W1
x only or x + 6 only.
x+6
x
x
x+6
Page 22 of 37
Part (c) (ii)
(ii)
*
Att 2
If the perimeter of the rectangle is 40 cm, write down an equation in x
to represent this information.
.
(ii)
*
*
5 marks
4 x + 12 = 40.
Accept either x + x + ( x + 6) + ( x + 6) = 40 or 2( x + x + 6) or 4 x + 12 = 40 for full marks.
If x + x + ( x + 6) + ( x + 6) = 40 present give full marks for this section, irrespective of any
subsequent errors within the section.
Accept candidate’s incorrect perimeter from (c) (i) = 40 for full marks.
Blunders (−3)
B1
x × ( x + 6) = 40 [if not the candidate’s expression above].
Slips (−1)
S1
An x or an x + 6 omitted in transcription from part (c) (i). ( max −3)
Part (c) (iii)
5 marks
(iii)
(iii)
Att 2
Solve the equation that you formed in part (ii) above, for x.
x + x + ( x + 6) + ( x + 6) = 40
4 x + 12 = 40
4 x + 12 = 40
4 x = 40 − 12
4 x = 28
4 x = 40 − 12
4 x = 28
x=7
x=7
x + x + x + x = 40 − 6 − 6
4x = 28
x=7
Candidate’s equation from (ii) must be progressed to the form " ax = ....." before any marks
can be earned for this section.
*
x = 284 ⇒ 4 marks.
*
Correct Equation for part (ii) may be embedded in this section and would earn full marks for
part (ii) if part (ii) is incorrect or vacant.
*
Correct answer without work ⇒ 2 marks.
*
Incorrect answer without work ⇒ 0 marks.
Blunders (−3)
B1
Error(s) in progressing equation (e.g. transposition).
B2
Adds ‘x s to ‘numbers’ and continues e.g. 4 x + 12 = 16x.
*
Slips (−1)
S1
Errors in addition (to max −3).
S2
Error in division e.g. x = 284 ⇒ x = 6 (say).
28
S3
4 and stops.
Attempts (2 marks)
A1
x + x + x + x = 40 − 6 − 6 and stops.
A2
4x only or 12 only appears and stops.
A3
Correct answer from arithmetical approach.
Page 23 of 37
Part (c) (iv)
5 marks
(iv) Find the area of the square with the same perimeter as the given rectangle.
Give your answer in cm2.
Perimeter of Square = 4l
(iv)
4l = 40
l = 10
l cm
*
*
⇒ Area of square = 10 × 10 = 100 cm 2 .
Accept 100 as answer (no need for units).
Correct answer without work ⇒ 2 marks.
Blunders (−3)
l ≠ 10 and continues.
B1
Slips (−1)
S1
Numerical errors within correct approach (max−3).
Attempts (2 marks)
x + x + ( x + 6) + ( x + 6)
4 x + 12
A1
or
and stops.
4
4
A2
Some use of 40 e.g. (40) 2 or 40 .
Worthless (0)
W1
7 × 13 or 7 × 13 = 91.
Page 24 of 37
Att 2
QUESTION 5
Part (a)
Part (b)
Part (c)
10 marks
20 (5, 5, 5, 5) marks
20 (5, 5, 10) marks
Part (a)
Att 3
Att (2, 2, 2, 2)
Att (2, 2, 3)
10 marks
Att 3
Solve the equation 5 x − 6 = 3( x + 4)
5 x − 6 = 3 x + 12
(a)
5 x − 3 x = 12 + 6
(4 marks)
2 x = 18
(7 marks)
x=9
*
*
*
(3 marks)
(10 marks)
x = 182 ⇒ 9 marks.
Correct answer without work ⇒ 7 marks.
Incorrect answer without work ⇒ 0 marks.
Blunders (−3)
B1
Error in distributive law and continues, e.g. 3x + 4 or x + 12 (once only).
B2
Each error in progressing equation (e.g. transposition).
Slips (−1)
S1
Error in division e.g. x = 182 ⇒ x = 8 (say).
S2
Numerical errors (to max −3).
18
S3
2 and stops.
Attempts (3 marks)
A1
Adds or subtracts‘x’s to ‘numbers’ and continues e.g. 5x − 6 = ± x or 3( x + 4) = 3(5 x) = 15 x.
