1
Workbook of
Solution 4
NUMBERS
Number
Chapter 1: Number
May / June 2010 Page 23
1
During one week in April, in Quebec, the daily minimum temperatures were
–5°C,
–1°C,
3°C,
–2°C,
2°C,
0°C,
6°C,
Write down
(a) the lowest of these temperatures,
Answer: (a)
-5°C
[1]
(b) the range of these temperatures.
range = highest – lowest
= 6 – (-5)
= 11
Answer: (b)
2
48%
23
53
11
4.80
53 = 4.818181
11
Answer: (b)
3
[1]
[First convert all the numbers
to decimal then compare and
write original numbers not
decimal ]
Write the numbers in order of size with the largest first.
23 = 4.795831523, 48% = 0.48, 11°C
53 > 4.80 >
23 > 48%
11
[1]
Ricardo changed $ 600 into pounds (£) when the exchange rate $ 1 = £ 0.60.
He later changed all the pounds back into dollars when the exchange rate was $ 1 = £ 0.72.
How many dollars did he receive?
$
£
$
£
1
0.60
1
0.72
600
x
x
360
x = 600 × 0.6 = £360
x = 360 = $500
0.72
Answer:
$500
[2]
3
Workbook of Variant 3
4
The maximum speed of car is 252 km/h.
Change this speed into metres per second.
= 252 ×
= 70 m/s
[to convert km to m → × 5 ]
hr
s
18
5
18
Answer:
5
70 m/s
[2]
Amalie makes a profit of 20% when she sells a shirt for $ 21.60.
[% is always 100]
Calculate how much Amalie paid for the shirt.
old (C.P.)
new (S.P.)
%
100
120 (100 + 20)
cost
x
21.60
x = 100 × 2.160
x = 18
100
Answer:
9
$18
[2]
1 second = 106 microseconds.
Change 3 × 1013 microseconds into minutes. Give your answer in standard form.
x=
seconds
microseconds
1
106
x
3 × 1013
3 × 1013
106
x = 30000000 sec
30000000 sec = 30000000 minute
60
= 500000
= 5 × 105
Answer:
4
5 × 105 min
[2]
Number
10The length of each side of an equilateral triangle is 74 mm, correct to the nearest millimetre.
Calculate the smallest possible perimeter of the triangle.
74 mm correct to nearest millimetre is half of 1 mm = 1
2
lower bound of length = 74 − 0.05
= 0.5
= 73.5
perimeter = 3(73.5)
= 220.5
Answer:
220.5 mm
[2]
May June 2010 Paper 43
1
Daniella is 8 years old and Edward is 12 years old.
(a) Their parents give them some money in the ratio of their ages.
(i) Write a ratio
Daniella’s age : Edward’s age
in the simplest form.
8 = 2
12
3
Answer: (a) (i)
2:3
[1]
(ii) Daniella receives $ 30.
Show that Edward receives $ 45.
Daniella
Edward
Ratio
2
3
Money
30
x
3 × 30
2
x =
x = $45
Answer: (a) (ii)
$ 45
[1]
(iii)What percentage of the total amount of money given by their parents does Edward
receive?
= Edward’s money × 100
total
=
45
× 100
45 + 30
= 60
5
Workbook of Variant 3
10
Diagram 1
Diagram 2
Diagram 3
Diagram 4
The diagrams show squares and dots on a grid.
Some dots are on the side of each square and other dots are inside each square.
The area of the square (shaded) in Diagram 1 is 1 unit2.
(a) Complete Diagram 4 by marking all the dots.
[1]
(b) Complete the column in the table below for diagram 4, 5 and n.
Diagram
1
2
3
4
5
------
n
Number of units of area
1
4
9
16
25
------
n2
Number of dots inside the square
1
5
13
25
41
------
(n−1)2 + n2
Number of dots on the sides of
the square
4
8
12
16
20
------
4n
Total number of dots
5
13
25
41
61
------
n2 + (n+1)2
For number of unit of area
1
4
9
12 22 92 42 52
16 25 -----------
n2
For number of dot inside the square :- nth term = (n − 1)2 + n2
So in 4th diagram : (4 − 1)2 + 42 = 25
in 5th diagram : (5 − 1)2 + 52 = 41
For number of dots on the sides of the square
1 × 4
2 × 4
3 × 4
4 × 4
5×4
4
8
12
16
20------
16 + 25
25 + 36
4n.
For total number of dots
1 + 4
4 + 9
9 + 16
513254161-----------
6
n2 + (n + 1)2
[7]
Number
(c) For Diagram 200, find the number of dots
(i) inside the square,
(200 − 1)2 + 2002 = 79601
Answer: (c) (i)
79601
[1]
(ii) on the sides of the square.
4 × 200 = 800
Answer: (c) (ii)
800
[1]
(d) Which diagram has 265 dots inside the square?
(n − 1)2 + n2 = 265
[ans through caculator]
n2 − 2n + 1 + n2 = 265
2n2 − 2n + 1 − 265 = 0
2n2 − 2n − 264 = 0
n = 12
Answer: (d)
12
[1]
October / November 2010 Paper 23
1
Write down the number which is 3.6 less than –4.7.
–4.7 – 3.6 = –8.3
Answer:
–8.3
[1]
2A plane took 1 hour and 10 minutes to fly from Riyadh to Jeddah.
The plane arrived in Jeddah at 2305.
At what time did the plane depart from Riyadh?
−
hour
minutes
23
05
1
10
Borrow 1hour from 23 and add 60 minutes to 05 minute.
7
Workbook of Variant 3
Answer: (a) (iii)
60 %
[2]
(b) Daniella invests her $ 30 at 3% per year, compound interest.
Calculate the amount Daniella has after 2 years.
Give your answer correct to 2 decimal places.
Amount = P(1 +
Amount = P(1 +
Amount = 31.827
r )2
100
3 )2
100
Answer: (b)
$31.83
[3]
(c) Edward also invests $ 30.
He invests this money at rate of r % per year, simple interest.
After 5 years he has a total amount of $ 32.25.
Calculate the value of r.
Amount = principal + simple interest
32.25 = 30 + simple interest
simple interest = 32.25 − 30
simple interest = 2.25
simple interest =
2.25 =
p×t×r
100
30 × 5 × r
100
2.25 × 100
=r
30 × 5
r = 1.5
Answer: (c)
8
r = 1.5
[2]
Number
[1 hour = 60 minutes]
hour
minutes
22
65
1
10
21
55
−
21:55
Answer:
3
[1]
Calculate 3 2.352 − 1.092
Give your answer correct to 4 decimal places.
3 2.352 − 1.092 = 1.63045875
1.6305
Answer:
5
Show that 3
[2]
3
1
1
+1
=5
4
3
12
Write down all the steps in your working.
Answer 3 3 + 1 1
4
3
=
15 + 4
4
3
=
15 × 3 + 4 × 4
12
=
61
12
=
5
[L.C.M. of 3 and 4 is 12]
1
12
5
Answer:
6
1
12
[1]
Write the following in order of size, smallest first.
20
41
80
161
↓
0.487804878
↓
04968944099
0.492
4.93%
↓
0.0493
4.93% <
Answer:
20
80
< 0.492 <
41
161
[2]
9
Workbook of Variant 3
7
In France, the cost of the one kilogram of apricots is € 3.38.
In the UK, the cost of one kilogram of apricots is £ 4.39.
£ 1 = € 1.04.
Calculate the difference between these prices.
Give your answer in pounds(£).
£
1
€
1.04
3.38
x
3.38
= £3.25
1.04
x=
= £ 4.39 − £ 3.25
= £1.14
Answer:
8
£ 1.14
[2]
A large rectangular card measures 80 centimetres by 90 centimetres.
Maria uses all this card to make small rectangular cards measuring 40 millimetres by
15 millimetres.
Calculate the number of small cards.
40 millimetres = 4 cm
15 milimetres = 1.5cm
90
1.5
= 60,
80
= 20
4
Number of small card = 60 × 20
= 1200
Answer:
1200
[2]
10 Nikhil invests $ 200 for 2 years at 4% per year compound interest.
Calculate the exact amount Nikhil has after 2 years.
x
Amount = P 1 +
100
Amount = 200 1 +
n
4
100
2
Amount = 216.32
Answer:
10
$216.32
[2]
Number
12 The side of a square is 6.3 cm, correct to the nearest millimetre.
The lower bound of the perimeter of the square is u cm and the upper bound of the perimeter
is v cm.
Calculate the value of
(a) u,
1 millimetre = 0.1cm
half of 0.1 =
lower bound of side = 6.3 − 0.05
0.1
= 0.05
2
= 6.25
lower bound of perimeter = 4 × 6.25
= 25
Answer: (a)
u = 25
[1]
(b) v − u.
upper bound of perimetre = 6.3 + 0.05
= 6.35
v = upper bound of perimetre = 4 × 6.35
= 25.4
v − u = 25.4 − 25
= 0.4
Answer: (b)
0.4
[1]
13 a × 107 + b × 106 = c × 106
Find c in terms of a and b
Give your answer in its simplest form.
a × 107 + b × 106 = c × 106
a × 10 × 106 + b × 106 = c × 106
[107 = 10 × 106 as am × an = am + n]
106 (10a + b) = c × 106
c = 10a + b
Answer: (b)
c = 10a + b
[2]
11
Workbook of Variant 3
14 Priyantha completes a 10 km run in 55 minutes 20 seconds.
Calculate Priyantha's average speed in km/h.
55 minute =
55
60
11
+
12
Average Speed =
=
11
hour
12
20
hour =
60×60
20 seconds =
Total time =
hour =
1
180
1
180
83
90
=
Total distance
Total time
10
83 ÷ 90
= 10.84337349
Answer:
10.8 km/h
[2]
October / November 2010 Paper 43
1
Thomas, Ursula and Vanessa share $ 200 in the ratio
Thomas : Ursula : Venessa = 3 : 2 : 5
(a) Show that Thomas receives $ 60 and Ursula receives $ 40.
Answer (a)
Thomas
Ursula
Vanessa
Total
Ratio
3
2
5
10
money
x
y
x =
3 × 200
= 60
10
200
[taking T's and total's column and cross multiply]
[taking U's and total's column and cross multiply]
y =
2 × 200
= 40
10
[2]
(b) Thomas buys a book for $ 21.
What percentage of his $ 60 does Thomas have left?
60 − 21 = 39 → left
39
× 100 = 65
60
Answer: (b)
12
65%
[2]
Number
(c) Ursula buys a computer game for $ 36.80 in a sale.
The sale price is 20% less than the original price.
Calculate the original price of the computer game.
x =
Old
New
%
100
80
Cost
x
36.80
100 × 36.80
80
x = 46
Answer:
$ 46
[3]
(d) Vanessa buys some books and some pencils.
Each book costs $ 12 more than each pencil.
The total cost of 5 books and 2 pencils is $ 64.20.
Find the cost of one pencil.
let cost of one pencil be x.
so cost of one book will be x+12
5 (x + 12) + 2x = 64.20
5x + 60 + 2x = 64.20
7x = 64.20 − 60
7x = 4.20
4.20
7
x = 0.6
x =
Answer:
$0.6
[3]
13
Workbook of Variant 3
11
Diagram 1
Diagram 2
Diagram 3
Diagram 4
The first four Diagrams in a sequence are shown above.
Each Diagram is made from dots and one centimetre lines.
The area of each small square is 1 cm2.
(a) Complete the table for Diagram 5 and 6.
Diagram
1
2
3
4
5
6
Area(cm2)
2
6
12
20
30
42
Number of dots
6
12
20
30
42
56
Number of one centimetre line
7
17
31
49
71
97
[4]
(b) The area of Diagram n is n(n + 1) cm2.
(i) Find the area of Diagram 50.
50 (50 + 1) = 2550
Answer: (b) (i)
2550 cm2
[1]
(ii) Which Diagram has an area of 930 cm2 ?
n (n + 1) = 930
n2 + n = 930
n2 + n − 930 = 0
∆ n = 30
Answer: (b) (ii)
Answer(c)
14
© UCLES 2010
0580/43/O/N/10
30
[1]
[1]
Number
(c) Find, in terms of n, the number of dots in Diagram n.
Area of diagram n is n (n + 1)
For number of dots in diagram , n will get replaced
By (n + 1) in area's nth term
(n + 1) (n + 1 + 1) = (n + 1) (n + 2)
Answer: (c)
(n + 1) (n + 2)
[1]
(d) The number of one centimetre lines in Diagram n is 2n2 + pn + 1
(i) Show that p = 4.
Answer (d)(i):
For diagram 1, number of one centimetre lines are 7
n = 1, 2(1)2 + p (1) + = 7
2 + p +1 = 7
p = 7 − 2 − 1
p = 4
Answer: (d) (i)
[1]
(ii) Find the number of one centimetre lines in Diagram 10.
Number of one centimetre lines in diagram n = 2n2 + 4n + 1
For diagram 100 : 2(10)2 + 4(10) + 1 = 241
Answer: (d) (ii)
p=4
241
[1]
(iii) Which Diagram has 337 one centimetre lines ?
2n2 + 4n + 1 = 337
2n2 + 4n + 1− 337 = 0
2n2 + 4n − 336 = 0
2(n2 + 2n − 168) = 0
n2 + 2n − 168 = 0
n2 + 14n − 12n − 168 = 0
n (n + 1) − 12 (n + 14) = 0
(n + 14) (n − 12) = 0
n − 12 = 0
n = 12
Answer: (b) (iii)
12
[3]
15
Workbook of Variant 3
(e) For each Diagram, the number of squares of area 1 cm2 is A, the number of dots is D and
the number of one centimetre lines is L.
Find a connection between A, D and L that is true for each Diagram.
Diagram
1
2
3
A
2
6
12
D
6
12
20
L
7
17
31
7=2+6−1
17 = 6 + 12 − 1
31 = 12 + 30 − 1
L = A + D − 1
Answer: (e)
L=A+D−1
[1]
May / June 2011 Paper 23
4Helen measures a rectangular sheet of paper as 197 mm by 210 mm, each correct to nearest
millimetre.
Calculate the upper bound for the perimeter of the sheet of paper.
1
= 0.5
2
= 197 + 0.5
Connect to nearest millimetre half of 1 millimetre =
Upper bound of width of rectangle
= 197.5
Upper bound of length of rectangle = 210 + 0.5
= 210.5
Upper bound of perimeter of sheet (ractangle)
= 2(l + b)
= 2(197.5 + 210.5)
= 816
Answer:(ii)
16
816 cm
[2]
Number
6
(a) Write 16 460 000 in standard form.
Answer:(a)
1.646 × 107
[1]
(b) Calculate 7.85 ÷ (2.366 × 102), giving your answer in standard form.
7.85 ÷ (2.336 × 102) = 0.0331783601
= 3.32 × 10-2
Answer: (b)
7
(a) Find the value of x when
x
27 × 24
18
=
=
36
(b) Show that
x=
36
[1]
2
1
4
÷1
=
.
3
6
7
Write down all the steps in your working.
2
3
÷1
=
2
3
÷
7
6
=
2
3
×
6
7
=
4
7
1
6
Answer: (b)
9
[2]
18
27
=
.
x
24
Answer: (a)
3.32 × 10-2
4
7
[2]
Eva invests $ 120 at a rate of 3% per year compound interest.
Calculate the total amount Eva has after 2 years.
Give your answer correct to 2 decimal places.
Amount
=
P 1+
Amount
=
120 1 +
Amount
=
127.308
n
t
100
3
100
2
Answer:
$127.31
[3]
17
Workbook of Variant 3
12Federico changed 400 euros(€) into New Zealand dollars (NZ$) at rate of € 1 = NZ$ 2.1.
He spent x New Zealand dollars and changed the rest back into euros at rate of € 1 = NZ$ d.
Find an expressions, in terms of x and d, for the number of euros Federico received.
€
1
400
NZ$
2.1
a
a = 400 × 2.1
a = 840 NZ$
He spend x NZ$
So he new have (840 − x) NZ$
b =
€
1
NZ$
b
840 − x
d
840 − x
d
€ 840 − x
Answer:
d
[3]
19 The scale of a map is 1 : 250 000.
(a) The actual distance between two cities is 80 km.
Calculate this distance on the map. Give your answer in centimetres.
80 km = 8000000 cm
Map
1
x
Actual
250000
8000000
8000000
250000
x = 32
x =
Answer: (a)
32 cm
(b) On the map a large forest has an area of 6 cm2.
Calculate the actual area of the forest. Give your answer in square kilometres.
18
Map
12
6
Actual
(250000)2
x
[2]
Number
x = 6 × (250000)2
x = 375000000000 cm2
375000000000 km2
1000002
x =
x = 37.5
Answer: (b)
37.5 km2
[2]
May / June 2011 Paper 43
1
Lucy works in a clothes shop.
(a) In one week she earned $ 277.20.
(i) She spent
1
of this on food.
8
Calculate how much she spend on food.
1
8
× 277.20 = 34.65
Answer: (a) (i)
$ 34.65
[1]
(ii) She paid 15% of the $ 277.20 in taxes.
Calculate how much she paid in taxes.
15
100
× 277.20 = 41.58
Answer: (a) (ii)
$ 41.58
[2]
(iii) The $ 277.20 was 5% more than Lucy earned in the previous week.
Calculate how much Lucy earned in the previous week.
Old
New
Ratio
100
105
Money
x
277.20
x =
100 × 277.20
105
x = 264
Answer: (a) (ii)
$ 264
[3]
19
Workbook of Variant 3
(b) The shop sells clothes for men, women and children.
(i) In one day Lucy sold colthes with a total value of $ 2200 in the ratio
men : women : children = 2 : 5 : 4
Calculate the value of the women's clothes she sold.
M
W
C
Total
2
5
4
11
x
2200
5 × 2200
11
x = 1000
x =
Answer: (b) (i)
(ii) The $ 2200 was
$ 1000
[2]
44
of the total value of the clothes sold in shop on this day.
73
Calculate the total value of the clothes sold in the shop on this day .
2200 =
44
73
×x
2200 × 73
=x
44
x = 3650
Answer: (b) (ii)
$ 3650
[2]
11 (a) (i) The first three positive integers 1, 2 and 3 have a sum of 6.
Write down the sum of the first 4 positive integers.
1 + 2 + 3 + 4 = 10
Answer: (a) (i)
10
[1]
(ii) The formula for the sum of the first n integers is n (n + 1) .
2
Show the formula is correct when n = 3.
3 (3 + 1)
=6
2
Answer: (a) (ii)
20
6
[2]
Number
(iii) Find the sum of the first 120 positive integers.
120 (120 + 1)
= 7260
2
Answer: (a) (iii)
7260
[1]
(iv) Find the sum of the integers
121 + 122 + 123 + 124 +
sum of first 200 positive integers − sum of first 120 positive integers
=
200(200+1)
− 7260
2
=
12840
+ 199 + 200.
Answer: (a) (iv)
12840
[2]
(v) Find the sum of the even numbers
2+4+6+
=
2 (1 + 2 + 3 ............... + 400)
=
2 (sum of first 400 positive integers)
=
=
+ 800.
2 ( 400(400+1) )
2
160400
Answer: (a) (v)
160400
[2]
(b) (i)Complete the following statements about the sums of cubes and the sums of integers.
13 = 11 = 1
13 + 23 = 9
1+2=3
13 + 23 + 33 = 36 (62)
13 + 23 + 33 + 43 = 100(102)
1+2+3=6
1 + 2 + 3 + 4 = 10
[2]
(ii) The sum of the first 14 integers is 105.
Find the sum of the first 14 cubes.
1052 = 11025
Answer: (b) (ii)
11025
[1]
(iii)Use the formula in part (a)(ii) to write down a formula for the sum of the first n cubes.
Answer: (b) (iii)
n (n +1)
2
2
[1]
21
Workbook of Variant 3
2
(iv) Find the sum of the first 60 cubes.
60 (60 + 1) = 3348900
2
2
Answer: (b) (iv)
3348900
[1]
(v) Find n when the sum of the first n cubes is 278 784.
[
n (n + 1)
2
]
2
= 278784
n (n + 1)
= 278784
2
n (n + 1)
= 528
2
n2 + n = 528 × 2
n2 + n = 1056
n2 + n − 1056 = 0
n (n − 32) + 33(n − 33) = 0
(n − 32) (n + 33) = 33
n − 32 = 0
n = 32
Answer: (b) (v)
32
[2]
October / November 2011 Paper 23
1
Martha divides $ 240 between spending and saving in the ratio
spending : saving 7:8
Calculate the amount Martha has for spending.
Spending
Saving
Total
ratio
7
8
15
Amount
x
x=
240
7 × 240
15
= 112
Answer:
22
112
[2]
Number
2210
211
212
213
214
215
216
Form the list of numbers, find
(a) a prime number,
Answer: (a)
211
[1]
(b) a cube number.
Answer: (b)
5
216
[1]
The population of the city is 128 000, correct to the nearest thousand.
(a) Write 128 000 in standard form.
Answer: (a)
1.28 × 105
[1]
(b) Write down the upper bound of the population .
1000
= 500
2
Upper bound of the population = 128000 + 500
half of thousand =
= 128500
Answer: (b)
128500
[1]
6Pedro invested $ 800 at a rate of 5% per year compound interest.
Calculate the total amount he has after 2 years.
Amount
=
=
=
x
P 1+
100
5
800 1 +
100
882
n
2
Answer:
$ 882
[1]
23
Workbook of Variant 3
7
Show that 3–2 + 2–2 =
13
36
Write down all the steps of your working.
3-2 + 2-2
1
32
1
9
1×9
4×1
+
4×9
4×9
+
+
1
22
1
4
4 + 9
36
13
36
Answer:
8
Find the value of
3 17.1 − 1.89
10.4 + 8.36
3 17.1
13
36
[2]
− 1.89
10.4 + 8.36
=
0.1864118168
Answer:
0.186
[2]
9In Vienna, the mid-day temperatures, in °C, are recorded during a week in December.
This information is shown below.
–221–3–1–20
Calculate
(a) the difference between the highest temperature and the lowest temperature,
2 − (−3) = 5
Answer: (a)
5°C
[1]
(b) the mean temperature.
−2 + 2 + 1 + (-3) + (-1) + (-2) + 0
7
= −0.7142857143
=
Answer: (b)
24
−0.714 °C
[2]
Number
10 Maria decides to increase her homework time of 8 hours per week by 15%.
Calculate her new homework time.
Give your answer in hours and minutes.
Old
New
%
100
115
Time
8
x
8 × 115
100
= 9.2 hours
x=
9.2 hours = (9 + 0.2) hour
0.2 hours = 0.2 × 60
= 12 min
9 hours and 12 minutes
Answer:
9 hours 12 minutes
[3]
12Alberto changes 800 Argentine pesos (ARS) into dollars ($) when the rate is $ 1 = 3.8235 ARS.
He spends $ 150 and changes the remaining dollars back into pesos when rate is
$ 1 = 3.8025 ARS.
Calculate the amount Alberto now has in pesos.
x=
$
ARS
1
3.8235
x
800
800
3.8235
x = $ 209.2323787
= $ 209.2323787 − $ 150
= $ 59.23237871
$
ARS
1
3.8235
59.2323787
x
x = 59.23237871 × 3.8025
x = 225.23112
Answer:
225.23
ARS
[3]
25
Workbook of Variant 3
13 During a marathon race an athelete loses 2% of his mass.
At the end of the race his mass is 67.13 kg.
Calculate his mass before the race.
Old
New
%
100
98
mass
x
67.13
100 × 67.13
98
x=
x = 68.5
68.5 kg
Answer:
[3]
October / November 2011 Paper 43
12 (a) The nth term of a sequence is n (n + 1).
(i) Write the two missing terms in the spaces 2, 6, 12, 20, 30,
3(3+1) = 12, 5(5+1) = 30
[2]
(ii) Write down an expression in terms of n of the (n + 1)th term.
n will get replaced by n + 1
(n + 1) (n + 1 + 1) = (n + 1) (n + 2)
Answer (a)(ii):
(n + 1) (n + 2)
[1]
(iii) The difference between the nth term and the (n + 1)th term is pn + q.
Find the values of p and q.
(n + 1) (n + 2) − [n (n + 1)]
n2 + 3n + 2 − [n2 − n]
n2 + 2n + n + 2 − n2 + n
2n + 2 (compose it with pn + q)
Answer (a)(iii):
p=
q=
26
2
2
[2]
Number
(iv) Find the positions of the two consecutive terms which have a difference of 140.
2n + 2 = 140
2n = 140 − 2
2n = 138
n = 69
n + 1 = 69 + 1
= 70
Answer (a)(iv):
69
and
70
[2]
(b) A sequence u1, u2, u3, u4, .............is given by the following rules.
u1 = 2,
u2 = 3
and
un = 2un−2 + un−1 for n ≥ 3.
For example, the third term is u3 and u3 = 2u1 + u2 = 2 × 2 + 3 = 7
So the sequence is 2, 3, 7, u4, u5,
(i) Show that u4 = 13.
Answer (b)(i):
u4 = 2u2 + u3 = 2(3) + 7 = 13
[1]
(ii) Find the value of u5
u5 = 2u3 + u4 = 2(7) + 13 = 27
Answer (b)(ii): u5 =
27
[1]
(iii) Two consecutive terms of the sequence are 3413 and 6827.
Find the term before and the term after these two given terms.
(Assume u1, u2, u3, u4 as shown below)
u3 = 2u1 + u2
6827 = 2u1 + 3413
6827 − 3413
2
u1 = 1707
u1 =
u4 = 2u2 + u3
u4 = 2(3413) + 6827
u4 = 13653
Answer (b)(iii):
1707,
, 3413, 6827,
13653
[2]
27
Workbook of Variant 3
May / June 2012 Paper 23
2
Hans invests $ 750 for 8 years at a rate of 2% per year simple interest.
Calculate the interest Hans receives.
SI =
p×t×r
100
SI =
750 × 8 × 2
100
SI = 120
120
Answer: $
3
[2]
(a) Calculate 3 71.5 + 220.9 and write down your full calculator display.
3.260770655
Answer (a):
[1]
(b) Write your answer to part(a) correct to 4 significant figures.
3.261
Answer (b):
5
[1]
9 cm
5 cm
5 cm
12 cm
The diagram shows a quadrilateral.
The lengths of the sides are given to the nearest centimetre.
Calculate the upper bound of perimeter of the quadrilateral.
half of 1 =
1
2
= 0.5
Upper bound of sides are 9 + 0.5 = 9.5,
5 + 0.5 = 5.5, 12 + 0.5 = 12.5
Upper bound of perimeter of quadrilateral.
= 9.5 + 5.5 + 12.5 + 5.5
= 33
Answer:
28
33
cm [2]
Number
8
During her holiday, Hannah rents a bike.
She pays a fixed cost of $ 8 and then a cost of $ 4.50 per day.
Hannah pays with a $ 50 note and receives $ 10.50 change.
Calculate for how many days Hannah rents the bike.
Total rent = 50 − 10.50
= 39.5
let hannah rent the bike for x days.
8 + 4.5x = 39.5
4.5x = 39.5 − 8
x=
39.5 − 8
4.5
7
Answer:
days
[3]
11 Boris invests $ 280 for 2 years at a rate of 3% per year compound interest.
Calculate the interest Boris receives at the end of the 2 years.
Give your answer correct to 2 decimal places.
x
Amount = P 1 +
100
Amount = 280 1 +
n
3
100
2
Amonts= 297.052
compound interest
= Amount − Principal
compound interest
= 297.052 − 280
compound interest
= 17.052
Answer:
$
17.05
[4]
12 Without using your calculator, work out the following.
Show all the steps of your working and give each answer as a fraction in its simplest form.
1
(a) 11 −
3
12
=
11
12
=
11−4
12
=
7
12
−
1×4
3×4
[L.C.M. 3 and 12 is 12]
Answer (a):
7
12
[2]
29
Workbook of Variant 3
1
÷
4
(b)
1
4
13
44
11
13
13
11
×
13
44
Answer (b):
[2]
May / June 2012 Paper 43
1
A train travels from Paris to Milan.
(a) The train departs from Paris at 2028 and the journey takes 9 hours 10 minutes.
(i) Find the time train arrives in Milan.
Hour
20
9
29
+
Minute
28
10
38
Hour
29
24
5
−
Minutes
38
38
Answer (a)(i):
5
38
[1]
(ii) The distance between Paris and Milan is 850 km.
Calculate the average speed of the train.
10 minute =
=
10
60
1
6
hour
hour
1
6
Total time = 9 +
=
55
6
hour
Average speed =
=
(9h 10m to
55
6
acn be done in calculator )
Total distance
Total time
850
55/6
= 92.727272
Answer (a) (ii):
30
92.7
km/h [2]
Number
(b) The total number of passengers on the train is 640.
(i) 160 passengers have tickets which cost $ 255 each,
330 passengers have tickets which cost $ 190 each,
150 passengers have tickets which cost $ 180 each.
Calculate the mean cost of a ticket.
160×225 + 330×190 + 150×180
640
Mean =
Mean =
203.90625
Answer (b)(i):
$
204
[3]
(ii) There are men, women and children on the train in the ratio
men : women : children
4 : 3 : 1.
Show that the number of women on the train is 240.
Answer (b)(ii):
ratio
M
W
C
Total
4
3
1
8
number
x =
x
640
640 × 3
= 240
8
[2]
(iii) 240 is an increase of 60% on the number of women on the train the previous day.
Calculate the number of women on the train the previous day.
Old
New
%
100
160
no. of women
x
240
100 × 240
160
x = 150
x =
Answer (b)(iii):
150
[3]
31
Workbook of Variant 3
(c) The length of the train is 210 m.
It passes through a station of length 340 m, at a speed of 180 km/h.
Calculate the number of seconds the train takes to pass completely through the station.
Speed :
Total distance = 210 + 340 = 550
time =
180 km/h =
180 ×
5
18
=
50 m/s
[as we need time in seconds]
[To connect km/h to km/s ×
5
]
18
Distance
550
=
= 11s
Speed
50
11
Answer (c):
s [3]
11
Diagram 1
Diagram 2
Diagram 3
The diagrams show a sequence of dots and circles.
Each diagram has one dot at the centre and 8 dots on each circle.
The radius of the first circle is 1 unit.
The radius of each new circle is 1 unit greater than the radius of the previous circle.
(a) Complete the table for diagrams 4 and 5.
Diagram
1
2
3
4
5
Number of dots
9
17
25
33
41
Area of the largest circle
2π
4π
9π
16π
25π
Total length of the circumferences of the circle
2π
6π
12π
20π
30π
[4]
(b) (i) Write down, in terms of n, the number of dots in diagram n.
here a = 9, d = 17−9 = 9
nth term = a + (n−1)d
= 9 + (n−1)8
= 9 + 8n − 8
= 8n + 1
Answer (b)(i):
32
8n + 1
[2]
Number
(ii) Find n, when the number of dots in diagram n is 1097.
8n + 1 = 1097
8n = 1097 − 1
8n = 1096
n = 1096
8
n
=
137
Answer (b)(ii):
n=
137
[2]
(c) Write down, in term of n and π , the area of the largest circle in
(i) diagram n,
π 4π
12π
9π
22π
3π
(ii) diagram 3n.
(3n)2π =
Answer (c)(i):
n 2π
[1]
Answer (c)(ii):
9n2π
[1]
9n2π
(d) Find, in terms of n and π, the total length of the circumferences of the circles in diagram n.
2π
6π
π( 2
6
π[ 1 × 2
12π
20π
30π
12 20 30 )
2×3
3×4
4×5
5×6
]
π[n(n+1)]
πn(n+1)
Answer (d):
[2]
October / November 2012 Paper 23
1
Samantha invests $ 600 at a rate of 2% per year simple interest.
Calculate the interest Samantha earns in 8 years.
simple interest =
=
p×t×r
8
600 × 8 × 2
100
= 96
Answer:
$
96
[2]
33
Workbook of Variant 3
2
2
1
Show that
10
2
2
+
5
= 0.17
Write down all the steps in your working.
Answer:
1
10
2
+
1
100
1 + 16
100
+
2
2
5
4×4
25×4
17
100
0.17
3
[2]
Jamie needs 300 g of flour to make 20 cakes.
How much flour does he need to make 12 cakes?
Flour
Cakes
300
20
x
12
x = 300 × 12
20
x = 180
180
Answer:
5
g [2]
Maria pays $ 84 rent.
The rent is increased by 5%.
Calculate Maria's new rent.
x=
Old
New
%
100
105
sent
84
x
84 × 105
100
x = 88.2
Answer:
34
$
88.2
[2]
Number
7
Find the value of
7.2
11.8 − 10.95
Give your answer correct to 4 significant figures.
7.2
= 8.470588235
11.8 − 10.95
8.471
Answer:
8
[2]
A carton contains 250 ml of juice, correct to the nearest millilitre.
Complete the statement about the amount of juice, j ml, in the carton.
1
= 0.5
2
lower bound of 250
= 250 − 0.5 = 249.5
upper bound of 250
= 250 + 0.5 = 250.5
half of 1 =
Answer:
249.5
≤j≤
250.5
[2]
15 A model of a ship is made to a scale of 1 : 200.
The surface area of the model is 7500 cm2.
Calculate the surface area of the ship, giving your answer in square metres.
Model
Actual
12
(200)2
7500
x
x = 2002 × 7500
x = 300000000 cm2
x=
30000000
m2
2
100
x = 30000 m2
Answer:
30000
m2 [3]
35
Workbook of Variant 3
October / November 2012 Paper 43
1
(a) The Martinez family travels by car to Seatown.
The distance is 92 km and the journey takes 1 hour 25 minutes.
(i) The family leaves home at 07 50.
Write down the time they arrive at Seatown.
Hour
Minute
07
50
1
25
8
75
+
75 − 60 = 15,
8+1 = 9
09:15
09:15
Answer (a)(i):
[1]
(ii) Calculate the average speed for the journey.
25 minute =
Total time = 1 +
25
60
=
17
12
Average speed
=
=
=
5
12
hour
5
12
hour
=
Total distance
Total time
92
17 ÷ 12
64.94117647
Answer (a)(ii):
64.9
km/h [2]
(iii) During the journey, the family stops for 10 minutes.
Calculate 10 minutes as a percentage of 1 hour 25 minutes.
1 hour 25 min = (60 + 25) min = 85 min
10
85
× 100 = 11.76470588
Answer (a)(iii):
36
11.8
% [1]
Number
(iv) 92 km is 15% more than the distance from Seatown to Deecity.
Calculate the distance from Seatown to Deecity.
Old
New
%
100
115
Distance
x
92
x =
100 × 92
x = 80
115
80
Answer (a)(iv):
km [3]
(b) The Martinez family spends $ 150 in the ratio
fuel : meals : gifts =
11 : 16 : 3
(i) Show that $ 15 is spend on gifts.
Answer (b)(i):
Ratio
F
M
G
Total
11
16
3
30
x
150
Money
[Taking G's column and Total's column and cross multiply]
x =
3 × 150
= 15
30
[2]
(ii) The family buys two gifts.
The first gift costs $ 8.25.
Find the ratio
Cost of first gift : cost of second gift.
Give your answer in its simplest form.
Cost of first gift
Cost of second gift
8.25
6.75
=
= 8.25
= 15 − 8.25 = 6.75
11
9
Answer (b)(ii):
11
:
9
[2]
37
Workbook of Variant 3
10 (a) Complete the table for the 6th term and the nth term in each sequence.
Sequance
11, 9, 7, 5, 3
1, 4, 9, 16, 25
2, 6, 12, 20, 30
3, 9, 27, 81, 243
1, 3, 15, 61, 213
A
B
C
D
E
A
a = 11
a = 9 − 11 = −2
a + (n + 1) d
11 + (n + 1) −2
11 − 2n + 2
13 − 2n
B
1
4
9
16
12 22 32 42
n2
C
2612
1×22×33×4
n (n + 1)
D
3927
31 32 33
3n
E
38
1=3−2
3=9−6
15 = 27 − 12
E=D−C
E = 3n − n (n + 1)
6th term
1
36
42
729
687
nth term
13 − 2n
n2
n (n + 1)
3n
3n − n (n + 1)
[12]
Number
(b) Find the value of the 100th term in
(i) Sequence A.
nth term
= 13 − 2n
100th term = 13 − 2(100)
100th term = -187
Answer (b)(i):
–187
[1]
(ii) Sequence C,
nth term
=
100 th term
100th term =
n (n + 1)
=
100 (100 + 1)
10100
Answer (b)(ii):
10100
[1]
(c) Find the value of n in sequence D when the nth term is equal to 6561.
Sequence D→ 3n = 6561
nth term3n = 38
Answer (c): n =
8
[1]
(d) Find the value of the 10th term in sequence E.
E → nth term
= 3n − n (n + 1)
10th term = 310 − 10 (10 + 1)
10th term = 58939
Answer (d):
58 939
[1]
39
Workbook of Variant 3
May / June 2013 Paper 23
1
Sheila can pay her hotel bill in Euros (€) or Pounds (£).
The bill was € 425 or £ 365 when the exchange rate was £ 1 = € 1.14.
In which currency was the bill cheaper?
Show all your working.
£
€
1
1.14
365
x
x = 365 × 1.14
x = € 416.1
£365 = €416.1 is cheaper.
Answer:
2
pounds(£)
[2]
The Ocean View Hotel has 300 rooms numbered from 100 to 399.
A room is chosen at random.
Find the probability that the room number ends in zero.
probabilioty =
3
30
300
Answer:
30
300
[2]
The time in Lisbon is the same as the time in Funchal.
A plane left Lisbon at 08 30 and arrives in Funchal at 10 20.
It then left Funchal at 12 55 and returned to Lisbon.
The return journey took 15 minutes more.
What time did the plane arrive in Lisbon?
Time taken for journey : 10 20 − 8 30 = 0150
Return Journey = 0150 + 0015 = 0205
Arrival time = 1255 + 0205 = 1500
Answer:
40
15:00
[2]
Number
4
Use a calculator to find
5 5
24
(a)
5 5
24
= 2.282177323
2.28
Answer (a):
cos 40°
7
(b)
cos40o
= 0.1094349204
7
0.109
Answer (b):
5
[1]
[1]
Write the following in the order of size, smallest first.
1.5
2
–1.5
−2 3
2
2
3
3
3
1.3103706970.5443310541.8371173070.7631428284
2
(1.5) 3
Answer:
9
2
3
1.5
<
−2
3
2
3
2
<
(1.5) 3
<
2
3
–1.5
[2]
Calculate, giving your answers in standard form,
(a) 2 × (5.5 × 104),
2 × (5.5 × 104)
110000
1.1 × 105
Answer:
1.1 × 105
[2]
(b) (5.5 × 104) − (5 × 104).
(5.5 × 104) − (5 × 104)
104 (5.5 −5)
104 (5.5 − 5)
104 (0.5)
= 5000
= 5 × 103
Answer:
5 × 103
[2]
41
Workbook of Variant 3
11 The sum of prime numbers less than 8 is equal to 17.
(a) Find the sum of the prime numbers less than 21.
Prime more than 8 and less than 21 are 11, 13, 17, 19
Sum of the prime number less than 21: 17+11+13+17+19 = 77
Answer (a):
77
[2]
(b) The sum of the prime numbers less than x is 58.
Find an integer value for x.
17 + 11 + 13 + 17 = 58 → This is sum of prime numbers less than 18.
Answer (b): x =
18
[2]
May / June 2013 Paper 43
1 (a) Ali and Ben receive a sum of money.
They share it in the ratio 5 : 1.
Ali receives $ 2345.
Calculate the total amount.
Ali
Ben
Total
Ratio
5
1
6
Money
2345
x =
x
2345×6
5
x = 2814
Answer (a)(ii): $
2814
[2]
(b) Ali uses 11% of his $ 2345 to buy a television.
Calculate the cost of the television.
11
100
× 2345 = 257.95
Answer (b)(ii): $
42
257.95
[2]
Number
(c) A different television costs $ 330.
(i) Ben buys one in a sale when this cost is reduced by 15%.
How much does Ben pay?
x =
Old
New
%
100
85
Cost
330
x
330×85
100
x = 280.5
Answer (c)(i): $
280.5
[2]
(ii) $ 330 is 12% less than the cost last year.
Calculate the cost last year.
x =
Old
New
%
100
88
Cost
x
330
100×330
88
x = 375
Answer (c)(ii): $
375
[3]
(d) Ali invests $ 1500 of his share in a bank account.
The account pays compound interest at a rate of 2.3% per year.
Calculate the total amount in the account at the end of 3 years.
Amount = P 1 +
x
n
100
2.3
100
Amount = 1500 1 +
Amount = 1605.898751
3
Answer (d): $
1605.90
[3]
43
Workbook of Variant 3
(e) Ali also buys a computer for $ 325.
He later sells this computer for $ 250.
Calculate Ali's percentage loss.
percentage loss =
Percentage loss =
Loss
× 100
Cost Price
325 − 250
× 100
325
percentage loss = 23.07692308
Answer (e):
23.1
% [3]
October / November 2013 Paper 23
1
Christa had a music lesson every week for one year.
Each of the 52 lessons lasted for 45 minutes.
Calculate the total time that Christa spent in music lessons.
Give your time in hours.
52 × 45
= 2340 minutes
2340 minute
=
2340
hour
60
= 39 hour
39
Answer:
4
h [2]
Write the following in order of size, smallest first.
1
100– 0.1
100
-0.1736451777 -5.671281820.010.6309573445
1
100
tan100
cos100
Answer:
<
<
cos100°tan100°
5
<
100 -01
[2]
Write
(a) 60 square metres in square centimetres,
60 square metres = 60 × 1002 cm2
= 600000
Answer (a):
44
600000
cm2
[1]
Number
(b) 22 metres per second in kilometres per hour.
22m/s = 22 ×
18
5
km/h = 79.2
Answer (b):
6
79.2
km/h [2]
In 2012 the cost of a ticket to an arts festival was $ 30.
This was 20% more than the ticket cost in 2011.
Calculate the cost of the ticket in 2011.
Old(2011)
New(2012)
%
100
120
Cost
x
30
x=
100 × 30
120
x = 25
Answer: $
25
[3]
12 Write the answer to the following calculations in standard form.
(a) 600 ÷ 8000
600
= 0.075
8000
= 7.5 × 10 -2
Answer (a):
7.5 × 10 -2
[2]
(b) 108 − 7 × 106
108 − 7 × 106 = 93000000 = 9.3 × 107
Answer (b):
9.3 × 107
[2]
45
Workbook of Variant 3
October / November 2013 Paper 43
1 (a) (i) In a camera magazine, 63 pages are used for adverts.
The ratio number of pages of adverts : number of pages of review = 7 : 5.
Calculate the number of pages used for reviews.
A
R
Ratio
7
5
Pages
63
x
x =
63 × 5
= 45
7
45
Answer (a)(ii) :
[2]
(ii) In another copy of the magazine, 56 pages are used for reviews and for photographs.
The ratio number of pages of reviews : number of pages of photographs = 9 : 5.
Calculate the number of pages used for photographs.
Ratio
R
P
Total
9
5
14
x
56
Pages
5 × 56
14
x = 20
x =
Answer (a)(iii) :
20
% [2]
(iii) One copy of the magazine cost $ 4.90.
An annual subcription costs $ 48.80 for 13 copies.
Calculate the percentage discount by having an annual subcription.
An annual subcription without discount cost : 490 × 13 = 63.7
Discount = 63.7 − 48.80 = 14.9
percentage discount =
=
=
14.9
63.7
discount
× 100
cost price
× 100
23.39089482
Answer (a)(iii) :
46
23.4
% [3]
Number
(b) In a car magazine, 25% of the pages are used for selling second-hand cars.
1
% of the remaining pages are used for features,
2
and the other 36 pages are used for reviews.
62
Work out the total number of pages in the magazine.
let the total number of pages be x
25
62.5
75
×x+
×
× x + 36 = x
100
100
100
0.25x + 0.46875x + 36 = x
36 = x − 0.25x − 0.46875x
36 = 0.28125x
36
= x
0.28125
x = 128
128
Answer (b) :
9
[4]
The first four diagrams in a sequence are shown below.
Diagram 1
Diagram 2
Diagram 3
The diagram are made from dots (•) and squares (
Diagram 4
) joined by lines.
(a) Complete the table.
Diagram
1
2
3
4
5
n
Number of dots
6
9
12
15
18
3n + 3
Number of squares
0
1
3
6
10
Number of triangle
4
9
16
25
36
Number of lines
9
18
30
45
63
1
n(n−1)
2
(n+1)
2
3
(n+1)(n+2)
2
[9]
47
Workbook of Variant 3
(b) Which diagram has 360 lines?
3
2
(n + 1) (n + 2) = 360
360 × 2
3
= 240
(n + 1) (n + 2) =
n2 + 2n + n + 2
n2 + 3n + 2 − 240
= 0 [Product = -238, sum = 3]
n2 + 3n − 238 = 0
n(n + 17) − 14(n + 17) = 0
(n + 17) (n − 14) = 0
n − 14 = 0
n = 14 [Number of lies cannot be negative number]
Answer (b) :
14
[2]
(c) The total number of lines in the first n diagrams is
1 3
n + pn2 + qn.
2
(i) When n = 1,
show that p + q = 8
Answer (c)(i) :
1
2
1
.
2
(1)3 + p(10)2 + q(1) = 9
1
2
+p+q=9
1
2
p+q=9−
p + q = 8
1
2
[1]
(ii)By choosing another value of n and using the equation in part (c) (i), find the values of p
and q.
When n = 2,
48
1
2
(2)3 + p(2)2 + q(2) = 9 + 18
4 + 4p + 2q = 27
Number
4p + 2q = 27 − 4
4p + 2q = 23
Solution first and this equation simultaneously.
1st equation : p + q = 8.5
2nd equation: 4p + 2q = 23
Multiplying 1st equation by −2
−2p − 2q = -17
4p + 2q = 23
→ adding 2 equation
2p = 6
6
2
p= 3
p =
Substituting value of p in first equation
p+q=8
1
2
3+q=8
1
2
q = 8
1
2
=
11
2
−3
Answer (c)(ii): p =
q=
3
11
2
[5]
May / June 2014 Paper 23
1
In March 2011, the average temperature in Kiev was 3 oC.
In March 2012, the average temperature in Kiev was 19 oC lower than in March 2011.
Write down the average temperature in Kiev in March 2012.
3 − 19 = –16
Answer :
–16oC
C
°
[1]
49
Workbook of Variant 3
3
Chris changes $ 1350 into euros (€) when € 1 = $ 1.313.
Calculate how much he receives.
x =
€
$
1
1.313
x
1350
1350
1.313
x = 1028.179741
1030
Answer: €
[2]
5 (a) Use your calculator to find the value of 7.5–0.4 ÷ 57 .
Write down your full calculator display.
7.5-0.4 ÷
57 = 0.0591613488
Answer (a) :
0.059161
[1]
(b) Write your answer to part(a) in standard form.
Answer (b) :
5.91613488 × 10–2
[1]
8Hans draws a plan of a field using a scale of 1 centimetres to represent 15 metres.
The actual area of the field is 10 800 m2.
Calculate the area of the field on the plan.
15 metres = 15 × 100cm = 1500 cm
Scale will be x =
1 : 1500
Map
Actual
12
15002
x
10800
3
10800
=
2
625
1500
× 100 m2 = 48 cm2
Answer :
50
48
cm2
[2]
Number
10 Without using a calculator, work out 1 1 − 7 .
4
9
Write down all the steps in your working.
1
1
7
−
4
9
7×4
9×5
−
9×4
9×4
45 − 28
36
17
36
Answer :
17
36
[3]
15A rectangle has length 127.3 cm and width 86.5 cm, both correct to 1 decimal places.
Calculate the upper bound and the lower bound for the perimeter of the rectangle.
Length and width both corrected to 1 decimal place.
0.1
= 0.05
2
Upper bound of length = 127.3 + 0.05 =
127.35
Upper bound of width =
86.55
half of 0.1 =
86.5 + 0.05
=
Upper bound of perimeter of rectangle = 2(127.35 + 86.55) =
Lower bound of length = 127.3 − 0.05 =
127.25
Lower bound of width = 86.5 − 0.05
86.45
=
Lower bound of perimeter of rectangle =
427.8
2(127.25 + 86.45) =
427.4
Answer: Upper bound =
Lower bound =
427.8
427.4
cm
cm [3]
May / June 2014 Pager 43
1
In July, a supermarket sold 45 981 bottles of fruit juice.
(a) The cost of a bottle of fruit juice was $ 1.35.
Calculate the amount received from the sale of the 45 981 bottles.
Give your answer correct to the nearest hundred dollars.
45981 × 1.35 = 62074.35 = 62100
Answer (a): $
62100
[2]
51
Workbook of Variant 3
(b) The number of bottles sold in July was 17% more than the number sold in January.
Calculate the number of bottles sold in January.
Old (January)
New(July)
%
100
117
Number of bottles
x
45981
x = 100×45981 117
x = 39300
39300
Answer (b):
[3]
(c) There were 3 different flavours of fruit juice.
The number of bottles sold in each flavour was in the ratio
Apple : orange : cherry = 3 : 4 : 2
The total number of bottles sold was 45 981.
Calculate the number of orange juice sold.
Ratio
Apple
Orange
Cherry
Total
3
4
2
9
x
Actual numbers
45981
4×45981
9
x =
x = 20436
Answer (c):
20436
[2]
(d) One bottle contains 1.5 litres of fruit juice.
Calculate the number of 330 ml glasses that can be filled completely from one bottle.
1.5 litre 1500
330
=
1.5 × 1000 ml
=
1500 ml
= 4.54545454
Answer (d):
Calculate the number of bottles that are recycled.
5
9
× 45981 = 25545
Answer (e):
52
[3]
5
of the 45 981 bottles are recycled.
9
(e)
4
25545
[2]
Number
11
Diagram 1
Diagram 2
Diagram 3
The first three diagrams in a sequence are shown above.
Diagram 1 shows an equilateral triangle with sides of length 1 unit.
In Diagram 2, there are 4 triangles with sides of length
In Diagram 3, there are 16 triangles with side of length
(a) Complete this table for Diagrams 4, 5, 6 and n.
Length of side
1
2
1
4
unit.
unit.
Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 Diagram 6 Diagram n
1
1
1
1
1
1
1
2
Length of sides 20
as a power of 2
4
2-1
8
2-2
16
2-3
32
2-4
2n−1
2-5
2-(n−1)
[6]
(b) (i)Complete this table for the number of the smallest of the triangles in Diagrams 4, 5
and 6.
Number of smallest
triangle
Number of smallest
triangle as a pwer of 0
Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 Diagram 6
1
4
16
64
256
1024
20
22
24
26
28
210
[2]
(ii)Find the number of the smallest triangles in Diagram n, giving your answer as a power
of 2.
0
2
4
6
8
a = 0, d = 2, nth term = a + (n − 1) d = 0 + (n −1)2 = 2n − 2.
Answer (b)(ii):
22n − 2
[1]
53
Workbook of Variant 3
(c) Calculate the number of the smallest triangles in the diagram where the smallest triangle
1
have sides of length
unit.
128
1
sides of length 128 unit.
1
128
1
27
= 2-(n-1)
=
1
2n-1
7 = n-1
n=8
Number of the smallest triangles when n=8
22(8)-2 = 16384
16384
Answer (c):
[2]
October / November 2014 Paper 23
1
$ 1 = 8.2 rand
Change $ 350 into rands.
$
Rand
1
8.2
350
x
x = 350 × 8.2
x = 2870
2870
Answer:
2
rand
[2]
Write the following in order of size, smallest first.
0.34 0.60.620.73
0.340.77459666920.360.343
Answer:
54
0.34
<
0.73
<
0.62
<
0.6
[2]
Number
3
Work out 4 × 10–5 × 6 × 1012
Give your answer in standard form.
4 × 10-5 × 6 × 1012
=
240000000
=
2.4 × 108
Answer:
5
2.4 × 108
[2]
A train takes 65 minutes to travel 52 km.
Calculate the average speed of the train in kilometres per hour.
65
60
65 minutes=
Average speed
=
Average speed
=
Average speed
=
hour
13
12
=
hour
Total distance
Total time
52
13/12 48
48
Answer:
km/h [2]
10 Maryah borrows $ 12 000 to start a business.
The loan is for 3 years at a rate of 5% per year compound interest.
The loan has to be paid back at the end of the 3 years.
n
Calculate the total amount to be paid back.
P1+
x
100
=
12000 1+
5
100
=
13891.5
Amount
=
3
Answer: $
13891.5
[3]
11 (a) Here are the first three terms of a sequence
U1 = 13
U2 = 13 + 23 U3 = 13 + 23 + 33
The nth term is given by Un =
Work out the value of U39.
U39 =
1
39
1 2
n (n + 1)2
4
(39)2 (39 + 1)2 =
608400
Answer (a):
U39 =
608400
[2]
55
Workbook of Variant 3
(b) Here are the first three terms of another sequence.
V1 = 23
V2 = 23 + 43 V3 = 23 + 43 + 63
By comparing this sequence with the sequence in part (a), find a formula for the nth term Vn.
U1 = 1
U2 = 9
U3 = 36
V1 = 8
V2 = 72
V3 = 288
Vn = 8Un
1 2
n (n + 1)2
4
Vn = 8
Vn = 2n2 (n + 1)2
Answer (b):Vn=
2n2 (n + 1)2
[1]
October / November 2014 Paper 43
2
There are three different areas A, B and C for seating in a theatre.
The numbers of seats in each area are in the ratio A : B : C = 11 : 8 : 7
There are 920 seats in area B.
(a) (i) Show that there are 805 seats in area C.
Answer (a)(i):
Ratio
Numbers of seats
x
=
x
=
A
11
B
8
920
C
7
Total
26
x
y
920 × 7
8
805
[1]
(ii)Write the number of seats in area B as a percentage of the total number of seats.
920 × 26
8
y = 2990
y =
=
920
× 100 = 30.76923077
2990
Answer (a)(ii):
56
30.8
% [2]
Number
(b) The cost of a ticket for a seat in each area of the theatre is shown in the table
Area A
$11.50
Area B
$15
Area C
$ 22.50
3
of area C tickets were sold.
5
The total amount of money taken from ticket sale was $ 35 834.
Calculate the number of area A tickets that were sold.
Let the Number of are tickets sold be x
For a concert 80% of area B tickets were sold and
80
× 920 × 15 +
100
3
5
× 805 × 22.50 + 11.50x = 35834
11040 + 10867.5 + 11.50x = 35834
11.50x = 35834−11040−10867.5
11.50x = 13926.5
x =
13926.5
11.50
= 1211
1211
Answer (b):
[5]
(c) The total ticket sales of $ 35 834 was 5% less than the ticket sales at the previous concert.
Calculate the ticket sales at the previous concert.
Old
New
%
100
95
Number of tickets
x
35834
x =
100 × 35834
95
x = 37720
Answer (c):
$
37720
[3]
57
Workbook of Variant 3
9
(a)Ricardo asks some motorists how many litres of fuel they use in one day.
The numbers of litres, correct to the nearest litre, are shown in the table.
Number of litres
16
17
18
19
20
Number of motorists
11
10
p
4
8
cumulative frequency
11
21
25
29
37
(i) For this table, the mean number of litres is 17.7.
Calculate the value of p.
16 × 11 + 17 × 10 + 18 × p + 19 × 4 + 20 × 8
= 17.7
11 + 10 + p + 4 + 8
176 + 170 + 18p + 76 + 160
33 + p
582 + 18p = 17.7(33 + p)
582 + 18p = 584.1 + 17.7p
18p − 17.7p = 584.1 − 582
= 17.7
0.3p = 2.1
p =
2.1
0.3
= 7
Answer (a)(i):
7
p=
[4]
(ii) Find the median number of litres.
Ramk =
n+1
37+1
=
= 19, Median = 17
2
2
17
Answer (a)(ii):
litres
[1]
(b) Manuel completed a journey of 320 km in his car.
The fuel for the journey cost $ 1.28 for every 6.4 km travelled.
(i) Calculate the cost of fuel for this journey.
Cost($)
KM
1.28
6.4
x
320
x =
1.28 × 320
6.4
x= 64
Answer (b)(i):
58
$
64
[2]
Number
(ii) When Manuel travelled 480 km in his car it used 60 litres of fuel.
Manuel's car used fuel at the same rate for the journey of 320 km.
Calculate the number of litres of fuel the car used for the journey of 320 km.
x =
km
litres of fuel
480
60
320
x
320×60
480
x = 40
Answer (b)(ii):
40
litres
[2]
(iii) Calculate the cost per litre of fuel used for the journey of 320 km.
x =
cost
litres of fuel
64
40
x
1
64
40
x = 1.6
Answer (b)(iii): $
1.6
[2]
(c) Ellie drives a car at a constant speed of 30 m/s correct to the nearest 5 m/s.
She maintains this speed for 5 minutes correct to the nearest 10 seconds.
Calculate the upper bound of the distance in kilometres that Ellie could have travelled.
Speed of 30m/s correct to nearest 5m/s
half of 5 = 5/2 = 2.5
upper bound of speed = 30+ 2.5 = 32.5 m/s
5 minutes correct to nearest 10 seconds
half of 10 = 10/2 =
5 minutes = 5 × 60 =
upper bound of time = 300 + 5 = 305 seconds
Distance = speed × time
[upper bound = upper bound × upper bound]
upper bound of distance = 32.5 × 305
5
300 seconds
= 9912.5 metre
=
9912.5
1000
= 9.9125
(Do not round off your final answer in bounds question)
Answer (c):
9.9125
km [5]
59
Workbook of Variant 3
May / June 2015 Paper 23
1
Ahmed and Babar share 240 g of sweets in the ratio 7 : 3.
Calculate the amount Ahmed receives.
x=
Ahmed
Babar
Total
Ratio
7
3
10
Sweets(g)
x
240
7 × 240
10
x= 168
168
Answer:
7
g [2]
James buys a drink for 2 euros (€).
Work out the cost of the drink in pounds (£) when £ 1 = € 1.252.
Give your answer correct to 2 decimal places.
Pounds(£)
euros(€)
1
1.252
x
2
x=
2
1.252
x = 1.597444089
Answer:
8
1.60
[3]
7
5
÷
.
8
9
Show all your working and give your answer as a fraction in its lowest terms.
Without using a calculator, work out 1
7
8
1
÷
5
9
15
18
÷
5
9
15
8
×
9
5
27
8
Answer:
60
£
27
8
[3]
Number
10 In a sale, the cost of a coat is reduced from $ 85 to $ 67.50.
Calculate the percentage reduction in the cost of the coat.
Difference
Original
Percentage reduction =
=
85 − 67.50
85
=
20.58823529
× 100
× 100
Answer:
20.6
% [3]
May / June 2015 Paper 43
2
(a) (i) Eduardo invests $ 640 at a rate of 2% per year compound interest.
Show that, at the end of 6 years, Eduardo has $ 721, correct to the nearest dollar.
Answer (a)(i):
Amount
=
P 1+
r
100
n
2 6
100
=
640 1 +
=
720.7439483
=
721
[2]
(ii) Manuela also invests $ 640.
At the end of 4 years, Manuela has $ 721.
Find the yearly compound interest rate.
n
Amount = P 1 r
100
4
r
721 = 640 1+
100
4
721 = 1+ r
100
640
4 721
r
=1+
640
100
4 721
640
100 (
−1=
r
100
4 721
− 1) = r
640
r = 3.024098432
Answer (a)(ii):
3.02
% [4]
61
Workbook of Variant 3
(b) Carlos buys a motor scooter for $ 1200.
Each year the value of the scooter decreases by 10% of his value at the beginning of
that year.
Find the value of the scooter after 3 years.
r
Amount = value of scooter = P 1 −
100
3
10
= 1200 1 −
100
= 874.8
n
Answer (b):
$
874.8
[2]
11 The first four terms of sequence A, B, C and D are shown in the table.
Sequence
A
1st term
2nd term
3rd term 4th term
1
3
2
4
3
5
B
C
D
3
-1
-3
4
0
0
5
1
5
5th term
nth term
4
6
5
7
n
n+2
6
2
12
7
3
21
n+2
n−2
n2−4
(a) Complete the table.
A
[Ideal sequence : 1, 2, 3, 4, .........n]
take numerator and deniminator separately
n
n+2
B
3
4
5
6
1
2
5
12 21
n+2
C
-1 0
n−2
D
-3 0
Sequence B × Sequence C
= (n+2) (n-2)
=n2 − 4
62
[8]
Number
(b) Which term in sequence A is equal to
n
=
n+2
36
37
37n = 36(n+2)
37n = 36n + 72
37n − 36n = 72
36
?
37
n = 72
Answer (b):
72
[2]
(c) Which term in sequence D is equal to 725?
n2 − 4 = 725
n2 = 725 + 4
n2 = 729
n = 729
n = 27
Answer (c):
27
[2]
October / November 2015 Paper 23
1
Write 168.9 correct to 2 significant figures.
Answer:
2
Calculate
[1]
2.07 − 1.89
5.71 − 3.92
2.07 − 1.89
5.71 − 3.92
= 0.1005586592
Answer:
3
170
0.101
[1]
Write 1.7 × 10–4 as an ordinary number.
Answer:
0.00017
[1]
63
Workbook of Variant 3
5 11 12 13 14 15 16
From the list of numbers, write down
(a) the factors of 60,
Answer (a):
12, 15
[1]
(b) the prime numbers.
Answer (b):
15 Work out
11, 13
[1]
2
1
1
+
–
, giving your answer as a fraction in its lowest terms.
3
6
4
Do not use a calculator and show all the steps of your working.
2×1
4×2
3×1
+
−
2×6
4×8
3×4
8+2−3
12
[L.C.M. of 3, 6 and 4 is 12]
7
12
Answer:
20 The volume of a cuboid is 878 cm3, correct to the nearest cubic centimetre.
The length of the base of the cuboid is 7 cm, correct to the nearest centimetre.
The width of the base of the cuboid is 6 cm, correct to the nearest centimetre.
Calculate the lower bound for the height of the cuboid.
Volume of cuboid =
length × base × height
volume
length×base
[lower bound = lower bound ÷ upper bound]
height =
Volume = 878 (correct to nearest cm3)
half of 1 =
1
2
= 0.5, lower bound of volume = 878 − 0.5 =
877.5
length = 7 (correct to nearest cm)
half of 1 =
base = 6
1
2
=
0.5,
(correct to nearest cm)
upper bound of base =
64
upper bound of length =
6 + 0.5 = 6.5
7 + 0.5 =
7.5
7
12
[3]
Number
lower bound for the height of the cuboid =
= 18
877.5
7.5 × 6.5
Answer:
18
cm [3]
October / November 2015 Paper 43
1 (a) Kolyan buys water for $ 2.60.
He also buys biscuits.
(i) The ratio cost of biscuits : cost of water
=
3:2
Find the cost of the biscuits.
x
=
x
=
Cost of biscuits
Cost of water
Ratio
3
2
Cost
x
2.60
3 × 2.60
2
3.9
Answer (a)(i): $
3.9
[2]
(ii) Kolyan has $ 9 to spend.
Work out the total amount Kolyan spends on water and biscuits as a fraction of the $ 9.
Give your answer in its lowest terms.
3.9 + 2.6
9
=
13
18
Answer (a)(ii):
13
18
[2]
(iii) The $ 9 is 62.5 % less than the amount Kolyan had to spend last week.
Calculate the amount Kolyan had to spend last week.
65
Workbook of Variant 3
Old
New
%
100
37.5
Amount
x
9
100 × 9
37.5
x =
x = 24
24
Answer (a)(iii):
[3]
(b) Priya buys a bicycle for $ 250
Each year the value of the bicycle decreases by 8% of its value at the beginning of that year.
Calculate the value of Priya's bicycle after 10 years.
Give your answer correct to the nearest dollar.
r
P1−
100
Amount
=
value of the bicycle
=
=
8
250 1 − 100
=
108.5971136
n
10
109
Answer (b): $
10 Complete the table for each sequence.
Sequeance
1st term
2nd
term
A
15
8
1
B
5
18
6
19
C
D
2
2
5
6
Sequence A :
1581-6
-7-7-7
It is linear sequence
nth term
a + (n−1)d
=
15 + (n−1)-7
=
15 − 7n + 7
=
66
=
-7n + 22
5th
term
6th term
nth term
-6
-13
-20
−7n+22
7
20
8
21
9
22
10
23
n+4
n+17
10
18
17
54
26
162
37
486
n2 + 1
2 × 3n-1
3rd term 4th term
[3]
Number
Sequence B :
7
n+4
.................
20
n+17
6
19
5
18
[take numerator and denominator separately ideal sequence is 1,2,3,4.......n ]
Sequence C :
Method 1
2
5
10 17
1+1
4+1
9+1
16+1
12+1
22+1
32+1
42+1
nth term is n2+1
Method 2
∆0
2
∆1
5
3
∆2
10
5
2
17
7
2
26
9
2
Its quadratic sequence
nth term
∆2 =
2a
2
=
2a
a
=
1
:
∆1 =
3a+b
3
=
3(1)+b
b
=
0
∆3 =
a+b+c
2
=
1+0+c
c
=
1
an2 + bn + 1
nth term ; n2 + 1
Sequence D :
2618 54162
x3
x3
x3
nth term
=
a × rn-1
nth term
=
2 × 3n-1
x3
[11]
67
Workbook of Variant 3
68
2
Workbook of
Solution 4
ALGEBRA 1
Algebra
Chapter 2: Algebra 1
May / June 2010 Paper 23
12 Expand and simplify
2(x − 3)2 − (2x − 3)2
=
2(x2 − 6x + 9) − (4x2 − 12x + 9)
=
2x2 − 12x + 18 − 4x2 + 12x − 9
=
2x2 − 4x2 − 12x + 12x + 18 − 9
=
−2x2 + 9
=
9 − 2x2
[(a2 − b)2 = a2 − 2ab + b2)]
[Gather all like terms]
Answer:
9 − 2x2
[3]
14 Solve the equation y
=9
2
y
2 × [3 (y − 4)]
+
= 97y = 18 + 24
2
2×1
3(y − 4) +
6 (y − 4) + y
= 97
y = 42
(common denomenator)
2
42
6y − 24 + y = 9 × 2y =
7
6y − 24 + y = 18y = 6
Answer: y =
6
[3]
71
Workbook of Variant 3
May / June 2010 Paper 43
9
(a) The cost of a bottle of water is $ w.
The cost of a bottle of juice is $ j.
The total cost of 8 bottles of water and 2 bottles of juice is $ 12.
The total cost of 12 bottles of water and 18 bottles of juice is $ 45.
Find the cost of a bottle of water and the cost of a bottle of juice.
1st equation: 8w + 2j = 12
2nd equation: 12w + 18j = 45
(8w + 2j = 12) × –9
[2 unknown simulttaneous equation]
[mulitiply entire 1st equation by -9]
–72w – 18j = –108
+
12w + 18j = 45
−60w = −63
w = −63
−60
w = 1.05
8(1.05) + 2j = 12
8.4 + 2j = 12
2j = 3.6
j = 3.6
2
j = 1.8
Answer: (a) cost of a bottle of water = $ =
cost of a bottle of water = $ =
$1.05
1.8
(b) Roshni cycles 2 kilometres at y km/h and then runs 4 kilometres at (y − 4) km/h.
The whole journey takes 40 minutes.
(i) Write an equation in y and show that it simplifies to y2 – 13y + 12 = 0.
Answer (b)(i):
72
Time = distance
speed
t1 = Time1 =
2
y
t2 = time2 =
4
y−4
[5]
Algebra
Total time = t1 + t2
Total time =
40
hour
60
[since speed is in km/h, time has to be in hours]
40
2 + 4 =
y
60
y−4
4
2(y − 4) + 4y
=
6
y (y − 4)
4
2y − 8 + 4y
=
2
6
y − 4y
6y − 8
4
=
y − 4y
6
6(6y − 8) = 4(y2 − 4y)
36y − 48 = 4y2 − 16y
0 = 4y2 − 16y − 36y + 48
0 = 4y2 − 52y + 48
0 = 4(y2 − 13y + 12)
y2 − 13y + 12 = 0
[4]
(ii) Factorise y2 − 13y + 12.
y2 − 12y − y + 12.[Sum is –13 and product is 12]
y(y − 12) − (y − 12)
(y − 12) (y − 1)
Answer: (b) (ii)
(y − 12) (y − 1)
[2]
(iii) Solve the equation y2 − 13y + 12 = 0.
(y − 12) (y − 1) = 0
Either y − 12 = 0
or
y−1=0
y = 12y = 1
Answer: (b) (iii)
y=
12
or y =
1
[1]
(iv) Work out Roshni’s running speed.
Roshni’s running speed
=y−4
= 12− 4
[1 will give negative speed]
= 8
Answer: (b) (iv)
8 km/h
[1]
73
Workbook of Variant 3
(c) Solve the equation
u2 − u − 4 = 0
Show all your working and give your answers correct to 2 decimal places.
a = 1, b = −1, c = −4
x =
−b ± b2 − 4ac
2a
x =
− (−1) ± (−1)2 − 4 (1) (−4)
2(1)
x = 1 ± 17
2
x = 1 + 17 or x = 1 − 17
2
2
x = 2.561552813 or x = −1.561552813
Answer: (c) u =
−1.56
2.56
or u =
[4]
October / November 2010 Paper 23
17 Solve the simultaneous equations.
5x − y = − 10
x + 2y = 9
[Multiply 1st equation by 2]
(5x − y = − 10) × 2
10x − 2y = − 20
+
x + 2y = 9
11x = −11
x = −1
substitute the value of x in 1st equation
5 (−1) − y= −10
−5 − y = −10
−5 + 10 = y
y=5
Answer: x =
y=
74
−1
5
[3]
Algebra
October / November 2010 Paper 43
3
(a) Expand the brackets and simplify.
x (x + 3) + 4x (x − 1)
= x2 + 3x + 4x2 − 4x
= x2 + 4x2 + 3x − 4x [collect all like terms]
= 5x2 − x
Answer: (a)
5x2 − x
[2]
(b) Simplify (3x3)3.
= 33x3×3
= 27x9
Answer: (b)
27x9
[2]
(c) Factorise the following completely.
(i) 7x7 + 14x14
= 7x7 (1 + 2x7)
[common factor]
Answer: (c) (i)
7x7 (1 + 2x7)
[2]
(ii) xy + xw + 2ay + 2aw
= x (y + w) + 2a (y + w)
[factorise by grouping]
= (y + w) (x + 2a)
Answer: (c) (ii)
(y + w) (x + 2a)
[2]
(iii) 4x2 − 49
= (2x)2 − 72
= (2x + 7) (2x − 7)
[a2 − b2 = (a + b) (a − b)]
Answer: (c) (iii)
(2x + 7) (2x − 7)
[1]
75
Workbook of Variant 3
(d) Solve the equation.
2x2 + 5x + 1 = 0
Show all your working and give your answers correct to 2 decimal places.
a = 2, b = 5, c = 1
x =
−b ± b2 − 4ac
2a
x =
−5 ± (5)2 − 4 (2) (1)
2(2)
x = −5 ± 17
4
x = −5 + 17 4
x = −5 − 17
or
4
x = −0.2192235936
or
x = −2.280776406
Answer: (d) x =
−0.22
−2.28
or x =
[4]
May / June 2011 Paper 23
5
y
0
x
The sketch shows the graph of y = axn where a and n are integers.
Write down a possible value for a and a possible value for n.
Answer: a =
n=
76
–1
2
[2]
Algebra
1
Factorise completely.
2xy − 4yz
2y (x − 2z)
[common factor]
Answer:
2
2y (x − 2z)
[2]
Make x the subject of the formula.
y=
x
+5
3
y−5=
x
3
3 (y − 5) = x
x = 3(y − 5)
Answer: x =
8
3 (y − 5)
[2]
Solve the simultaneous equations.
x + 2y = 3
2x − 3y = 13
[multiply 1st equation by -2]
−2x − 4y = −6
+
2x − 3y = 13
−7y = 7
y =
7
−7
y = −1
subtitute the value of y in 1st equation
x + 2 (-1) = 3
x-2=3
x = 3 + 2
x = 5
Answer: x =
5
y=
−1
[3]
77
Workbook of Variant 3
10 The cost of a cup of tea is t cents.
The cost of a cup of coffee is (t + 5) cents.
The total cost of 7 cups of tea and 11 cups of coffee is 2215 cents.
Find the cost of one cup of tea.
7t + 11 (t + 5) = 2215
7t + 11t + 55 = 2215
18t + 55 = 2215
18t = 2215 − 55
18t = 2160
t =
2160
18
t = 120
Answer:
120
cent [3]
May / June 2011 Paper 43
3
x cm
2x cm
(x + 5) cm
The diagram shows a square of side (x + 5) cm and rectangle which measures 2x cm by x cm.
The area of the square is 1 cm2 more than area of the rectangle.
(a) Show that x2 − 10x − 24 = 0
Answer (a):
Area of sequnce − area of rectangle = 1
side2 − l × b = 1
(x + 5)2 − (2x) (x) = 1
x2 + 10x + 25 – 2x2 = 1
78
Algebra
0 = 2x2 − x2 − 10x − 25 + 1
0 = x2 − 10x − 24
x2 − 10x − 24 = 0 [3]
(b) Find the value of x.
x2 − 10x − 24 = 0
[Product = − 24, Sum = − 10]
x2 − 12x + 2x − 24 = 0
[Splitting middle term]
x (x − 12) + 2 (x − 12) = 0
(x − 12) (x + 2) = 0
x = 12[since x is width of rectangle, x cannot be negative]
12
Answer: (b) x =
[3]
(c) Calculate the acute angle between the diagonals of the rectangle.
A
x 12
O
Consider
triangle OAX
O
O
A
tan θ =
6
12
θ = tan-1
6 x 6
12
o
6
(o)
x
( 12 )
θ = 26.56505118
12
A
(A)
tan θ =
B
OX is half of 24
A
B
Angle AOB is double of AOX
26.56505118 × 2 = 53.13010235
Answer: (c)
53.1
[3]
79
Workbook of Variant 3
October / November 2011 Paper 23
3
Solve the simultaneous equations.
x + 5y = 22
x + 3y = 12
[multiply 1st equation by –1]
− x − 5y = − 22
+
x + 3y = 12
− 2y = − 10
y=
−10
−2
y=5
subtitute the value of y in 1st equation
x + 5 (5) = 22
x + 25 = 22
x = 22 − 25
x = −3
Answer: x =
y=
−3
5
[2]
11 Factorise completely.
p2x − 4q2x
x(p2 − 4q2)
[common factor]
x(p2 − (2q)2)
[a2 − b2 = (a + b)(a − b)]
x(p + 2q) (x − 2q)
Answer:
80
x (p + 2q) (x − 2q)
[3]
Algebra
October / November 2011 Paper 43
5
(a)The cost of a bottle of juice is 5 cents more than the cost of a bottle of water. Mohini buys 3
bottles of water and 6 bottles of juice. The total cost is $ 5.25.
Find the cost of a bottle of water.
Give your answer in cents.
let x → cost of a bottle of water
cost of a bottle of juice = x + 5
3x + 6(x + 5) = 525
3x + 6x + 30 = 525
[convert in cents]
9x + 30 = 525
9x = 525 − 30
9x = 495
x =
495
9
x = 55
Answer: (a)
55
cent [4]
(b) The cost of a biscuit is x cents.
The cost of a cake is (x + 3) cents.
The number of biscuits Roshni can buy for 72 cents is 2 more than the number of cakes she
can buy for 72 cents.
(i) Show that x2 + 3x − 108 = 0
Answer (b)(i):
72
x
Number of biscuits =
Number of cake =
Number of biscuts − Number of cake = 2
72
72
−
=2
x+3
x
72 (x + 3) − 72x
=2
x2 + 3x
72
x+3
72x + 216 − 72x = 2(x2 + 3x)
216 = 2x2 + 6x
0 = 2x2 + 6x − 216
81
Workbook of Variant 3
0 = 2(x2 + 3x − 108)
x2 + 3x − 108 = 0 [3
(ii) Solve the equation x2 + 3x − 108 = 0.
[product = –108, sum = 3]
x2 + 3x − 108 = 0
x2 + 12x − 9x − 108 = 0
[splitting middle term]
x (x + 12) − 9 (x + 12) = 0
(x + 12) (x − 9) = 0
x + 12 = 0
or
x−9=0
x = −12
or
x=9
Answer: (b) (ii) x =
−12
or x =
9
[3]
(iii) Find the total cost of 2 biscuits and 1 cake.
2x + x + 3
2 (9) + 9 + 3 = 30
30
Answer: (b) (iii)
cents [1]
May / June 2012 Paper 23
15 Solve the equation 2x2 + 6x − 3 = 0
Show your working and give your answers correct to 2 decimal places.
a = 2,
b = 6,
c = −3
2
x = −b ± b − 4ac
2a
2
x = −6 ± (6) − 4 (2) (–3)
2(2)
x = −6 ± 60
4
x = −6 + 60 4
x = 0.4364916731
or
or
x = −6 − 60
4
x = -3.436491673
Answer: (b) x =
82
0.44
or x =
-3.44
[4]
Algebra
May / June 2012 Paper 43
10 (a) Rice costs $ x per kilogram.
Potatoes cost $ (x + 1) per kilogram.
Total cost of 12 kg of rice and 7 kg of potatoes is $ 31.70.
Find the cost of 1 kg of rice.
12x + 7 (x + 1) = 31.70
12x + 7x + 7 = 31.70
19x + 7 = 31.70
19x = 31.70 − 7
19x = 24.7
24.7
19
x =
x = 1.3
Answer: (a) $
1.3
[3]
(b) The cost of a small bottle of juice is $ y.
The cost of a large bottle of juice is $ (y + 1)
When Catriona spends $ 36 on small bottles only, she receives 25 more bottles than when
she spends $ 36 on large bottles only.
(i) Show that 25y2 + 25y − 36 = 0.
Answer (b)(i):
Number of small bottle of juice =
36
y
Number of large bottle of juice =
36
y+1
Number of small bottle of juice − Number of large bottle of juice = 25
36
36
−
= 25
y+1
y
36 (y + 1) − 36y
y (y + 1)
= 25
36y + 36 − 36y
= 25
y2 + y
36 = 25 (y2 + y)
36 = 25y2 + 25y
0 = 25y2 + 25y − 36
25y2 + 25y − 36 = 0
[4]
(ii) Factorise 25y2 + 25y − 36.
83
Workbook of Variant 3
[Product = -900, sum = 25]
25y2 + 25y − 36
= 25y2 − 20y + 45y +36
= 5y (5y − 4) + 9(5y − 4)
= (5y − 4) (5y + 9)
(splitting middle term)
Answer: (b) (ii)
(5y − 4) (5y + 9)
[2]
(iii) Solve the equation 25y2 + 25y − 36 = 0.
(5y − 4) (5y + 9) = 0
Either 5y − 4 = 0
or
5y + 9 = 0
4
= 0.8
5
or
y=
y =
–9
= –1.8
5
Answer: (b) (iii) y =
0.8
–1.8
or y =
[1]
(iv) Find the total cost of 1 small bottle of juice and 1 large bottle of juice.
y + y + 1 = 0.8 + 0.8 + 1
Answer: (b) (iv) $
2.6
[1]
October / November 2012 Paper 23
4
Expand the brackets.
y (3 – y3)
3y − y4
Answer:
3y − y4
October / November 2012 Paper 43
5 (a) Marcos buys 2 bottles of water and 3 bottles of lemonade.
The total cost is $ 3.60.
The cost of one bottle of lemonade is $ 0.25 more than the cost of one bottle of water.
Find the cost of one bottle of water.
Let x → cost of one bottle of water.
cost of one bottle of lemonade = x + 0.25
2x + 3(x + 0.25) = 3.60
2x + 3x + 0.75 = 3.60
84
[2]
Algebra
5x + 0.75 = 3.60
5x = 3.60 − 0.75
5x = 2.85
2.85
5
x =
x = 0.57
Answer: (a) $
0.57
[4]
(b)
5 cm2
y cm
x cm
6 cm2
Y cm
(x + 2) cm
The diagram shows two rectangles.
The first rectangle measures x cm by y cm and has an area of 5 cm2.
The second rectangle measures (x + 2) cm by Y cm has an area of 6 cm2.
(i) When y + Y = 1 show that x2 − 9x − 10 = 0
Answer (b)(i):
Area of 1st rectangle : xy = 5
y =
5
x
[Make y the subject, because what we have to prove is in terms of x]
Area of 2nd rectangle: (x + 2) Y = 6
Y =
6
x+2
y + Y = 1
[Make Y the subject]
[given]
6
5
+
=1
x+2
x
5(x + 2) + 6x
=1
x (x + 2)
5x + 10 + 6x
=1
x2 + 2x
11x + 10 = x2 + 2x
0 = x2 + 2x − 11x − 10
0 = x2 − 9x − 10
x2 − 9x − 10 = 0 [4]
(ii) Factorise x2 − 9x − 10.
85
Workbook of Variant 3
product = −10,
sum − 9
x2 − 9x − 10
= x2 − 10x + x − 10
(splitting middle term)
= x(x − 10) + 1 (x − 10)
= (x − 10) (x + 1)
Answer: (b) (ii)
(x − 10) (x + 1)
[2]
(iii) Calculate the perimeter of the first rectangle.
x = 10,
y=
5
= 0.5
10
y = 0.5
perimeter = 2 (x + y)
= 2 (10 + 0.5)
= 21
Answer: (b) (iii)
21
cm [2]
(c)
(2x + 3) cm
5 cm
(x + 3) cm
The diagram shows a right-angled triangle with sides of length 5 cm,
(x + 3) cm and (2x + 3) cm.
(i) Show that 3x2 + 6x – 25 = 0.
Answer (c)(i):
(2x + 3)2 = (x + 3)2 + 52(Pythagoras theorem)
4x2 + 12x + 9 = x2 + 6x + 9 + 25
4x2 + 12x + 9 − x2 − 6x − 9 − 25 = 0
4x2 − x2 + 12x − 6x + 9 − 9 − 25 = 0
3x + 6x − 25 = 0
(ii) Solve the equation 3x2 + 6x – 25 = 0
86
[4]
Algebra
Show all your working and give your answers correct to 2 decimal places.
a = 3, b = 6, c = -25
x =
x =
−b ± b2 − 4ac
2a
−6 ± (6)2 − 4 (3) (−25)
2(3)
x = −6 ± 336
6
x = −6 + 336 6
x = 2.055050463
or
or
x = −6 − 336
6
x = –4.055050463
Answer: (c) (ii) x =
2.06
or x =
–4.06
[4]
(iii) Calculate the area of the triangle.
Area of triangle =
1
× base × height
2
=
1
× (x + 3) (5)
2
=
1
× (2.6 + 3) (5)
2
= 12.65
Answer: (c) (iii)
12.7
cm2
[2]
May / June 2013 Paper 23
10 Find the value of 2x + y for the simultaneous equations.
3x + 5y = 48
2x − y = 19
[Multiply 2nd equation by 5]
3x + 5y = 48
+
10x − 5y = 95
13x = 143
x =
143
13
x = 11
Substitute the value of x in 2nd original equation
87
Workbook of Variant 3
22 − y = 19
22 − 19 = y
y = 3
2x + y = 2(11) + 3
= 22 + 3
= 25
Answer:
2x + y = 25
[4]
May / June 2013 Paper 43
10 (a) (i) Solve 2(3x − 7) = 13.