A2
5 x − 6 = 3 x + 12 and stops.
A3
3x appears and stops.
A4
5 x = 3( x + 4) + 6 and stops.
Worthless (0)
W1
Adds or subtracts‘x’s to ‘numbers’ and stops.
Page 25 of 37
Part (b) (i)
(i)
5 marks
Factorise
Att 2
4ab + 8b
4b(a + 2)
(i)
*
Accept 4(ab + 2b) or b(4a + 8) or 2(2ab + 4b) for 5 marks.
Blunders (−3)
B1
An incorrect common factor.
B2
Stops after some correct effort at factorisation e.g. 4b(a ) + 4b(2) or similar.
Slips (−1)
S1
Numerical errors when taking out a factor e.g. 4b(a + 4) .
Attempts (2 marks)
A1
4(ab) and /or + 8(b) or effort at brackets.
A2
Indication of common factor e.g. 4ab + 8b .
Part (b) (ii)
(ii)
*
*
5 marks
Factorise:
ab + 2ac + 5b + 10c
ab + 2ac + 5b + 10c
(ii)
Att 2
ab + 5b + 2ac + 10c
a (b + 2c) + 5(b + 2c)
(b + 2c)(a + 5)
b(a + 5) + 2c(a + 5)
(a + 5)(b + 2c)
Correct answer without work ⇒ 2 marks.
Incorrect answer without work ⇒ 0 marks.
Blunders (−3)
B1
Stops after first line of correct factorisation.
B2
Error(s) in factorising any pair of terms.
B3
Incorrect common factor and continues e.g. a (b + 2c) + 5(b + c) = (b + 2c)(a + 5) .
An instance of correct answer from incorrect work.
Slips (−1)
S1
(b + 2c) ± (a + 5) .
S2
Correct second line of factorisation but gives 5a (b + 2c).
Attempts (2 marks)
A1
Pairing off, or indication of pairing off, and stops.
A2
Correctly factorises any pair and stops.
Page 26 of 37
Part (b) (iii)
(iii)
5 marks
Att 2
Factorise: x 2 + 2 x − 15
(iii)
x 2 + 2 x − 15
= x 2 + 5 x − 3x − 15
= x( x + 5) − 3( x + 5) (2 marks)
= ( x + 5)( x − 3)
x
+5
x
(2 marks)
*
*
*
*
⇒ ( x + 5)( x − 3)
−3
Quadratic equation formula method is subject to Slips and Blunders.
Accept (with or without brackets) for 5 marks any of the following
(x + 5) and (x − 3) . ( The word and is written down.)
(x + 5) or (x − 3) .
( The word or is written down.)
Accept ( x + 5), ( x − 3) for 5 marks.
Correct answer without work ⇒ 5 marks.
Blunders (−3)
B1
Incorrect two term linear factors of x 2 + 2 x − 15 formed from correct, but not applicable,
factors of x 2 and ± 15 .
B2
Correct cross method but factors not written.
B3
x( x + 5) − 3( x + 5) or x( x − 3) + 5( x − 3) and stops.
B4
Incorrect common factor and continues (applies to guide number method).
Slips (−1)
S1
Uses quadratic equation formula, but has wrong signs in factors (once only).
S2
Uses quadratic equation formula to find x = −5 and x = 3 and stops.
S3
( x + 5) ± ( x − 3) .
Attempts (2 marks)
A1
Correct factors of x 2 only.
A2
Correct factors of –15 or +15 only.
A3
5 x − 3 x only appears.
A4
Correct quadratic equation formula with or without substitution and stops.
Page 27 of 37
Part (b) (iv)
(iv)
(iv)
*
*
5 marks
Att 2
Factorise: x 2 − y 2
( x + y )( x − y )
Accept (with or without brackets) for 5 marks any of the following
(x + y ) and (x − y ) . [The word and is written down.]
(x + y ) or (x − y ) . [The word or is written down.]
Accept ( x + y ), ( x − y ) for 5 marks.
Blunders (−3)
B1
Incorrect two term linear factors of x 2 − y 2 formed from correct, but not applicable, factor
of x 2 and ± y 2 .
B2
( y + x)( y − x) .
Slips (−1)
S1
Solves x 2 = y 2 to give x = y and x = − y and stops.
S2
( x + y) ± ( x − y) .
Attempts (2 marks)
A1
Correct factors of x 2 only.
A2
Correct factors of y 2 or − y 2 only.
A3
x or ± y appears.
A4
x 2 − y 2 = x.x − y. y and stops.
A5
( xy )( xy ).
A6
Mention of the difference of two squares.
Page 28 of 37
Part (c) (i)
(i)
(i)
*
*
*
*
*
*
5 marks
Att 2
x+5 x+2
+
as a single fraction.
4
3
Give your answer in its simplest form.
Express
3( x + 5) + 4( x + 2)
12
3x + 15 + 4 x + 8
=
12
7 x + 23
=
12
(5 marks)
x + 5 x + 2 2x + 7
+
=
⇒ 0 marks.
4
3
7
All Blunders and Slips in the simplification of the numerator subject to a max. deduction (−3).
3( x + 5) + 4( x + 2)
and stops ⇒ 2 marks. (B3)
12
3 x + 15 + 4 x + 8
and stops ⇒ 4 marks. (S3)
12
14 x + 46
⇒ 4 marks. (S3)
24
14 x + 46 ⇒ 3 marks. (S1and S3)
Adds numerators and then denominators i.e.
Blunders (−3)
B1 Incorrect common denominator and continues.
B2
Incorrect numerator from candidate’s common denominator. e.g.
B3
B4
B5
No simplification of numerator.
Errors in distributive law. [See * 2]
Errors in sign when multiplying. [See * 2]
Slips (−1)
S1 Correct common denominator implied.
S2 Numerical errors in arithmetical operations.
S3 Not in simplest form. [See * 4].
Attempts (2 marks)
A1 12 only or a multiple of 12 only appears.
5 x 2 x 15 x + 8 x
A2
+
=
.
4
3
12
Worthless (0)
 x + 5  x + 2 
W1 

 and stops.
 4  3 
6 x 3x
x + 5 x + 2 5x 2x 7 x
W2
+
and stops or
+
=
+
=
= x.
4
3
4
3
4
3
7
Page 29 of 37
4( x + 5) + 3( x + 2)
.
12
Part (c) (ii)
5 marks
Att 2
Hence, or otherwise, solve the equation
(ii)
x+5 x+2 5
+
=
4
3
2
(ii)
7 x + 23 5
=
12
2
14 x + 46 = 60
14 x = 60 − 46
14 x = 14
x = 1.
*
*
*
*
*
*
*
7 x + 23 5
− =0
12
2
2(7 x + 23) − 12(5)
=0
12
14 x + 46 − 60 = 0
14 x − 14 = 0
14 x = 14
x =1
12(7 x + 23) 12(5)
=
12
2
7 x + 23 = 6(5)
7 x + 23 = 30
7x = 7
x =1
Candidate’s equation must be of the form
3( x + 5) + 4( x + 2) = 30
12
3 x + 15 + 4 x + 8 = 30
7 x + 23 = 30
7 x = 30 − 23
7x = 7
x =1
ax + b 5
= if full marks are to be earned for this
c
2
section.
Correct trial and error ⇒ Att mark only.
x + 5 x + 2 5x 2x 7 x
7x 5
5
+
=
+
=
= x from (i) and then
= or x = ⇒ 2 marks.
4
3
4
3
7
7
2
2
5 x 2 x 15 x + 8 x
15 x + 8 x 5
23 x 5
+
=
from (i) and then
= or
= ⇒ 2 marks.
4
3
12
12
2
12
2
x + 5 x + 2 2x + 7
2x + 7 5
+
=
from (i) and then
= etc can gain full marks.
4
3
7
7
2
x+5 x+2 5
+
= ⇒ 4 x + 20 + 3 x + 6 = 10 etc ⇒ B2.
4
3
2
x+5 x+2 5
+
= ⇒ x + 1 ⋅ 25 + x + 0 ⋅ 66 = 2 ⋅ 5 etc ⇒ B2.
4
3
2
Blunders (−3)
B1
Error(s) in establishing an equation without fractions and continues.
B2
Error(s) in progressing equation (e.g. transposition).
Slips (−1)
S1 Error in division in final step to find x.
Attempts (2 marks)
A1 Adding unlike terms in progressing equation.