6x − 14 = 13
6x = 13 + 14
6x = 27
6x =
27
6
x = 4.5
4.5
Answer: (a) (i) x
[3]
(ii) Solve by factorising x2 − 7x + 6 = 0
Product = 6, sum = –7
x2 − 7x + 6 = 0
x2 − 6x − x + 6 = 0
[Spliting middle term]
x(x − 6) −1 (x − 6) = 0
(x − 6) (x − 1) = 0
Either x − 6 = 0
x = 6
or
or
x−1=0
x=1
Answer: (a) (ii) x =
x+2
3x − 2
+
=4
10
5
2 × (3x − 2)
+ x + 2 = 4
2×5
10
1
(iii) Solve
88
2 (3x − 2) + x + 2
=4
10
[LCM of 5 and 10 is 10]
or x =
6
[3]
Algebra
6x − 4 + x + 2 = 40
6x + x − 4 + 2 = 40
7x − 2 = 40
7x = 40 + 2
7x = 42
x =
42
7
x = 6
Answer: (a) (iii) x =
(b) 12 =
=
5
12 + 22 + 32
=
14
12 + 22 + 32 + 42
=
30
+ n2 = an3 + bn +
Work out the values of a and b.
[take any value of n and substitute]
n
6
an3 + bn2 +
When n = 1
→
When n = 2
1
6
2
=5
6
1
=5
3
8a + 4b = 5 −
8a + 4b =
1
=1
6
5
1st equation
6
a(2)3 + b(2)2 +
8a + 4b +
6
1
=1
6
a + b = 1 −
a + b =
n
given
a(1)3 + b(1)2 +
a + b +
[4]
1
12 + 22
12 + 22 + 32 + 42 +....................
6
1
3
14
2nd equation
3
89
Workbook of Variant 3
[Solve 2 equtaions simultaneously]
[Multiply 1st equation by −4]
-10
−4a − 4b =
3
+
8a + 4b = 14
3
4a =
a =
4
3
4
3×4
1
3
(Substitute value of a in 1st equation)
a =
a + b =
5
6
1
5
+b=
3
6
b =
5
1
−
6
3
b =
1
2
Answer: (b) a =
b=
1
3
1
2
[6]
October / November 2013 Paper 23
9
(a) Expand and simplify (a + b)2.
(a + b)2
= (a + b) (a + b)
= a (a + b) + b (a + b)
= a2 + ab + ab + b2
= a2 + 2ab + b2
Answer: (a)
90
a2 + 2ab + b2
[2]
Algebra
(b) Find the vaue of a2 + b2 when a + b = 6 and ab = 7.
(a + b)2 = a2 + 2ab + b2 (from given part a)
62 = a2 + 2(7) + b2[a + b = 6, ab = 7] given
36 = a2 + 14 + b2
36 − 14 = a2 + b2
a2 + b2 = 22
Answer: (b)
22
[1]
7The solutions of the equation x2 − 6x + d = 0 are both integers. d is a prime number.
Find d.
Sum = –6
Product = d
Sum
Product
–6 + 0
0
–5 – 1
5
–4 – 2
8
–3 – 3
9
→
Prime
Answer: (d)
3
5
[3]
Solve the equation 1 + 2x = –15.
2x = −15 − 1
x =
–16
2
x = –8
Answer: x
–8
[2]
91
Workbook of Variant 3
October / November 2013 Paper 43
10 (a) Simplify.
x2 − 3x
x2 − 9
x (x − 3)
[a2 − b2 = (a + b) (a − b)]
(x + 3) (x − 3)
x
x+3
x
x+3
Answer: (a)
[3]
(b) Solve
20
15
−
=2
x
x+1
15 (x + 1) − 20x
=2
x (x + 1)
15x + 15 − 20x
=2
x2 + x
15 − 5x = 2 (x2 + x)
15 − 5x = 2x2 + 2x
0 = 2x2 + 2x + 5x − 15
0 = 2x2 + 7x − 15
2x2 + 7x − 15 = 0
2x2 − 3x + 10x − 15 = 0
[Sum = 7, Product = –30]
x(2x − 3) + 5(2x − 3) = 0
(2x − 3) (x + 5) = 0
Either 2x − 3 = 0
2x = 3
x =
or
or
x+5=0
x = –5
3
2
Answer: x =
92
3
2
or x =
–5
[7]
Algebra
May / June 2014 Paper 23
4
Factorise completely.
15a3 − 5ab
= 5a (3a2 − b)
[common factor]
Answer:
5a (3a2 − b)
[2]
12 (a) Factorise 3x2 + 2x − 8.
[product = −24, sum = 2]
3x2 + 2x − 8
= 3x2 − 4x + 6x − 8
[splitting middle term]
= x(3x − 4) + 2 (3x − 4)
= (3x − 4) (x + 2)
Answer: (a)
(3x − 4) (x + 2)
[2]
(b) Solve the equation 3x2 + 2x − 8 = 0.
(3x − 4) (x + 2) = 0
Either 3x − 4 = 0
3x = 4
x =
or
or
x+2=0
x = –2
4
3
Answer: (b) x =
4
3
or x =
–2
[1]
19 Robbie pays $ 10.80 when he buys 3 notebooks and 4 pencils.
Paniz pays $14.50 when she buys 5 notebooks and 2 pencils.
Write down simultaneous equations and use them to find the cost of a notebook and the cost of
a pencil.
let x
y
→ cost of a notebook.
→ cost of pencil.
1st equation : 3x + 4y = 10.80
2nd equation : 5x + 2y = 14.50
[Multiply 2nd equation by –2]
93
Workbook of Variant 3
3x + 4y = 10.80
+
–10x + 4y = –29
→ (1st equation)
→ (2nd equation after multiplication)
–7x = –18.2
–18.2
–7
x=
x = 2.6
Substitute value of x in 1st equation
3(2.6) + 4y = 10.80
7.8 + 4y = 10.80
4y = 10.80 − 7.8
4y = 3
4y =
3
4
y = 0.75
Answer: cost of a notebook = $
cost of pencil = $
2.6
0.75
[5]
May / June 2014 Paper 43
8
2x – 3
7
(a) (i)Show that the equation
+
= 1 can be simplified to
2
x+4
2x2 + 3x − 6 = 0.
Answer (a)(i):
7 × 2 + (2x − 3) (x + 4)
=1
2 (x + 4)
14 + 2x (x + 4) −3 (x + 4) = 2 (x + 4)
14 + 2x2 + 8x − 3x − 12 = 2x + 8
2x2 + 8x − 3x − 2x + 14 − 12 − 8 = 0
2x2 + 3x − 6 = 0
(ii) Solve the equation 2x2 + 3x − 6 = 0
Show all your working and give your answers correct to 2 decimal places.
a = 2, b = 3, c = -6
94
[3]
Algebra
x =
−b ± b2 − 4ac
2a
x =
−3 ± (3)2 − 4 (2)(-6)
2(2)
x = −3 ± 57
4
x = −3 + 57 4
or
x = −3 + 57
x = 1.137458609
or
x = –2.637458609
4
Answer: (a) (ii) x =
1.14
or x =
–2.64
[4]
(b)The total surface area of a cone with radius x and slant height 3x is equal to the area of a
circle with radius r.
Show that r = 2x.
[The curved surface area, A, of a cone with radius r and slant height l is A = πrl.]
Answer (b):
The total surface area of cone = curved surface of cone + area of circle
[area of circle = πr2]
πr2 = π(x) (3x) + πx2
πr2 = 3 πx2 + πx2
πr2 = 4 πrx2
r2 =
4πx2
π
r2 = 4x2
r = 4x2
r = 2x [4]
October / November 2014 Paper 23
6
Solve the equation.
2x + 5
=8
3
2x + 5 = 8 × 3
2x + 5 = 24
95
Workbook of Variant 3
2x = 24 − 5
2x = 19
19
2
x = 9.5
x=
Answer: x =
9.5
[3]
October / November 2014 Paper 43
8
(a) A straight line joins the points (−1, −4) and (3, 8).
(i) Find the midpoint of the this line.
( x +2 x , y +2 y )
= ( –1 + 3 , –4 + 8 )
2
2
=
1
2
1
2
= (1, 2)
Answer: (a) (i)
(1, 2)
[2]
(ii) Find the equation of this line.
Give your answer in the form y = mx + c
m =
y2 − y2
x2 − x1
m = 8 − (−4)
3 − (−1)
m =
8+4
3+1
m =
12
4
m = 3
Substitute value of m in general equation of line
y = mx + c
8 = 3(3) + c
8−9=c
→ (general equation of line)
c = -1
Answer: (a) (ii) y =
96
3x − 1
[3]
Algebra
(b) (i) Factorise x2 + 3x − 10.
[product = −10, sum = 3]
x2 − 2x + 5x − 10
[splitting middle term]
x(x − 2) + 5(x − 2)
(x − 2) (x + 5)
(x − 2) (x + 5)
Answer: (b) (i)
[2]
(ii) The graph of y = x2 + 3x − 10 is sketched below.
Line of symmetry x = - 1.5
y
(a, 0)
0
x
(b, 0)
(0, c)
Write down the values of a, b and c.
(a, 0) and (b, 0) are x - intercept.
[to find x - intercept, substitute y = 0]
x2 + 3x − 10 = y
(x − 2) (x + 5) = 0
x = 2,
x = −5
a = −5, b=2
(0, c) is y - intercept
[To find y - intercept, substitute x = 0]
y = 02 + 3(0) − 10
y = −10
Answer: (b) (ii) a
–5
b
2
c
–10
[3]
97
Workbook of Variant 3
(iii) Write down the equation of the line of symmetry of the graph of y = x2 + 3x −10.
x =
2−5
2
x = −1.5
–1.5
Answer: (b) (iii)
[1]
(c) Sketch the graph of y = 18 + 7x − x2 on the axes below.
Indicate clearly the values where the graph crosses the x and y axes.
y
(0,18)
(-2,0)
To find x - inrecept put y = 0
0 = 18 + 7x − x2
0
(9,0)
x
x2 − 7x − 18 = 0
x(x − 9) + 2(x − 9) = 0
(x − 9) (x + 2) = 0
x − 9 = 0, or x + 2 = 0
x = 9,
x = −2
to find y-intercept put x = 0
y = 18 + 7(0) − 02
y = 18
Answer: (c)
(d) (i) x2 + 12x − 7 = (x + p)2 − q
Find the value of p and value of q.
Method: 1
x2 + 12 − 7
(x + 6)2 − 36 − 7
(x + 6)2 − 43.
98
(now compare)
18
[4]
Algebra
Method: 2
(expand right hand side and compare)
x2 + 12x − 7 = (x + p)2 − q
x2 + 12x − 7 = x2 + 2px + p2 − q [(a + b)2 = a2 + 2ab + b2]
After comparing
12x = 2px
12x
=p
2x
p = 6
p2 − q = −7
62 − q = −7
36 − (−7) = q
q = 43.
6
Answer: (d) (i) p =
43
or q =
[3]
(ii) Write down the minimum value of y for the graph of y = x2 + 12x − 7.
Answer: (d) (ii)
–43
[1]
October / November 2014 Paper 43
10 (a)
(3x – 5) cm
(2x – 3) cm
(15 – 2x) cm
(2x + 7) cm
(i)Write an expression, in term of x, for the perimeter of the quadrilateral. Give your answer
in its simplest form.
Perimeter = 3x − 5 + 2x − 3 + 2x + 7 + 15 − 2x [Perimeter means boundry]
Perimeter = 3x + 2x + 2x − 2x − 5 − 3 + 7 + 15
Perimeter = 5x + 14
Answer: (a) (i)
5x + 14
cm [2]
99
Workbook of Variant 3
(ii) The perimeter of the quadrilateral is 32 cm.
Find the length of the longest side of the quadrilateral.
5x + 14 = 32
5x = 32 − 14
5x = 18
x =
18
5
x = 3.6
3x − 5 = 3(3.6) − 5 = 5.8
2x − 3 = 2(3.6) − 3 = 4.2
2x + 7 = 2(3.6) + 7 = 14.2
→ (longest side)
15 − 2x = 15 − 2(3.6) = 7.8
14.2
Answer: (a) (ii)
(b)
(6b – a) m
(5a – 2b) m
14 m
(7a – 6b) m
am
(3b + a) m
The triangle has a perimeter of 32.5 m.
The quadrilateral has a perimeter of 39.75 m.
Write two equtions in terms of a and b and simplify them.
Use an algebraic method to find the values of a and b.
Show all your working,
14 + 7a − 6b + 3b + a = 32.5
7a + a − 6b + 3b = 32.5 − 14
8a − 3b = 18.5
100
→ 1st equation
5a − 2b + 6b − a + 13.5 + a = 39.75
5a − a + a − 2b + 6b = 39.75 − 13.5
5a + 4b = 26.25
→ 2nd equation
13.5 m
cm [3]
Algebra
Muyltiply 1st equation by 4 and 2nd equation by 3
32a − 12b = 74
15a + 12b = 78.75
47a = 152.75
(add equations)
a = 152.75
47
a = 3.25
(Substitute value of a in 1st equation)
8(3.25) − 3b = 18.5
26 − 3b = 18.5
26 − 18.5 = 3b
7.5 = 3b
3b = 7.5
b =
7.5
3
b = 2.5
Answer: (b) (ii) a
b
3.25
2.5
[6]
May / June 2015 Paper 23
2
Factorise completely.
9x2 − 6x
= 3x (3x − 2)
[common factor]
Answer:
5
3x (3x − 2)
[2]
Factorise 2x2 – 5x − 3.
[product = −6, sum −5]
2x2 − 5x − 3
= 2x2 − 6x + x − 3
[splitting middle term]
= 2x (x − 3) + 1(x − 3)
= (x − 3) (2x + 1)
Answer:
(x − 3) (2x + 1)
[2]
101
Workbook of Variant 3
9
Solve the equation.
3(x + 4) = 2(4x − 1)
(3x + 12) = (8x – 2)
12 + 2 = 8x − 3x
14 = 5x
5x = 14
x =
14
5
x = 2.8
2.8
Answer: x =
[3]
14 Solve the equation.
2x2 + x − 2 = 0
Show your working and give your answers correct to 2 decimal places.
a = 2, b = 1, c = −2
2
x = −b ± b − 4ac
2a
2
x = −1 ± (1) − 4 (2) (−2)
2(2)
x = −1 ± 17
4
x = −1 + 17 4
or
x = −1 − 17
x = 0.7807764064
or
x = –1.280776406
4
Answer: (c) (ii) x =
0.78
or x =
–1.28
[4]
May / June 2015 Paper 43
7
(a) The total surface area of a cone is given by the formula A = πrl + πr2
(i) Find A when r = 6.2 cm and l = 10.8 cm.
A = π (6.2) (10.8) + π (6.2)2 [Hint : type this in calculator as it is]
a = 331.1238657
Answer: (a) (i)
102
331
cm2
[2]
Algebra
(ii) Rearrange the formula to make l the subject.
A = πrl + πr2
A − πr2 = πrl
2
A − πr = l
π 2r
A − πr2
π 2r
Answer: (a) (ii) l
[2]
(b) (i) Irina walks 10 km at 4 km/h and then a further 8 km at 5 km/h.
Calculate Irina's average speed for the whole journey.
Time =
Time1 = t1 =
t1
=
Distance
Speed
2.5 hours
Time2 = t2 =
10
4
8
5
t2 = 1.6 hours
Total time = T1 + T2 = 2.5 + 1.6 = 4.1 hours
Total distance = 10 + 8 = 18 km.
Average Speed =
=
18
4.1
Total Distance
Total Speed
= 4.390243902
Answer: (b) (i)
4.39
(ii)
Dariella walks x km at 5 km/h and then runs (x + 4) km at 10 km/h.
The average speed of this journey is 7 km/h.
Find the value of x.
Show all your working.
Time1 = t1 =
x
5
Time = t2 =
x+4
10
Total time = t1 + t2
=
x
x+4
+
5
10
=
10x + 5(x + 4)
5(10)
km/h [3]
103
Workbook of Variant 3
= 10x + 5x + 20
50
=
15x + 20
50
Total distance = x + x + 4 = 2x + 4
Average speed =
Total Time =
Total distance
Total time
Total Distance
Average Speed
2x + 4
15x + 20
=
7
50
∴
7(15x + 20) = 50(2x + 4)
105x + 140 = 100x + 200
105x − 100x = 200 − 140
5x = 60
x = 12
Answer: (b) (ii) x =
12
[5]
(c) (i) Priyantha sells her model car for $19.80 at a profit of 20%.
Calculate the original price of the model car.
%
x =
Old
New
C. P.
S. P.
100
120
x
19.80
100 × 19.80
120
x = 16.5
Answer: (c) (i) $
104
16.5
[3]
Algebra
(ii) Dev sells his model car for $ x at a profit of y %.
Find an expression, in terms of x and y, for the original price of this model car.
Write your answer in single fraction.
Let a be original price.
%
Old
New
C. P.
S. P.
100
100 + y
a
x
100 x = a(100 + y)
100 x
=a
100 + y
100 x
100 + y
a =
Answer: (c) (ii) $
100 x
100 + y
[3]
October / November 2015 Paper 23
6
Simplify
1 − 2u + u + 4
1 + 4 −2u + u
5−u
Answer:
7
5−u
[2]
Factorise completely.
2x − 4x2
2x(1 − 2x)
[Common factor]
Answer:
2x(1 − 2x)
[2]
21 Solve the equation 3x2 + 4x − 5 = 0.
Show all your working and give your answer correct to 2 decimal places.
a = 3, b = 4, c = -5
2
x = −b ± b − 4ac
2a
105
Workbook of Variant 3
2
x = −4 ± (4) − 4 (3) (–5)
2(3)
x = −4 ± 76
6
x = −4 + 76 or x = −4 − 76
x = 0.7862996478
or
x = −2.119632981
6
6
Answer: x =
0.79
or x =
–2.12
[4]
October / November 2015 Paper 43
7 (a) The cost of a loaf of bread is x cents.
The cost of a cake is (x − 5) cents.
The total cost of 6 loaves of bread and 11 cakes is $ 13.56.
Find the value of x.
6x + 11(x − 5) = 1356
6x + 11x − 55 = 1356
[convert $13.56 in cents as x is in cent]
17x = 1336 + 55
17x = 1411
x = 1411
17
x = 83
83
Answer: (a) x =
(b)
y+1
y
y+3
The area of the rectangle and the area of the triangle are equal.
Find the value of y.
Area of rectangle = area of triangle
l × b =
106
1
×b×h
2
2y + 1
[4]
Algebra
y(y + 3) =
1
(2y + 1)(y + 1)
2
2y(y + 3) = (2y + 1)(y + 1)
2y2 + 6y = 2y(y + 1) + 1(y + 1)
2y2 + 6y = 2y2 + 2y + y + 1
2y2 + 6y = 2y2 + 3y +1
2y2 + 6y − 2y2 − 3y = 1
3y = 1
y =
1
3
Answer: (b) y =
(c)
The cost of a bottle of water is (w − 1) cents.
The cost of a bottle of milk is (2w − 11) cents.
A certain number of bottles of water costs $ 4.80.
The same number of bottles of milk cost $ 7.80.
Find the value of w.
Number of bottles of water = Number of bottles of milk.
1
3
[4]
480
780
=
w−1
2w − 11
480 (2w − 11) = 780(w − 1)
960w − 5280 = 780w − 780
960w − 780w = 5280 − 780
180w = 4500
w =
4500
180
w = 25
Answer: (c) w =
25
[4]
(d)
u cm
t
(3u – 2) cm
The area of the triangle is 2.5 cm2.
107
Workbook of Variant 3
(i) Show that 3u2 − 2u − 5 = 0
area of trangle =
2.5 =
2 × 2.5 = 3u2 − 2u
5 = 3u2 − 2u
0 = 3u2 − 2u − 5
1
× base × height
2
1
(3u − 2) (u)
2
Answer: (d) (i)
0 = 3u2 − 2u − 5
[2]
(ii) Factorise 3u2 − 2u − 5.
[product = −15, sum = −2]
3u2 − 2u − 5
3u2 + 3u − 5u − 5
3u (u + 1) − 5 (u + 1)
(3u − 5) (u + 1)
(u + 1) (3u − 5)
Answer: (d) (ii)
[2]
(iii) Find the size of angle t.
3u2 − 2u − 5 = 0
(u + 1) (3u − 5) = 0
3u − 5 = 0
u =
or
5
u = –1
3
TOA
u+1=0
5
3
t
3=
tan t =
O
A
5
–2
3
=3
( 53 ÷ 3)
5
t = tan (
÷ 3)
3
tan t =
-1
t = 29.0546041
Answer: (d) (iii)
108
29.1
[3]
3
Workbook of
Solution 4
MENSURATION
Mensuration
Chapter 3: Mensuration
May / June 2010 Paper 23
17
K
O
40°
5.6 cm
L
OKL is a sector of a circle, center O, radius 5.6 cm. Angle KOL = 40o
Calculate
(a) the area of the sector,
area of the sector =
=
θ
× πr 2
360
40
× π(5.6)2
360
= 10.94670507
Answer: (a)
10.9
cm2
[2]
(b) the perimeter of the sector.
Perimeter of sector = Arc length + 2r
=
θ
× 2πr + 2r
360
=
40
× 2π(5.6) + 2(5.6)
360
= 15.10953752
Answer (b):
15.1
cm [2]
111
Workbook of Variant 3
May / June 2010 Paper 43
8
3cm
6cm
10 cm
A solid metal cuboid measures 10 cm by 6 cm by 3 cm.
(a) Show that 16 of these solid metal cuboids will fit exactly into a box which has internal
measurements 40 cm by 12 cm by 6 cm.
Answer (a):
40
= 4;
10
12
= 2;
6
6
=2
3
4 × 2 × 2 = 16
16
Answer: (a)
[2]
(b) Calculate the volume of one metal cuboid.
Volume of cuboid = l × b × h
= 10 × 6 × 3
= 180
Answer: (b)
(c) One cubic centimetre of the metal has a mass of 8 grams.
The box has a mass of 600 grams.
Calculate the total mass of the 16 cuboids and the box in
(i) grams,
Volume of one metal cuboid = 180 cm3
Volume of 16 metal cuboids = 16 × 180
= 2880 cm3
112
180
cm3
[1]
Mensuration
cm3
gram
1
8
2880
x
x = 2880 × 8
x = 23040
Total mass of 16 cuboid and box = 23040 + 600
= 23640
Answer: (c) (i)
23640
g [2]
(ii) kilograms.
23640
= 23.640
1000
[To convert grams to kilograms ÷ 1000]
Answer: (c) (ii)
23.6
kg [1]
(d) (i) Calculate the surface area of one of the solid metal cuboids.
Surface area of cuboid = 2(lb + bh + hl)
= 2(10 × 6 + 6 × 3 + 3 × 10)
= 216
Answer: (d) (i)
216
cm2
[2]
(ii) The surface of each cuboid is painted. The cost of the paint is $ 25 per square metre.
Calculate the cost of painting all 16 cuboids.
Surface area of one cuboid = 216 cm2
Surface area of 16 cuboid = 3456 cm2
3456 cm2 =
3456 m2
1002
= 0.3456 m2
m2
$
1
25
0.3456
x
x = 0.3456 × 25
= 8.64
Answer: (d) (ii) $
8.64
[3]
113
Workbook of Variant 3
(e) One of the solid metal cuboids is melted down. Some of the metal is used to make 200
identical solid spheres of radius 0.5 cm.
Calculate the volume of metal from this cuboid which is not used.
4
π r 3]
3
Let volume of metal from this cuboid which is not used be x
Volume of metal cuboid = 200 (volume of sphere) + x
[The volume, V, of a sphere of radius r is V =
( 43 πr ) + x
4
180 = 200( π (0.5) ) + x
3
180 = 200
3
3
180 = 104.7197551 + x
x = 180 − 104.7197551
x = 75.28024488
Answer: (e)
75.3
cm3
[3]
(f) 50 cm3 of metal is used to make 20 identical solid spheres of radius r.
Calculate the radius r.
50 = 20 (volume of one sphere)
50 = 20 ×
4
π r3
3
50 × 3
= r3
20 × 4 × π
3
50 × 3
=r
20 × 4 × π
r = 0.8419451505
Answer: (f) r =
114
0.842
cm [3]
Mensuration
October / November 2010 Paper 23
18
x°
8 cm
The diagram shows a sector of a circle of radius 8 cm.
The angle of the sector is x°.
The perimeter of the sector is (16 + 14π) cm.
Find the value of x.
The perimeter of the sector is (16 + 14x) cm.
Find the value of x
Perimeter of sector = Arc length + 2r
16 + 14π =
x
× 2π(8) + 2(8)
360
16 + 14π =
x
× 2(π)8 + 16
360
16 + 14π − 16 =
14π =
[Arc length =
θ
× 2πr]
360
x
× 2π(8)
360
x
× 2π(8)
360
14π × 360 = x
2π(8)
315 = x
n = 315
Answer: x =
315
[3]
115
Workbook of Variant 3
October / November 2010 Paper 43
8
3 cm
12 cm
The diagram shows a solid made up of a hemisphere and a cylinder.
The radius of both the cylinder and the hemisphere is 3 cm.
The length of the cylinder is 12 cm.
(a) (i) Calculate the volume of the solid.
4
π r 3]
3
Volume of this solid = volume of cylinder + volume of hemisphere
= π r 2h +
[The volume, V, of the sphere with radius r is V =
( 43 × πr )
1 4
= π(3) (12) +
× π(3) )
2 (3
1
2
3
2
3
= 395.8406744
Answer: (a) (i)
396
cm3
[4]
(ii) The solid is made of steel and 1 cm3 of steel has a mass of 7.9 g.
Calculate the mass of the solid.
Give your answer in kilograms.
cm3
gram
1
7.9
395.8406744
x
x = 7.9 × 395.8406744
x = 3127.141328 g
31227.141328g =
3127.141328
kg
1000
[to convert gram to kg ÷ 1000]
= 3.127141328
Answer: (a) (ii)
116
3.13
kg [2]
Mensuration
(iii) The solid fits into a box in the shape of a cuboid, 15 cm by 6 cm by 6 cm.
Calculate the volume of the box not occupied by the solid.
volume of box not occupied by solid = volume of cuboid − Volume of solid
= 15 × 6 × 6 − 395.8406744
[volume of cuboids = l × b × h ]
= 144.1593256
Answer: (a) (iii)
144
cm2
[2]
cm2
[5]
(b) (i) Calculate the total surface area of the solid.
You must show your working.
[The surface area, A, of a sphere with radius r is A = 4πr2]
Total surface area = A (circle) + CSA (cylinder) + A (hemisphere)
Total surface area = πr2 + 2πrh + 2πr2
= π(3)2 + 2π(3)(12) + 2π(3)2
= 311.0176727
[surface area of hemisphere =
4πr2
= 2πr2]
2
Answer: (b) (i)
311
(ii) The surface of the solid is painted.
The cost of the paint is $ 0.09 per millilitre.
One millilitre of paint covers an area of 8 cm2.
Calculate the cost of painting the solid.
x =
millilitre
cm2
1
8
x
311.0176727
311.0176727
8
x = 38.87720909
millilitre
cost
1
0.09
38.87720909
y
y = 38.87720909 × 0.09
y = 3.4989488181
Answer: (b) (ii) $
3.50
[2]
117
Workbook of Variant 3
May / June 2011 Paper 23
15 A cylinder has a height of 12 cm and a volume of 920 cm3.
Calculate the radius of the base of the cylinder.
volume of cylinder = πr2h
920 = π × r2 × 12
920
= r2
π × 12
r2 =
920
π × 12
r =
920
π × 12
r = 4.940015986
Answer:
4.94
May / June 2011 Paper 43
4
A
8 cm
6 cm
O
B
9 cm
C
The circle, centre O, passes through the points A, B and C.
In the triangle ABC, AB = 8 cm, BC = 9 cm and CA = 6 cm.
(a) Calculate angle BAC and show that it rounds to 78.6o, correct to 1 decimal place.
118
cm [3]
Mensuration
cos A =
b2 + c2 − a2
2bc
cos A =
82 + 62 − 92
2(8)(6)
A = cos-1
(
A = 78.58484226
82 + 62 − 92
2(8)(6)
)
x = 78.6o
(b) M is the midpoint of BC.
(i) Find angle BOM.
[Angle at centre is double of angle on circcumference]
Angle BOC = 78.6 × 2 = 157.2
Angle BOM = angle BOC ÷ 2 = 78.6
[Isosceles triangle]
78.6
Answer: (b) (i) Angle BOM =
[1]
(ii)Calculate the radius of the circle and show that it rounds to 4.59 cm, correct to 3
significant figure.
Answer (b)(ii):
sin 78.6 =
D
O
H
78.6
sin 78.6 = 4.5
r
r =
r
4.5
sin 78.6
= 4.590566485
B
4.5
M
4.5
C
≈ 4.59
(c) Calculate the area of the triangle ABC as a percentage of area of the circle.
Area of triangle ABC × 100
Area of circle
1
× 8 × 6 × sin 78.6
2
π × (4.59)2
35.54530377 %
× 100
[Area of triangle =
1
× a × b × sin C]
2
Answer: (c)
35.5
% [4]
119
Workbook of Variant 3
October / November 2011 Paper 23
14
r
2r
The sphere of radius r fits exactly inside the cylinder of radius r and height 2r.
Calculate the percentage of the cylinder occupied by the sphere.
[The volume, V, of a sphere with radius r is V =
Area of sphere
× 100
Area of cylinder
4
π × r3
3
π × r2 × 2r
4
3
2
[
4
π r 3]
3
× 100
× 100
4
÷ 2] × 100
3
= 66.66666667
Answer:
19
66.7
A
O
50°
9 cm
B
The diagram shows a sector AOB of circle, centre O, radius 9 cm with angle AOB = 50°.
Calculate the area of the segment shaded in the diagram.
120
% [3]
Mensuration
Area of shaded segment = area of segment − area of triangle
=
θ
× πr 2
360
−
1
× a × b × sin C
2
=
50
× π ×92
360
−
1
× 9 × 9 × sin 50
2
= 4.318117407
4.32
Answer:
cm2
[4]
October / November 2011 Paper 43
1
0.8 m
0.5 m
1.2 m
1.2 m
d
0.4 m
A rectangular tank measures 1.2 m by 0.8 m by 0.5 m .
(a) Water flows from the full tank into a cylinder at rate of 0.3 m3/min.
Calculate the time it takes for the full tank to empty.
Give your answer in minutes and seconds.
Volume = Rate × Time × something
[something depends on unit of rate]
Volume = rate × time × __________
m3 = m3/min × min
[equation balanced]
1.2 × 0.8 × 0.5 = 0.3 × t
1.2 × 0.8 × 0.5
=t
0.3
121
Workbook of Variant 3
t = 1.6 min.
1.6 min = (1 + 0.6) min
0.6 min = 0.6 × 60 sec
= 36 sec
[To convert min to sec.]
Answer: (a)
1
min,
36
[3]
(b) The radius of the cylinder is 0.4 m.
Calculate the depth of water, d, when all the water from the rectangular tank is in the
cylinder.
Volume of rectangular tank = Volume of cylinder
1.2 × 0.8 × 0.5 = πr2d
1.2 × 0.8 × 0.5 = π(0.4)2d
1.2 × 0.8 × 0.5
=d
π(0.4)2
0.9549296586 = d
d = 0.9549296586
Answer: (b) d =
0.954
m. [3]
(c) The cylinder has a height of 1.2 m and is open at the top.
The inside surface is painted at cost of $ 2.30 per m2.
Calculate the cost of painting the inside surface.
Inside surfce area of cylinder = πr2 + 2πrh
= π(0.4)2 + 2π(0.4)(1.2)
= 3.518583772 m2
m2
$
1
2.30
3.518583772
x
x = 3.518583772 × 2.30
= 8.092742676
Answer: (c) $
122
8.09
[4]
Mensuration
May / June 2012 Paper 23
16
4 cm
15 cm
The diagram shows a solid prism of length 15 cm.
The cross section of the prism is a semi circle of radius 4 cm.
Calculate the total surface area of the prism.
area of cross section (semi circle) =
=
πr2
2
π(4)2
2
= 8π
perimeter of cross section =
2πr
+r+r
2
=
2π(4)
+4+4
2
4π + 8
=
Total surface area of prism = 2(area of cross section) + perimeter × length
= 2(8π) + (4π + 8) × 15
= 358.7610417
Answer:
359
cm2
[4]
123
Workbook of Variant 3
May / June 2012 Paper 43
5
3 cm
3 cm
6 cm
8 cm
The diagram shows two solid spheres of radius 3 cm lying on the base of cylinder of radius 8 cm.
Liquid is poured into the cylinder until the spheres are just covered.
[The volume, V, of a sphere with radius r is V =
4
π r 3]
3
(a) Calculate the volume of liquid in the cylinder in
(i) cm3,
Volume of liquid in the cylinder = volume of cylinder with height 6 − volume of 2 spheres.
( 43 πr )
4
= π(8) (6) − 2(
×π×3)
3
= π r 2h − 2
2
3
3
= 980.1769079
Answer: (a) (i)
980
cm2
[4]
litres
[1]
(ii) litres.
980 cm3 =
980
= 0.980
1000
[1000 cm3 = 1 litre]
Answer: (a) (ii)
124
0.98
Mensuration
(b) One cubic centimetre of the liquid has a mass of 1.22 grams.
Calculate the mass of the liquid in the cylinder.
Give your answer in kilograms.
cm3
grams.
1
1.22
980.1769079
x
x = 980.1769079 × 1.22
x = 1195.815825 gram
= 1195.815828 gram
= 1195.815828
1000
= 1.195815828 kg
Answer: (b)
1.2
kg [2]
(c) The spheres are removed from the cylinder.
Calculate the new height of the liquid in the cylinder.
METHOD 1:-
( 43 πr ) = πr h
4
2(
π × 3 ) = π (8) h
3
2
3
2
3
2
72π = π82h
72π
=h
π82
h = 1.125
new height = 6 − 1.125
= 4.875
METHOD 2 :-
980.1769079 = πr2h
980.1769079 = π(8)2h
980.1769079
=h
π(8)2
h = 4.875
Answer: (c)
4.88
cm [2]
125
Workbook of Variant 3
October / November 2012 Paper 23
17
B
5r
4r
O
A
The diagram shows a sector of a circle, centre O, radius 5r.
The length of the arc AB is 4r.
Find the area of the sector in terms of r, giving your answer in its simplest form.
arc length =
4r =
θ
× 2πr
360
θ
× 2π(5r)
360
4r × 360
=θ
2π(5r)
144
=θ
π
area of sector =
θ × πr 2
360
=
144 ÷ π
× π(5r)2
360
=
144 ÷ π
× π × 25 × r2
360
= 10r2
Answer:
October / November 2012 Paper 43
3
A metal cuboid has a volume of 1080 cm3 and a mass of 8 kg.
(a) Calculate the mass of one cubic centimetre of the metal.
126
10r2
[3]
Mensuration
Give your answer in grams.
cm2
gram
1080
8000
1
x
8000
1080
x =
= 7.4074070407
Answer: (a)
7.41
g [1]
(b) The base of the cuboid measures 12 cm by 10 cm.
Calculate the height of the cuboid.
Volume of cuboid = l × b × h
1080 = 12 × 10 × h
1080
=h
12 × 10
9 = h
Answer: (b)
9
cm [2]
(c) The cuboid is melted down and made into a sphere with radius r cm.
(i) Calculate the value of r.
[The volume, V, of a sphere with radius r is V =
4
π r 3]
3
Volume of cuboid = volume of sphere
1080 =
4
π r3
3
1080 × 3
=r
4×π
1080 × 3
=r
4×π
6.364706508 = r
r = 6.364706508
Answer: (c) (i) r =
6.36
[3]
(ii) Calculate the surface area of the sphere.
[The surface area, A, of a sphere with radius r is A = 4 πr2]
Surface area of sphere = 4πr2
127
Workbook of Variant 3
= 4π(6.364706508)2
= 509.0572513
Answer: (c) (ii)
509
cm2
[3]
(d) A larger sphere has a radius R cm.
The surface area of this sphere is double the surface area of the sphere with radius r cm in
part (c).
Find the value of
METHOD 1
R
.
r
As surface area of sphere with radius R is double the surface area of the radius r in part(c),
area factor of these 2 spheres is 2:1
( ) = 21
R
r
2
R
=
r
R
= 2
r
2
1
METHOD 2
( ) = 2(509.0572513)
509.0572513
R
2
(
=
)
r
1
R
r
2
2
R
= 2
r
Answer: (d)
128
2
[2]
Mensuration
May / June 2013 Paper 23
4 cm
7
5 cm
10 cm
18 cm
4 cm
The shaded shape has rotational symmetry of order 2. Work out the
shaded area.
5 cm
= 2[10 × 4 + 18 × 5]
10 cm
5 cm
= 260
18 cm
Answer:
260
cm2
[3]
cm2
[4]
18
B
A
5 cm
120°
O
A and B lie on a circle, centre O, radius 5 cm. Angle AOB = 120o.
Find the area of the shaded segment.
Area of shaded segment = area of sector − area of triangle
=
θ × πr2 − 1 × a × b × sin C
2
360
1
= 120 × π(5)2 −
× 5 × 5 × sin 120
2
360
= 15.35462123
Answer:
15.4
129
Workbook of Variant 3
May / June 2013 Paper 43
I
4
H
J
F
40 cm
7 cm
E
G
22 cm
EFGHIJ is solid metal prism of length 40 cm.
The cross section EFG is a right angled triangle. EF = 7 cm and EG = 22 cm.
(a) Calculate the volume of the prism.
volume of prism = area of cross section × length
=
1
× b × h × length
2
=
1
× 22 × 7 × 40
2
= 3080
3080
Answer: (a)
cm3
[2]
(b) Calculate the length FJ.
EJ2 = EG2 + GJ2 [By pythagoras theorem]
EJ2 = 222 + 402
4
J
F
EJ = 222 + 402
Base
EJ = 45.65084884
40cm
FJ2 = EJ2 + EF2
7cm
FJ2 = 45.650848842 + 72
FJ = 45.650848842 + 72
FJ = 46.18441296
E
22cm
G
Answer: (b) FJ =
130
J
E
46.2
cm [4]
Mensuration
(c) Calculate the angle between FJ and the EGJH of the prism.
sin θ =
O
H
sin θ =
7
46.2
θ = sin–1
θ
=
(
F
7
46.2
(0)
46.2
7cm
)
8.714742019
(4)
J
E
Answer: (c)
8.7
[3]
(d) The prism is melted and made into spheres.
Each sphere has a radius 1.5 cm.
Work out the greatest number of spheres that can be made.
[The volume, V, of a sphere with radius r is V =
number of spheres =
4
π r 3]
3
volume of cuboid
volume of sphere
3080
=
4
× π 1.53
3
= 217.8654332
Answer: (d)
217
[3]
(e) (i) A right angled triangle is the cross section of another prism.
This triangle has height 4.5 cm and base 11.0 cm.
Both measurements are correct to 1 decimal place.
Calculate the upper bound for the area of this triangle.
1 decimal place = 0.1
0.1 ÷ 2 = 0.05
4.5 + 0.05 = 4.55
(upper bound of height)
11.0 + 0.05 = 11.5
(upper bound of base)
upper bound for area =
=
1
×b×h
2
1
4.55 × 11.05
2
= 25.13875
Answer: (e) (i)
25.13875cm2
cm2
[2]
131
Workbook of Variant 3
(ii) Write your answer to part(e)(i) correct to 4 significant figures.
25.14 cm2
Answer: (e) (ii)
cm2
[1]
)
[2]
October / November 2013 Paper 23
2
Three of the vertices of a parallelogram are at (4, 12), (8, 4) and (16, 16).
y
24
20
16
12
8
4
–8
–4
0
4
8
12
16
20
24
x
Write down the co-ordinates of two possible positions of the fourth vertex.
(–4, 0) is also possible.
Answer:
(
20
8
2
) and (
16 The diagram shows the entrance to a tunnel.
The circular arc has a radius of 3 m and centre O.
AB is horizontal and angle AOB = 120o.
O
3m
A
132
120°
3m
B
24
Mensuration
During a storm the tunnel filled with water, to the level shown by the shaded area in the diagram.
(a) Calculate the shaded area.
Shaded area = area of triangle + 2(area of sector)
( 360θ × πr )
30
1
=
× 3 × 3 × sin 120 + 2 (
× π(3) )
360
2
=
1
× a × b × sin C + 2
2
2
2
= 8.609503297
Answer: (a)
8.61
m2
[4]
m3
[1]
(b) The tunnel is 50 m long.
Calculate the volume of water in the tunnel.
Volume of water in the tunnel = area of cross section × length
= 8.609503297 × 50
= 430.4751649
Answer: (b)
430
October / November 2013 Paper 43
6
Sandra has designed this open container.
The height of the container is 35 cm.
35 cm
The cross section of the container is designed from three semi-circles with diameters 17.5 cm,
6.5 cm and 24 cm.
17.5 cm
6.5 cm
133
Workbook of Variant 3
6.5 cm
17.5 cm
(a) Calculate the area of the cross section of the container.
1
17.5 cm
→ area of shape 1 =
π (8.75)2
= 120.2640938
2
→ area of shape 2 =
π (12)2
π (3.25)2
–
= 209.6031349
2
2
6.5 cm
2
1
Area of cross section = 120.2640938 + 209.631349
3
= 329.8672287
6.5 cm
2
17.5 cm
[r = 8.75]
Answer: (a)
330
cm2
[3]
cm2
[4]
litres
[3]
(b) Calculate the external surface area of the container, including the base.
1
perimeter
=
3
2
2π(8.75)
2π(12)
2π(3.25)
+
+
2
2
2
perimeter = 24π
External surface area = perimeter × height + area of base
= 24π × 35 + 329.8672287
= 2968.805058
Answer: (b)
2970
(c) The container has a height of 35 cm.
Calculate the capacity of the container.
Give your answer in litres.