A2 Some effort at removal of fractions.
A3 Oversimplification e.g. 7 x + 23 = 5 and continues.
A4 Oversimplification as a result of errors in part (i).
A5 Trial and error.
Page 30 of 37
Part (c) (iii)
(iii)
10 marks
Att 3
Solve for x and for y:
3x − y = 8
x + 2y = 5
(iii)
3x − y = 8
x + 2y = 5
3x − y = 8
x + 2y = 5
y = 3x − 8
x + 2(3x − 8) = 5
6 x − 2 y = 16
3x − y = 8
3 x + 6 y = 15
x + 6 x − 16 = 5
7 x = 5 + 16
− 7 y = −7
y =1
x=3
7 x = 21
x=3
x + 2y = 5
7 x = 21
x=3
y =1
*
*
y =1
Apply only one blunder deduction (B1 or B2) to any error(s) in establishing the first
equation in terms of x only or the first equation in terms of y only.
Finding the second variable is subject to a maximum deduction (−3).
Blunders (−3)
B1
Error(s) in establishing the first equation in terms of x only [ 7 x = 21 ] or the first equation in
terms of y only [ − 7 y = −7 ] through elimination by cancellation.
B2
Error(s) in establishing the first equation in terms of x only or the first equation in terms of y
only through elimination by substitution.
B3
Errors in transposition in solving the first one variable equation.
B4
Errors in transposition when finding second variable.
B5
Incorrect substitution when finding second variable.
B6
Finds one variable only.
Slips (−1)
S1
Numerical errors (max −3) in solving first one variable equation and when finding second
variable.
Attempts (3 marks)
A1
Attempt at transposition and stops.
A2
Multiplies either equation by some number and stops.
A3
Correct answers without algebraic work.
Page 31 of 37
QUESTION 6
Part (a)
Part (b)
Part (c)
10 (5, 5) marks
30 (20, 10) marks
10 (5, 5) marks
Part (a) (i)
5 marks
(i)
f ( x ) = 5 x − 6.
(i)
*
*
*
*
Att 2
Find: f (3)
f (3) = 5(3) − 6
= 15 − 6
=9
*
Att (2,2)
Att (7,3)
Att (2,2)
(4 marks)
(5 marks)
Function concept correct:
f (3) = 5(3) − 6 or f (2) = 15 − 6 i.e. multiplication of 3 by 5
is clearly indicated or is implied by subsequent work.
Completion of f (3) subject to maximum deduction of –1.
Correct function concept i.e. 5(3) − 6 and stops ⇒ 4 marks.
Ignores x giving 5−6 = −1 ⇒ 0 marks.
3[ f ( x)] = 15 x − 18 ⇒ 0 marks.
Correct answer without work ⇒ 2 marks.
Blunders (−3)
B1
f (3) incorrect: misunderstanding of the concept of a function.
Misreadings (−1)
M1
f (−3) instead of f (3).
Slips (−1)
S1
Numerical errors (to max –1).
Attempts (2 marks)
A1
Treats as equation and continues or stops.
Page 32 of 37
Part (a) (ii)
(ii)
5 marks
f ( x ) = 5 x − 6.
(ii)
Att 2
f (−2)
f (−2) = 5(−2) − 6
= −10 − 6
= −16
*
*
*
*
*
(5 marks)
Function concept correct:
f (−2) = 5(−2) − 6 or f (−2) = −10 − 6 i.e. multiplication of −2 by 5
is clearly indicated or is implied by subsequent work.
Completion of f (−2) subject to maximum deduction of –1.
Correct function concept i.e. 5(−2) − 6 and stops ⇒ 4 marks.
Ignores x giving 5−6 = −1 ⇒ 0 marks.
− 2[ f ( x)] = −10 x + 12 ⇒ 0 marks.
Correct answer without work ⇒ 2 marks.
Blunders (−3)
B1
f (−2) incorrect: misunderstanding of the concept of a function.
Misreadings (−1)
M1
f (2) instead of f (−2).
Slips (−1)
S1
Numerical errors and sign errors (to max –1).
Attempts (2 marks)
A1
Treats as equation and continues or stops.
Page 33 of 37
Part (b)
20 marks
Att 7
Draw the graph of the function
f : x → x2 + x − 3
in the domain −3 ≤ x ≤ 2,
f (-3)
f (-2)
f (-1)
f (0)
f(1)
f(2)
=
=
=
=
=
=
where x ∈ R.