Capacity of the container = area of cross section × height
= 329.8672287 × 35
= 11545.353 cm3
11545.353 cm3 =
11545.353
litre
1000
= 11.545353
Answer: (c)
134
11.5
Mensuration
(d) Sandra’s container is completely filled with water.
All the water is then poured into another container in the shape of a cone.
The cone has radius 20 cm and height 40 cm.
20 cm
r
40 cm
h
(i) The diagram shows the water in the cone.
Show that r =
h
.
2
Answer (d)(i):
h
r
=
40
20
r =
h
× 20
40
r =
h
2
(ii) Find the height, h, of the water in the cone.
[The volume, V, of a cone with radius r and height h is V =
1
πr2h]
3
Water of old container = Water in this cone
11545.353 =
1
π r 2h
3
11545.353 =
1
h
π
3
2
( )h
2
11545.353 × 22 × 3
= h3
π
h =
3
11545.353 × 22 × 3
π
h = 35.33020818
Answer: (d) (ii) h =
35.3
cm [3]
135
Workbook of Variant 3
May / June 2014 Paper 23
21
The diagram shows two concentric circles and three radii.
The diagram has rotational symmetry of order 3.
A club uses the diagram for its badge with some sections shaded.
The radius of the large circle is 6 cm and the radius of the small circle is 4 cm.
Calculate the total perimeter of the shaded area.
1
perimeter =
=
2
3
4
θ
2πr
× 2πr +
+r+r
360
2
240
2π(6)
× 2π(6) +
+6+6
360
2
= 62.26548246
Answer:
136
62.3
cm [5]
Mensuration
May / June 2014 Paper 43
10
8 cm
r cm
(a)The three sides of an equilateral triangle are tangents to a circle of radius r cm.
cmthe triangle are 8 cm long.
The sides8of
Calculate the value of r.
cm correct to 1 decimal place.
Show that it rounds tor 2.3,
Answer (a):
O
A
tan 30 =
tan 30 = r
4
rcm
30
4 × tan 30 = r
4cm
2.309401077 = r
r = 2.309401077
8 cm
≈ 2.3
Answer: (a)
2.3
[3]
12 cm
(b)
8 cm
12 cm
The diagram shows a box in the shape of a triangular prism of height 12 cm.
The cross section is an equilateral triangle of side 8 cm.
137
Workbook of Variant 3
Calculate the volume of the box.
Volume of box = area of cross section × height
( 12 × a × sin 60) × height
1
= (
×8 × sin 60) × 12
2
=
2
2
= 332.5537551
Answer: (b)
333
cm3
[4]
(c) The box contains biscuits.
Each biscuit is a cylinder of radius 2.3 centimetres and height 4 millimetres.
Calculate
(i) the largest number of biscuits that can be placed in the box,
no. of biscuits =
height of box
height of each biscuit
=
120
= 30
4
30
Answer: (c) (i)
[3]
(ii) the volume of one biscuit in cubic centimetres,
4 millimetres = 0.4 centimetres
volume of one biscuit = πr2h
= π(2.3)2(0.4)
= 6.647610055
Answer: (c) (ii)
6.65
cm2
[2]
%
[3]
(iii) the percentage of the volume of the box not filled with biscuits.
volume of box not filled with biscuits = 332.5537551 − 30(6.64761005)
= 133.1254535
percentage of the volume of box not filled with biscuits
=
133.1254535
× 100
332.5537551
= 40.03125854
Answer: (c) (iii)
138
40
Mensuration
October / November 2014 Paper 23
18
6 cm
15 cm
The diagram shows a glass, in the shape of a cone, for drinking milk.
The cone has a radius of 6 cm and height 15 cm.
A bottle of milk holds 2 litres.
(a) How many times can the glass be completely filled from the bottle?
[The volume, V, of a cone with radius r and height h is V =
2 litres = 2000cm3
Number of glass can be filled completely =
1
πr2h.]
3
volume of bottle of milk
volume of one glass
2000
=
1
× π(6)2 × 5
3
= 3.536776513
3
Answer: (a)
[4]
(b) Calculate the volume of milk left in the bottle.
Give your answer in cm3.
volume of milk left = Total volume of milk in bottle − volume of 3 glass
( 13 πr h)
1
= 2000 − 3 (
× π (6) × 12)
3
= 2000 − 3
2
2
= 303.5399671
Answer: (b)
304
cm3
[3]
139
Workbook of Variant 3
October / November 2014 Paper 43
1
(a) ABCD is a trapezium.
11 cm
A
B
4.7 cm
D
C
2.6 cm
17 cm
(i) Calculate the length of AD.
AD2 = 2.62 + 4.72
AD = 2.62 + 4.72
AD = 5.371219601
5.37
Answer: (a) (i) AD =
cm [2]
(ii) Calculate the size of angle BCD.
3.4 = 17 − 2.6 – 11
tan θ =
O
A
tan θ =
O
A
tan θ =
4.7
3.4
B
(0)
4.7 cm
3.4
A
4.7
θ = tan
3.4
C
θ = 54.11786275
Answer: (a) (ii) angle BCD =
54.1
[3]
(iii) Calculate the area of the trapezium ABCD.
Area of trapezium =
=
1
(a + b) × h
2
1
(17 + 11) × 4.7
3
= 65.8
Answer: (a) (iii)
140
65.8
cm2
[2]
Mensuration
(b) A similar trapezium has perpendicular height 9.4 cm.
Calculate the area of this trapezium.
( )=
9.4
4.7
2
x
65.8
2
x
9.4 =
2
65.8
4.7
2
65.8 × 9.4 = x
4.72
263.2 = x
x = 263.2
263
Answer: (b)
cm2
[3]
7
P
A
6.5 cm
56°
8 cm
8 cm
O
C
x
B
Q
The diagram shows a triangle and a sector of a circle.
In triangle ABC, AB = AC = 8 cm and angle BAC = 56o.
Sector OPQ has centre O, sector angle x and radius 6.5 cm.
(a) Show that the area of triangle ABC is 26.5 cm2 correct to 1 decimal place.
Answer (a):
Area of triangle =
=
1
× a × b × sin C
2
1
× 8 × 8 × sin 56
2
= 26.52920232
Answer: (a)
26.52920232
[2]
141
Workbook of Variant 3
(b) The area of sector OPQ is equal to the area of triangle ABC.
(i) Calculate the sector angle x.
Area of sector = area of triangle
θ
× πr2 = 26.52920232
360
x
× π6.52 = 26.52920232
360
x = 360 × 26.52920232
π × 6.52
x = 71.9531989
72
Answer: (b) (i)
[3]
(ii) Calculate the perimeter of the sector OPQ.
Perimeter of the sector = Arc length + r + r
=
θ
× 2πr + r+ r
360
=
71.9531989
× 2π(6.5) + 6.5 + 6.5
360
= 21.16283148
21.1
Answer: (b) (ii)
(c) The diagram shows a sector of a circle, radius r cm.
r cm
30°
(i) Show that the area of the shaded segment is
142
(
)
1 2 1
r
π − 1 cm2
4
3
cm [3]
Mensuration
Answer (c)(i):
area of shaded segment =
area of sector − area of triangle
=
θ
1
× π× r2 −
× a × b × sin C
360
2
=
30
1
× πr 2 −
r × r × sin 30
360
2
=
1
1 2
1
× πr 2 −
r ×
12
2
2
=
1
1
× r2
π−1
4
3
(
)
Answer: (c) (i)
1 2 1
π−1
r
4
3
(
)
[4]
(ii) The area of the segment is 5 cm2.
Find the value of r.
Find the value of r.
(
)
1 2 1
r
π-1 = 5
4
3
r2 =
5×4
( 13 π −1)
r =
5×4
( 13 π−1)
r = 20.58520824
Answer: (c) (ii) r =
20.6
[3]
May / June 2015 Paper 23
15 The circumference of a circle is 30 cm.
(a) Calculate the radius of the circle.
circumference of circle = 2πr
30 = 2πr
30
=r
2π
r =
30
2π
r = 4.774648293
Answer: (a)
4.77
cm [2]
143
Workbook of Variant 3
(b)
The length of the arc of the semi-circle is 15 cm.
Calculate the area of the semi-cicle.
Arc length of the semi-circle is 15, so circumference of a circle is 30 cm
∴ r = 4.774648293
Arc of semi circle =
π r2
2
2
= π (4.774648293)
2
= 35.8098622
Answer: (b)
May / June 2015 Paper 43
35.8
cm2
[2]
45 cm
8 (a) A cylinder tank contains 180 000 cm3 of water.
The radius of the tank is 45 cm.
Calculate the height of water in tank.
Volume of cylinder tank = πr2h
180000 = π(45)2 h
45 cm
180000
=h
π(45)2
h = 28.29421211
Answer: (a)
(b)D
C
70 cm
40 cm
A
D
B
50 cm
40 cm
C 150 cm
70 cm
150 cm
A
144
50 cm
B
28.3
cm [2]
Mensuration
The diagram shows an empty tank in the shape of a horizontal prism of length 150 cm.
The cross section of the prism is an isoscels trapezium ABCD.
AB = 50 cm, CD = 70 cm and the vertical height of the trapezium is 40 cm.
(i) Calculate the volume of the tank.
Volume of prism = area of cross section × length
= [
1
(a + b) × h] × length
2
=
1
(50 + 70) × 40 × 150
2
= 360000
360000
Answer: (b) (i)
cm3
[3]
litre
[1]
(ii) Write your answer to part (b)(i) in litres.
360000 cm3 =
360000
1000
= 360 litre
360
Answer: (b) (ii)
(c) The 180 000 cm3 of water flows from the tank in part (a) into the tank in part (b) at rate of
15 cm3/s.
Calculate the time this takes.
Give answer in hours and minutes.
Volume = Rate x time x something.
[Something depends on unit of rate]
Volume = Rate x time x __________
cm3 = cm3/s x s
[equation balanced]
180000 = 15 x t
12000 = t
t = 12000 seconds
12000 s =
3
12000
10
1
=
=3
hours
60 x 60
3
3
1
1
hours = (3 +
) hours
3
3
1
hours = 20 min
3
3 hours, 20 min.
Answer: (c)
3
h 20
min
[3]
145
Workbook of Variant 3
(d)
70 cm
D
x cm
F
40 cm
C
E
h cm
A
B
50 cm
The 180 000 cm3 of water reached the level EF as shown above.
EF = x cm and the height of the water is h cm.
(i) Using the properties of similar triangles, show that h = 2 (x − 50).
Answer (d)(i):
h
40
=
h =
h =
C
1
(x − 50)
2
10
1
(x − 50)
2
10
E
40
h
× 40
B
1
(x − 50) × 4
2
h = (x − 50) × 2
1 (x - 50)
2
10
h = 2(x − 50)
Answer: (d) (i)
2(x − 50)
cm3
[2]
(ii) Using h = 2(x − 50), show that the shaded area, in cm2, is x2 − 2500.
Answer (d)(ii):
Shaded area =
1
(a + b) h
2
1
(x + 50) × 2 × (x − 50)
2
= (x + 50) (x − 50)
=
= x2 − 502
= x2 − 2500
Answer: (d) (ii)
146
x2 − 2500
[1]
Mensuration
(iii) Find the value of x.
Volume = area of cross section × length
180000 = (x2 − 2500)150
180000
= x2 − 2500
150
1200 = x2 − 2500
1200 + 2500 = x2
x = 1200 +2500
x = 60.8276253
Answer: (b)(iii) x =
60.8
Answer: (d)(iv) h =
21.7
[2]
(iv) Find the value of h.
h = 2(x − 50)
h = 2(60.8276253 − 50)
h = 21.65525061
[1]
October / November 2015 Paper 23
25
B
8 cm
O
30°
C
A
OAB is the sector of a circle, centre O, with radius 8 cm and sector angle 30o.
BC is perpendicular to OA.
Calculate the area of the region shaded on the diagram.
147
Workbook of Variant 3
B
cos 30 = OC 8 × cos 30 = OC
OB
COS θ = A
H
cos 30 = OC OC = 6.92820323
8
shaded area =
area of sector −
(4)
8cm
area of triangle.
=
θ
1
× π× r2 −
× a × b × sin C
360
2
=
30
1
× π × (8)2 −
× (6.92820323) × (8) × sin 30
360
2
O
30
C
(A)
= 2.898754359
Answer:
2.9
cm2
October / November 2015 Paper 43
3
The diagram shows a horizontal water trough in the shape of a prism.
35 cm
12 cm
6 cm
120 cm
25 cm
The cross section of this prism is a trapezium.
The trapezium has parallel sides of length 35 cm and 25 cm and a perpendicular height of 12 cm.
The length of the prism is 120 cm.
(a) Calculate the volume of the trough.
Volume of prism = area of cross section × length
=
1
(a + b) × h × length
2
=
1
(25 + 35) × 12 × 120
2
= 43200
148
[5]
Mensuration
Answer: (a)
43200
cm2
[3]
cm3
[2]
(b) The trough contains water to a depth of 6 cm.
(i) Show that volume of water is 19 800 cm3.
Answer (b)(i):
here b =
25 + 35
= 30
2
Volume of the water = (area of cross section of water) × length
=
1
(25 + 30) × 6 × 120
2
= 19800 cm3
Answer: (b) (i)
19800
(ii) Calculate the percentage of the trough that contains water.
=
volume of water
× 100
volume of trough
=
19800
× 100
43200
= 45.83333333
Answer: (b) (ii)
45.8
$ [1]
(c) The water is drained from the trough at rate of 12 litres per hours.
Calculate the time it takes to empty the trough.
Give your answer in hours and minutes.
Volume = Rate x time x something.
[Something depends on unit of rate]
Litre = litre/hour x hour
(Never change unit of rate, instead change unit of valueme accordingly)
19800 cm3 =
19800
litre
1000
= 19.8 litre
Volume = rate x time
19.8 = 12 x t
[equation balanced]
19.8
=t
12
t = 1.65 hours
149
Workbook of Variant 3
1.65 = 1 + 0.65 hours
0.65 hours = 0.65 x 60 min
39 min
Answer: (c)
1
39
h min
[4]
(d) The water from the trough just fills a cylinder of radius r cm and height 3r cm.
Calculate the value of r.
Volume of cylinder = πr2h
19800 = π(r)2(3r)
19800 = 3πr3
19800
= r3
3π
r =
3
19800
3π
r = 12.800750953
Answer: (d) r =
12.8
[3]
(e) The cylinder has a mass of 1.2 kg.
1 cm3 of water has mass of 1 g.
Calculate the total mass of the cylinder and the water.
Give your answer in kilograms.
1 cm3 of water has mass of 1 g
19800 cm3 of water will have mass of 19800 gram.
19800
= 19.8 kg.
1000
19800 grams =
Total mass of cylinder and water = 1.2 + 19.8
= 21
Answer: (e)
150
21
kg [2]
4
Workbook of
Solution 4
GEOMETRY
Geometry
Chapter 4: Geometry
May / June 2010 Paper 23
13 (a) Write down the number of lines of symmetry for the diagram below.
Answer (a):
0
[1]
2
[1]
(b) Write down the order of rotational symmetry for the diagram below.
Answer (b):
(c) The diagram shows a cuboid which has no square faces.
Draw one of the planes of symmetry of the cuboid on the diagram.
[1]
153
Workbook of Variant 3
22
B
A
C
The diagram shows a farmer’s field ABC.
The farmer decides to grow potatoes in the region of the field which is
•
nearer to A than to C [Draw perpendicular bisector of AC]
and
•
nearer to AB than to AC
[Draw angle bisector of A]
Using a straight edge and compasses only, construct two loci accurately and
shade this region on the diagram.
154
[5]
Geometry
May / June 2010 Paper 43
5
(a)
A
3 cm
Q
4 cm
P
B
3.6 cm
C
The diagram shows two triangles ACB and APQ.
Angle PAQ = angle BAC and angle AQP = angle ABC.
AB = 4 cm, BC = 3.6 cm and AQ = 3 cm.
(i) Complete the following statement.
Triangle ACB is
Similar
to triangle APQ [1]
(ii) Calculate the length of PQ.
[Triangle ACB is similar to APQ, so sides of triangle will be in propotion]
AQ
PQ
=
AB
BC
PQ
3
=
3.6
4
PQ =
3 × 3.6
4
PQ = 2.7
Answer: (a)(ii) PQ =
2.7
cm [2]
155
Workbook of Variant 3
(iii) The area of triangle ACB is 5.6 cm2.
Calculate the area of triangle APQ.
( AQ
)
AB
( 34 )
2
2
=
=
Area of triangle APQ
Area of triangle ACB
x
5.6
32 × 5.6
=x
42
x = 3.15
Answer: (a) (iii)
(b)
H
31.5
cm2 [2]
S
R
O
U
M
61°
T
N
R, H, S, T and U lie on a circle, centre O.
HT is a diameter and MN is a tangent to the circle at T.
Angle RTM = 61o.
Find
(i) angle RTH,
Angle HTM = 90,
∴ angle RHT = 90 − 61 = 29
Answer: (b) (i) Angle RHT =
29
[1]
Answer: (b) (ii) Angle RHT =
61
[1]
(ii) angle RHT,
Angle HRT = 90,
∴ Angle RHT + 90 + 29 = 180
∴ angle RHT = 61
156
Geometry
(iii) angle RST,
Angle RHT = angle RST (angle subtended by same arc)
Answer: (b) (iii) Angle RST =
61
[1]
(iv) angle RUT,
Angle RUT = 180 − 61 = 119
(Opposite angle of cyclic quadrilateral add upto 180)
Answer: (b) (iv) Angle RUT =
119
[1]
(c) ABCDEF is hexagon.
The interior angle B is 4° greater than interior angle A.
The interior angle C is 4° greater than interior angle B, and so on, with each of the next
interior
angles 4° greater than the previous one.
(i) By how many degrees is interior angle F greater than interior angle A?
Angle A
=
x
Angle B
=
x+4
Angle C
=
x+8
Angle D
=
x + 12
Angle E
=
x + 16
Angle F
=
x + 20
Answer: (c) (i)
20
[1]
(ii) Calculate interior angle A.
sum of all interior angles of hexagon = (6 − 2) 180 [Sum of intenior angle of polygon is
= 720(n - 2) 180]
x + x + 4 + x + 8 + x + 12 + x + 16 + x + 20 = 720
6x + 60 = 720
6x = 720 − 60
6x = 660
x =
660
6
x = 110
Answer: (c) (ii)
110
[3]
157
Workbook of Variant 3
October / November 2010 Paper 23
9
A
8 cm
Q
P
10 cm
C
12 cm
B
APB and AQC are straight lines. PQ is parallel to BC.
AP = 8 cm, PQ = 10 cm and BC = 12 cm.
Calculate the length of AB.
AB
= BC
AP
PQ
12
AB
=
10
8
AB =
12 × 8
10
AB = 9.6
Answer: AB =
9.6
[2]
19 A model of a car is made to a scale of 1 : 40.
The volume of the model is 45 cm3.
Calculate the volume of the car.
Give your answer in m3.
( 401 )
3
=
45
[In similar triangles, (length ratio)3 = Volume ratio]
x
13
45
=
3
x
40
x = 45 × 403
x = 2880000 cm3
2880000 cm3 =
2880000
m3
3
100
Answer:
158
2.88
m3
[3]
Geometry
23
A
D
50°
O
86°
30°
C
B
The points A, B, C and D lie on the circumference of the circle, centre O.
Angle ABD = 30o, angle CAD = 50o and angle BOC = 86o.
(a) Give the reason why angle DBC = 50o.
Answer: (a)
Angle substended by same arc
[1]
(b) Find
(i) angle ADC,
Angle ACD = 30°[Angle subtended by same arc]
In triangle ACD; 50 + 30 + x = 180
x = 100
Answer: (b) (i) Angle ADC
100
[1]
(ii) angle BDC,
Angle BDC = 86 ÷ 2 [The angle at the centre of a circle is twice the angle on circimference]
= 43
Answer: (b) (ii) Angle BDC =
43
[1]
159
Workbook of Variant 3
(iii) angle OBD.
Angle OBC =
(180 − 86)
= 47[Isosceles triangle]
2
Angle DBC = Angle OBD + Angle OBC
50 = Angle OBD + 47
Angle OBD = 50 − 47
= 3
Answer: (b) (iii) Angle OBD =
3
[2]
October / November 2010 Paper 43
5
C
D
SHOP
B
A
The diagram shows an area of land ABCD used for a shop, a car park and gardens.
(a) Using a straight edge and compasses only, construct
(i) the locus of points equidistant from C and from D,
[Draw perpendicular bisector of line CD][2]
(ii) The locus of points equidistant from AD and from AB.
[Draw angle bisector of angle DAB][2]
(b) The shop is on the land nearer to D than to C and nearer to AD than to AB.
160
Write the word SHOP in this region on the diagram.
[1]
Geometry
(c) (i) The scale of the diagram is 1 centimetre to 20 metres.
The gardens are the part of the land less than 100 m from B.
Draw the boundary for the gardens.
cm
1
x
x =
m
20
100
100
= 5 cm
20
[Draw arc with centre B, radius 5 cm inside ABCD][1]
(ii)The car park is the part of the land not used for the shop and not used for the gardens.
Shade the car park region on the diagram.
[1]
May / June 2011 Paper 23
20
V
U
W
g°
70°
O
h°
X
e°
f°
A
B
T
The diagram shows a circle, centre O.
VT is a diameter and ATB is a tangent to the circle at T.
U, V, W and X lie on the circle and angle VOU = 70o.
Calculate the value of
(a) e,
Angle UOT = 180 − 70 = 110
Angle e =
[Angle VOU and angle VOT are supplementory]
180 − 110
= 35
2
[Triangle VOT is Isosceles triangle]
Answer: (a) e =
35
[1]
161
Workbook of Variant 3
(b) f,
Angle f = 90 − e = 90 − 35 = 55
Answer: (b) f =
55
[1]
(c) g,
Angle UOT = 110
∴ angle UWT =
110
= 55 [Angle at centre of a circle is twice the angle on circumference]
2
55
Answer: (c) g =
[1]
(d) h,
h + g = 180
[Opposite angles of a cyclic quadrilateral add upto 180]
h + 55 = 180
h = 180 − 55
h = 125
Answer: (d) h =
125
[1]
17
20 cm
10 cm
9 cm
d cm
The diagrams show two mathamatically similar containers.
The larger container has a base with diameter 9 cm and height 20 cm.
The smaller container has base with diameter d cm and height 10 cm.
(a) Find the value of d.
10
d
=
[In similar shapes, sides are in proportion]
20
9
d =
10 × 9
20
d = 4.5
Answer: (a) d =
162
4.5
[1]
Geometry
(b) The larger container has a capacity of 1600 ml.
Calculate the capacity of the smaller container.
x
( ) = 1600
10
20
3
103
x
=
203
1600
103 × 1600
=d
203
[Capacity refers to volume. (length ratio)3 = volume ratio]
200 = d
d = 200
Answer: (b)
200
ml
[2]
May / June 2011 Paper 43
8
D
C
Q
A
B
(a)Draw accurately the locus of points, inside the quadrilateral ABCD, which are
6 cm from the point D.
[Draw arc with radius 6cm, centre D]
[1]
(b) Using a straight edge and compasses only, construct
(i) the perpendicular bisector of AB,
[2]
(ii) the locus of points, inside the quadrilateral, which are equidistant from AB and from BC.
[Draw angle bisector of angle ABC]
[2]
(c) The point Q is equidistant from A and from B and equidistant from AB and from BC.
(i) Label the point Q on the diagram.
[Point of intersection of Perpendicitor bisector and angle bisector]
[1]
163
Workbook of Variant 3
(ii) Measure the distance of Q from the line AB
Answer: (c) (ii)
2.7
cm [1]
(d) On the diagram, shade the region inside the quadrilateral which is
• less than 6 cm from D.
and
• nearer to A than to B.
and
• nearer to AB than to BC.[1]
October / November 2011 Paper 23
22
A
B
4.40 cm
X
3.84 cm
9.40 cm
D
C
A, B, C and D lie on a circle.
AC and BD intersect at X.
(a) Give reason why angle BAX is equal to angle CDX.
Answer: (a)
Angles subtended by same arc.
[1]
(b) AB = 4.40 cm, CD = 9.40 cm and BX = 3.84 cm.
(i) Calculate the length of CX.
CX
CD
=
BX
AB
9.40
CX
=
4.40
3.84
9.40 × 3.84
= CX
4.40
CX = 8.203636364
Answer: (b) (i) CX =
164
8.2
cm [2]
Geometry
(ii) The area of triangle ABX is 5.41 cm2.
Calculate the area of triangle CDX.
[In similar triangles, (length ratio)2 = area ratio]
Area of triangle CDX
=
( CD
)
AB
Area of triangle ABX
2
9.402 × 5.41
=x
4.402
x = 24.69150826
24.7
Answer: (b) (ii)
cm2
[2]
October / November 2011 Paper 43
3
(a)
D
C
A
40°
B
30°
E
ABCD is a quadrilateral with angle BAD = 40o.
AB is extended to E and angle EBC = 30°.
AB = AD and BD = BC.
(i) Calculate angle BCD.
Angle DBA = Angle ADB(Isosceles triangle)
∴ Angle DBA =
180 − 40
= 70
2
Ange DBA + angle DBC + angle CBE = 180
70 + angle DBC + 30 = 180
Angle DBC = 180 − 70 − 30
Angle DBC = 80
DB = CD
Angle BCD =
180 − 80
2
Angle BCD = 50
Answer: (a) (i) angle BCD =
50
cm [3]
165
Workbook of Variant 3
(ii) Give a reason why DC is not parallel to AE.
Answer: (a) (ii)
Angle BCD is not equal to angle CBE
[1]
(b) A regular polygon has n sides.
5n
degrees.
2
Each exterior angle is
Find the value of n.
[Each extarior angle of regular polygon =
5n
360
=
2
n
360 × 2 = 5n × n
360 × 2 = 5n2
360
]
n
360 × 2
= n2
5
n =
360 × 2
5
n = 12
Answer: (b) n =
(c)
O
A
The diagram shows a circle, centre O.
A, B, and C are points in the circumference.
OC is parallel to AB.
Angle OCA = 25°.
166
C
25°
B
12
[3]
Geometry
Calculate angle OBC.
Angle CAB = 25
[OC is parallel to AB
Angle OAC = 25
[Triangle OAC is an isosceles triangle]
angle OBA = 50 [Triangle AOB is an isosceles triangle]
∴ Alternate angles]
angle AOB = 180 − (50 + 50)
angle AOB = 80
angle AOC =
80
= 40
2
[Angle at centre of a circle is twice the angle at circumfernce]
Let angle OBC = x
Consider triangle ABC,
Angle CAB + angle ABC + angle BCA = 180
25 + 50 + x + 40 = 180
x = 180 − 25 − 50 − 40
x = 65
Answer: (c) angle OBC =
65
[3]
May / June 2012 Paper 23
1
A
73°
B
C
120°
82°
x°
D
E
The diagram shows a quadrilateral ABCD.
CDE is a straight line.
167
Workbook of Variant 3
Calculate the value of x.
82 + 120 + 73 + Angle ADC = 360
Angle ADC = 360 − 82 − 120 − 73
Angle ADC = 85
Angle ADC + x = 180
85 + x = 180
x = 180 − 85
x = 95
Answer: x =
May / June 2012 Paper 43
8
(a)
B
C
88°
A
u°
Y
68°
w°
O
v°
D
E
A, B, C, D and E lie on the circle, centre O.
CA and BD intersect at Y.
Angle DCA = 88o and angle CYD = 68o.
Angle BAC = uo, angle AED = vo and reflex angle AOD = wo.
Calculate the value of u, v and w.
88 + 68 + Angle CDY = 180
Angle CDY = 180 − 88 − 68
Angle CDY = 24
Angle CDY = u [Angle subtended by same arc]
88 + v = 180
v = 180−88
v = 92
168
S
R
X
Q
95
[2]
v°
D
E
Geometry
DOA = 88 × 2 = 176
W = 360 − 176 = 184
Answer: (a) u =
24
v=
92
w=
184
[4]
(b)
S
R
X
Q
P
P, Q, R and S lie on the circle. PR and QS intersect at X.
The area of triangle RSX = 1.2 cm2 and PX = 3 SX.
Calculate the area of triangle PQX.
PX = 3SX
3
PX
=
1
SX
[In similar triangle, length ratio2 = area ratio]
of triangle PQX
( PXSX ) = Area
Area of triangle RSX
3
x
( ) =
1
1.2
2
2
1.2 × 32 = x
x = 10.8
Answer: (b)
10.8
cm2
[2]
169
Workbook of Variant 3
(c)
F
G 4x°
J
x°
2x°
O
H
I
K
GI is a diameter of the circle.
FGH is a tangent to the circle at G.
J and K also lie on the circle.
Angle JGI = xo, angle FGJ = 4xo and angle KGI = 2xo.
Find
(i) the value of x,
4x + x = 90
5x = 90
x =
90
5
x = 18
Answer: (c) (i) x =
18
Answer: (c) (ii) angle JKG =
72
[2]
(ii) the size of angle JKG,
Angle JKG = 4x
x
= 4(18)
= 72
x
[2]
(iii) the size of angle GJK.
72 + 2x + x + Angle GJK = 180
72 + 2(18) + 18 + Angle GJK = 180
Angle GJK = 180 − 72 − 2(18) − 18
= 54
Answer: (c) (iii) angle GJK =
170
54
[1]
Geometry
October / November 2012 Paper 23
6
T
X
R
X
Using a straight edge and compasses only, construct the locus of points which are
equidistant from R and from T.
[2]
October / November 2012 Paper 43
8
D
A
28°
52°
X
C
B
(a) A, B, C and D lie on a circle.
The chords AC and BD intersect at X.
Angle BAC = 28° and angle AXD = 52°.
Calculate angle XCD.
Angle AXB + 52 = 180
Angle AXB = 180 − 52
Angle AXB = 128
Angle XAB + angle AXB + angle XBA = 180
S
171
C
Workbook of Variant 3
B
28 + 128 + XBA = 180
XBA = 180 – 28 – 128
XBA = 24
Answer: (a) angle XCD =
24
[3]
(b)
S
P
22x° O
R 25x°
Q
PQRS is a cyclic quadrilateral in the circle, centre O.
Angle QOS = 22xo and angle QRS = 25xo.
Find the value of x.
Reflex angle SOQ = 2(SRQ) [Angle at centre of a circle is twice the angle on circumerence]
= 2(25x)
= 50x
22x + 50x = 360
[Angle around a point add upto 360]
72x = 360
360
72
x = 5
x =
Answer: (b) x =
172
5
[3]
Geometry
(c)
L
8 cm
O
44°
K
M
In the diagram OKL is a sector of a circle, centre O and radius 8 cm.
OKM is a straight line and ML is a tangent to the circle at L.
Angle LOK = 44o.
Calculate the area shaded in the diagram.
ML is tangent to the circle at L.
Angle OLM = 90°
tan θ =
tan 44 =
O
A
LM
8
8 × tan 44 = LM
LM = 7.725510198
Shaded area = area of triangle − area of sector
=
1
θ
×b×h−
× π r2
2
360
=
44
1
(7.725510198)8 −
× (π)(8)2
360
2
= 6.327804924
Answer: (c)
6.33
cm2
[5]
May / June 2013 Paper 23
6
The volumes of two similar cones are 36 π cm3 and 288 π cm3.
The base radius of the smaller cone is 3 cm.
Calulate the base radius of the larger cone.
[In similar shapes, (length ratio)3 = volume ratio]
173
Workbook of Variant 3
(
( 3x ) =
radius of large cone
radius of small cone
3
volume of large cone
volume of small cone
288 π
36 π
3
x=
)=
3
288 π x 33
36 π
x = 6
Answer:
6
cm [3]
May / June 2013 Paper 43
2
(a)In this question show all your construction arcs and use only a ruler and compasses to draw
the boundaries of your region.
This scale drawing shows the positions of four towns P, Q, R and S, on a map where 1 cm
represents 10 km.
North
P
Bearing of S from P.
Q
Scale: 1cm to 10 km
S
R
174
Geometry
A nature reserve lies in the quadrilateral PQRS.
The boundaries of the nature reserve are :
•
•
•
•
equidistant from Q and from R [Draw perpendicular bisector of QR]
equidistant from PS and from PQ [Draw Angle brisector of angle SPQ]
60 km from R[Draw arc with radius 6 cm centre R]
along QR[One of the boundaries of nature reserve should be QR]
(i) Shade the region which represents the nature reserve.
[7]
(ii) Measure the bearing of S from P.
[Its easier to measure acute angle and subract from 360, 360 - 141 = 219]
219
Answer: (a) (ii)
[1]
(b) A circular lake in the nature reserve has a radius of 45 m.
(i) Calculate the area of the lake.
Area of circular tank = πr2
= π (45)2
= 6361.725124
Answer: (b) (i)
6360
m2
[2]
(ii)
A fence is placed along part of the circumference of the lake.
This arc subtends an angle of 210° at the centre of the circle.
Calculate the length of the fence.
length of the fence = Arc length
=
=
θ
×2πr
360
210°
210
× 2 π (45)
360
= 164.9336143
Answer: (b) (ii)
165
m [2]
175
Workbook of Variant 3
8
(a)
B
27°
C
A
O
E
D
A, B, C, D and E are points on the circle, centre O.
Angle ABD = 27o.
Find
(i) angle ACD,
Angle ACD = Angle ABD
[Angle subtended by same arc]
Answer: (a) (i) angle ACD =
27
[1]
(ii) angle AOD,
Angle AOD = 2(ABD)
[Angle at centre is twice the angle on circumference]
= 2(27)
M
= 54
L
100°
45 cm
(iii) angle AED.
Angle AED = 180 − 27
K
= 153
67°
Answer: (a) (ii) angle AOD =
[1]
32 cm
[Opposite angles of a cyclic quadrilatral add upto 180]
Answer: (a)N (iii) angle AED =
176
54
153
[1]
Geometry
(b)
M
L
100°
67°
45 cm
32 cm
K
N
The diagram shows quadrilateral KLMN.
KL = 45 cm, LN = 32 cm, angle KLN = 100o and NLM = 67o.
(i) Calculate the length KN.
a2 = b2 + c2 − 2bc cos A
KN2 = 452 + 322 − 2(45) (32) cos 100
KN = 452 + 322 − 2(45) (32) cos100
KN = 59.57437999
Answer: (b) (i) KN =
59.6
cm [4]
(ii) The area of triangle LMN is 324 cm2.
Calculate the length LM.
Area of triangle =
324 =
1
× a × b × sin C
2
1
× 32 × LM × sin 67
2
324 × 2
= LM
32 × sin 67
LM = 21.99879764
Answer: (b) (ii) LM =
22
cm [3]
(iii) Another triangle XYZ is mathematically similar to triangle LMN.
177
Workbook of Variant 3
M
L
Y
X
Z
N
XZ = 16 cm and the area of triangle LMN is 324 cm2.
Calculate the area of triangle XYZ.
In similar triangles, (length ratio)2 = Area ratio
area of trangle XYZ
( ) = area
of triangle LMN
x
16
(
=
)
324
32
XZ
LN
2
2
x
162
=
2
324
32
324 × 162
=x
322
x = 81
Answer: (b) (iii)
October / November 2013 Paper 23
13
M
D
x
42°
C
O
A
B
N
178
81
cm2
[2]
Geometry
The vertices of the rectangle ABCD lie on a circle, centre O.
MN is a line of symmetry of the rectangle.
AC is a diameter of the circle and angle ACD = 42o.
Calculate
(a) angle CAM,
MN is line of symmetry of the rectangle
∴ Angle OXC = 90°
In triangle OXC
Angle OXC + angle COX + angle XCO = 180
90 + angle COX + 42 = 180
angle COX = 180 − 90 − 42
angle COX = 48
angle CAM =
48
2
= 24
Answer: (a) Angle CAM =
24
Answer: (b) Angle DCM =
24
[2]
(b) angle DCM.
Angle AMC = 90°
let angle DCM = x
In triangle CAM
24 + 90 + x + 42 = 180
x = 180 − 24 − 90 − 42
x = 24
[2]
179
Workbook of Variant 3
15
A
(a) Construct the locus of all the points which are 3 cm from vertex A and
outside the rectangle.
[2]
[draw circle with centre A radius 3 cm outside the rectangle]
(b)Construct, using a straight edge and compasses only, one of the lines of
symmetry of the rectangle.
[Draw perendicular bisector for either length or breadth, Here both are drawn. ]
[2]
October / November 2013 Paper 43
4
(a) One angle of an isosceles triangle is 48o.
Write down the possible pairs of value for the remaining two angles.
Either 48 + 48 + x = 180
x = 180 − 48 − 48
x = 84
48 and 84
or
180 − 48
= 66
2
66 and 66
Answer: (a)
48
66
180
and
and
84
66
[2]
Geometry
(b) Calculate the sum of the interior angles of a pentagon.
Sum of interior angle of a polygon = (n − 2) × 180
Sum of interior angles of a pentagon = (5 − 2) × 180 = 540.
Answer: (b)
540
[2]
(c) Calculate the sum of the angles a, b, c, d, e, f and g shown in this diagram.
a
g
b
f
c
e
d
[Angle around a point adds upto 360. Sum of interior angles of heptagon is 900]
a + b + c + d + e + f + g = 7(360) − (900) = 1620
Answer: (c)
1620
[2]
(d) The trapezium ABCD, has four angles as shown.
All the angles are in degrees.
B
3y – 20
A
C
4x – 5
2x + 5
x + y – 10
D
(i) Show that 7x + 4y = 390.
Answer (d)(i):
[Sum of interior angels of quadrilateral in 360]
2x + 5 + 3y − 20 + 4x − 5 + x + y − 10 = 360
181
Workbook of Variant 3
2x + 4x + x + 3y + y = 360 – 5 + 20 + 5 + 10
7x + 4y = 390
Answer: (d) (i)
7x + 4y = 390
[1]
(ii) Show that 2x + 3y = 195
Answer (d)(ii):
[Co-interior angels sum up to 180°]
2x + 5 + 3y − 20 = 180
2x + 3y = 180 − 5 + 20
2x + 3y = 195
Answer: (d) (ii)
2x + 3y = 195
[1]
(iii) Solve these simultaneous equations.
(Multiply (d) (i)'s equation by 3 and (d) (ii) by –4)
21x + 12y = 1170
+
–8x + –12y = –780
13x = 390
x =
390
13
x = 30
Substitute the value of x in d (i)'s equation
7(30) + 4y = 390
210 + 4y = 390
4y = 390 − 210
y = 390 −
y = 45
182
210
4
Answer: (d) (iii) x =
30
y=
45
[4]
Geometry
(iv) Use your answer to part (d)(iii) to find the size of all four angles of trapezium.
2x + 5 = 2(30) + 5 = 65
3y − 5 = 2(45) − 5 = 115
4x − 5 = 4(30) − 5 = 115
x + y − 10 = 30 + 45 −10 = 65
Answer: (d) (iv)
65
,
115
,
115
,
65
[1]
May / June 2014 Paper 23
7
46°
8.69 cm
74°
9.65 cm
9.65 cm
60°
x°
46°
7.22 cm
y cm
These two triangles are congruent.
Write down the value of
(a) x,
In first diagram side opposite to angle 74o is 9.65 cm
In second diagram, side opposite to angle x is 9.65 cm
Answer: (a) x =
74
[1]
(b) y.
In first diagram,adjacent side of angle 46o are 8.69 cm and 9.65 cm
In second diagram, adjacent side of angle 46o are y and 9.65 cm
Answer: (b) y =
8.69
[1]
183
Workbook of Variant 3
May / June 2014 Paper 43
7
(a)
E
x°
t°
D
x° C
F
q°
Y
A
p° 32°
B
X
ABCDEF is a hexagon.
AB is parallel to ED and BC is parallel to FE.
YFE and YABX are straight lines.
Angle CBX = 32o and angle EFA = 90o.(dotted lines are extensions)
Calculate the valune of
(i) p,
p + 32 = 180
(Supplementary angles)
p = 180 − 32
p = 148
Answer: (a)(i) p =
148
[1]
(ii) q,
Angle AYF = 32
Angle AFY = 90
∴ 32 + 90 + angle AFY = 180
(angles in a triangle)
Angle YAF = 180 − 32 − 90 = 58
Angle YAF + q = 180
58 + q = 180
q = 180 − 58
q = 122
Answer: (a)(ii) q =
184
122
[2]
Geometry
(iii) t,
Angle AYF + t = 180
32 + t = 180
t = 180 − 32
t = 148
148
Answer: (a)(iii) t =
[1]
(iv) x.
ABCDEF is hexagon, sum of all integer angles of hexagon
= (6 − 2) × 180 = 720
q + 90 + t + x + x + p = 720
122 + 90 + 148 + 2x + 148 = 720
2x = 720 − 122 − 90 − 148 − 148
2x = 212
x =
212
2
x = 106
106
Answer: (a)(iv) x =
(b)
[3]
R
x°
S
T
Q
y°
63°
P
U
P, Q, R and S are points on a circle and PS = SQ.
PR is a diameter and TPU is the tangent to the circle at P.
Angle SPT = 63o.
Find the value of
185
Workbook of Variant 3
(i) x,
Angle SPR = 90 − 63 = 27
angle RSP = 90
Considering triangle RSP,
27 + 90 + x = 180
x = 180 − 90 − 27
x = 63
Answer: (b) (i) x =
63
Answer: (b)(ii) y =
54
[2]
(ii) y.
Angle SRP = Angle SQP
Angle SQP = x
Considering triangle SQP
x + x + y = 180
63 + 63 + y = 180
y = 180 − 63 − 63
y = 54
[2]
October / November 2014 Paper 23
7
Find the interior angle of a regular polygon with 18 sides.
Interior angle of a regular polygon =
(n − 2) × 180
n
Interior angle of a regular polygon with 18 sides =
(18 − 2) × 180
18
= 160
Answer:
186
160
[3]
Geometry
12
D
C
E
A
B
(a) Draw the locus of the points which are 3 cm from E.
[1]
(b) Using a straight edge and compasses only, construct the bisector of angle DCB.
[2]
(c) Shade the region which is
• less than 3 cm from E[Draw radius of 3 cm with centre E]
and
nearer to CB than to CD. [Draw angle bisector of angle BCD]
•
[1]
October / November 2014 Paper 43
3
A
B
52°
D
O
C
56°
E
A, B, C and D are points on a circle, centre O.
CE is a tangent to the circle at C.
187
Workbook of Variant 3
(a) Find the size of the following angles and give a reason for each answer.
52
Angle subtended by same
(i) Angle DAC =
, because,
arc.
[2]
104
(ii) Angle DOC =
twice angle at circumference.
[2]
(iii) Angle BCO =
and radius is 90°
, because,
34
, because,
angle at centre is
Angle between tangent
[2]
(b) CE = 8.9 cm and CB = 7 cm.
(i) Calculate the length of BE.
B
a2 = b2 + c2 − 2bc cos A
7 cm
BE = 7 + 8.9 − 2(7) (8.9) cos 56
2
2
2
BE = 72 + 8.92 − 2(7) (8.9) cos 56
C
BE = 7.650788471
56°
8.9 cm
Answer: (b) (i) BE =
7.65
Answer: (b) (ii) angle BEC =
49.3
E
[4]
(ii) Calculate angle BEC.
sin B
sin A =
b
a
sin E
sin 56
=
7
7.65078847
sin E =
7 x sin 56
7.65078847
E = sin–1
7 x sin 56
( 7.65078847
)
E = 49.33374739
188
[3]
Geometry
May / June 2015 Paper 43
6
(a)
E
X
(Assume)
D
140°
120°
F
C
A
B
y(Assume)
In the hexagon ABCDEF, AB is parallel to ED and AF is parallel to CD.
Angle ABC = 90o, angle CDE = 140o and angle DEF = 120o.
Calculate angle EFA.
[AYDX is a parallel to gram]
Calculate angle EFA.
Angle EXF + angle EDC = 180
Angle EXF + 140 = 180
Angle EXF = 180 − 140
= 40
D
Angle XEF = 180 − 120
Angle XEF = 60
C
30°
X
Angle XFE = 180 − (60 + 40)
Angle XFE = 80
Angle EFA = 180 − 80
=
100°
A
B
100
Answer: (a) Angle EFA =
100
[4]
189
Workbook of Variant 3
(b)
D
C
30°
X
100°
A
B
In the cyclic quadrilateral ABCD, angle ABC = 100° and angle BDC = 30°.
The diagonals intersect at X.
(i) Calculate angle ACB.
Angle BAC = Angle BDC = 30
[Angle subtended by same arc]
In triangle ABC,
Angle BAC + Angle ABC + Angle ACB = 180
30 + 100 + Angle ACB = 180
Angle ACB = 180 − 100 − 30
= 50
50
Answer: (b) (i) Angle ACB =
[2]
(ii)Angle BXC = 89°.
Calculate angle CAD.
Here Angle CAD is same as XAD
Angle ADX = Angle ACB = 50
[Angle subtended by same arc]
In triangle ADX,
Angle ADX + Angle AXD + Angle XAD = 180
50 + 89 + Angle XAD = 180
Angle XAD = 180 − 50 − 89
= 41
Answer: (b) (ii) CAD =
41
[2]
(iii) Complete the statement.
Triangles AXD and BXC are
190
similar
[1]
Geometry
(c)
R
S
Y
Q
P
P, Q, R and S lie on a circle.
PR and QS intersect at Y.
PS = 11 cm, QR = 10 cm and area of triangle QRY = 23 cm2.
Calculate the area of triangle PYS.
[In similar triangle, (length ratio)2 = Area ratio]
PS
( QR
)=
11
(
=
20 )
2
area of triangle PYS
area of trangle QRY
2
x
23
112 × 23
=x
102
x = 27.83
Answer: (c)
(d) A regular polygon has n sides.
Each exterior angle is equal to
(i) Find the value of n.
n
360
=
10
n
cm2
[2]
n
degrees.
10
Each exterior angle of rectangle polygon =
27.8
360
n
n × n = 360 × 10
n2 = 3600
n = 3600
n = 60
Answer: (d) (i) n =
60
[3]
191
Workbook of Variant 3
(ii) Find the size of an interior angle of this polygon.
Each interior angle of rectangle polygon =
=
(n − 2) × 180
n
(60 − 2) × 180
60
= 174
174
Answer: (d) (ii)
[2]
October / November 2015 Paper 23
8
Find the sum of the interior angles of a 25 sided polygon.
Sum of interior angle of polygon = (n − 2) × 180
Sum of interior angle of 25 sided polygon = (25 − 2) × 180
= 4140
4140
Answer:
[2]
14 Two containers are mathematically similar.
Their volumes are 54 cm3 and 128 cm3.
The height of the smaller container is 4.5 cm.
Calculate the height of the larger container.
In similar shapes, [(length ratio)3 = volume ratio]
(
( 4.5x ) =
height of large container
height of small container
3
volume of large container
) = volume
of small container
3
128
54
x3
128
=
3
54
4.5
128 × 4.53
54
x3 =
x=
3
128 × 4.53
54
x=6
Answer:
192
6
cm [3]
Geometry
October / November 2015 Paper43
8
(a)
C
D
A
B
In the diagram, D is on AC so that angle ADB = angle ABC.
(i) Show that angle ABD is equal to angle ACB.
Answer: (a)(i)
Angle A is common to both triangles
So, Angle BAD = Angle BAC
Angle ABD = Angle ACB [given]
So, Third angle of triangles are equal.
Hence Angle ABD = Angle ACB
[2]
(ii) Complete the statement.
similar
Triangle ABD and ACB are
[2]
(iii) AB = 12 cm, BC = 11 cm and AC = 16 cm.
Calculate the length of BD.
AB
BD
=
[In similar triangles, sides are in proportion]
AC
BC
12
BD
=
16
11
12 × 11
= BD
16
BD = 8.25
E
D
102°
u°
Answer: (a) (iii) BD =
A
v°
38°
x°
w°
C
8.25
cm [2]
193
Workbook of Variant 3
(b)
E
D
102°
u°
A
v°
38°
x°
C
w°
B
A, B, C, D and E lie on the circle.
Angle AED = 102° and angle BAC = 38°.
BC = CD
Find the value of
(i) u,
Angle BDC = Angle BAC
[Angle subtended by same arc]
u = 38
Answer: (b) (i) u =
38
Answer: (b) (ii) v =
38
[1]
(ii) v,
Angle CBD = Angle CDB = u
Angle DAC = Angle DBC = u
v = u = 38
[1]
(iii) w,
w = 180 − 102 = 78
[ABDE is cyclic quadrilatre]
Answer: (b) (iii) w =
78
Answer: (b) (iv) x =
26
[1]
(iv) x.
In triangle ABC.
38 + w + 38 + x = 180
38 + 78 + 38 + x = 180
x = 26
194
[1]
Geometry
(c)
P
m°
O
2m°
Q
R
In the diagram P, Q and R lie on the circle, centre O.
PQ is parallel to OR.
Angle QPO = m° and angle QRO = 2m°.
Find the value of m.
Angle RQP = 180 − 2m [Angle ORQ and angle RQP are co-interior]
Angle ROP = 180 − m[Angle QPO and angle ROP are co-interior]
Angle ROP = 2(180 − 2m)[Angle at centre in twice angle on circumference]
180 − m + 2(180 − 2m) = 360
180 − m + 360 − 4m = 360
180 + 360 − 360 = m + 4m
[Angle around a point is 360°]
180 = 5m
180
=m
5
m = 36
Answer: (c) m =
36
[5]
195
5
Workbook of
Solution 4
ALGEBRA 2
Algebra 2
Chapter 5: Algebra 2
May / June 2010 Paper 23
6
3x × 94 = 3n.
Find n in terms of x.
3x × 94 = 3n
3x × (32)4 = 3n[32 = 9]
3x × 38 = 3n[(an)n = amn]
3x + 8 = 3n[am × an = am+n]
∴ n=x+8
Answer: n =
8
x+8
[2]
Write as a single fraction in its simplest from.
x + x–1
3
2
x + x–1
3
2
= 2x + 3x – 3
6
= 5x – 3
6
Answer:
5x – 3
6
[2]
16 Make y the subject of the formula. A = r (y + 2)
5
r
(y
+
2)
A=
5
5A = r (y + 2)
5A = ry + 2r
5A – 2r = ry
199
Workbook of Variant 3
∴ 5A – 2r = y
r
∴ y = 5A – 2r
r
Answer: y =
5A – 2r
r
[3]
20
Line 1
y
4
3
Line 2
2
1
0
1
2
3
4
x
Line 3
Find the three inequalties which define the shaded region on the grid.
First find equation of line surrounded by region .
It will be easy if we will name the line. Region is triangle so we will have 3 lines. Naming of line
is done in diagram.
Line 1 : x = 0 ,
For line 2 : (Pick any 2 point which are on line 2)
(2, 1) (4, 2)
m = y2 - y1 = 2 - 1 = 1 x2 - x1
4-2
2
[y - intercept = c = 0 (from diagram)]
Line 2 , y = 1 x[y = mx + c]
2
For line 3 : (pick any 2 point which are on line 3)
(4, 0) (0, 4)
200
Algebra 2
m = y2 - y1 = 4 - 0 = -1
x2 - x1
0-4
[y - intercept = c = 4 (from diagram)]
line 3, y = -x + 4[y = mx + c]
x + y = 4
Here
x ≥ 0,
For line 2: We need the region towards y-axis
∴y≥ 1 x
2
For line 3: We need the region below the line
∴x+y≤4
Answer:
x≥o
y≥ 1 x
2
x+y≤4
[5]
October / November 2010 Paper 23
16
g
=
2
h
i
Find i in terms of g and h.
g
=
2
h
i
( ) = hi
g
2
2
g2 = h
i
4
i = 4h
g2
Answer: i =
24 (a) Write
4h
g2
[3]
1
2
–
as a single fraction in its lowest terms.
y
x
x – 2y [L.C.M is xy]
xy
Answer: (a)
x – 2y
xy
[2]
201
Workbook of Variant 3
2
(b) Write x + x in its lowest terms.
3x + 3
= x (x + 1) [factorise numerator and denominator]
3 (x + 1)
= x
3
Answer: (b)
22
y
Line 2
Line 3
9
8
7
6
5
4
3
2
Line 1
1
–6
–5
–4
–3
–2
–1
0
1
2
3
4
x
–1
Find the three inequalities which define the shaded triangle in the diagram.
First find equation of line surounded by region.
Region is triangle so have 3 lines,
line 1 : y = 1 , line 2 : x = 3
For line 3, (take any two points which lie on the line)
(1,6) and (2,7)
m = y2 – y1 = 7 – 6 = 1
x2 – x1
2–1
[y - intercept = c = 5 (from diagram)]
∴ line 3: y = x + 5[y = mx + c]
Here
202
y≥1
and
x≤3
x
3
[3]
Algebra 2
For line 3, we need the region
towards x - axis
∴ y ≤ x + 5
(≤ as it is solid line)
Answer:
y≥1
x≤3
y≤x+5
[5]
May / June 2010 Paper 23
11 The volume of a solid varies directly as the cube of its length.
When the length is 3 cm, the volume is 108 cm3.
Find the volume when the length is 5 cm.
V α L3V = 4 L3
V = KL3V = 4 (5)3
108 = K(3)3V = 500
108 = K
33
K=4
Answer:
13
500
[3]
y
y= 4-x
y = x+4
4
R
y=1
0
x
The diagram shows the lines y = 1, y = x + 4 and y = 4 – x.
On the diagram, label the region R where y ≥ 1, y ≥ x + 4 and y ≤ 4 – x.
203
Workbook of Variant 3
[Here, to find region R, lets shade unwanted region] .
identify the lines , y = 1 → horizontal line.
y = x + 4 → line with positive slope and positive y - intercept
y = 4 – x → line with negative slope and positive y - intercept [3]
18 Simplify the following
(a) (3x3)3
33 x3 × 3
27x9
Answer: (a)
(b) (125x6)
27x9
[2]
2
3
2
= (53 x6) 3
2
= 53 x 3 2
x6 x 3
= 52 x2 x 2
= 25x4
Answer: (b)
16 Write
25x4
[2]
2
3
+
as a single fraction.
x–2
x+2
Give your answer in its simplest from.
2
3
+
x–2
x+2
= 2(x + 2) + 3 (x – 2)
(x – 2) (x + 2)
= 2x + 4 + 3x – 6
(x – 2) (x + 2)
= 2x + 3x + 4 – 6
(x – 2) (x + 2)
=
5x – 2
(x – 2) (x + 2)
Answer:
204
5x – 2
(x – 2) (x + 2)
[3]
Algebra 2
October / November 2011 Paper 23
4
Find the value of
(
27
8
)
–4
3
Give your answer as an exact fraction.
=
=
=
–4
3
(
27
8
(
8
27
) 23
33
4
3
)
4
3
( )
= 2
3
( )
3x
[( ab ) = ( ba ) ]
-n
n
4
3
4
= 2
34
= 16
81
Answer:
16
81
[2]
15 ap = px + c
Write p in terms of a, c and x.
ap = px + c
ap – px = c
p (a – x) = c
p=
c
a–x
Answer: p =
c
a–x
[3]
18 Write as a single fraction, in its simplest from.
1–x
2+x
–
x
1 – 2x
= (1 – x) (1 – 2x) – x(2 + x)
x (1 – 2x)
= 1(1 – 2x) – x(1 – 2x) – x(2 + x)
x (1 – 2x)
2
2
= 1 – 2x – x + 2x – 2x – x
x(1 – 2x)
205
Workbook of Variant 3
2
2
= 1 – 2x – 2x – x – 2x – x
x(1 – 2x)
2
= 1 – 5x + x
x(1 – 2x)
Answer:
1 – 5x + x2
x(1 – 2x)
[4]
16The time, t, for a pendulum to swing varies directly as the square root of its length, l.
When l = 9, t = 6.
(a) Find a formula for t in terms of l.
t α l
t = k l
6 = k 9
6 = k (3)
k = 2
∴ t = 2 l
Answer: (a) t =
2 l
[2]
(b) Find t when l = 2.25.
t = 2 2.25
t = 3
Answer: (b) t =
October / November 2011 Paper 43
10 H
assan stores books in large boxes and small boxes.
Each large box holds 20 books and each small box holds 10 books.
He has x large boxes and y small boxes.
(a) Hassan must store at least 200 books.
Show that 2x + y ≥ 20.
Answer (a):
20x + 10y ≥ 200
10 (2x + y) ≥ 200
2x + y ≥ 200
206
3
[1]
Algebra 2
(b) Hassan must not use more than 15 boxes.
He must use at least 3 small boxes.
The number of small boxes must be less than or equal to the number of large boxes.
Write down three inequalities to show this information.
[not more than implies less than or equal to]
[At least means greates than or equal to]
Answer: (b)
x + y ≤ 15
y≥3
y≤x
[3]
(c) O
n the grid, show the information in part (a) and part (b) by drawing for straight lines and
shading the unwanted regions.
y
2x+y=20
y=x
20
18
16
14
12
10
8
6
4
y=3
2
x+y=15
0
2
4
6
8
10
12
14
16
18
20
x
[6]
(d) A large box costs $ 5 and a small box costs $ 2.
(i) Find the least possible total cost of the boxes.
[Pick co-ordinates from corner of quadrilateral. Since x and y are number of boxes, we
will have to pick nearest integer co-ordinate inside wanted region]
(12, 3) → 5(12) + 2(3) = 66
207
Workbook of Variant 3
(8, 7) → 5(8) + 2(7) = 54
(7, 6) → 5(7) + 2(6) = 47 → least cost.
(9, 3) → 5(9) + 2(3) = 51
Answer: (d) (i) $
47
[1]
(ii) F
ind the number of large boxes and the number of small boxes which give this least
possible cost.
(7, 6) least cost
Answer: (b) (ii) Number of large boxes =
Number of small boxes =
7
6
[2]
May / June 2012 Paper 23
13 (a) Find the value of 7p – 3q when p = 8 and q = –5.
7p – 3q
7(8) – 3(–5)
= 71
Answer: (a)
71
[2]
(b) Factorise completely.
3uv + 9vw
3u (v + 3w)[common factor]
Answer: (b)
4
3u (v + 3w)
[2]
Solve the inequality.
3y + 7 ≤ 2 – y
3y + y ≤ 2 – 7
4y ≤ –5
y ≤ –5
4
y ≤ –1.25
Answer:
208
y ≤ –1.25
[2]
Algebra 2
10The periodic time, T, of a pendulum varies directly as the square root of its length, l.
T = 6 when l = 9.
Find T when l = 25.
Tα l
T=k l
6= k 9
6 = k (3)
k=2
∴T=2 l
T = 2 25
T = 10
Answer: T =
10
[3]
14 Simplify the following.
(a) (4pq2)3
43p3q2 x 3
64p3q6
Answer: (a)
(b) (16x8)
=
[2]
–1
4
1
(16 x )
8
[its better to get rid of negative powers, a–1 = 1 ]
a
1
4
1
1
(24 x8) 4
1
=
2
=
64p3q6
1
4x
4
1
x8 x 4
1
2x2
Answer: (b) w =
1
2x2
[3]
209
Workbook of Variant 3
9
Make w the subject of formula.
t = 2 – 3w
a
3w = 2 – t
[Keeping 'w' positive]
a
3w = a(2 – t)
w = a(2 – t)
3
Answer: w =
a(2 – t)
3
[3]
20 Simplify fully.
x2 – x – 20
x3 – 10x2 + 25x
[Factorise numerator and denominator]
2
= x – 5x + 4x – 20 [product = –20, sum = –1]
x (x2 – 10x + 25)
=
x(x – 5) + 4(x – 5) [product = 25, sum = –10]
x(x2 – 5x – 5x + 25)
=
(x – 5) (x + 4)
x[x(x – 5) – 5(x – 5)]
= (x – 5) (x + 4)
x(x – 5) (x – 5)
=
x+4
x(x – 5)
Answer:
x+4
x(x – 5)
[5]
October / November 2012 Paper 23
16 Make y the subject for the formula.
A = πx2 − πy2
πy2 = πx2 − A
y2 =
y=
πx2 − A
π
πx2 − A
π
Answer: y =
210
πx2 − A
π
[3]
Algebra 2
21 Simplify the following.
h2 – h – 20
h2 – 25
[Factorise numerator and denominator]
2
= h − 5h + 4h − 20
h2 − 52
= h(h − 5) + 4(h − 5)
(h + 5) (h − 5)
= (h − 5) (h + 4)
(h + 5) (h − 5)
= (h + 4)
(h + 5)
Answer: y =
(h + 4)
(h + 5)
[4]
14
y varies inversely as the square root of x.
When x = 9, y = 6.
Find y when x = 36.
1
x
y=k
6=
k
9
6=
k
3
k = 18
y=
18
x
y = 18
36
y = 18
6
y = 18
6
y=3
Answer:
3
[3]
211
Workbook of Variant 3
13 Write the following as a single fraction in its simplest form.
x+2
2x – 1
−
+1
3
4
4(x + 2) − 3(2x − 1) + 12
12
4x + 8 − 6x + 3 + 12
12
8 + 3 + 12 + 4x − 6x
12
23 − 2x
12
Answer:
( )
49
10 Without using calculator, show that
16
Write down all the steps in your working.
=
=
=
( )
( )
49
16
16
49
( )
42
72
–3
2
=
23 − 2x
12
[3]
64
.
343
–3
2
3
2
3
2
3
42 x 2
=
3
72 x 2
=
43
73
=
64
343
Answer:
(256w256)
11 Simplify
64
343
[2]
1
4
1
(44 w256) 4 (256 = 44)
1
1
44 x 4 × w256 x 4
4 × w64
Answer:
212
4w64
[2]
Algebra 2
May / June 2013 Paper 23
13 Write as a single fraction in its simplest form
x+3
x−1
−
x−3
x+1
=
(x + 3) (x + 1) − [(x − 1) (x − 3)]
(x − 3) (x + 1)
= x(x + 1) + 3(x + 1) − [x(x − 3) −1(x − 3)]
(x − 3)(x + 1)
=
x2 + x + 3x +3 − [x2 − 3x − x + 3]
(x − 3) (x + 1)
=
x2 + 4x + 3 − [x2 − 4x + 3]
(x − 3) (x + 1)
=
x2 + 4x + 3 − x2 + 4x − 3
(x − 3) (x + 1)
=
8x
(x − 3) (x + 1)
Answer:
8x
(x − 3) (x + 1)
[4]
14 (a) Solve 3n + 23 < n + 41
3n − n < 41 − 23
2n < 18
n<
18
2
n<9
Answer: (a)
(b) Factorise completely.
n<9
[2]
ab + bc + ad + cd
b(a + c) + d(a + c)
[Factorising by grouping]
(a + c) (b + d)
Answer: (b)
(a + c) (b + d)
[2]
213
Workbook of Variant 3
20 (a) y =
8+
4
x
Find y when x = 2
Give your answer correct to 4 decimal places.
y =
8+
4
2
y = 8 + 2
y = 10
y = 3.16227766
Answer: (a) y =
(b) Rearrange y =
y2 = 8 +
4
x
y2 – 8 =
4
x
x =
8+
[2]
4
to make x the subject.
x
4
y –8
2
Answer: (b) x =
8
3.16
4
y2 – 8
[4]
The mass, m, of a sphere varies directly with the cube of its radius, r.
m = 160 when r = 2.
Find m when r = 5.
m α r3
m = kr3
160 = k(2)3
160 = k(8)
160
=k
8
k = 20
m = 20r3
m = 20(5)3
m = 2500
Answer: m =
214
2500
[3]
Algebra 2
May / June 2013 Paper 43
3
(a) Luk wants to buy x goats and y sheep.
(i) He wants to buy at least 5 goats.
Write down an inequality in x to represent this condition.
[At least means greates than or equal to]
Answer: (a) (i)
x≥5
[1]
(ii) He wants to buy at least 11 sheep.
Write down an inequality in y to represent this condition.
Answer: (a) (ii)
y ≥ 11
[1]
(iii) He wants to buy at least 20 animals.
Write down an inequality in x and y to represent this condition.
AAnswer: (a) (iii)
x + y ≥ 20
[1]
(b) Goat costs $ 4 and sheep costs $ 8.
The maximum Luk can spend is $ 160.
Write down an inequality in x and y and show that it simplifies to x + 2y ≤ 40
Answer (b):
4x + 8y ≤ 160
4(x + 2y) ≤ 160
x + 2y ≤
160
4
x + 2y ≤ 40
215
Workbook of Variant 3
(c) (i)On the grid below, draw four lines to show the four inequalities and shade the unwanted
regions.
y
40
x=5
35
30
25
20
15
y=11
10
5
0
5
10
15
20
25
30
35
40
x
x+2y=40
x+y=20
[7]
(ii) Work out the maximum number of animals that Luk can buy.
[pick co-ordinates from corner of quadrilateral Since x and y are animal we will have to
pick nearest integer co - ordinates inside wanted region ]
(5, 15) → 5 + 15 = 20
(5, 17) → 5 + 17 = 22
(18, 11) → 18 + 11 = 29
→ Maximum
(9, 11) → 9 + 11 = 20
Answer: (c) (ii)
216
29
[2]
Algebra 2
October / November 2013 Paper 23
8
m varies directly as the cube of x.
m = 200 when x = 2
Find m when x = 0.4.
m α x3
m = kx3
200 = k(2)3
200 = k(8)
200
=k
8
k = 25
m = 25x3
m = 25(0.4)3
m = 1.6
Answer: m =
14 (a) Simplfy (64q–2)
1.6
[3]
1
2
1
= (64q–2) 2
1
= (82q–2) 2
1
1
= 82 × 2 q–2 × 2
= 8 q–1
=
8
q
Answer: (a)
8
q
[2]
(b) 57 ÷ 59 = p2
Find p.
57 ÷ 59 = p2
57 – 9 = p2
5–2 = p2
(5–1)2 = p2
217
Workbook of Variant 3
p = 5–1
p =
1
5
p = 0.2
Answer: (b) p =
0.2
[2]
May / June 2014 Paper 23
6
Simplify.
3x2y3 × x4y
3x2+4
3x6
y3+1
y4
Answer:
9
3x6y4
[2]
Solve the inequality.
5t + 23 < 17 − 2t
5t + 23 < 17 − 2t
5t + 2t < 17 − 23
7t < –6
t < −
6
7
t<−
Answer:
11 y varies as the cube root of (x + 3).
When x = 5, y = 1.
Find value of y when x = 340.
yα3x+3
y=k3x+3
1=k3x+3
1=k38
1 = k(2)
k=
218
1
2
6
7
[2]
Algebra 2
1
2
1
y=
2
y=
3
x+3
3
340 + 3
3
343
y=
1
2
y=
1
×7
2
y = 3.5
Answer: y =
3.5
[3]
October / November 2014 Paper 23
8
Make x the subject of the formula.
y = 2 +
x−8
y=2+ x−8
y−2= x−8
8 + (y − 2)2 = x
x = 8 + (y − 2)2
Answer: x =
9
8 + (y − 2)2
[3]
y varies inversely as (x + 5).
y = 6 when x = 3
Find y when x = 7.
yα
1
x+5
y=
k
x+5
6=
k
3+5
6= k
8
6×8=k
k = 48
y=
48
x+5
219
Workbook of Variant 3
48
7+5
y=
y = 48
12
y=4
Answer: y =
4
[3]
13 Write as a single fraction, in its simplest form.
3
2x
+
+ 3 + 2x
2x
3
=
3(3) + 2x(2x) + 6x(3) + 6x(2x)
6x
[LCM is 6x]
2
2
= 9 + 4x + 18x + 12x
6x
=
4x2 + 12x2 + 18x + 9
6x
=
16x2 + 18x + 9
6x
Answer:
16x2 + 18x + 9
6x
[3]
October / November 2014 Paper 43
6
(a) Simlpify
3
x5
3
x3 ÷ 5 [am × an = am+n]
x
(i) x3 ÷
= x3 ×
=
x3 + 5
3
=
x8
3
x5
3
Answer: (a) (i)
220
x8
3
[1]
Algebra 2
(ii) 5xy8 × 3x6y–5
5xy8 × 3x6y–5
= 15x1 + 6 y8 + (–5)
= 15x7
y3
15x7y3
Answer: (a) (ii)
[2]
2
(iii) (64x12)
3
2
(64x12) 3
2
x12) 3
= (43
2
= (43 x 3
= 42
2
x12 3 )
x8
= 16x8
16x8
Answer: (a) (iii)
[2]
(b) Solve 3x2 − 7x − 12 = 0.
Show your working and give answers correct to 2 decimal places.
a = 3, b = -7, c = -12
x =
− b ± b 2 − 4ac
2a
x =
−(−7) ± (− 7)2 − 4(3)(−12)
2(3)
x =
7 ± 193
6
x =
7 + 193
x = 7 − 193
6
6
x = 3.482073998 or x = -1.148740665
Answer: (b) x =
3.48
or x =
-1.15
[4]
221
Workbook of Variant 3
(c) Simpify.
x2 − 25
x3 − 5x2
x2 − 25
x3 − 5x2
[a2 – b2 = (a + b) (a – b)]
x2 − 52
x2 (x − 5)
(x + 5) (x − 5)
x2(x − 5)
x+5
x2
Answer: (c)
x+5
x2
[3]
May / June 2015 Paper 23
13 Simplify
(a) 12x12 ÷ 3x3
12x12
3x3
= 4x12 − 3
= 4x9
Answer: (a)
4x9
[2]
1
(b) (256y256)
8
1
(256y256) 8
1
= (28 y256) 8
1
1
= 28 × 8 y256 × 8
= 2y32
Answer: (b)
222
2y32
[2]
Algebra 2
October / November 2015 Paper 23
10 Find the value of
(a) ( 5 )8
(5)8
1
= (5 2 )8
1
= 5 2 × 8
= 54
= 625
Answer: (a)
625
[1]
–2
( 271 )
27
= (
1 )
3
(b)
2
3
2
= 33 × 3
= 32
= 9
Answer: (b)
9
[1]
11 Write the following as single fraction.
x
2
x
x +
2
(a) x +
=
2x + x
2
=
3x
2
Answer: (a)
3x
2
[1]
223
Workbook of Variant 3
2
x
x2 + 2
=
x
(b) x +
Answer: (b)
16 Make a the subject of the formula
s = ut +
x2 + 2
x
[1]
1 2
at
2
1 2
at
2
s − ut =
2(s − ut) = at2
2(s − ut)
=a
t2
a = 2(s − ut)
t2
Answer: a =
2(s − ut)
t2
[3]
17 Simplify.
(
(
(
x64
16y16
x64
16y16
)
)
1
4
1
4
1
x64 x 4
1
24 x 4
1
y16 x 4
)
( )
x16
2y4
Answer:
19 y is inversely proportional to (x + 2)2
When x = 1, y = 2,
Find y in term of x.
224
yα
1
(x + 2)2
y=
k
(x + 2)2
2=
k
(1 + 2)2
2=
k
32
x16
2y4
[3]
Algebra 2
2=
k
9
k = 18
y=
18
(x + 2)2
Answer:
18
(x + 2)2
[2]
22 Simplify.
4 + 10w
8 – 50w2
[factorise numerator and demominator]
2(2 + 5w)
2(4 − 25w2)
=
2 + 5w
2 − (5w)2
=
2 + 5w
(2 + 5w) (2 − 5w)
=
1
(2 − 5w)
2
Answer:
1
(2 − 5w)
[4]
225
6
Workbook of
Solution 4
TRIGONOMETRY
228
Trigonometry
Chapter 6: Trigonometry
May / June 2010 Paper 23
11
P
cliff
beach
A
F
55 m
The diagram shows a point P at the top of a cliff.
The point F is on the beach and and vertically below P.
The point A is 55 m from F, along the horizontal beach.
The angle of elevation of P from A is 17o.
Calculate PF, the height of the cliff.
tan 17 =
tan 17 =
O
P
A
PF
(0)
55
17°
55m
55 × tan 17 = PF
PF = 16.81518748.
Answer: PF =
21
(A)
F
16.8
m [3]
North
Q
North
140°
North
50 m
R
100 m
P
229
Workbook of Variant 3
The diagram shows three points P, Q and R on horizontal ground.
PQ = 50 m, PR = 100 m and angle PQR = 140o.
(a) Calculate angle PRQ.
[here 2 sides and 2 angles are involyed ∴ sine rule]
sin A
a
sin R
50
=
=
sin B
b
sin 140
100
sin R = 50 ×
R = sin-1
(
sin 140
100
50 × sin 140
100
)
= 18.74723725
Answer: (a) Angle PRO =
18.7o
[3]
(b) The bearing of R from Q is 100o.
Find the bearing of P from R.
=
360 − (80 + 18.7472)
Q 100
80
=
261.2528
18.7
R
P
Answer: (b)
May / June 2010 Paper 43
2
C
B
8 cm
5 cm
3 cm
A
11 cm
D
In the quadrilateral ABCD, AB = 3 cm, AD = 11 cm and DC = 8 cm.
The diagonal AC = 5 cm and angle BAC = 90o.
Calculate
230
261o
[2]
Trigonometry
(a) the length of BC,
[by pythagoras theorem]
C
BC2 = BA2 + CA2
2
5 cm
BC
= 32 + 52
BC = 32 + 52
B
A
3 cm
BC = 5.83095895
Answer: (b) Angle BC =
5.83
cm [2]
(b) angle ACD,
b2 + c2 − a2
cos A =
cos C =
C = cos-1
C = 113.5781785
2bc
52 + 82 − 112
2(5)(8)
52 + 82 − 112
2(5)(8)
Answer: (b) Angle ACD =
113.6
[4]
(c) the area of the quadrilateral ABCD.
Area of quadrilateral ABCD = Area of triangle ABC + area of triangle ACD
=
=
1
2
1
2
×b×h+
×3×5+
1
2
1
2
× a × b × sin c
× 5 × 8 × sin 113.5781
= 25.83031374
Answer: (c)
25.8
cm2
[3]
231
Workbook of Variant 3
October / November 2010 Paper 43
2
R
4 km
Q
7 km
85°
S
4.5 km
40°
P
The diagram shows five straight roads.
PQ = 4.5 km, QR = 4 km and PR = 7 km.
Angle RPS = 40o and angle PSR = 85o.
(a) Calculate angle PQR and show that it rounds to 110.7o.
Answer (a)
b2 + c2 − a2
cos A =
cos Q =
Q = cos-1
Q = 110.74238
2bc
42 + 4.52 − 72
2(4)(4.5)
42 + 4.52 − 72
2(4)(4.5)
≈ 110.7
Answer: (a)
110.7
[4]
(b) Calculate the length of the road RS and show that it rounds to 4.52 km.
Answer (b)
sin A
a
sin 40
RS
RS =
sin B
=
b
=
sin 85
7
7 × sin 40
sin 85
R = 4.516700678
≈ 4.52 km.
Answer: (b)
232
4.52
[3]
Trigonometry
(c) Calculate the area of the quadrilateral PQRS.
[Use the value of 110.7o for angle PQR and the value of 4.52 km for RS.]
angle SRP = 180 − (85 + 40)
= 55
area of quadrilateral PQRS = area triangle SRP + area of triangle QRP
1
=
2
1
=
2
× a × b + sin C +
1
2
× 4.52 × 7 × sin55 +
× a × b × sin C
1
2
× 4 × 4.5 × sin110.7
= 21.37798162
Answer: (c)
21.4
km2
[5]
October / November 2011 Paper 23
21
North
C
km
North
2.7
4.5 km
A
5 km
B
The diagram shows 3 ships A, B and C at sea.
AB = 5 km, BC = 4.5 km and AC = 2.7 km.
(a) Calculate angle ACB.
Show all your working.
[here 3 sides and one angle is involved ∴ cosine rule]
cos A =
cos C =
b2 + c2 − a2
2bc
2.72 + 4.52 − 52
2(2.7)(4.5)
233
Workbook of Variant 3
2.72 + 4.52 − 52
C = cos-1
C = 84.00009876
2(2.7)(4.5)
≈ 84o
84o
Answer: (a) Angle ABC =
[4]
(b) The bearing of A from C is 220o.
Calculate the bearing of B from C.
220 − 84 = 136o
C
0
220
0
84
A
B
136
Answer: (b)
October / November 2011 Paper 43
6
Q
P
3cm
D
C
4 cm
A
12cm
B
The diagram shows a triangular prism of length 12 cm.
The rectangle ABCD is horizontal and rectangle DCPQ is vertical.
The cross-section is triangle PBC in which angle BCP = 90o, BC = 4 cm and CP = 3 cm.
(a) (i) Calculate the length of AP.
2
AC
= AB2 + BC2
C
D
AC2 = 122 + 42
4 cm
AC = 122 + 42
A
B
A
12 cm
234
P pythagoras theorem]
[By
3
C
[1]
Trigonometry
AP2 = AC2 + PC2[By pythagoras theorem]
P
C
AP2 = 122 + 42 + 32
3
4 cm AP
= 122 + 42 + 32
cm
B
AP = 13
A
C
Answer: (a) (i) AP =
13
cm [3]
(ii) Calculate the angle of elevation of P from A.
P
(4)
O
(0)
sin
A=
m
c
4
13
3 cm
sin A =
A
C
A = sin-1
3
13
3
13
A = 13.3423638
Answer: (a) (ii)
13.3
[2]
(b) (i) Calculate angle PBC.
P
tan B =
(0)
3 cm
tan B =
B
C
4 cm
B
= tan-1
(A)
O
A
3
4
3
4
PF = 36.86989765
Answer: (b) (i) Angle PBC =
36.9
[2]
(ii) X is on BP so that angle BXC = 120o.
Calculate the length of XC.
P
Considering triangle B × C
X
3 cm
0
120
B
36.869
4 cm
sin A
a
=
sin B
b
C
235
Workbook of Variant 3
sin 36.869
xc
XC = 4 ×
=
sin 120
4
sin 36.869
sin 120
XC = 2.771223402
2.77
Answer: (a) (ii) XC =
cm [3]
May / June 2012 Paper 23
6
C
9 cm
A
B
28°
15 cm
Calculate the area of triangle ABC.
Area of triangle ABC =
=
1
2
1
2
× a × b × sin C
× 15 × 9 × sin 28
= 31.68933049
Answer:
236
31.7
cm2
[2]
Trigonometry
21
P
5 cm
D
C
8 cm
M
B
8 cm
The diagram shows a pyramid on a square base ABCD.
The diagonals of the base, AC and BD, intersect at M.
The sides of the square are 8 cm and the vertical height of the pyramid, PM, is 5 cm.
Calculate
(a) the length of the edge PB,
DB2 = DA2 + AB2
2
D
C
DB
= 82 + 82
DB2 = 82 + 82
8 cm
DB = 8 2
MB
is half of DB
A
B
8 cm
MB =
MB =
P
DB
2
8 2
2
MB = 4 2
5 cm
PB2 = PM2 + MB2
[By pythagores theorem]
PB2 = (4 2 )2 + 52
M
42
B
PB = (4 2 )2 + 52
PB = 7.549834435
Answer: (a) PB =
7.55
cm [3]
237
Workbook of Variant 3
(b) the angle between PB and the base ABCD.
P
sin A =
7.549 cm
5 cm
(0)
sin
A=
(4)
B
M
A = sin-1
O
A
5
7.549
(
5
7.549
)
A = 41.4785324
Answer: (b)
May / June 2012 Paper 43
2
North
D
95°
10 km
40°
A
30°
12 km
C
17 km
B
The diagram shows straight roads connecting the towns A, B, C and D.
AB = 17 km, AC = 12 km and CD = 10 km.
Angle BAC = 30o and angle ADC = 95o.
(a) Calculate angle CAD.
Considering triangle ADC.
238
sin A
a
sin A
10
=
=
sin B
b
sin 95
12
41.5
[3]
Trigonometry
10 × sin 95
sin A =
A = sin-1
A = 56.11540844
12
(
10 × sin 95
12
)
Answer: (a) Angle CAD =
56.1
[3]
(b) Calculate the distance BC.
Considering triangle ABC.
a2 = b2 + c2 − 2bc cos A
CB2 = 122 +172 − 2(12)(17) cos 30
CB = 122 + 172 − 2(12)(17) cos 30
CB = 8.925336703
Answer: (b) BC =
8.93
km [4]
(c) The bearing of D from A is 040o.
Find the bearing of
D
(i) B from A
40 + 56.115 + 30
126.115
40°
A
56.115°
30°
C
B
(ii) A from B.
Answer: (c) (i)
126o
[1]
180 − 126.115 = 53.885
126.115°
angle to be found
360 − 53.885 = 306.115
Answer: (c) (ii)
306°
[1]
239
Workbook of Variant 3
(d) Angle ACB is obtuse.
Calculate angle BCD.
Considering triangle ABC.
sin A
a
sin C
17
sin B
=
b
sin 30
=
8.925
17 × sin 30
sin C =
C = sin-1
C = 72.24720984
8.925
(
17 × sin 30
8.925
)
Since Angle ACB is obtuse (given)
Angle ACB = 180 − 72.247
= 107.753
Angle DCA = 180 − (95 + 56.115)
[Angle in a triangle]
= 28.885
Angle BCD = angle ACB + angle DCA
= 107.753 + 28.885
= 136.638
Answer: (d) Angle BCD =
240
137
[4]
Trigonometry
October / November 2012 Paper 23
18
C
23°
A
13 cm
6 cm
B
In triangle ABC, AB = 6 cm, BC = 13 cm and angle ACB = 23o.
Calculate angle BAC, which is obtuse.
sin A
a
sin A
13
=
=
sin B
b
sin 23
6
13 × sin 23
sin A =
A = sin-1
A = 57.84205967
6
(
13 × sin 23
6
)
Angle BAC is obtuse (given)
obtuse angle of BAC = 180 − 57.842
= 112.158
Answer: Angle BAC =
122.2
[4]
241
Workbook of Variant 3
24
Q
P
6 cm
C
B
5 cm
D
A
10 cm
The diagram shows a triangular prism.
ABCD is a horizontal rectangle with DA = 10 cm and AB = 5 cm.
BCQP is vertical rectangle and BP = 6 cm.
Calculate
(a) the length of DP,
DB2 = DA2 + AB2
C
DB = 10 + 5
2
2
2
6cm
5cm
DB = 102 +52
P
B
D
DB = 5 5 cm
A
10 cm
D
5 5cm
B
DP2 = DB2 + BP2
P
2
C
DP
= (5 5 )2 B
+ 62
6cm
DP = (5 5 )25cm
+ 62
D DP10=cm
12.68857754
A
D
5 5cm
B
Answer: (a) DP =
242
12.7
cm [3]
Trigonometry
(b) The angle between DP and the horizontal rectangle ABCD.
sin D =
sin D =
O
A
6
12.6889
(
688
12.
)
6
D = sin-1
D = 28.22191168
12.6889
D
(4)
cm
P
6cm
(0)
B
Answer: (b)
28.2
[3]
October / November 2012 Paper 43
6
A
16 cm
B
25 cm
C
The area of triangle ABC is 130 cm2.
AB = 16 cm and BC = 25 cm.
(a) Show clearly that angle ABC = 40.5o, correct to one decimal place.
Answer(a)
Area of triangle =
130 =
2 × 130
16 × 25
sin C =
1
2
1
2
× a × b × sin C
× 16 × 25 × sin C
= sin C
2 × 130
16 × 25
243
Workbook of Variant 3
C = sin-1
2 × 130
16 × 25
= 40.54160187
= 40.5
40.5
Answer: (a)
[3]
(b) Calculate the length of AC
a2 = b2 + c2 − 2bc cos A
AC2 = 162 + 252 − 2(16)(25) cos 40.542
AC = 162 + 252 − 2(16)(25) cos 40.542
AC = 16.52441367
16.5
Answer: (b)
cm [4]
(c) Calculate the shortest distance from A to BC.