(-3)2 + (-3) −3 = 3
(-2)2 + (-2) −3 = −1
(-1)2 + (-1) −3 = −3
(0)2 + (0) − 3 = −3
(1)2 + (1) − 3 = −1
(2)2 + (2) − 3 = 3
or
x
x2
x
-3
f(x)
-3
9
-3
-3
3
-2
4
-2
-3
−1
-1
1
-1
-3
-3
0
0
0
-3
-3
1
1
1
-3
-1
2
4
2
-3
3
Table
20 marks
Att 7
*
Each individual error in the rows other than the f(x) row, apart from Blunders below, attracts a
deduction of −1 subject to a maximum deduction of −3 per row. [f(x) max(−6)]
Blunders (−3)
x 2 taken as 2x all the way. [In row headed x 2 by candidate]
B1
B2
x taken as −x all the way. [In row headed x by candidate]
B3
-3 calculated as –3x all the way. [In row headed − 3 by candidate]
B4
Adds in top row when evaluating f (x) .
B5
Omits ‘−3’ row or omits ‘x’ row.
B6
Omits a value in the domain each time to max of −9 (5 values missing ⇒ Att 7).
B7
Each incorrect image without work.
Slips (−1)
S1
Numerical Slips (to max −3) in any row other than f (x) row.
S2
Misreads ‘−3’ as ‘+3’ and places ‘+3’ in the table or ‘+x’ as ‘−x’ and places ‘−x’ in the table.
S3
Each incorrect f (x) value calculated by addition within columns in student’s table (no limit).
But note B4.
Attempts (7 marks)
A1
Omits x 2 row from table or treats x 2 as x.
A2
Table with only f (x) = x2.
A3
Any effort at calculating point(s).
A4
One point only calculated and nothing else.
Page 34 of 37
Graph
*
*
*
10 Marks
Att 3
Att 7 + Att 3 ⇒ one point only calculated and graphed correctly.
Correct graph but no table ⇒ full marks, i.e. 30 marks.
Accept reversed co-ordinates (i) if axes not labelled or (ii) if axes are reversed to compensate
(see B1 below).
Blunders (−3)
B1
Reversed co-ordinates plotted against non-reversed axes (once only) [See 3rd *].
B2
Axes not graduated uniformly (once only).
B3
Points not joined or joined in incorrect order (once only).
Slips (−1)
S1
Each point of candidate graphed incorrectly.
S2
Each point from table not graphed (subject to N1).
Attempts (3 marks)
A1
Graduated axes only (need not be labelled).
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Part (c) (i)
5 marks
Att 2
Use the graph drawn in 6(b) to estimate:
(i)
the values of x for which f ( x) = 0.
(i)
Work to be shown on graph and answers written here.
x = −2 ⋅ 3 or x = 1 ⋅ 3
*
*
Correct answer (clearly consistent with graph) inside tolerance without graphical indication
⇒ 2 marks.
A candidate’s incorrect graph can earn up to full marks for this section. [Use similar
tolerances]
Blunders (−3)
B1
Answer on diagram but outside of tolerance ( ± 0 ⋅ 25 ).
B2
Fails to write down the answers.
B3
Only one answer or indication.
Attempts (2 marks)
A1
Algebraic evaluation.
Worthless (0)
W1
Answers outside of tolerance without graphical indication.
W2
f (0) giving −3 as answer.
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Part (c) (ii)
5 marks
Att 2
Use the graph drawn in 6(b) to estimate:
(ii)
the value of f ( x) when x = 0 ⋅ 5.
(ii)
Work to be shown on graph and answers written here
f (0 ⋅ 5) = −2 ⋅ 25
*
*
Correct answer (clearly consistent with graph) inside tolerance without graphical indication
⇒ 2 marks.
A candidate’s incorrect graph can earn up to full marks for this section.
(Use similar tolerances)
Blunders (−3)
B1
Answer on diagram but outside of tolerance ( ± 0 ⋅ 25 ).
B2
Fails to write down the answer.
Attempts (2 marks)
A1
Algebraic evaluation or calculator.
Worthless (0)
W1
Answer outside of tolerance without graphical indication.
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