[shortest distance is always perpendicular distance]
Considering triangle ABX
sin B =
sin B =
O
A
A
(4)
16 cm
AX
AB
sin 40.5 =
AX
B
40.5
(0)
x
C
16
AX = 16 × sin 40.5
AX = 10.39116877
Answer: (c)
244
10.4
cm [2]
Trigonometry
October / November 2013 Paper 23
10
(4)
(o)
628 m
h
15°
Calculate the length h.
Give your answer correct to 2 significant figures.
sin 15 =
sin 15 =
O
A
h
628
h = 628 × sin 15
h = 162.5383603
Answer: h =
160
m [3]
October / November 2013 Paper 43
2
A field, ABCD, is in the shape of a quadrilateral.
A footpath crosses the field from A to C.
C
26°
B
55 m
65°
A
32°
62 m
122°
D
(a) Use the sine rule to calculate the distance AC and show that it rounds to 119.9 m, correct to
1 decimal place.
245
Workbook of Variant 3
Using sine rule, considering triangle ACD
sin A
a
=
sin 122
AC
AC =
sin B
b
=
sin 26
62
62 × sin 122
sin 26
AC = 119.9417032
≈ 119.9
Answer: (a)
119.9
[3]
(b) Calculate the length of BC.
a2 = b2 + c2 − 2bc cos A
BC2 = 552 + 119.92 − 2(55)(119.9) cos 65
BC =
552 + 119.92 − 2(55)(119.9) cos 65
BC = 108.7524609
Answer: (b) BC =
109
m [4]
(c) Calculate the area of triangle ACD.
Area of triangle ACD =
=
1
2
1
2
× a × b × sin C
× 62 × 119.942 × sin 32
= 1970.346868
Answer: (c)
1970
m2
(d) The field is for sale at $ 4.50 per square metre.
Calculate the cost of the field.
Area of triangle ABC =
=
1
2
1
2
× a × b × sin 65
Area Cost / m2
14.50
4959.717004
× 55 × 119.942 × sin 65
= 2989.370136
[Cross Multiply]
x = 4959.717004 × 4.50
= 22318.72652
246
x
[2]
Trigonometry
Total area of field = Area of triangle ABC + area of triangle ACD
= 2989.370136 + 1970.346868
= 4959.717004
Answer: (d) $
22300
[3]
May / June 2014 Paper 23
14
P
66°
77°
37°
Q
12.5 cm
R
Calculate PR.
sin A
=
a
sin 37
b
=
PR
PR =
sin B
sin 66
12.5
12.5 × sin 37
sin 66
A = 8.234606966
Answer: PR =
16
H
8.23
cm [3]
G
F
E
12 cm
C
D
A
3 cm
4 cm
B
247
Workbook of Variant 3
ABCDEFGH is a cuboid.
AB = 4 cm, BC = 3 cm and AG = 12 cm.
Calculate the angle that AG makes with the base ABCD.
AC2 = AB2 + CB2
AC2 = 42 + 32
2
12 cm
3 cm
AC = 5 cm
A
B
4 cm
A
5 cm
(A)
C
A
cos A =
D A= 5
cos
H
G
C
12
12 cm
3 cm
5
A = cos-1
A
12
4 cm
A = 65.37568165
B
A
5 cm
(A)
C
Answer:
248
G
C
D
AC = 4 +3
2
65.4
[4]
Trigonometry
May / June 2014 Paper 43
3
P
12 cm
X
17 cm
Q
R
(a) The diagram shows triangle PQR with PQ = 12 cm and PR = 17 cm.
The area of triangle PQR is 97 cm2 and angle QPR is acute.
(i) Calculate angle QPR.
Area of triangle PQR =
97 =
1
2
1
2
× a × b × sin C
× 12 × 17 × sin P
97 x 2
= sin P
12 x 17
P = sin-1
( 1297xx172 )
P = 71.98589094
72
Answer: (a) (i) Angle QPR =
(ii) The midpoint of PQ is X.
Use the cosine rule to calculate the length of XR.
P
Using cosine rule,
6 cm
a2 = b2 + c2 − 2ab cos A
XR = 62 + 172 − 2(6)(17) cos 72
72
X
XR2 = 62 + 172 − 2(6)(17) cos 72
[3]
17 cm
6 cm
Q
XR = 16.18519488
R
Answer: (a) (ii) XR =
16.2
cm [4]
249
Workbook of Variant 3
(b)
42°
9.4 cm
a cm
37°
Calculate the value of a.
sin θ =
O
H
(4)
9.4 cm
x
sin 37 =
9.4
x = 9.4 × sin 37
x 42°
(0)
(A)
x
37°
a
x = 5.657061218
A
cos θ =
H
(4)
x
cos 42 = 9.4 cm
a
cos 42 =
a =
37°
5.657061218
x 42°
(0)
a
(A)
x
(4)
a
5.657061218
cos 42
a = 7.612326728
7.61
Answer: (b) a =
[4]
(c) sin x = cos 40o, 0o ≤ x ≤ 180o
Find the two values of x.
sin x = cos 40
x = sin-1(cos 40)
x = 50
sin θ = sin (180 − θ)
180 − 50 = 130o
Answer: (c) x =
250
50o
or
x=
130o
[2]
(4)
Trigonometry
May / June 2015 Paper 23
3
5 cm
x°
2 cm
Calculate the value of x.
A
cos x =
H
2
cos x =
5
x = cos-1
( )
2
5
x = 66.42182152
Answer: x =
11
66.4
[2]
C
100°
30°
A
24 cm
B
Use the sine rule to calculate BC.
Using sine rule ,
sin A
a
sin 30
BC
=
=
sin B
b
sin 100
24
251
Workbook of Variant 3
BC =
24 × sin 30
sin 100
BC = 12.18511934
Answer: BC =
12.2
[3]
20
R
9 cm
P
10 cm
Q
The area of triangle PQR is 38.5 cm2.
Calculate the length QR.
1
× a × b × sin C
Area of triangle =
Area of triangle PQR =
38.5 =
2 × 38.5
9 × 10
1
2
2
1
2
× 9 × 10 × sin P
× 9 × 10 × sin P
= sin P
sin P =
P = sin-1
2 × 38.5
9 × 10
2 × 38.5
9 × 10
P = 58.82116446
a2 = b2 + c2 − 2ab cos A
QR2 = 92 + 102 − 2(9)(10) cos 58.82116446
QR = 92 + 102 − 2(9)(10) cos 58.82116446
QR = 9.370806686
Answer: QR =
252
9.37
cm [6]
Trigonometry
October / November 2015 Paper 23
9
5 cm
2 cm
x°
Calculate the value of x.
sin x =
O
sin x =
2
H
x = cos-1
5
( )
2
5
x = 23.57817848
Answer: x =
23.6
[2]
October / November 2015 Paper 43
5
K
40°
65°
680 km
D
North
2380 km
M
1560 km
C
The diagram shows some distances between Mumbai (M), Kathmandu (K), Dhaka (D) and
Colombo (C).
(a) Angle CKD = 65o
253
Workbook of Variant 3
Use the cosine rule to calculate the distance CD
Considering triangle CDK,
Using cosine rule,
a2 = b2 + c2 − 2ab cos A
QR2 = 6802 + 23802 − 2(680)(2380) cos 65
QR = 6802 + 23802 − 2(680)(2380) cos 65
QR = 2181.483259
Answer: (a) CD =
2180
km [4]
(b) Angle MKC = 40o.
Use the sine rule to calculate the acute angle KMC.
Considering triangle KMC,
Using sine rule
sin A
a
=
sin 30
2380
sin M =
=
sin B
b
sin 40
1560
2380 × sin 40
sin M = sin-1
1560
2380 × sin 40
1560
A = 78.71418742
Answer: (b) Angle KMC =
78.7
[3]
(c) The bearing of K from M is 050o.
Find the bearing of M from C.
=
180 − (50 + 78.7)
North
[Co-interior angle]
K
=
51.3
North
50
M
= 360 − 51.3
[Angle around a point add upto 360°]
78.7
= 308.7
C
Answer: (c)
254
309
[2]
Trigonometry
(d) A plane from Colombo to Mumbai leaves at 21:15 and the journey takes 2 hours 24 minutes.
(i) Find the time the plane arrives at Mumbai.
hours Minutes
+
21
15
2
24
23
39
23:39
Answer: (d) (i)
[1]
(ii) Calculate the average speed of the plane.
Distance between Colombo to Mumbai is 1560 km
Time taken : 2 hours 24 minutes
2+
Average speed = total distance/total time
24
60
=
= 2.4
1560
2.4
= 650
Answer: (d) (ii)
650
km/h [2]
255
7
Workbook of
Solution 4
GRAPHS
258
Graphs
Chapter 7: Graphs
May / June 2010 Paper 23
19
v
3
Speed
(m / s)
0
4
Time (seconds)
15
t
The diagram shows the speed-time graph for 15 seconds of the journey of a cyclist.
(a) Calculate the acceleration of the cyclist during the first 4 seconds.
[Acceleration is gradient when speed in increasing]
acceleration =
3
rise
=
= 0.75
4
run
Answer (a):
0.75
m/s2
[1]
m/s2
[3]
(b) Calculate the average speed for the first 15 seconds.
here total distance = area covered under the graph
=
1
(a + b) h
2
=
1
(11 + 15) 3
2
= 39
average speed =
=
total distance
total time
39
15
= 2.6
Answer (b):
2.6
259
Workbook of Variant 3
May / June 2010 Paper 43
6
(a) Complete the table of value for
x
y
-4
-3
-2
-4.3 -3.3 -2.5
-1
-2
y=x+
-0.5
-2.5 1
.
x
0.5
2.5
1
2
2
2.5
3
3.3
4
4.3 [2]
(b)
y
y=2x+1
y= x+ 1
x
4
3
2
1
–4
–3
–2
–1
0
1
2
3
4
x
–1
–2
–3
–4
y= x+1
On the grid, draw the graph of y = x +
1
for –4 ≤ x ≤ –0.5 and 0.5 ≤ x ≤ 4.
x
Six of the points have been plotted for you.[3]
260
Graphs
(c) There are three integer values of k for which the equation x +
Write down these three values of k.
-1
Answer: (c) k =
1
= k has no solutions.
x
0
or k =
(d) Write down the ranges of x for which the gradient of the graph of y = x +
Answer (d):
(e) To solve the equation x +
1
or k =
[2]
1
is positive.
x
x < -1 and x > 1
[2]
1
= 2x + 1, a straight line can be drawn on the grid.
x
(i) Draw this line on the grid for –2.5 ≤ x ≤ 1.5.
For y = 2x + 1
x
1
2
3
y
3
5
7
[2]
(ii) On the grid, show how you would find the solutions.
Point of intersection gives solution
(iii) Show how the equation x +
[1]
1
= 2x + 1 can be rearranged into the form x2 + bx + c = 0
x
and find the values of the b and c.
1
= 2x + 1
x
x +
x2 +1
x
= 2x + 1
x2 + 1 = x(2x + 1)
x2 + 1 = 2x2 + x
0 = 2x2 + x − x2 − 1
x2 + x − 1 = 0
[Compalre it with x2 + bx + c]
Answer: (e) (iii) b =
1
c=
-1
[3]
261
Workbook of Variant 3
October / November 2010 Paper 23
15Find the equation of the straight line which passes through the point (0, 8) and (3, 2).
y = mx + c
m =
y2 − y1
x2 − x1
2−8
3−0
m = -2
m =
To find c, take any one point and put in the equation
y = -2x + c
8 = -2(0) + c
c = 8
y = -2x + 8
Answer: y =
21
20
18
16
14
1
12
2
Speed
10
(m / s)
8
6
4
3
2
0
10
20
Time (seconds)
262
30
40
-2x + 8
[3]
Graphs
The graph shows 40 seconds of a car journey.
The car travelled at a constant speed of 20 m/s, decelerated to 8 m/s2 then accelerated back to
20 m/s2.
Calculate
(a) the deceleration of the car,
[Deceleration is gradient where speed is decreasing]
Deceleration =
rise
20 − 8
=
= 2.4
run
5
Answer: (a)
2.4
m/s2
[1]
(b) the total distance travelled by car during the 40 seconds.
Method 1
Total distance = area under the graph
Total distance = area rectangle − area of triangle
= (40 × 20) −
= 680
( 12 × 20 × 12)
Method 2
Total distance = shape 1 + shape 2 + shape 3
= [
1
1
(10 + 15) × 12] + [
(10 + 25) × 12] + [8 × 40]
2
2
= 680
Answer: (b)
680
m [3]
October / November 2010 Paper 43
7
(a) Complete the table for the function f(x) =
x
y
-4
-3
x3
+ 1.
10
-2
-1
0
1
2
3
-5.4 -1.7 0.2
0.9
1
1.1
1.8
3.7
[2]
(b) On the grid, draw the graph of y = f(x) for –4 ≤ x ≤ 3.
263
Workbook of Variant 3
y
g (x)
4
f (x)
3
2
1
–4
–3
–2
–1
0
g (x)
1
2
3
x
–1
–2
–3
–4
–5
–6
–7
–8
4
, x ≠ 0.
x
(c) Complete the table for the fuction g(x) =
[4]
x
-4
-3
-2
-1
1
2
3
g(x)
-1
-1.3
-2
-4
4
2
1.3
[2]
(d) On the grid, draw the graph of y = g(x) for –4 ≤ x ≤ –1 and 1 ≤ x ≤ 3.
(e) (i) Use your graphs to solve the equation
x3
4
+1 =
.
x
10
[Find point of intersection of both curves]
Answer: (e) (i) x =
264
[3]
-2.9
or x =
2.1
[2]
Graphs
(ii) The equation
x3
4
+1=
can be written as x4 + ax + b = 0.
x
10
Find the values of a and b.
x3
4
+1=
x
10
x3 + 10
4
=
x
10
x(x3 + 10) = 4 × 10
x4 + 10x = 40
x4 + 10x = 40
x4 + 10x − 40 = 0
[compare this with x4 + ax = b]
a = 10, b = -40
Answer: (d) (ii) a =
10
b=
-40
[2]
May / June 2011 Paper 23
14
y
13
1
0
3
x
The diagram shows the straight line which passes through the points (0,1) and (3,13).
Find the equation of the straight line.
y = mx + c
m=
y2 − y1
x2 − x1
265
Workbook of Variant 3
m=
13 − 1
3−0
m=4
line is passing through 1 on y - intercept
c=1
y = 4x + 1
Answer:
y = 4x + 1
[3]
22
3
Speed
(km / min)
0
4
50
Time (min)
60
A train journey takes one hour.
The diagram shows the speed-time graph for this journey.
(a) Calculate the total distance of the journey.
Give your answer in kilometres.
Total distance = A (trapezium)
Total distance =
Distance = 159
1
(46 + 60) × 3
2
Answer: (a)
159
km [3]
(b) (i) Convert 3 kilometres/minute into metres/second.
3 × 1000 metre
3 kilometres
=
= 50
60 seconds
minute
Answer: (b) (i)
266
50
m/s [2]
Graphs
(ii) Calculate the acceleration of the train during the first 4 minutes.
Give your answer in metres / second2.
4 minutes = 4 × 60 second = 240 second.
rise
50
5
=
=
= 0.2083333333
run
240
24
Acceleration =
0.208
Answer: (b) (ii)
m/s2
[2]
May / June 2011 Paper 43
5(a)Complete the table of values for the function f(x), where f(x) = x2 +
1
,x≠0
x2
x
-3
-2.5
-2
-1.5
-1
-0.5
0.5
1
1.5
2
2.5
3
f(x)
9.11
6.41
4.25
2.69
2
4.25
4.25
2
2.69
4.25
6.41
9.11
[3]
(b) On the grid, draw the graph of y = f (x) for –3 ≤ x ≤ –0.5 and 0.5 ≤ x ≤ 3.
y
10
f (x)
f (x)
9
8
7
y=2x
6
5
4
y=3
3
2
1
–3
–2
–1
0
1
2
3
x
–1
–2
[5]
267
Workbook of Variant 3
(c) (i) Write down the equation of the line of symmetry of the graph.
[Line of symmetry for this graph is y - axis]
x=0
Answer: (c) (i)
[1]
(ii) Draw the tangent to the graph of y = f(x) where x = –1.5
Use the tangent to estimate the gradient of the graph of y = f(x) where x = −1.5
After drawing tangent take any 2 points which are on tangent and find gradient.
(-1.5, 2.69)(-3, 6.2)
m =
y2 − y1
x2 − x1
m =
6.2 − 2.69
-3 − (-1.5)
m = -2.34
-2.34
Answer: (c) (ii)
(iii) Use your graph to solve the equation x2 +
x2 +
[3]
1
= 3.
x2
1
= f(x). Draw line y = 3, point of intersection will give solution]
x2
Answer: (c) (iii) x =
-1.6
or x =
0.7
or x =
0.7
or x =
1.6
(iv)Draw a suitable line on the grid and use your graphs to solve the equation x2 +
[Draw line y = 2x, point of intersection with graph f(x) will give solution]
[2]
1
= 2x.
x2
For y = 2x
x
1
2
3
y
2
4
6
Answer: (c) (iv) x =
268
1
or x =
1.8
[3]
Graphs
October / November 2011 Paper 23
23
40
Speed
(m / s)
0
5
10
Time (minutes)
15
he diagram shows the speed-time graph for the first 15 minutes of a train journey. The train
T
accelerates for 5 minutes and then continues at a constant speed of 40 metres/second.
(a) Calculate the acceleration of the train during the first 5 minutes.
Give your answer in m/s2.
Acceleration =
rise
40
2
=
=
= 0.1333333333.
run
5 × 60
15
Answer: (a)
2
or 0.133
15
m/s2
[2]
(b) Calculate the average speed for the first 15 minutes of the train journey.
Give your answer in m/s.
Total distance = Area of trapezium.
Total distance =
=
1
(a + b) h
2
1
(600 + 900) × 40
2
= 30000
Average speed =
=
Total distance
Totale time
30000
900
= 33.3333333333
Answer: (b)
33.3
m/s [3]
269
Workbook of Variant 3
October / November 2011 Paper 43
2
(a) Complete the table of values for y = 2x.
x
-2
-1
0
1
2
3
y
0.25
0.5
1
2
4
8
[2]
(b) On the grid, draw the graph of y = 2x for –2 ≤ x ≤ 3.
y
10
9
y=2x+2
y=2x
8
7
6
5
targent at x =1
4
3
2
1
–2
0
–1
1
2
3
x
–1
(c) (i) On the grid, draw the straight line which passes through the points (0, 2) and (3, 8).
[3]
[1]
(ii) The equation of this line is y = mx + 2.
Show that the value of m is 2.
Method 1
m=+
m=
6
3
m = 2
270
rise
run
Method 2
m=
y2 − y1
x2 − x1
m=
8−2
3−0
m=2
[1]
Graphs
(iii) One answer to the eqution 2x = 2x + 2 is x = 3
Use your graph to find the other answer.
[point of intersect will give another answer]
For y = 2x + 2
x
1
2
3
y
4
6
8
-0.75
Answer: (c) (iii) x =
[1]
(d) Draw the tangent to the curve at the point where x = 1.
Use this tangent to calculate an estimate of the gradient of y = 2x when x = 1.
Take 2 any points which are on tangent and find gradient
m =
y2 − y1
x2 − x1
m =
4.6 − 2
3−1
(x1 y1)
(1, 2)
(x2 y2)
(3, 4.6)
m = 1.3
1.3
Answer: (d)
[3]
May / June 2012 Paper 23
19
120
100
2
1
80
Speed
(km / h) 60
3
40
20
0
0.5
1
1.5
2
2.5
3
3.5
Time (minutes)
The diagram shows the speed-time graph for part of a car journey.
The speed of the car is shown in kilometres/hour.
271
Workbook of Variant 3
Calculate the distance travelled by the car during the 3.5 minutes shown in the diagram.
Give your answer in kilometres.
Total distance = shape 1 + shape 2 + shape 3
(shape 1 and 2 have same area)
Total distance = 2(
= 2
( 12 ×
1
× b × h) + l × b
2
) ( 1207
1
× 20 +
120
× 100
)
= 6
6
Answer:
km [4]
May / June 2012 Paper 43
7
f(x) = 2x
(a) Complete the table.
x
f(x)
0
1
0.5
1.4
1
2
1.5
2.8
(b) Draw the graph of y = f(x) for 0 ≤ x ≤ 4.
272
2
4
2.5
5.7
3
8
3.5
11.3
4
16
[3]
Graphs
y
f (x)
16
15
14
13
y= 3x
12
11
targent
10
9
8
7
6
y=5
5
4
3
2
1
0
0.5
1
1.5
2
2.5
3
3.5
4
x
[4]
(c) Use your graph to slve the equation 2x = 5
[Draw line y = 5 point of intersection with f(x) will give solution]
Answer: (c) x =
2.3
[1]
273
Workbook of Variant 3
(d) Draw a suitable straight line and use it to solve the equation 2x = 3x.
[Draw line y = 3x and point of intersection with f(x) will give solution]
For y = 3x
x
1
2
3
y
3
6
9
0.4
Answer: (d) x =
or x =
3.25
[3]
(e) Draw a suitable tangent and use it to find the co-ordinates of the point on the graph of
y = f(x), where the gradient of the graph is 3.
[draw a parallel line with line y = 3x such that new drawn line should be tangent of f(x)]
Answer: (e)
2.25
(
,
4.8
)
[3]
October / November 2012 Paper 23
19
5
4
3
Speed
(metres per
second)
2
2
1
1
0
2
4
6
8
10
12
14
16
18
Time (seconds)
The diagram shows the speed time graph for the last 18 seconds of Roman’s cycle journey.
(a) Calculate the deceleration.
[Deceleration is gradient where speed is decreasing]
rise
5
=
= 0.625
8
run
Deceleration =
Answer: (a)
274
5
/ 0.625
8
m/s2
[1]
Graphs
(b) Calculate the total distance Roman travels during the 18 seconds.
distance = shape 1 + shape 2
distance =
1
1
(a + b) h +
(a + b) h
2
2
distance = [
1
1
(1 + 5) × 4] + [
(6 + 14) × 5]
2
2
distance = 62
62
Answer: (b)
m [3]
October / November 2012 Paper 43
4
f(x) =
2
− 3x, x ≠ 0.
x2
(a) Complete the table.
x
-3
-2.5
-2
-1.5
-1
-0.5
0.5
1
1.5
f(x)
9.2
7.8
6.5
5.4
5
9.5
6.5
-1
-3.6
2
2.5
3
-5.5 -7.2 -8.8
[2]
275
Workbook of Variant 3
(b) On the grid, draw the graph of y = f(x), for –3 ≤ x ≤ –0.5 and 0.5 ≤ x ≤ 3.
[5]
y
10
f (x)
y=3x
9
8
7
6
5
y=4
4
3
2
1
–3
–2
–1
0
1
2
3
x
–1
–2
–3
–4
–5
–6
–7
–8
–9
276
f (x)
Graphs
(c) Use your graph to solve the equations.
(i) f(x) = 4
[draw line y = 4]
Answer: (c) (i) x =
0.6
Answer: (c) (ii) x =
0.75
[1]
(ii) f(x) = 3x
[draw line y = 3x]
[point of intersection is answer]
[2]
(d) The equation f(x) = 3x can be written as x3 = k.
Find the value of k.
f(x) = 3x
2
− 3x = 3x
x2
2
= 3x + 3x
x2
2
= 6x
x2
2
= x3
6
x3 =
1
3
Answer: (d) k =
1
3
[2]
277
Workbook of Variant 3
(ii) Find the equation of this line.
y = mx + c
m =
m =
(y2 − y1)
(x2 − x1)
−9 − 5
3 − (-1)
m = −3.5
y = mx + c
y = −3.5 + c
5 = −3.5(-1) + c
5 = 3.5 + c
c = 1.5
y = −3.5x + 1.5
y = -3.5x + 1.5
Answer: (e) (ii) y =
[3]
(iii) Complete the statement.
tangent
The straight line in part (e) (ii) is a
to the graph of y = f(x)[1]
May / June 2013 Paper 43
5
(a) Complete this table of values for the fuction f(x) =
278
1
− x2, x ≠ 0.
x
x
-3
-2
-1
-0.5
-0.2
0.2
0.5
1
2
3
f(x)
-9.33
-4.5
-2
-2.25
-5.04
4.96
1.75
0
-3.5
-8.67
[3]
Graphs
(b) Draw the graph of f(x) =
1
− x2 for –3 ≤ x ≤ –0.2 and 0.2 ≤ x ≤ 3.
x
y
5
x = -2
y =2x -2
4
3
2
1
–3
–2
0
–1
1
2
x
3
–1
–2
y = -3
–3
–4
–5
–6
–7
–8
f (x)
–9
f (x)
–10
[5]
(c) Use your graph to solve f(x) = –3.
y = -3 [draw y = -3]
Answer: (c) x =
-1.55
or x =
-0.35
or x =
1.8
[3]
279
Workbook of Variant 3
(d) By drawing a suitable line on your graph, solve the equation f(x) = 2x – 2.
For y = 2x - 2
x
1
2
3
y
0
2
4
-2.6
Answer: (c) x =
or x =
-0.4
1
or x =
[3]
(e) By drawing a suitable tangent, work out an estimate of the gradient of the curve at the point
where x = –2.
You must show your working.
(After drawing tangent at x = -2 take any 2 points which are on tangent and find gradient)
(-2,
m =
-5.04)
(0,
2.8)
2.8 − (-5.04)
0 − (-2)
= 3.92
Answer: (e)
3.92
October / November 2013 Paper 43
3
A rectangular metal sheet measures 9 cm by 7 cm.
A square, of side x cm, is cut from each corner.
The metal is then folded to make an open box of height x cm.
9 cm
7 cm
x cm
x cm
(a) Write down, in terms of x, the length and width of the box.
280
x cm
[3]
Graphs
Length = 9 − x − x = 9 − 2x
Width = 7 − x − x = 7 − 2x
Answer: (a) Length =
Width =
9 − 2x
7 − 2x
[2]
(b) Show that the volume, V, of the box is 4x3 − 32x2 + 63x.
Volume of cuboid = l × b × h
= (9 − 2x) (7 − 2x)x
= (9 − 2x) (7x − 2x2)
= 9(7x − 2x2) − 2x (7x − 2x2)
= 63x − 18x2 − 14x2 + 4x3
= 4x3 − 18x2 − 14x2 + 63x
= 4x3 − 32x2 + 63x
Answer: (b)
4x3 − 32x2 + 63x
[2]
(c) Complete this table of values for V = 4x3 − 32x2 + 63x
x
0
0.5
1
1.5
2
2.5
3
3.5
V
0
24
35
36
30
20
9
0
4(0.5)3 − 32(0.5)2 + 63(.05) = 24
4(2.5)3 − 32(2.5)2 + 63(2.5) = 20
[2]
(d) On the grid, draw the graph of V = 4x3 − 32x2 + 63x for 0 ≤ x ≤ 3.5
281
Workbook of Variant 3
Three of the points have been plotted for you.
V
40
35
30
25
20
15
10
5
0
0.5
1
1.5
2
2.5
3
3.5
x
[3]
(e) The volume of the box is at least 30 cm3.
Write down, as an inequality, the possible values of x.
Answer: (e)
0.73 ≤ x ≤ 2
[2]
(f) (i) Write down the maximum volume of the box.
Answer: (f) (i)
(ii) Write down the value of x which gives the maximum volume.
282
37
cm3
[1]
Graphs
[corresponding value of x where volume is maximum]
Answer: (f) (ii)
8
(a) Rearrange s = ut +
s − ut =
1.25
[1]
1 2
at to make a the subject.
2
1 2
at
2
2(s − ut) = at2
2(s − ut)
a =
t2
=a
2(s − ut)
t2
2(s − ut)
Answer: (a) a =
t2
[3]
(b) The formula v = u + at can be used to calculate the speed, v of a car.
u = 15, a = 2 and t = 8, each correct to the nearest integer.
Calculate the upper bound of the speed v.
u, a and t correct to nearest integer.
1
= 0.5
2
half of 1 =
upper bound of u = 150 + 0.5 = 15.5
upper bound of a = 2 + 0.5 = 2.5
upper bound of t = 8 + 0.5 = 8.5
upper bund of speed V = 15.5 + 2.5 × 8.5
= 36.75
Answer: (b)
36.75
[3]
(c) The diagram shows the speed-time graph for a car travelling between two sets of
283
Workbook of Variant 3
traffic lights.
16
Speed
(m/s)
0
10
Time (seconds)
20
25
(i) Calculate the deceleration of the car for the last 5 seconds of the journey.
deceleration =
rise
16
=
= 3.2
run
5
Answer: (c) (i)
3.2
m/s2
[1]
(ii) Calculate the average speed of the car between the two sets of traffic lights.
Total distance = area of trapezium
Distance =
=
1
(a + b) h
2
1
(10 + 25) × 16
2
= 280 m
Average speed =
=
280
25
Total distance
Total time
= 11.2
Answer: (c) (ii)
May / June 2014 Paper 23
284
11.2
m/s [4
Graphs
13 Find the equation of the line passing through the point with co-ordinates (5, 9) and (–3, 13).
y = mx + c
m=
y2 − y1
x2 − x1
m=
13 − 9
-3 − 5
m=
-1
2
Take any one point to find value of C
y=x+c
9=
-1
(5) + c
2
c=9+
5
2
c=
23
2
y=
-1
23
x+
2
2
y=
Answer:
-1
23
x+
2
2
[3]
May / June 2014 Paper 43
4
The table shows some values for the function y =
x
y
-3
-2
-2.89 -1.75
-1
-0.5
0.5
1
0
3.5
4.5
2
1
+ x, x ≠ 0
x2
2
3
4
2.25 3.11 4.06
(a) Complete the table of values.
(b) On the grid, draw the graph of y =
[3]
1
+ x for –3 ≤ x ≤ –0.5 and 0.5 ≤ x ≤ 4.
x2
285
Workbook of Variant 3
y
5


4
y = 1-x
y=1
x +x
2
targent at x = 2


3
2
y=3

1
–3
–2
–1
0
1
2
3
4
x
–1

–2
y = 1x + x
2

–3
[5]
(c) Use your graph to solve the equation
1
+x−3=0
x2
1
+x=3
x2
1
+x−3=0
x2
[Make changes such that left hand side becomes original fraction]
Draw line y = 3
y = 3
Answer: (c) x =
286
-0.5
or x =
0.6
or x =
2.9
[3]
Graphs
(d) Use your graph to solve the equation
Draw a line y = 1 − x
Point of intersection will give solution
1
+x=1−x
x2
For y=1 − x
x
1
2
3
y
0
-1
-2
-0.65
Answer: (d) x =
[3]
(e) By drawing a suitable tangent, find an estimate of the gradient of the curve at the point
where x = 2.
x = 2
After drawing tangent, take any 2 point witch are on tangent and find gradient
(2, 2.25) and (4, 3.6)
m =
y2 − y1
x2 − x1
m =
3.6 − 2.25
4−2
m = 0.675
0.675
Answer: (e)
(f) Using algebra, show that you can use the graph at y = 0 to find
3
[3]
−1
Answer (f):
y =
1
+x
x2
0 =
1
+x
x2
0 =
1 + x3
x2
0 = 1 + x3
-1 = x3
x =
3
−1
3
Answer:
−1
[3]
287
Workbook of Variant 3
May / June 2015 Paper 23
12
10
Speed (m/s)
0
u
3u
Time (seconds)
A car starts from rest and accelerates for u seconds until it reaches a speed of 10 m/s.
The car then travels at 10 m/s for 2u seconds.
The diagram shows the speed-time graph for this journey.
The distance travelled by the car in the first 3u seconds is 125 m.
(a) Find the value of u.
Distance =
125 =
125 ×
1
2
1
(a + b) h
2
[(3u − u) + 3u] 10
2
= 2u + 3u
10
125 = 5u
u = 25
25
Answer: (a) u =
[3]
(b) Find the acceleration in the first u seconds.
Acceleration =
rise
10
=
=2
run
5
Answer: (b)
288
2
m/s2
[1]
Graphs
May / June 2015 Paper 43
3
f(x) =
8
x
+
,x≠0
2
x
2
(a) Complete the table of value for f(x).
x
-5
-4
-3
-2
-1.5
1.5
2
2.5
3
3.5
4
5
f(x)
-2.2
-1.5
-0.6
1
2.8
4.3
3
2.5
2.4
2.4
2.5
2.8
[3]
(b) On the grid, draw the graph of y = f(x) for –5 ≤ x ≤ –1.5 and 1.5 ≤ x ≤ 5.
y
Y=1-x
X= -3
5
x
4
3
x
x
x
x
x
f (x)
x
2
x
–5
–4
–3
–2
1
–1
0
1
2
3
4
5
x
x
–1
x
f (x) x
–2
–3
[5]
289
Workbook of Variant 3
(c) Solve f(x) = 0.
f(x) means y.
y = 0
is x axis.
Intersection of f(x) with x axis is solution
Answer: (c) x =
-2.6
[1]
(d) By drawing a suitable line on the grid, solve the equation f(x) = 1 − x.
For y = 1 − x
x
1
2
3
y
0
-1
-2
[draw line y = 1 − x]
Answer: (d) x =
-1.6
[3]
(e) By drawing a tangent at the point (–3, –0.6), estimate the gradient of the graph of
y = f(x) when x = –3
After during tangent at x = -3 take 2 any point which are on tangent and find gradient.
(3, 0.6)(0, 2.8)
m =
y2 − y1
x2 − x1
m =
2.8 − (-0.6)
0 − (-3)
m = 1.133333333
Answer: (e)
290
1.13
[3]
Graphs
October / November 2015 Paper 23
26
45
Speed
(km/h)
0
10
20
Time (seconds)
30
The diagram shows the speed-time graph of a car.
The car travels at 45 km/h for 20 seconds.
The car then decelerates for 10 seconds until it stops.
(a) Change 45 km/h into m/s.
[to convert km/h to m/s → ×
5
]
18
5
m/s = 12.5
18
45 km/h = 45 ×
Answer: (a)
12.5
m/s [2]
(b) Find the deceleration of the car, giving your answer in m/s2.
deceleration =
rise
12.5
=
= 1.25
run
10
Answer: (b)
1.25
m/s2
[1]
(c) Find the distance travelled by the car during the 30 seconds, giving your answer in metres.
distance = Area of trapezium
distance =
distance =
1
(a + b) h
2
1
(20 + 30) × 12.5
2
distance = 312.5
Answer: (c)
312.5
m [3]
291
Workbook of Variant 3
October / November 2015 Paper 43
4
f(x) = x −
1
,
2x2
x≠0
(a) Complete the table of values.
x
-3
-2
-1.5
-1
-0.5
-0.3
0.3
0.5
1
1.5
2
f(x)
-3.1
-2.1
-1.7
-1.5
-2.55
-5.9
-5.3
-1.5
0.5
1.3
1.9
[2]
(b) On the grid, draw the graph of y = f(x) for –3 ≤ x ≤ –0.3 and 0.3 ≤ x ≤ 2.
[5]
y
5
y=2-x
4
3
2x
–3
–2
–1
x
1
x
0
1
x
f (x)
y=1
2
x
–1
x
x
x
–2
x
x
f(x)
x
–3
–4
–5
x
x–6
(c) Use your graph to solve the equation f(x) = 1.
f(x) means y
draw line y = 1 and point of intersection will be solution.
Answer: (c) x =
292
1.3
[1]
Graphs
(d) There is only one negative integer value, k, for which f(x) = k has only one solution for all real
x.
Write down this value of k.
y = -1 is only horizontal line which negative integer and will have only one solution
Answer: (d) k =
-1
[1]
1
(e) The equation 2x −
− 2 = 0 can be solved using graph of y = f(x) and a straight
2x2
line graph.
(i) Find the equation of this straight line.
[take this equation and keep doing algebric changes till left hand becomes exactly same
1
as origin f(x) = x −
]
2x2
2x −
1
−2=0
2x2
1
=2
2x2
subtract x both side
2x −
1
=2−x
2x2
2x − x −
x −
1
=2−x
x2
y=2-x
Answer: (e) (i)
(ii) On the grid, draw this straight line and solve the equation 2x −
y=2−x
[1]
1
− 2 = 0.
2x2
For y = 2 − x
x
0
1
2
y
2
1
0
[Point of intersection gives solution]
Answer: (e) (ii)
1.2
[3]
293
Workbook of Variant 3
294
8
Workbook of
Solution 4
SETS, VECTORS,
AND FUNCTIONS
296
Sets, Vectors and Functions
Chapter 8: Sets, Vectors and Functions
May / June 2010 Paper 23
7
Shade the required regions in the Venn diagram below.
A
B
A
B
C
C
[2]
15
G
g
O
N
h
H
In triangle OGH, the ratio GN : NH = 3 : 1.
OG = g and OH = h.
Find the following in terms of g and h, giving your answers in their simplest form.
297
Workbook of Variant 3
(a) HG
HG = HO + OG
= -h + g = g − h
Answer: (a) HG
g−h
[1]
(b) ON
GN : NH = 3:1
HN =
1
of HG
4
ON = OH + HN
= h +
1
(g − h)
4
= h +
1
1
g−
h
4
4
= h −
1
1
h+
g
4
4
ON =
1
3
g+
h
4
4
Answer: (b) ON =
18 f (x) = x2 + 2
g(x) = (x + 2)2
1
3
g+
h
4
4
[2]
h(x) = 3x – 5
Find
(a) g f (–2),
f(-2) = (-2)2 + 2
f(-2) = 4 + 2
f(-2) = 6
gf(-2) = g(6)
= (6 + 2)2
= 82
= 64
Answer: (a)
298
64
[2]
Sets, Vectors and Functions
(b) h–1 (22).
h-1(22) = x
22 = h(x)
22 = 3x − 5
22 + 5 = 3x
3x = 27
x = 27/3
x = 9
9
Answer: (b)
[2]
October / November 2010 Paper 23
4
Shade the required region on each Venn diagram.
P
A
Q
B
R
A ∩ B'
(P ∪ Q) ∩ R'
[2]
11 In a group of 24 students, 21 like football and 15 like swimming.
One student does not like football and does not like swimming.
Find the number of students who like both football and swimming.
21 − x + x + 15 − x + 1 = 24
=24
21 + 15 + 1 − x = 24
21 + 15 + 1 − 24 = x
S
F
21-x
x
15-x
13 = x
x = 13
1
Answer:
13
[2]
299
Workbook of Variant 3
25 f : x → 2x − 7
g:x→
Find
(a) fg
( 12 ),
g(
1
)=
2
g(
1
)=2
2
1
x
1
1
2
1
1
g( ) =
2
0.5
1
) = f(2)
2
= 2(2) − 7
fg(
= -3
Answer: (a)
-3
[2]
(b) g f (x),
gf(x) =
1
2x − 7
Answer: (b) gf(x) =
1
2x − 7
[1]
(c) f –1(x).
f(x) = 2x − 7
y = 2x − 7
[Replaced f(x) with y]
x = 2y − 7
[Inter changed ‘x’ and ‘y’]
x + 7 = 2y[Make y the subject]
x +7
=y
2
f-1(x) =
x+7
[repleced y with f-1(x)]
2
Answer: (c) f-1(x) =
300
x+7
2
[2]
Sets, Vectors and Functions
October / November 2010 Paper 43
9
y
3
A
2
1
–4
–3
–2
C
–1 0
–1
1
2
3
4
5
x
–2
–3
–4
B
(a) The points A(5, 3), B(1, –4) and C(–4, –2) are shown in the diagram.
(i) Write CA as a column vector.
[9 unit right and 5 unit up]
Answer: (a) (i) CA =
( )
[1]
(7)
[2]
BA
[1]
9
5
(ii) Find CA − CB as a single column vector.
( -25 )
9
CA − CB = ( ) − ( 5 )
5
-2
=( 9-5 )
5 - (-2)
4
=( )
7
CB =
4
Answer: (a) (ii)
(iii)Complete the following statement.
BA =
( 47 )
Answer: (a) (iii) CA − CB
301
Workbook of Variant 3
(iv) Calculate CA .
CA  = 92 + 52
CA  = 10.29563014
10.3
Answer: (a) (iv)
[2]
(b)
D
u
C
t
M
A
B
1
ABCD is a trapezium with DC parallel to AB and DC =
AB.
2
M is the midpoint of BC.
AD = t and DC = u.
Find the following vectors in terms of t and / or u.
Give each answer in its simplest form.
(i) AB
DC =
1
AB
2
2DC = AB
AB = 2u
Answer: (b) (i) AB =
2u
[1]
(ii) BM
BC = BA + AD + DC
BC = -2u + t + u
BC = t − u
M is midpoint of BC
BM =
1
BC
2
BM =
1
(t − u)
2
Answer: (b) (ii) BM =
302
1
(t − u)
2
[2]
Sets, Vectors and Functions
(iii) AM
AM = AB + BM
= 2u +
1
(t − u)
2
= 2u +
1
t− 1 u
2
2
= 2u −
1
u+ 1 t
2
2
=
3
u+ 1 t
2
2
Answer: (b) (iii) AM =
3
u+ 1 t
2
2
[2]
May / June 2011 Paper 23
3
(a)
A
B
A
B
Shade the region A ∩ B'
A
(b)
[1]
B
7
4
5
7
4
5
3
A
B
3
This Venn diagram shows the number of elements in each region.
Write down the value of n(A ∪ B' ).
n (A U B’) = 7 + 4 + 3
= 14
Answer: (b) n(A U B’)=
14
[1]
303
Workbook of Variant 3
May / June 2011 Paper 43
9
f (x) = 3x + 1
g(x) = (x + 2)2.
(a) Find the value of
(i) g f (2)
f(2) = 3(2) + 1
f(2) = 7
gf(2) = g(7)
= (7 + 2)2 =
92 = 81
Answer: (a) (i)
81
[2]
(ii) f f (0.5)
f(0.5) = 3(0.5) + 1
f(0.5) = 2.5
ff(0.5) = f(2.5)
= 3(2.5) + 1 = 8.5
Answer: (a) (ii)
8.5
[2]
(b) Find f–1 (x), the inverse of f(x).
f(x) = 3x + 1
y = 3x + 1
x = 3y + 1
x − 1 = 3y
x−1 =y
3
f-1(x) =
x−1
3
x−1
Answer: (b)
(c) Find fg(x).
Give your answer in its simplest form.
fg(x) = 3(x + 2)2 + 1
= 3(x + 2)(x + 2) + 1
= 3[x(x + 2) + 2(x + 2)] + 1
= 3[x2 + 2x + 2x + 4] + 1
304
3
[2]
Sets, Vectors and Functions
= 3x2 + 12x + 12 + 1
= 3x2 + 12x + 13
3x2 + 12x + 13
Answer: (c)
x2 + f(x) = 0
(d) Solve the equation [2]
Show all your working and give answers correct to 2 decimal places.
x2 + f(x) = 0
x2 + 3x + 1 = 0
a = 1, b = 3, c = 1
x =
−b ± b2 − 4ac
2a
2
x = −3 ± 3 − 4(1)(1)
2(1)
x =
-3 ± 5
2
x =
–3 + 5
2
x = -0.3819660113
x=
–3 − 5
2
or x = -2.618033989
-0.38
Answer: (d) x =
-2.62
or x =
[4]
10 (a)
C
L
D
N
M
q
A
p
B
ABCD is a parallelogram.
L is the midpoint of DC, M is the midpoint of BC and N is the midpoint of LM.
AB = p and AD = q.
(i) Find the followings in terms of p and q, in their simplest form.
305
Workbook of Variant 3
(a) AC
AC = AD + DC
= q + p
= p + q
p+q
Answer: (a) (i) (a) AC =
[1]
(b) LM
LM = LC + CM
=
1
1
(DC) +
(CB)
2
2
=
1
1
(p) +
(-q)
2
2
=
1
1
p−
q
2
2
Answer: (a) (i) (b) LM =
1
1
p−
q
2
2
[2]
(c) AN
AN = AD + DL + LN
AN = q +
1
1
DL +
LM
2
2
AN = q +
1
1
p+
2
2
AN = q +
1
1
1
p+
p−
q
2
4
4
( 12 p − 12 q)
AN =
1
1
1
p+
p+q−
q
2
4
4
AN =
3
3
p+
q
4
4
Answer: (a) (i) (c) AN =
3
3
p+
q
4
4
[2]
(ii) Explain why your answer for AN shows that the point N lies on the AC.
Answer(a)(ii)AN =
AN =
3
(p+q)
4
AN =
Answer: (a) (ii)
306
3
3
p+
q
4
4
3
AC, AN is multiple of AC
4
[1]
Sets, Vectors and Functions
(b)
F
G
2x°
(x + 15)°
H
J
75°
E
EFG is a triangle.
HJ is parallel to FG.
Angle FEG = 75o.
Angle EFG = 2xo and angle FGE = (x + 15)o.
(i) Find the value of x.
considering triangle FEG
2x + 75 + x + 15 = 180
2x + x = 180 − 75 − 15
3x = 90
x = 30
Answer: (b) (i) x =
30
[2]
(ii) Find angle HJG.
Angle HJG = 180 − Angle JGF [angle HJG and angle JGF are co-interior angles]
Angle HJG = 180 − (x+15)
Angle HJG = 180 − (30 + 15)
Angle HJG = 135
Answer: (b) (ii) Angle HJG =
135
[1]
307
Workbook of Variant 3
October / November 2011 Paper 23
17
R
F
13-x
x
11-x
In the Venn diagram = {students in a survey}, R = {student who like rugby} and
F = {student who like football}.
n ( ) = 20
n (R ∪ F) = 17
n (R) = 13
n(F) = 11
(a) Find
(i) n (R ∩ F),
13 − x + x + 11 − x = 17
13 + 11 − x = 17
13 + 11 − 17 = x
x = 7
Answer: (a) (i)
7
[1]
(ii) n(R' ∩ F),
n(R’ ∩ F) = 11 − x
n(R’ ∩ F) = 11 − 7
n(R’ ∩ F) = 4
Answer: (a) (ii)
4
[1]
(b) A student who likes rugby is chosen at random.
Find the probability that this student also likes football.
[Number student who like rugby = 13]
[ from rugby, who like football = 7]
Answer: (b)
308
7
13
[1]
Sets, Vectors and Functions
October / November 2011 Paper 43
8
f(x) = x2 + x − 1
g(x) = 1− 2x h(x) = 3x
(a) Find the value of hg(–2).
g(-2) = 1 − 2(-2)
g(-2) = 5
hg(-2) = h(5)
= 35
= 243
243
Answer: (a)
[2]
(b) Find g–1(x).
g(x) = 1 − 2x
y = 1 − 2x
x = 1 − 2y
2y = 1 − x
y =
1−x
2
1−x
2
Answer: (b) g-1(x) =
[2]
(c) Solve the equation f(x) = 0.
Show all your working and give your answer correct to 2 decimal places.
f(x) = 0
x2 + x − 1 = 0
a = 1, b = 1, c = -1
x =
−b ± b2 − 4ac
2a
x =
−1 ± (1)2 − 4(1)(-1)
2(1)
x =
−1 ± 5
2
x =
−1 + 5
2
x = −1 − 5
2
x = 0.6180339887orx = 1.618033989
Answer: (c) x =
0.62
or x =
-1.62
[4]
309
Workbook of Variant 3
(d) Find fg(x).
Give your answer in its simplest form.
fg(x) = (1 − 2x)2 + (1 − 2x) − 1
fg(x) = (1 − 2x)(1 − 2x) + (1 − 2x) − 1
fg(x) = 1(1 − 2x) − 2x(1 − 2x) + 1 − 2x − 1
fg(x) = 1 − 2x − 2x + 4x2 + 1 − 2x − 1
fg(x) = 4x2 − 2x − 2x − 2x + 1 + 1− 1
fg(x) = 4x2 − 6x + 1
4x2 − 6x + 1
Answer: (d) fg(x) =
[3]
(e) Solve the equation
h–1(x) = 2
x = h(2)
x = 32
x = 9
9
Answer: (e)
[1]
11 (a)
y
5
R
4
3
Q
2
P
–3
1
–2
–1
0
1
2
3
4
5
x
The points P and Q have co-ordinates (–3, 1) and (5, 2).
(i) Write PQ as a column vector.
PQ = 8 units right, 1 unit up
Answer: (a) (i) PQ =
310
( 81 )
[1]
Sets, Vectors and Functions
(ii) QR = 2
( –11 ) = ( –22 )
Mark the point R on the grid.
[QR = from Q, 2 unit to left and 2 unit up]
(iii) Write down the position vector of the point P.
[position vector means from O to P]
[From origin , 3 unit left and 1 unit up]
Answer: (a) (iii)
(b)
( -31 )
[1]
U
L
u
O
M
K
V
v
In the diagram, OU = u and OV = v.
K is on UV so that UK =
M is the midpoint of KL.
2
3
UV and L is on OU so that OL =
OU.
3
4
Find the following in term of u and v, giving your answers in their simplest form.
(i) LK
UV = -u + v
UK =
2
(UV)
3
UK =
2
(-u + v)
3
UK =
-2
2
u+
v
3
3
LU =
1
3
OU [As OL =
OU]
4
4
LU =
1
u
4
LK = LU + UK
311
Workbook of Variant 3
=
1
2
2
u−
u+
v
4
3
3
2
= -5 u +
v
3
12
Answer: (b) (i) LK =
-5 u + 2 v
3
12
[4]
(ii) OM
M is midpoint of LK
∴ LM =
OM = OL + LM
OM =
3
1
OU +
LK
4
2
OM =
3
1
2
u+
( -5 u +
v)
4
2
3
12
OM =
3
2
u − -5 u +
v
4
6
24
1
LK
2
1
OM = 13 u +
v
3
24
Answer: (b) (ii) OM =
May / June 2012 Paper 23
18
Q
R
M
q
O
p
X
P
O is the origin and OPRQ is a parallelogram.
The position vectors of P and Q are p and q.
X is on PR so that PX = 2XR.
312
13 u + 1 v
3
24
[2]
Sets, Vectors and Functions
Find in terms of p and q, in their simplest forms
(a) QX,
Method 1:
PX = 2 X R
PX =
2
PR
3
PX =
2
q
3
OX = QO + OP + PX
2
q
3
= -q + p +
=p+
2
q−q
3
=p−
1
q
3
Method 2 :
QX = QR + RX
=p+
1
(RP)
3
=p+
1
(-q)
3
=p−
1
q
3
p−
Answer: (a) QX =
1
q
3
[2]
(b) the position vector of M, the midpoint of QX.
[position vector means OM , position vector always from O]
OM = OQ + QM
OM = q +
1
QX
2
OM = q +
1
1
(p −
q)
2
3
OM = q +
1
1
p−
q
2
6
OM =
1
1
p+q−
q
2
6
OM =
1
5
p+
q
2
6
313
Workbook of Variant 3
Answer: (b)
1
5
p+
q
2
6
[2]
May / June 2012 Paper 43
9
f(x) = 1 − 2x g(x) =
(a) Find the value of
1
, x ≠0
x
h(x) = x3 + 1
(i) gf(2)
f(2) = 1 − 2(2)
f(2) = -3
gf(2) = g(-3)
= 1
-3
= -1
3
Answer: (a) (i)
-1
3
[2]
(ii) h(–2)
h(-2) = (-2)3 + 1
h(-2) = -8 + 1
h(-2) = -7
Answer: (a) (ii)
-7
[1]
(b) Find fg(x).
Write your answer as a single fraction.
fg(x) = 1 − 2
fg(x) = 1 −
( 1x )
2
x
fg(x) = x − 2
x
Answer: (b) fg(x) =
(c) Find h–1(x), the inverse of h(x).
h(x) = x3 + 1
y = x3 + 1
x = y3 + 1
x − 1 = y3
y = 3 x − 1
314
x−2
x
[2]
Sets, Vectors and Functions
Answer: (c) h-1(x) =
3
x−1
[2]
(d) Write down which of these sketches shows the graph of each of
y = f(x), y = g(x) and y = h(x).
y
0
y
0
x
Graph A
Graph D
x
Graph B
y
0
y
x
Graph C
y
y
0
x
0
x
Graph E
0
x
Graph F
[f(x) is straight line with position y =- intersect and negative gradient]
[g(x) is inverse fuction where y ≠ 0]
[h(x) is cubic function with positive y - intersect]
Answer: (b) y = f(x) Graph =
A
y = g(x) Graph =
F
315
Workbook of Variant 3
y = h(x) Graph =
D
Answer: (e) x =
29
[3]
(e) k(x) = x5 − 3
Solve the equation k –1 (x) = 2
x = k(2)
x = (2)5 − 3
x = 32 − 3
x = 29
[2]
October / November 2012 Paper 23
9
Shade the required region in each of the Venn diagrams.
P
A
Q
B
R
A'
(P ∩ R) ∪ Q
[2]
20
D
E
d
O
c
In the diagram, O is the origin.
316
C
Sets, Vectors and Functions
OC = c and OD = d.
E is on CD so that CE = 2ED.
Find, in term of c and d, in their simplest forms,
(a) DE,
DC = DO + OC
DC = -d + c
DC = c − d
CE = 2 ED
1
DC
3
1
DE =
(c - d)
3
DE =
Answer: (a) DE =
1
(c - d)
3
[2]
(b) the position vector of E.
[position vector of E means OE]
OE = OD + DE
OE = d +
1
(c − d)
3
OE = d +
1
1
c−
d
3
3
OE =
1
1
c+d−
d
3
3
OE =
1
2
c+
d
3
3
Answer: (b)
23 f(x) = 3x + 5
1
2
c+
d
3
3
[2]
g(x) = 4x − 1
(a) Find the value of gg(3).
g(3) = 4(3) − 1
g(3) = 11
gg(3) = g(11)
= 4(11) − 1
= 43
317
Workbook of Variant 3
Answer: (a)
43
[2]
(b) Find fg(x), giving your answer in its simplest form.
fg(x) = 3(4x − 1) + 5
fg(x) = 12x − 3 + 5
fg(x) = 12x + 2
Answer: (b) fg(x) =
12x + 2
[2]
(c) Solve the equation.
f–1(x) = 11
f-1(x) = 11
x = f(11)
x = 3(11) + 5
x = 38
Answer: (c) x =
May / June 2013 Paper 23
15
A
B
12 – x
13
14
x
20 – x
C
15 – x
8
y
The Venn diagram shows the number of elements in sets A, B and C.
(a) n(A ∪ B ∪ C) = 74
318
38
[1]
Sets, Vectors and Functions
Find x.
13 + 12 − x + 14 + 20 − x + x + 15 − x + 8 = 74
13 + 12 + 14 + 20 + 15 + 8 − x − x + x − x = 74
82 − 2x = 74
82 − 74 = 2x
8 = 2x
8
=x
2
x = 4
Answer: (a) x =
4
Answer: (b) y =
26
[2]
(b) n(%) = 100
Find y.
74 + y = 100
y = 100 − 74
y = 26
[1]
(c) Find the value of n((A ∪ B)' ∩ C).
A
B
C
Answer: (c)
8
[1]
319
Workbook of Variant 3
19
C
D
B
c
b
E
A
O
OABCDE is a regular polygon.
(a) Write down the geometrical name for this polygon.
Answer: (a)
Hexagon
[1]
(b) O is the origin. OB = b and OC = c.
Find, in terms of b and c, in their simplest form,
(i) BC ,
BC = BO + OC
BC = -b + c
Answer: (b)(i) BC =
-b + c
[1]
(ii) OA,
BA =
1
CO
2
BA =
1
(-c)
2
BA = -
[property of hexagon]
1
c
2
OA = OB + BA
OA = b −
1
c
2
b−
Answer: (b)(ii) OA =
320
1
c
2
[2]
Sets, Vectors and Functions
(iii) the position vector of E.
position vector means OE,
OE is parallel to BC
OE = BC
Answer: (b)(iii)
16 f(x) = x +
Find
2
− 3, x ≠ 0
x
g(x) =
−b + c
[1]
x
−5
2
(a) fg(18),
g(18) = 18 − 5
2
g(18) = 4
fg(18) = f(4)
= 4 +
=
2
−3
4
3
2
= 1.5
Answer: (a)
1.5
[2]
(b) g–1(x).
g(x) =
x
−5
2
y =
x
−5
2
x =
y
−5
2
x + 5 =
y
2
2(x + 5) = y
y = 2x + 10
Answer: (b) g-1(x) =
2x + 10
[2]
321
Workbook of Variant 3
October / November 2013 Paper 23
19 f(x) = 2x + 3
g(x) = x2
(a) Find fg(6).
g(6) = 62
g(6) = 36
fg(6) = f(36) = 2(36) + 3 = 75
75
Answer: (a)
[2]
(b) Solve the equation gf(x) = 100
gf(x) = (2x + 3)2
gf(x) = 100
(2x + 3)2 = 100
2x + 3 = ± 100
2x + 3 = 100or2x + 3 = − 100
2x + 3 = 10
2x + 3 = -10
2x = 10 − 3
2x = -10 − 3
x = 3.5x = -6.5
Answer: (b) x =
3.5
-6.5
or x =
[3]
(c) Find f –1(x)
f(x) = 2x + 3
y = 2x + 3
x = 2y + 3
x − 3 = 2y
y = x − 3
2
Answer: (c) f-1(x) =
x−3
2
[2]
(d) Find ff –1(5)
ff-1(5) = 5 [ff-1(x) = x]
Answer: (d)
322
5
[1]
Sets, Vectors and Functions
October / November 2013 Paper 43
7
(a) The co-ordinates of P are (–4, –4) and the co-ordinates of Q are (8, 14).
(i) Find the gradient of the line PQ.
Gradient = y2 − y1
x2 − x1
= 14 − (-4)
8 − (-4)
3
= 18 =
= 1.5
2
12
Answer: (a) (i)
1.5
[2]
(ii) Find the equation of the line PQ.
y = mx + c
y = 1.5 x + c
14 = 1.5(8) + c
14 = 12 + c
c = 14 − 12
c = 2
y = 1.5 x + 2
y =
3x
+2
2
Answer: (a) (ii) y =
3x
+2
2
[2]
(iii) Write PQ as a column vector.
PQ =
( 44++148 ) = ( 1218 )
Answer: (a) (iii) PQ =
( 1218 )
[1]
(iv) Find the magnitude of PQ.
PQ = 122 + 182
PQ = 21.63330765
Answer: (a) (iv)
21.6
[2]
323
Workbook of Variant 3
(b)
T
A
R
4a
O
B
3b
In the diagram, OA = 4a and OB = 3b.
R lies on AB such that OR =
1
(12a + 6b).
5
T is the point such that BT =
3
OA.
2
(i) Find the following in terms of a and b, giving each answer in its simplest form.
(a) AB
AB = AO + OB
= -4a + 3b
= 3b − 4a
Answer: (b) (i) (a) AB =
3b − 4a
[1]
(b) AR
AR = AO + OR
AR = -4a +
1
(12a + 6b)
5
6
AR = -4a + 12 a +
b
5
5
AR =
6
8
b−
a
5
5
=
1
(6b − 8a)
5
Answer: (b) (i) (b) AR =
324
1
(6b − 8a)
5
[2]
Sets, Vectors and Functions
(c) OT
OT = OB + BT
OT = 3b +
3
OA
2
OT = 3b +
3
(4a)
2
OT = 3b + 12 a
2
OT = 6a + 3b
Answer: (b) (i) (c) OT =
6a + 3b
[1]
(ii) Complete the following statement.
The points O, R and 7 are in a straight line because
[
1
(12a + 6b)
5
OR =
=
]
6
(2a + b)
5
OT is multiple of OR and point O is common.
[1]
(iii) Triangle OAR and triangle TBR are similar.
Find the value of
BT =
3
OA
2
BT =
3
x 4a
2
area of triangle TBR
.
area of triangle OAR
BT = 6a
( )=
6a
4a
36 = 9
4
16
Answer: (b) (iii)
9
4
[2]
May / June 2014 Paper 23
17
= {x : 1 ≤ x ≤ 10, where x is an integer}
A = {square numbers}
B = {1, 2, 3, 4, 5, 6}
325
Workbook of Variant 3
=
A =
{1,2,3,....10}
{1,4,9}
(a) Write all the elements of
in their correct place in the Venn diagram.
A
B
9
1
2
3
4
5
6
7
8
10
[2]
(b) List the elements of (A ∪ B)'.
Answer: (b)
7, 8, 10
[1]
(c) Find n(A ∩ B' )
[shading of A B’]
A
B
Answer: (c)
326
1
[1]
Sets, Vectors and Functions
May / June 2014 Paper 43
5
(a)
y
5
A
4
3
2
B
1
0
1
2
3
4
5
6
8
7
x
(i) Write down the position vector of A.
OA =
( 24 )
[2 unit right, 4 unit up]
Answer: (a) (i)
( )
2
4
[1]
(ii) Find AB, the magnitude of AB.
AB = (-3)2 + 52
= 5.830951895
Answer: (a) (ii)
(b)
5.83
[2]
S
Q
M
q
O
p
P
R
O is origin, OP = p and OQ = q.
OP is extended to R so that OP = PR.
OQ is extended to S so that OQ = QS.
327
Workbook of Variant 3
(i) Write down RQ in term of p and q.
RQ = RP + PO + OQ
= −p − p + q
= −2p + q
−2p + q
Answer: (b) (i)
[1]
(ii) PS and RQ intersect at M and RM = 2MQ.
Use vectors to find the ratio PM : PS, showing all your working.
PM = PR + RM
PM = P +
2
RQ
3
PM = p +
2
(−2p + q)
3
PM = p −
4
2
p+
q
3
3
PM = −
PM =
1
2
p+
q
3
3
1
(−p + 2q)
3
PS = PO + OQ + QS
= −p + q + q
= −p + 2q
PM =
PS
1
(−p + 2q)
3
(−p + 2q)
1
PM =
3
PS
Answer: (b) (ii) PM : PS =
9
1
:
3
[4]
f(x) = 4 − 3x g(x) = 3–x
(a) Find f(2x) in terms of x.
f(2x)
=
4 − 3(2x)
f(2x)
=
4 − 6x
Answer: (a) f(2x) =
328
4 − 6x
[1]
Sets, Vectors and Functions
(b) Find ff(x) in its simplest form.
ff(x) = 4 − 3(4 − 3x)
= 4 − 12 + 9x
= 9x − 8
Answer: (b) ff(x) =
9x − 8
[2]
(c) Work out gg(–1).
Give your answer as a fraction.
g(-1) = 3-(-1)
gg(-1) = g(3) = 3-3 =
1
= 1
3
3
27
Answer: (c)
1
27
[3]
(d) Find f –1(x), the inverse of f(x).
f(x) = 4 − 3x
y = 4 − 3x
x = 4 − 3y
3y = 4 − x
y = 4 − x
3
Answer: (d) f-1(x)
4−x
3
[2]
(e) Solve the equation gf(x) = 1.
gf(x) = 3-(4 − 3x)
gf(x) = 1
3 − (4 − 3x) = 1
3 − (4 − 3x) = 30
−(4 − 3x) = 0
−4 + 3x = 0
3x = 4
x =
4
.
3
Answer: (e) x =
4
3
[3]
329
Workbook of Variant 3
October / November 2014 Paper 23
15 The lights and brakes of 30 bicycles are tested.
The table shows the results.
Lights
Brakes
Fail test
3
9
Pass test
27
21
The lights and brakes both failed on one bicycle only.
= {30 bicycles}
Complete the Venn diagrams.
(a)
hts fail
Lig
Brakes f
ail
19
..........
2
..........
hts fail
Lig
1
..........
8
rakes f
B..........
ail
19
..........
2
..........
1
..........
8
..........
[lights and brakes both failed on one bicycle only]
hts pass
Lig
Brakes p
ass
8
19
..........
..........
hts pass
g
i
L
ak2es p
Br..........
ass
[2]
(b)
1
..........
1
..........
8
..........
19
..........
2
..........
[In previous Venn diagram 19 were either light failed not brakes fail, it means 19 were both pass]
[2]
330
Sets, Vectors and Functions
16 f(x) = (x − 3)2g(x) =
Find
x−1
h(x) = x3
4
(a) hf(1),
f(1) = (1 − 3)2 = (-2)2 = 4
hf(1) = h(4) = 43 = 64
Answer: (a)
64
[2]
(b) g–1(x),
g(x) = x − 1
4
y = x − 1
4
x = y − 1
4
4x = y − 1
4x + 1 = y
f-1(x) = 4x + 1
Answer: (b) g-1(x)
4x + 1
[2]
(c) gh(x)
3
gh(x) = x − 1
4
Answer: (c) fh (x)
x3 − 1
4
[1]
(d) the solution to the equation f(x) = 0.
f(x) = 0
(x − 3)2 = 0
x − 3 = 0
x = 3
Answer: (d) (x)
3
[1]
331
Workbook of Variant 3
14
P
A
a
Q
B
b
O
The diagram shows two points, P and Q on a straight line AB.
P is the midpoint of AB and Q is the midpoint of PB.
O is the origin, OA = a and OB = b.
Write down, in terms of a and b, in its simplest form
(a) AP,
P is midpoint of AB
AP =
1
AB
2
AB = −a + b
AP =
1
(-a+b)
2
1
1
a+
b
2
2
AP = −
AP =
1
1
b−
a
2
2
Answer: (b) AP =
(b) the position vector of Q.
AP = PB
PQ =
1
PB
2
PQ =
1
1
1
(
b−
a)
2
2
2
PQ =
1
1
b−
a
4
4
OQ = OA + AP + PQ
1
1
1
1
b−
a+
b−
a
2
2
4
4
1
1
1
1
OQ = a −
a−
a+
b+
b
2
4
2
4
OQ = a +
332
1
1
b−
a
2
2
[2]
Sets, Vectors and Functions
OQ =
1
3
a+
b
4
4
Answer: (b)
1
3
a+
b
4
4
[2]
May / June 2015 Paper 23
16 (a) In this part, you may use this Venn digram to help you answer the questions.
F
S
In a class of 30 students, 25 study French (F), 18 study Spanish (S).
One student does not study French or Spanish.
(i) Find the number of students who study French and Spanish.
25 − x + x + 18 − x + 1 = 30
F
S
25 + 18 + 1 − x + x − x = 30
44 − 30 = x
x
25-x
18-x
x = 14
1
Answer: (a) (i)
14
[2]
(ii) One of the 30 students is chosen at random.
Find the probability that this student studies French but not Spanish.
Number of student studies french but spanish = 25 − x = 25 − 14 = 11
Total Number of Student = 30
Answer: (a) (ii)
11
30
[1]
(iii) A student who does not study Spanish is chosen at random.
Find the probability that this student studies French.
Number of student who does not study spanish = 25 − x + 1 = 25 − 14 + 1 = 12
From then who study french = 25 − x = 25 − 14 = 11
P
Q
Answer: (a) (iii)
11
12
[1]
333
Workbook of Variant 3
(b)
P
Q
R
On this Venn diagram, shade the region R ∩ (P ∪ Q)'
19
M
B
b
O
[1]
P
X
a
A
OAPB is a parallelogram.
O is the origin, OA = a and OB = b
M is the midpoint of BP.
(a) Find, in terms of a and b, giving your answer in its simplest form,
(i) BA,
BA = BO + OA
BA = -b + a
Answer: (a) (i) BA =
-b + a
[1]
(ii) the position vector of M.
OM = OB + BM
OM = b +
1
(BP)
2
OM = b +
1
a
2
b+
Answer: (a) (ii)
334
1
a
2
[1]
Sets, Vectors and Functions
(b) X is on BA so that BX : XA = 1 : 2
Show that X ties on OM.
Answer: (b)
OX = OB + BX
OX = b +
1
(BA)
3
OX = b +
1
(-b + a)
3
OX = b −
1
1
b+
a
3
3
OX =
2
1
b+
a
3
3
OX =
2
1
(b +
a)
3
2
OX =
2
1
(b +
a)
3
2
OX =
2
OM[OM = b + 1 a]
3
2
X lies on OM
[4]
May / June 2015 Paper 43
10 f(x) = 2x − 1
g(x) = x2 + xh(x) =
(a) Find ff(3).
f(3) = 2(3) − 1 = 5
ff(3) = f(5) = 2(5) − 1 = 9
2
,x≠0
x
Answer: (a)
9
[2]
(b) Find gf(x), giving your answer in its simplest form.
gf(x) = (2x − 1)2 + (2x − 1)
gf(x) = (2x − 1)(2x − 1) + 2x − 1
gf(x) = 2x(2x − 1) − 1(2x − 1) + 2x − 1
gf(x) = 4x2 − 2x − 2x + 1 + 2x − 1
gf(x) = 4x2 − 2x
Answer: (b)
4x2 − 2x
[3]
335
Workbook of Variant 3
(c) Find f–1(x).
f(x) = 2x − 1
y = 2x − 1
x = 2y − 1
x + 1 = 2y
x+1 =y
2
f-1(x) =
x+1
2
x+1
Answer: (c) f-1(x) =
2
[2]
(d) Find h(x) + h(x + 2), giving your answer as a single fraction.
h(x) + h(x + 2)
2
2
+
x
x+2
= 2(x + 2) + 2(x)
x(x + 2)
= 2x + 4 + 2x
x2 + 2x
= 4x + 4
x2 + 2x
Answer: (d)
October / November 2015 Paper 23
12
P
Q
3
5
10
9
The Venn diagram shows the number of elements in each set.
336
4x + 4
x2 + 2x
[4]
Sets, Vectors and Functions
(a) Find n(P' ∩ Q)
P
Q
Shading of (P’ ∩ Q)
Answer: (a)
10
[1]
(b) Complete the statement
Here 3 + 5 + 9 = 17 wich is represented by (P ∪ Q’)
Answer: (b)
n(
(P ∪ Q’)
) = 17 [1]
23 B
b
C
O
a
A
In the diagram, O is the origin, OA = a and OB = b.
C is on the line AB so that AC : CB = 1 : 2.
Find, in terms of a and b, in its simplest form,
(a) AC,
AC = 1/3 (AB)
AB = AD + OB
AB = -a + b
AC =
1
(-a + b)
3
Answer: (a) AC =
1
(-a + b)
3
[2]
337
Workbook of Variant 3
(b) the position vector of C.
OC = OA + AC
OC = a +
1
(-a + b)
3
OC = a −
1
1
a+
b
3
3
OC =
[Position vector of C means OC ]
2
1
a+
b
3
3
Answer: (b)
2
1
a+
b
3
3
[2]
October / November 2015 Paper 43
9
f(x) = 2x − 1
g(x) =
(a) Find h(3).
1
, x ≠ 0,
x
h(x) = 2x
h(3) = 23 = 8
Answer: (a)
8
[1]
(b) Find fg(0.5).
g(0.5) =
1 =2
0.5
fg(0.5) = f(2) = 2(2) -1 = 3
Answer: (b)
3
[2]
(c) Find f –1(x).
f(x) = 2x − 1
y = 2x − 1
x = 2y − 1
x + 1 = 2y
f-1(x) = x + 1
2
Answer: (c) f-1(x) =
x+1
2
[2]
(d) Find ff(x), giving your answer in its simplest form.
ff(x) = 2(2x − 1) − 1
ff(x) = 4x − 2 − 1
ff(x) = 4x − 3
Answer: (d)
338
4x − 3
[2]
Sets, Vectors and Functions
(e) Find (f(x))2 + 6, giving your answer in its simplest form.
(f(x))2 + 6
(2x − 1)2 + 6
(2x − 1)(2x − 1) + 6
2x(2x−1) − 1(2x − 1) + 6
4x2 − 2x − 2x + 1 + 6
4x2 − 4x + 7
Answer: (e)
4x2 − 4x + 7
[2]
(f) Simplify hh–1(x).
hh-1(x) = x
Answer: (f)
x
[1]
(g) Which of the following statements is true?
f–1(x)
=
f(x)
g–1(x) =
g(x)
h–1(x) =
h(x)
f(x) = 2x − 1
f-1(x) = x + 1
2
g(x) = 1
x
g-1(x) = 1
x
g-1(x) = g(x)
Answer: (g) g-1(x) =
g(x)
[1]
(h) Use two of the functions f(x), g(x) and h(x) to find the composite function which is equal to
2x+1 − 1.
fh(x) = 2(2x)
fh(x) = 2x + 1 − 1
[am × an = am + n]
Answer: (h)
fh(x)
[1]
339
9
Workbook of
Solution 4
MATRICES AND
TRANSFORMATIONS
342
Matrices and Transformations
Chapter 9: Matrices and Transformations
May / June 2010 Paper 23
23 A = (1
4)
B=
Find
( –23
–1
2
)
(a) AB,
( -23
AB = (1 4)1 × 2
-1
2
AB = (1 × 3 + 4 × −2
AB = (3 -8 Ab = (-5 7)
)
2×2
1 × -1 + 4 × 2)1 × 2
-1 + 8)
(-5
Answer: (a) AB =
7)
[2]
(b) the inverse matrix B–1,
B-1 =
B-1 =
(
2
1
2
3 × 2 − (-1 × 2)
(
2
1
2
6−2
(
2
B-1 = 1 2
4
1
3
1
3
1
3
)
→ B-1 =
1
ab - bc
( -cd
BB-1=
)
)
)
Answer: (b) B-1 =
(c) BB–1
-b
a
1
4
(2
2
1
3
)
[2]
=
( 10
0
1
)
Answer: (c) BB-1 =
( 10
0
1
)
[1]
343
Workbook of Variant 3
May / June 2010 Paper 43
4
y
12
11
10


A
9
B
8



7
6
y=6

5
P
4
3
T
Q
2
1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
x
–1
–2
–3
–4
(a) Draw the reflection of triangle T in the line y = 6.
Label the image A.
[2]
(b) Draw the translation of triangle T by the vector
Label the image B.
( –46 ).
[2]
(c) Describe fully the single transformation which maps triangle B onto triangle T.
Answer: (c)
( )
Transformation by vector -4
6
[2]
(d) (i) Describe fully the single transformation which maps triangle T onto triangle P.
Scale factor = new length = 1
old length
2
Answer: (d) (i)
344
Enlargement, s.f. = 1 , centre (4, 6)
2
[3]
Matrices and Transformations
(ii) Complete the following statement.
Answer: (d) (ii) Area of triangle P =
1
4
× area of triangle T
[1]
October / November 2010 Paper 43
4
(a) A =
( 24
3
B =
5
)
( 27 )
C = (1 2)
Find the following matrices.
(i) AB
( 24 35 ) ( 27 )
AB = ( 2 × 2 + 7 × 3 )
4×2+5×7
AB =
2×2
2×1
2×1
( 43 )
25
Answer: (a) (i)
[2]
(ii) CB
CB =
( 27 )
(1 2)1 × 2 CB = (1 × 2 + 2 × 7)1 × 1
CB = (2 + 14)
2×1
CB = (16)
(16)
Answer: (a) (ii)
[2]
(iii) A–1, the inverse of A
(
A-1 =
1
5
2×5−3×4
-4
A-1 =
1
5
10 − 12
-4
(
1 5
-2
-4
(
-3
)
2
-3
2
-3
2
)
)
Answer: (a) (iii)
1
-2
(
5
-4
-3
2
)
[2]
345
Workbook of Variant 3
(b) Describe fully the single transformation represented by the matrix
x
1
+y
0
(
y
0
1
0 .
–1
)
y comparing the given matrix with
B
identity matrix
)
x
( 10
Reflection over x - axis
Answer: (b)
[2]
-y
(c) Find the 2 by 2 matrix that represents an anticlockwise rotation of 90o about the origin.
x
1
y
0
(
-x
x
y
+ x moves to + y and
0 + y moves to - x.
1
)
Resultant matrix
(
y
0
1
-x
-1
0
)
Answer: (c)
346
(
0
1
-1
0
)
[2]
Matrices and Transformations
2
y
y= -x
6
X
5
4
3
b (i)
2
1
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
6
x
–1
–2
–3
–4
Z
Y
–5
–6
(a) (i) Draw the reflection of shape X in the x-axis. Label the image Y.[2]
(ii) Draw the rotation of shape Y, 90o clockwise about (0, 0). Label the image Z.[2]
(iii) Describe fully the single transformation that maps shape Z onto shape X.
Answer: (a) (ii)
Reflection over line y = -x
[2]
1
.[2]
2
1
(ii) Find the matrix which represents an enlargement, centre (0, 0), scale factor
.
2
(b) (i) Draw the enlargement of shape X, centre(0,0), scale factor
( )
1
2
Answer: (b) (i)
0
0
1
2
[2]
347
Workbook of Variant 3
October / November 2011 Paper 23
( 26 ). The order of the matrix N is 2 × 1.
20 (a) N =
P = (1 3). The order of the matrix P is 1 × 2.
(i) Write down the order of the matrix NP.
order of N
order of P
2×1
1×2
same
2 × 2
2×2
Answer: (a) (i)
[1]
(ii) Calculate PN.
PN = (1 3)1 × 2
( 26 )
2×1
PN = (1 × 2 + 3 × 6)1 × 1
PN = (2 + 18)1 × 1
PN = (20)1 × 1
(20)
Answer: (a) (ii)
(b) M =
( 22
3
4
[1]
)
Find M–1, the inverse of M.
(
M-1 =
1
4
2×4−2×3
-2
M-1 =
1 4
8−6
-2
(
(
M-1 = 1 4
2
-2
-3
2
-3
2
-3
2
)
)
)
Answer: (b) M-1
348
1
2
( -2
4
-3
2
)
[2]
Matrices and Transformations
October / November 2011 Paper 43
4
y
y=x
6
4
(b)
P
–6
–4
–2
2
W
0
2
4
x
6
(a)
–2
–4
–6
(a) Draw the reflection of shape P in the line y = x.
(b) Draw the translation of shape P by the vector
[2]
( –21 ).
[2]
(c) (i) Describe fully the single transformation that maps shape P onto shape W.
Answer: (c) (i)
Rotation 90o clockwise centse (0, 0)
[3]
(ii) Find the 2 by 2 matrix which represents this transformation.
x
1
+y
0
(
x
y
0
1
)
Resultant matrix
(
-y
0
-1
x
1
0
)
-y
( -1 0 )
0
Answer: (c) (ii)
1
[2]
349
Workbook of Variant 3
May / June 2012 Paper 23
17 A =
( 21
4
B = (1
3
)
2)
(a) Calculate BA.
BA = (1
2)1 × 2
( 21 43 )
2×2
BA = (1 × 2 + 2 × 1 1 × 4 + 2 × 3) 1 × 2
BA = (2 + 2 4 + 6) 1 × 2
BA = (4
10) 2 × 2
(4
Answer: (a)
10)
[2]
(b) Find A–1, the inverse of A.
(
A-1 =
1
3
2×3−1×4
-1
A-1 =
1
3
6−4
-1
(
(
= 1 3
2
-1
-4
2
-4
2
-4
2
)
)
)
Answer: (b)
350
1
2
( -1
3
-4
2
)
[2]
Matrices and Transformations
3
y
x=6
11
10
9
8
7
6
5
(a)
4
Q
3
2
P
1
–3
–2
–1
0
1
2
3
(a) Draw the translation of triangle P by
( b)
4
5
6
7
8
9
10
11
12
x
( 53 ).
[2]
(b) Draw the reflection of triangle P in the line x = 6.
[2]
(c) (i) Describe fully the single transformation that maps triangle P onto triangle Q.
Answer: (c) (i)
Rotation anticlockwise centre (0, 0)
[3]
(ii) Find the 2 by 2 matrix which represents the transformation in part (c)(i).
x
1
y
0
(
-x
x
y
0
1
)
Resultant matrix
(
y
0
1
-x
-1
0
)
(1
0
Answer: (c) (ii)
-1
0
)
[2]
351
Workbook of Variant 3
October / November 2012 Paper 23
22 (a) M =
( –13
2
1
)
Find M–1, the inverse of M.
M-1 =
M-1 =
(
1
1
1
3 × 1 − 2 × -1
(
1
1
1
3+2
(
1
= 1 1
5
-2
3
-2
3
-2
3
)
)
)
Answer: (a)
1
5
( 11
-2
3
)
[2]
(b) D, E and X are 2 × 2 Matrices.
I is the identity 2 × 2 matrix.
(i) Simplify DI.
DI = D Answer: (b) (i)
D
[1]
(ii) DX = E
Write X in terms of D and E
DX = E
X = D-1 E
Answer: (b) (ii) x =
352
D-1 E
[1]
Matrices and Transformations
October / November 2012 Paper 43
2
(a)
y
8
7
X
6
a(i)
5
4
3
2
1
–8
–7
–6
–5
–4
–3
–1 0
–2
1
2
3
4
5
6
7
8
x
–1
Y
a(ii)
(-6, -4)
–2
–3
–4
–5
–6
–7
–8
–9
(i) Draw the translation of triangle X by the vector
( –11
).
–1
(ii) Draw the enlargement of triangle Y with centre (–6, –4 ) and scale factor
[2]
1
.
2
[2]
353
Workbook of Variant 3
(b)
y
8
7
W
6
X
5
4
3
2
y=1
1
–8 –7 –6 –5 –4 –3 –2 –1 0
–1
Y
1
2
3
4
5
6
7
8
x
–2
–3
–4
Z
–5
–6
–7
–8
–9
Describe fully the single tranformation that maps
(i) Triangle X onto triangle Z,
Answer: (b) (i)
Reflection over line y = 1
[2]
(ii) Triangle X onto triangle Y.
Answer: (b) (ii)
354
Rotation 180, centre (3, 2)
[3]
Matrices and Transformations
May / June 2013 Paper 23
( 23 36 )
17 M =
( 21
N =
1
7
5
2
)
(a) Work out MN.
( 23 36 ) ( 21 17 52 )
2×1+3×7
MN = ( 2 × 2 + 3 × 1
3×2+6×1
3×1+6×7
2 + 21
10 + 6
MN = ( 4 + 3
)
6+6
3 + 42
15 + 12
23
16
MN = ( 7
)
12
45
27
MN =
2×2
2×3
2×5+3×2
3×5+6×2
)
2×3
2×3
2×3
Answer: (a)
(
23
45
16
27
)
[2]
( -3
-3
2
)
[2]
7
12
(b) Find M–1, the inverse of M.
(
M-1 =
1
6
2×6−3×3
-3
=
1
6
12 - 9
-3
(
(
= 1 6
3
-3
-3
2
-3
2
-3
2
)
)
)
Answer: (b)
1
3
6
October / November 2013 Paper 23
11 A =
( 34
–1 2
)
I= 1 0
(0 1)
Work out the following.
(a) AI
AI = A
(4
3
Answer: (a) AI =
-1
2
)
[1]
355
Workbook of Variant 3
(b) A–1
(
1
3
1
2
3 × 2 − 4 × -1
-4
A-1 =
(
A-1 =
1
2
6+4
-4
A-1 =
1 2
10
-4
(
1
3
1
3
)
)
)
Answer: (b) A-1 =
1
10
(
2
-4
1
3
)
[2]
17 (p, q) is the image of the point (x, y) under this combined transformation.
( pq ) = ( –10
0
1
) ( xy ) + ( 32 )
(a) Draw the image of the triangle under the combined transformation
(
-1
0
0
1
)
Represents reflection in the y axis so reflect the given triangle over y-axis
(represented with dotted line) and then translation it by vector
y
( 32 ).
4
3
2
1
4
–3
–2
–1
0
1
2
3
4
x
–1
–2
–3
–4
356
[3]
Matrices and Transformations
(b) Describe fully the single transformation represented by
x
1
y
0
y
0
1
(
-x
)
( –10
0
1
)
By comparing it with identity matrix
x
Reflection over y-axis
Answer: (b)
[2]
May / June 2014 Paper 23
22
y
7
6
5
4
A
3
2
1
–7
–6
–5
–4
–3
–2
–1 0
–1
1
2
(a)
3
4
5
6
7
x
–2
–3
B
–4
–5
–6
357
Workbook of Variant 3
(a) Draw the image of triangle A after a translation by the vector
( –43 ).
[2]
(b) Describe fully the single transformation which maps triangle A onto triangle B.
Answer: (b)
18 A =
Rotation 180o, centre (1, 0)
[3]
( 54 23 )
(a) Calculate A2
A2 = AA
( 54 23 ) ( 54 23 )
5×2+2×3
A = ( 5 × 5 + 2 × 4
)
4×5+3×4
4×2+3×3
A = ( 25 + 8 10 + 6 )
20 + 12
8+9
A = ( 33 16 )
32
17
A2 =
2×2
2×2
2
2×2
2
2
Answer: (a)
(
33
32
16
17
)
[2]
(b) Calculate A–1, the inverse of A.
(
A-1 =
1
3
5×3−4×2
-4
A-1 =
1
3
15 - 8
-4
(
(
A-1 = 1 3
7
-4
-2
5
-2
5
-2
5
)
)
)
Answer: (b)
358
1
7
(
3
-4
-2
5
)
[2]
Matrices and Transformations
October / November 2014 Paper 23
19 (a) N =
1
0
( –10
)
Describe fully the single transformation represented by N.
x
1
+y
0
y
0 1
(
)
x
-y
Rotation 90° clockwise centre (0, 0).
Answer: (a)
[3]
(b) Find the matrix which represents the single transformation that maps triangle A onto
triangle B.
y
y =x
7
6
5
4
A
3
2
B
1
0
1
2
3
4
5
6
7
8
9
10
x
Triangle A onto triangle B represents reflection in the line y = x.
Answer: (b)
(
0
1
1
0
)
[2]
359
Workbook of Variant 3
October / November 2014 Paper 43
5
P=
( 01
–1 0
)
Q=
( 10
–2 1
)
R = –3
(
5
)
(a) Work out.
(i) 4P,
4P = 4
( 01
-1
0
) = ( 44 ×× 01 44 ×× -10 ) = ( 04 -40 )
Answer: (a) (i)
( 04 -40 )
[1]
(ii) P − Q,
)
= ( 0 − 1 -1 − (-2) ) = ( -1
1−0
0−1
1
P−Q=
( 01
-1
0
) − ( 10
-2
1
1
-1
)
(1
1
-1
)
[1]
(0
0
-1
)
[2]
-1
Answer: (a) (ii)
(iii) P2,
P2 = P.P
( 01 -10 ) − ( 01 -11 )
P = ( 0 × 0 + 1 × -1 ) ( 0 × -1 + -1 × 0 )
1×0+0×1
1×1+0×0
P = ( 0 − 1 ) ( 0 + 0 )
4×1
0−1
= ( -1 0 )
0 -1
P2 =
2
2
Answer: (a) (iii)
-1
(iv) QR.
( 10 -21 ) ( -35 )
QR = ( 1 × -3 + -2 × 5 )
0 × -3 + 1 × 5
QR = ( -3 − 10 )
0+5
QR = ( -13 )
5
QR =
-13
Answer: (a) (iv)
360
(5)
[2]
Matrices and Transformations
(b) Find the matrix S, so that QS =
(
1
0
0
1
)=I
( 10 01 ).
QS = I
S = Q-1 I
S = Q-1 ( 10
1
1
=
(
0
1+0
1
2
= (
0
1 )
S = Q-1 =
1
1×1+2×0
)
2
1 )
2
1
Answer: (b)
(
)
[3]
( -1 0 )
[2]
1
0
2
1
May / June 2015 Paper 23
6
Find the 2 × 2 matrix that represents a rotation through 90o clockwise about (0, 0).
x
+y
1
0
(
+x
y
0
1
)
Resultant matrix
(
y
0
-1
-x
1
0
)
-y
0
Answer:
1
361
Workbook of Variant 3
May / June 2015 Paper 43
1
y
6
U
5
4
3
a(i)
(0,3)
T
2
V
1
–5
–4
–3
–2
1
–1 0
2
3
4
5
6
7
x
–1
–2
a(ii)
–3
–4
–5
–6
(a) On the grid, draw the image of
(i) tringle T after a reflection in the line x = –1,
[2]
(ii) triangle T after a rotation through 180° about (0, 0).
[2]
(b) Describe fully the single transformation that maps
(i) triangle T onto triangle U,
Translation by vector
Answer: (b) (i)
( )
-2
2
[2]
(ii) triangle T onto triangle V.
Scale factor = new length = 3
old length
1
o find centre join corrosponding points of triangle. (2 lines are enough) Point of
T
intersection gives center.
Answer: (b) (ii)
362
Enlargement, centre(0, 3), Scale factor = 3
[3]
Matrices and Transformations
9
( 21
P=
3
Q =
4
)
( 10
2
3
)
R=
( 01
u
S =
v
)
w
(8
3
2
)
(a) Work out PQ
( 21 34 ) ( 10 23 )
2×2+3×3
PQ = ( 2 × 1 + 3 × 0
)
1×1+4×0
1×2+4×3
PQ = ( 2 + 0 4 + 9 )
1+0
2 + 12
PQ = ( 2 13 )
1
14
PQ =
2×2
2×2
2×2
Answer: (a)
( 21
13
14
)
[2]
(b) Find Q–1.
(
Q-1 =
3
1
0
1×3−0×2
Q-1 =
3
1
0
3−0
(
(
3
Q-1 = 1 0
3
-2
1
-2
1
-2
1
)
)
)
Answer: (b)
(
1
3
3
0
-2
1
)
[2]
(c) PR = RP
Find the value of u and the value of v.
PR = RP
( 21 34 ) ( 01 uv ) = ( 01 uv ) ( 21 34 )
2 × u + 3v
= ( 2 × 0 + 3 × 1
=( 0×2+u×1
)
0×1+×4×1
1×u+4×v
1×2+v×1
( 3 2u + 3v ) = ( u 4u )
4
u + 4v
2+v
3 + 4v
2×2
2×2
2×2
2×2
0×3+u×4
1×3+v×4
)
By comparing,
u = 3
v + 2 = 4
v = 4 − 2
v = 2
Answer: (c) u =
3
v=
2
[3]
363
Workbook of Variant 3
(d) The determinant of S is 0.
Find the value of w.
Find value of w.
w × 2 − 8 × 3 = 0
2w − 24 = 0
2w = 24
w = 24
2
w = 12
Answer: (d) w =
12
Answer: u =
10
[2]
October / November 2015 Paper 23
13 M =
( 72
u
3
) and
M = 1.
Find the value of u.
Determinant = 1
7×3−2×u=1
21 − 2u = 1
21 − 1 = 2u
20 = 2u
2u = 20
u = 20
2
u = 10
364
[2]
Matrices and Transformations
October / November 2015 Paper 43
2
y
8
7
6
a(i)
5
4
3
2
T
1
(-1,0)
–8
–7
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
x
y= -1
–1
U
6
–2
–3
a(ii)
–4
W
–5
–6
(a) On the grid, draw the image of
(i) triangle T after a translation by the vector
( –44 ).[2]
(ii) triangle T after a reflection in the line y = –1
[2]
(b) Describe fully the single transformation that maps triangle T onto triangle U.
Answer: (b)
Rotation 180, centre (-1, 0)
[3]
365
Workbook of Variant 3
(c) (i) Describe fully the single transformation that maps triangle T onto triangle W.
Reflection in line y = -x
Answer: (c) (i)
[3]
(ii) Find the 2 × 2 matrix that represents the transformation in part (c)(i).
x
1
+y
0
y=x
(
y
0
1
)
+x moves yo -y
+x
-x
+
y moves to -x
-y
Resulted matrix
(
y
0
-1
-x
-1
0
)
Answer: (c) (ii)
366
(
0
-1
-1
0
)
[2]
10
Workbook of
Solution 4
STATISTICS AND
PROBABILITY
368
Statistics and Probability
Chapter 10: Statistics and Probability
October / November 2010 Paper 43
3
2
1
4
4
1
1
3
2
4
1
The diagram shows a circular board, divided into 10 numbered sectors.
When the arrow is spun it is equally likely to stop in any sector.
(a) Complete the table below which shows the probability of the arrow stopping at each number.
Number
1
2
3
4
Probability
0.4
0.2
0.1
0.3
4
10
1
10
[1]
(b) The arrow is spun once.
Find
(i) the most likely number,
Answer: (b) (i)
1
[1]
369
Workbook of Variant 3
(ii) the probability of a number less than 4.
possible outcomes
1
or 2
or 3
probability → 0.4 + 0.2 + 0.1 = 0.7
Answer: (b) (ii)
0.7
[1]
(c) The arrow is spun twice.
Find the probability that
(i) both numbers are 2,
possible outcomes 2 and 2
probability
0.2 × 0.2 = 0.04
Answer: (c) (i)
0.04
[1]
(ii) the first number is 3 and the second number is 4,
possible outcomes 3 and 4
probability
0.1 × 0.3 = 0.03
Answer: (c) (ii)
0.03
[2]
(iii) the two numbers add up to 4.
Possible outcomes
(2 and 2) or (1 and 3) or (3 and 1)
Probability = 0.2 × 0.2 + 0.4 × 0.1 + 0.1 × 0.4
= 0.12
Answer: (c)(iii)
0.12
[3]
(d) The arrow is spun several times until it stops at a number 4.
Find the probability that this happens on the third spin.
possible outcome
probability
not 4
and
not 4
and
4
(1−0.3) × (1−0.3) × (0.3) = 0.147
[Probability of not getting 4 = 1 − getting 4 = 1−0.3 = 0.7]
Answer: (d)
370
0.147
[2]
Statistics and Probability
7
(a)The table shows how many books were borrowed by the 126 members of a library group in
a month.
Number of book
11
12
13
14
15
16
Number of members
35
28
22
18
14
9
Cumulative frequency
35
63
85
103 117 126
Find the mode, the median and the mean for the number of books borrowed.
Mode is most occuring number
Median =
126 + 1
2
Mode = 11
= 63.5 term
63.5 term is in between 12 and 13
Median =
12 + 13
2
= 12.5
mean = ∑fx
∑f
mean =
11 × 35 + 12 × 28 + 13 × 22 + 14 × 28 + 15 × 14 + 16 × 9
126
mean = 12.8015873
Answer: (a) mode =
median =
mean =
11
12.5
12.8
[6]
371
Workbook of Variant 3
(b) The 126 members record the number of hours they read in one week.
The histogram shows the results.
Frequency
density
15
10
5
0
5
8
10
Time (hours)
12
16
20
h
(i) Use the information from the histogram to complete the frequency table.
Number of hours (h)
0<h≤5
5<h≤8
15
27
Frequency
8 < h ≤ 10 10 < h ≤ 12 12 < h ≤ 16 16 < h ≤ 20
30
20
24
10
[3]
(ii)Use the information in this table to calculate an estimate of the mean number of hours.
show your working
mean =
∑fx
∑f
mean =
2.5 × 15 + 6.5 × 27 + 9 × 30 + 11 × 20 + 14 × 24 + 18 × 10
126
mean = 9.674603175
Answer: (b) (ii)
372
9.67
hours
[4]
Statistics and Probability
October / November 2010 Paper 43
6
Sacha either walks or cycles to school.
On any day, the probability that he walks to school is
3
.
5
(a) (i) A school term has 55 days.
Work out the expected number of days Sacha walks to school.
55 ×
3
= 33
5
Answer: (a) (i)
33
[1]
(ii) Calculate the probability that Sacha walks to school on the first 5 days of the term.
1st day
3
5
(
)×(
2nd day
3
5
)×(
3rd day
3
5
)×(
4th day
3
5
)×(
5th day
3
5
Answer: (a) (ii)
(b) When Sacha walks to school, the probability that he is late is
When he cycles to school, the probability that he is late is
243
) = 3125
243
3125
[2]
1
.
4
1
.
8
(i) Complete the tree diagram by writing the probabilities in the four spaces provided.
3
5
2
5
1
4
late
3
4
not late
walks
1
8
late
cycles
7
8
[Each branch of tree diagram, adds upto 1]
not late
[3]
373
Workbook of Variant 3
(ii) Calculate the probability that Sacha cycles to school and is late.
possible outcomes are cycles
and
late
2
1
1
×
=
5
8
20
Answer: (b) (ii)
1
20
[2]
(iii) Calculate the probability that Sacha is late to school.
possible outcomes:
probability:
walks
and
late
or
cycles and late
3
1
1
1
×
+
=
5
4
20
5
Answer: (b) (iii)
1
5
[2]
10 (a)For a set of six integers, the mode is 8, the median is 9 and the mean is 10.
The smallest integer is greater than 6 and the largest integer is 16.
Find the two possible sets of six integers.
First set
7
8
8
10
x
16
1
2
3
4
5
6
Median = 9
Median is 9,So 3rd and 4th term will be 8 and 10
Mode is 8,So there will be one more 8 so second term is 8.
7 + 8 + 8 + 10 + x + 16
= 10
6
x = 11
Second set
8
8
8
10
y
16
1
2
3
4
5
6
2nd, 3rd, and 4th terms will remain same as given small integer is greater than 6 in first set
we took 7 here we will take 8
374
8 + 8 + 8 + 10 + y + 16
= 10
6
Statistics and Probability
y = 10
Answer: (a) First set
Second set
7
8
,
,
8
8
,
,
8
8
,
,
10
,
10
,
11
10
,
,
16
16
,
, [5]
(b) One day Ahmed sells 160 oranges.
He records the mass of each orange.
The results are shown in the table.
Mass (m grams) 50 < m ≤ 80 80 < m ≤ 90 90 < m ≤ 100 100 < m ≤ 120 120 < m ≤ 150
Frequency
30
35
40
40
15
(i) Calculate an estimate of the mean mass of the 160 oranges.
mean =
∑fx
∑f
mean =
65 × 30 + 85 × 35 + 95 × 40 + 110 × 40 + 135 × 15
160
[Here 'x' is taken on mid values of class]
mean = 94.6875
Answer: (b) (i)
94.7
g [4]
(ii) On the grid complete the histogram to show the information in the table.
375
Workbook of Variant 3
5
4
3
Frequency
density
2
1
0
50
60
70
80
90
100
110
120
130
140
150
m
Mass (grams)
Frequency density =
frequency
class width
for 90 < m ≤ 100 → fd =
40
10
=
4
for 100 < m ≤ 120
→ fd =
40
20
=
2
for 120 < m ≤ 150
→ fd =
15
30
=
0.5
[4]
May / June 2011 Paper 43
376
Statistics and Probability
6
200
180
160
UQ
140
120
Median
Cumulative
100
frequency
80
60
LQ
40
20
0
1
2
3
4
5
6
7
8
9
10
m
Mass (kilograms)
The masses of the 200 parcels are recorded.
The results are shown in the cumulative frequency diagram above.
(a) Find
(i) the median,
( 2n )
[
th
term given median]
n
200
=
= 100th term
2
2
Answer: (a) (i)
5.8
kg [1]
(ii) the lower quartile,
( 4n )
[
th
term given lower median]
n
200
=
= 50th term
4
4
Answer: (a) (ii)
4.6
kg [1]
(iii) the inter-quartile range,
377
Workbook of Variant 3
3
3
(n) =
(200) = 150
4
4
upper quartile = 7
2.4
Answer: (a) (iii)
kg [1]
(iv) the number of parcels with a mass greater than 3.5 kg.
Total number of parcels − mark on 3.5 kg
200 − 28
172
172
Answer: (a) (iv)
[2]
(b) (i)Use the information from the cumulative frequency diagram to complete the grouped
frequency table.
[Mark on 6 - Mark on 4]
[Mark on 7 - Mark on 6]
110 - 36 = 74150 - 110 = 40
Mass (m kg)
0< m ≤ 4
4<m≤6
6<m≤7
7 < m ≤ 10
Frequency
36
74
40
50
[2]
(ii) Use the grouped frequency table to calculate an estimate of the mean.
mean =
∑fx
∑f
mean =
2 × 36 + 5 × 74 + 6.5 × 40 + 8.5 × 50
200
mean = 5.635
Answer: (b) (ii)
5.64
(iii) Complete the frequency density table and use it to complete the histogram.
378
kg [4]
Statistics and Probability
Mass (m kg)
0<m≤4
4<m≤6
6<m≤7
7 < m ≤ 10
Frequency density
9
37
40
16.7
40
35
30
25
Frequency
20
density
15
10
5
0
1
2
3
4
5
6
7
8
9
10
m
Mass (kilograms)
[4]
7
Katrina puts some plants in her garden.
7
.
10
If there is a flower, it can only be red, yellow or orange.
2
1
When there is a flower, the probability it is red is
and the probability it is yellow is
.
3
4
The probability that a plant will produce a flower is
(a) Draw a tree diagram to show all this information.
Label the diagram and write the probability on each branch.
Answer: (a)
379
Workbook of Variant 3
Red
2
3
1
4
Flower
Yellow
2
3
7
10
1
12
Flower
Red
1
4
Orange
Yellow
7
10
3
10
1
12
Not Flower
Orange
[5]
3
(b) A plant is chosen at random.
10
Find the probability that it will not produce a yellow flower.
Not Flower
probability of not producing yellow = 1 − probability of producing yelloow
=1−
=
1
7
×
4
10
33
40
33
40
Answer: (b)
[3]
(c) If Katrina puts 120 plants in her garden, how many orange flowers would she expect?
120 times probability of producing orange flowers.
120
7
( 107
×
1
12
)
7
Answer: (c)
October / November 2011 Paper 43
7The times, t minutes, taken for 200 students to cycle one kilometre are shown in the table.
380
Time (t minutes)
0< t ≤ 2
2<t≤3
3<t≤4
4<t≤8
Frequency
24
68
72
36
cumulative frequency
24
92
164
200
[2]
Statistics and Probability
(a) Write down the class interval that contains the median.
200 + 1
= 100.5th term
2
100.5th term coming in interval 3 < t ≤ 4
Answer: (a)
3<t≤4
[1]
(b) Calculate an estimate of the mean.
Show all your working.
mean =
∑fx
∑f
Here 'x' is taken as mid value of class]
0+2
2
2+3
2
3+4
2
4+8
2
mean = 1 × 24 + 2.5 × 68 + 3.5 × 72 + 6 × 36
200
mean = 3.31
3.31
Answer: (b)
min
[4]
(c) (i)Use the information in the table opposite to complete the cumulative frequency table.
Time (t minute)
t≤2
t≤3
t≤4
t≤8
cumulativeFrequency
24
92
164
200
24 + 68
92 + 72
[1]
381
Workbook of Variant 3
(ii) On the grid, draw a cumulative frequency diagram.
200
180
Cumulative frequency
160
UQ
140
120
Median
100
80
60
LQ
40
20
0
1
2
3
4
Time (minutes)
5
6
7
8
t
[3]
(iii) Use your diagram to find the median, the lower quartile and the inter quartile range.
Median:
n
200
=
= 100
2
2
Median = 3.1
LQ
n
200
=
= 50
4
4
LQ = 2.45
UQ =
3
3
(n) =
(200) = 150
4
4
UQ = 3.75
IQR = UQ − LQ
3.75 − 2.45 = 1.3
Answer: (c) (iii) Median =
Lower quartile =
Inter-quartile range =
382
3.1
2.45
1.3
min
min
min
[3]
Statistics and Probability
9
Set A
S
U
M
S
Set B
M
I
N
U
S
The diagram shows two sets of cards.
(a) One card is chosen at random from Set A and replaced.
(i) Write down the probability that the card chosen shows the letter M.
1
4
Answer: (a) (i)
[1]
(ii)If this is carried out 100 times, write down the expected number of times the card chosen
shows the letter M.
100
( 14 ) = 25
25
Answer: (a) (ii)
[1]
(b) Two cards are chosen at random, without replacement, from Set A.
Find the probability that both cards show the letter S.
possible outcomes
S
and S
2
probability 4
×
1
1
2
=
=
6
3
12
1
6
Answer: (b)
[2]
(c) One card is chosen at random from Set A and one card is chosen at random from set B.
Find the probability that exactly one of the two cards shows the letter U.
possible outcomes
set A[U] and set B[not U] or set A[not U] and set B[U]
Probability
1
4
=
7
20
×
4
+
5
3
4
Answer: (c)
×
1
5
7
20
[3]
383
Workbook of Variant 3
(d) A card is chosen at random, without replacement, from Set B until the letter shown is either
I or U.
Find the probability that this does not happen until the 4th card is chosen.
1st card
not I or U
3
5
×
=
6
60
=
1
10
and
2nd card
and
3rd card
not I or U
2
4
×
1
3
×
and
not I or U
4th card
I or U
2
2
1
10
Answer: (d)
[2]
May / June 2012 Paper 23
7
Height (h cm)
0 < h ≤ 10
10 < h ≤ 15
15 < h ≤ 30
Frequency
25
u
9
Frequency density
2.5
4.8
v
The table shows information about the heights of some flowers.
Calculate the values of u and v.
[Class width = upper class limit - cover class limit]
frequency density =
4.8 =
frequency
class width
u
5
u = 4.8 × 5
= 24
v =
384
9
15
v = 0.6
Answer: u =
24
v=
0.6
min
min
[2]
Statistics and Probability
May / June 2012 Paper 43
4
(a)In a football league a team is given 3 points for a win, 1 point for a draw and 0 points for
a loss.
The table shows the 20 results for Athletico Cambridge.
Points
3
1
0
Frequency
10
3
7
Frequency density
10
13
20
(20-10-3)
(i) Find the median and the mode.
For mdian
n+1
20 + 1
=
= 10.5th term
2
2
10.5th term is comming between points 3 and 1 median = 2
2
Answer: (a) (i) Median =
3
Mode =
[3]
(ii) Thomas wants to draw a pie chart using the informatiom in the table.
Calculate the angle of the sector which shows the number of times Athletico
Cambridge were given 1 point.
20
360
x
3
3 × 360
= 54
20
x =
54
Answer: (a) (ii)
[2]
(b) Athletico Cambrdge has 20 players.
The table shows information about the height (h centimetres) of the players.
Height (h cm)
170 < h ≤180
180 < h ≤ 190
190 < h ≤ 200
Frequency
5
12
3
Calculate an estimate of the mean height of the players.
mean =
∑fx
[Here 'x' is taken as mid value of class]
∑f
mean =
175 × 5 + 185 × 12 + 195 × 3
20
mean = 184
385
Workbook of Variant 3
Answer: (b)
184
cm [4]
6
H
C
30
150
20
40
=
{240 passengers who arrive on a flight in Cyprus}
H = {passengers who are on holiday}
C = {passengers who hire a car}
(a) Write down the number of passengers who
(i) are on holiday,
30 + 150 = 180
Answer: (a) (i)
180
[1]
(ii) hire a car but are not on holiday.
Answer: (a) (ii)
20
[1]
(b) Find the value of n(H ∪ C' ).
30 + 150 + 40 = 220
Answer: (b)
220
[1]
(c) One of the 240 passengers is chosen at random.
Write down the probability that this passenger
(i) hires a car,
Number of passengers hires car = 150 + 20
= 170
170
17
Probability =
=
240
24
Answer: (c) (i)
17
24
[1]
(ii) is on holiday and hires a car.
Number of passengers who are on holiday and hires car = 150
Probability =
150
240
5
=
8
Answer: (c) (ii)
386
5
8
[1]
Statistics and Probability
(d) Give your answers to this part correct to 4 decimal places.
Two of the 240 passengers are chosen at random.
Find the probability that
(i) they are both on holiday,
Possible outcome = Holiday and holiday
Probability
180
179
×
= 0.5617154812
240
239
0.5617
Answer: (d) (i)
[2]
(ii) exactly one of the two passengers is on holiday.
(Holiday and not holiday) or (not holiday and holiday)
180
60
60
180
×
+
×
= 0.3765690377
240
239
240
239
0.3766
Answer: (d) (ii)
[3]
(e) Give your answer to this part correct to 4 decimal places.
Two passengers are chosen at random from those on holiday.
Find the probability that they both hire a car.
180 passengers are on holiday out of them 150 hire a car,
150
149
×
= 0.6936685289
180
179
0.6937
Answer: (e)
[3]
October / November 2012 Paper 23
12
Mass of parcel (m kilograms)
0 < m ≤0.5
0.5 < m ≤ 1.5
1.5 < m ≤ 3
Frequency
20
18
9
The table above shows information about parcel in a delivery van.
John wants to draw a histogram using this information.
Complete the table below.
Mass parcel (m kilogram)
0< m ≤0.5
0.5 < m ≤ 1.5
1.5 < m ≤ 3
Frequency
40
18
6
Frequency density =
frequency
class width
20
= 40
0.5
9
=6
1.5
[2]
387
Workbook of Variant 3
October / November 2012 Paper 43
7
1
(a)
2
2
3
4
Two discs are chosen at random without replacement from the five discs shown in the
diagram.
(i) Find the probability that both discs are numbered 2.
possible outcomes
2
probability
and
2
2
1
2
= 1
×
=
5
4
20
10
1
10
Answer: (a) (i)
[2]
(ii) Find the probability that the numbers on the two discs have a total of 5.
(1, 4) or (4, 1) or (2, 3) or (3, 2)
1
1
1
1
2
1
1
2
3
6
×
+
×
+
×
+
×
=
=
5
4
5
4
5
4
5
4
10
20
3
10
Answer: (a) (ii)
[3]
(iii) Find the probability that the numbers on the two discs do not have a total of 5.
1−
6
14
7
=
=
20
20
10
7
10
Answer: (a) (iii)
[1]
(b) A group of international students take part in a survey on the nationality of their parents.
E =
{student with an English parent}
F =
{student with a French parent}
n( ) = 50, n(E) = 15, n(F) = 9 and n(E ∪ F)' = 33
E
F
15-x
x
9-x
(i) Find n(E ∩ F).
33
15 − x + x + 9 − x + 33 = 50
x = 7
Answer: (b) (i)
(ii) Find n(E ' ∪ F).
x + 9 − x + 33
7 + 9 − 7 + 33 = 42
388
7
[1]
Statistics and Probability
Answer: (b) (ii)
42
[1]
(iii) A student is chosen at random.
Find the probability that this student has an English parent and a French parent.
Answer: (b) (iii)
7
50
[1]
(iv)A student who has a French parent is chosen at random.
Find the probability that this student also has an English parent.
9 students have a french parent out of them 7 have an english parent
Answer: (b) (iv)
7
9
[1]
389
Workbook of Variant 3
9 200 students take a Mathematics examination.
The cumulative frequency diagram shows information about the times taken, t minutes, to
complete the examination.
200
190
180
170
160
UQ
150
140
130
120
110
Cumulative
frequency
Median
100
90
80
70
60
LQ
50
40
30
20
10
0
30
40
50
60
Time (minutes)
390
70
80
90
t
Statistics and Probability
(a) Find
(i) the median,
n
200
=
= 100th term
2
2
72
Answer: (a) (i)
min
[1]
min
[1]
min
[1]
(ii) the lower quartile,
n
200
=
= 50th term
4
4
68
Answer: (a) (ii)
(iii) the inter quartile range,
3
3
(n) =
(200) = 150th term
4
4
Upper qurtile = 76
IQR = UQ − LQ
= 76 − 68
= 8
8
Answer: (a) (iii)
(iv) the number of students who took more than 1 hour.
164
Answer: (a) (iv)
[2]
(b) (i) Use the cumulative frequency diagram to complete the grouped frequency table.
Time t minute
30 < t ≤ 40 40 < t ≤ 50 50< t ≤ 60 60 < t ≤ 70 70 < t ≤ 80 80 < t ≤ 90
Frequency
9
11
16
28
108
28
[1]
(ii) Calculate an estimate of the mean time taken by the 200 students to complete the
examination.
Show all your working.
mean =
mean =
∑fx
[Here 'x' will be taken as mid value of class]
∑f
35 × 9 + 45 × 11 + 55 × 16 + 65 × 28 +75 × 108 + 85 × 28
200
mean = 69.95
Answer: (b) (ii)
70
min
[4]
391
Workbook of Variant 3
May / June 2013 Paper 23
12 Two spinners have sections numbered from 1 to 5.
Each is spun once and each number is equally likely.
The possibility diagram is shown below.
5
4
2
1
3
1
4
3
2
2
5
Second
3
spinner
5
1
1
2
3
4
First spinner
4
5
Find the probability that
(a) both spinners show the same number,
(1,1) (2,2) (3,3) (4,4) (5,5)
=
5
25
1
5
Answer: (a)
1
5
[2]
(b) the sum of the numbers shown on the two spinners is 7.
(2,5) (3,4) (4,3) (5,2)
4
25
Answer: (b)
392
4
25
[2]
Statistics and Probability
May / June 2013 Paper 43
9
Sam asked 80 people how many minutes their journey to work took on one day.
The cumulative frequency diagram shows the times taken (m minutes).
80
70
UQ
60
50
Median
Cumulative
40
frequency
30
20
LQ
10
0
10
20
30
40
50
m
Time (minutes)
(a) Find
(i) the median,
80
= 40th term
2
Answer: (a) (i)
14
min
[1]
393
Workbook of Variant 3
(ii) the lower quartile,
80
= 20th term
4
8
Answer: (a) (ii)
min
[1]
min
[1]
(iii) the inter-quartile range.
3
(80) = 60th term,
4
UQ = 30
IQR = UQ − LQ = 30 − 8 = 22
22
Answer: (a) (iii)
(b) One of the 80 people is chosen at random.
Find the probability that their journey to work took more than 35 minutes.
Give your answer as a fraction.
80 − mark on 35
80 − 69 = 11
p=
11
80
11
80
Answer: (b)
[2]
(c) Use the cumulative frequency diagram to complete this frequency table.
Time m minute
0 < m ≤ 10 10 < m ≤ 15 15< m ≤ 30 30 < m ≤ 40
Frequency
30
12
18
40 < m ≤ 50
16
4
76 - 16 = 16
80 - 76 = 4
[2]
(d) Using mid-interval values, calculate an estimate of the mean journey time for the 80 people.
mean = ∑fx
∑f
mean =
5 × 30 + 12.5 × 12 + 22.5 × 18 + 35 × 16 +45 × 4
80
mean = 18.0625
Answer: (d)
394
18.1
min
[3]
Statistics and Probability
(e) Use the table in part (c) to complete the histrogram to show the times taken by the 80
people. One column has already been completed for you.
4
3
Frequency
2
density
1
0
10
for 10 < m ≤ 15, fd =
12
= 2.4
5
for 15 < m ≤ 30, fd =
18
= 1.2
15
for 30 < m ≤ 40, fd =
16
= 1.6
10
for 40 < m ≤ 50, fd =
4
= 0.4
10
20
30
Time (minutes)
40
50
m
[5]
395
Workbook of Variant 3
6
In a box there are 7 red cards and 3 blue cards.
A card is drawn at random from the box and is not replaced.
A second card is then drawn at random from the box.
(a) Complete this tree diagram
First card
Second card
6
9
7
10
3
10
Red
Red
3
9
Blue
7
9
Red
Blue
2
9
Blue
[3]
(b) Work out the probability that the two cards are of the different colours.
Give your answer as a fraction.
red and blue or blue and red
3
7
7
3
42
7
× + × =
=
9
9
10
10
90
15
Answer: (b)
396
7
15
[3]
Statistics and Probability
October / November 2013 Paper 23
18 A gardener measured the lengths of 50 green beans from his garden.
The results have used to draw this cumulative frequency diagram.
50
40
UQ
30
Median
Cumulative
frequency
20
LQ
10
0
5
10
15
20
25
30
35
Length of green bean (cm)
Work out
(a) the median,
n
50
=
= 25th term
2
2
25th term will give us median median = 19
Answer: (a)
19
cm [1]
(b) the number of green beans that are longer than 26 cm,
Total green beans − mark on 26 cm
50 − 47
= 3
Answer: (b)
3
[2]
397
Workbook of Variant 3
(c) the inter-quartile range,
n
50
=
= 12.5th term
4
4
12.5 term gives LQ, LQ = 16.5
3
3
(n) =
(50) = 37.5th term
4
4
IQR = UQ − LQ = 22 − 16.5 = 5.5
5.5
Answer: (c)
cm [2]
(d) the probability that a green bean chosen at random is more than 14 cm long.
Total green beans − mark on 14 cm = 50 − 5 = 45
45
50
probability =
45
50
Answer: (d)
[2]
October / November 2013 Paper 43
5
(a)80 students were asked how much time they spend on the internet in one day.
This table shows the result.
Time (t hours)
0<t≤1
1<t≤2
2<t≤3
3<t≤5
Number of students
15
11
10
19
5 < t ≤ 7 7 < t ≤ 10
13
12
(i) Calculate an estimate of the mean time spent on the internet by the 80 students.
mean =
∑fx
∑f
mean =
0.5 × 15 + 1.5 × 11 + 2.5 × 10 + 4 × 19 + 6 × 13 + 8.5 × 12
80
mean = 3.8125
Answer: (a) (i)
398
3.81
hour
[4]
Statistics and Probability
(ii) On the grid, complete the histogram to show this information.
16
14
12
10
Frequency
density 8
6
4
2
0
1
2
3
4
5
6
7
8
9
10
t
Time (hours)
For intervel 3 < t ≤ 5
fd =
19
= 9.5
2
For intervel 5 < t ≤ 7
fd =
13
= 6.5
2
For intervel 7 < t ≤ 10
fd =
12
=4
3
[4]
399
Workbook of Variant 3
(b) The probability that Chaminda uses the internet on day is
3
.
5
The probability that Niluka uses the internet on any day is
3
.
4
(i) Complete the tree diagram.
Chaminda
Niluka
Uses the
internet
3
5
3
4
Uses the
internet
1
4
Does not
use the
internet
3
4
2
5
Does not
use the
internet
1
4
Uses the
internet
Does not
use the
internet
[2]
(ii)Calculate the probability, that on any day, at least one of the two students uses the
internet.
1 − does not uses the internet on both days
1−
=
=
2
1
×
5
4
18
20
9
10
Answer: (b) (ii)
9
10
[3]
(iii) Calculate the probability that Chaminda uses the internet on three consecutive days.
3
3
3
27
×
×
=
5
5
5
125
Answer: (b) (iii)
400
27
125
[2]
Statistics and Probability
May / June 2014 Paper 23
2
Michelle sells icecream.
The table shows how many of the different flavours she sells in one hour.
Flavour
Vanilla
Number sold
6
Strawberry Chocolate
8
Mango
9
7
Michelle wants to show this information in a pie chart.
Calculate the sector angle for mango.
7
× 360 = 84
6+8+9+7
84
Answer:
[2]
20Jenna draws a cumulative frequency diagram to show information about the scores of 500
people in a quiz.
500
400
UQ
300
Median
Cumulative
frequency
40th percentile
200
LQ
100
0
10
20
30
40
50
60
Score
Use the diagram to find
401
Workbook of Variant 3
(a) the median score,
n
500
=
= 250th term
2
2
250 th gives us median median = 34
Answer: (a)
34
[1]
(b) the inter quartile range,
125 term gives lower quartile LQ = 24
n
500
=
= 125th term
4
4
3
3
(n) =
(500) = 375th term
4
4
375 term gives upper quartile, UQ = 40
IQR = UQ − LQ
= 40 − 24
= 16
Answer: (b)
16
[2]
(c) the 40th percentile,
40
× 500 = 200th
100
200th term will give 40th percentile,
Answer: (c)
30
[1]
(d) the number of people who scored 30 or less but more than 20.
Mark on 30 score − mark on 20 score
200 − 80
120
Answer: (d)
402
120
[1]
Statistics and Probability
May / June 2014 Paper 43
2
4
3
Frequency
density
2
1
0
10
20
30
40
50
60
Amount ($x)
A survey asked 90 people how much money they gave to charity in one month.
The histogram shows the results of the survey.
(a) Complete the frequency table for the six columns in the histogram.
Amount
Frequency
0 < t ≤ 10
10 < t ≤ 25
25< t ≤ 30
30 < t ≤ 35
35 < t ≤ 50
50 < t ≤ 60
13
33
19
4
15
6
10 x 1.3 = 13
15 x 2.2 = 33
5 x 3.8 = 19
5 x 0.8 = 4
15 x 1 = 15
10 x 0.6 = 6
[5]
(b) Use your frequency table to calculate an estimate of the mean amount these 90 people
gave to charity.
mean = ∑fx ∑f
mean =
5 × 13 + 17.5 × 33 + 27.5 × 19 + 32.5 × 4 + 42.5 × 15 + 55 × 6
90
mean = 25.13888889
Answer: (b) $
25.1
[4]
403
Workbook of Variant 3
6
In this question, give all your answer as fractions.
N
A
T
I
O
N
The letters of the word NATION are printed on 6 cards.
(a) A card is a chosen at random.
Write down the probability that
(i) it has the letter T printed on it,
Answer: (a) (i)
1
6
[1]
(ii) it does not have the letter N printed on it,
4
2
=
6
3
Answer: (a) (ii)
2
3
[1]
(iii) the letter printed on it has no line of symmetry.
Letter N has no line of symmetry
2
1
=
3
6
Answer: (a) (iii)
1
3
[1]
(b) Lara chooses a card at random, replaces it, then chooses a card again.
Calculate the probability that only one of the cards she chooses has the letter N printed on it.
Possible outcomes
N and not N or not N and N
Probability
2
4
4
2
× + × 6
6
6
6
=
16
36
=
4
9
Answer: (b)
4
9
[3]
(c) Jacob chooses a card at random and does not replace it.
He continues until he chooses a card with the letter N printed on it.
Find the probability that this happens when he chose the 4th card.
Probabilitynot N ×
4
6
×
3
5
×
2
4
not N ×
×
2
3
not N ×
N
48
360
=
=
2
15
Answer: (c)
404
2
15
[3]
Statistics and Probability
October / November 2014 Paper 23
4
The four sector angles in a pie chart are 2xo, 3xo, 4xo and 90o.
Find the value of x.
2x + 3x + 4x + 90 = 360
9x + 90 = 360
9x = 360 − 90
x =
360 − 90
9
30
Answer: x =
[2]
17 The mass, m grams, of cornflakes in each of 200 boxes is recorded
The cumulative frequency diagram shows the results.
200
180
UQ
160
140
120
Cumulative
frequency 100
80
LQ
60
40
20
0
494
496
498
500
502
504
506
508
510
m
Mass (grams)
(a) Use the diagram to estimate the inter quartile range.
LQ = 502.6
n
200
=
= 50th term
4
4
3
3
(n) =
(200) = 150th term
4
4
UQ = 505.8
Answer: (a)
3.2
g [2]
405
Workbook of Variant 3
(b) Find the probability that a box chosen at random has a mass of 500 grams or less.
Mark on 500 gram = 16
Probability =
16
200
16
200
Answer: (b)
(c)
Mass (m gram)
[2]
496 < m ≤ 500 500 < m ≤ 504 504 < m ≤ 508 508 < m ≤ 510
Frequency
16
74
104
The data in this frequency table is to be shown in a histogram.
Complete the frequency density table below.
mass (m gram)
6
496 < m ≤ 500 500 < m ≤ 504 504 < m ≤ 508 508 < m ≤ 510
Frequency
4
18.5
26
3
74 ÷ 4 = 18.5
104 ÷ 4 = 26
6÷2=3
[2]
October / November 2014 Paper 43
4
Yeung and Ariven compete in a triathlon race.
The probability that Yeung finishes this race is
3
.
5
The probability that Ariven finishes this race is
2
.
3
(a) (i) Which of them is more likely to finish this race?
Give a reason for your answer.
[
3
= 0.6,
5
2
= 0.666666667]
3
Answer: (a) (i)
Ariven
beacuse
probability that Ariven finishes this race is more than Yeung
[1]
(ii) Find the probability that they both finish this race.
Yeung and ariven
3
2
6
×
=
5
3
15
Answer: (a) (ii)
406
2
6
/
5
15
[2]
Statistics and Probability
(iii) Find the probability that only one of them finishes this race.
Probability that yeung does not finishes this race
3
2
=
5
5
Probability that Ariven does not finishes this race
=1−
2
1
=
3
3
Yeung finishes, Ariven does not finished
=1−
(
or Ariven does not finishes
3
1
2
2
7
×
+
×
=
5
3
3
5
15
)
7
15
Answer: (a) (iii)
[3]
(b) After the first race, Yeung competes in two further triathlon races.
(i) Complete the tree diagram.
First race
Second race
6
7
Third race
1
7
6
7
2
5
Does not
finish
1
7
Finishes
3
10
Does not
finish
7
10
Finishes
3
10
Does not
finish
7
10
Finishes
3
10
Does not
finish
7
10
Finishes
3
10
Does not
finish
Finishes
Finishes
3
5
7
10
Does not
finish
Finishes
Does not
finish
[3]
407
Workbook of Variant 3
(ii) Calculate the probability that Yeung finishes all three of his races.
=
3
6
7
×
×
5
7
10
126 =
9
350
25
9
25
Answer: (b) (ii)
[2]
(iii) Calculate the probability that Yeung finishes at least one of his races.
1 − P [does not finishes any of his race]
1−
=
2
1
3
344
×
×
=
5
7
10
350
344
172
=
350
175
172
175
Answer: (b) (iii)
May / June 2015 Paper 23
17
200
UQ
150
Median
Cumulative
frequency 100
LQ
50
0
1
2
3
4
5
6
Time (seconds)
200 students take a reaction time test.
The cumulative frequency diagram shows the result.
Find
408
7
8
9
10
[3]
Statistics and Probability
(a) the median,
n
200
=
= 100th term
2
2
Median = 6
6
Answer: (a)
s [1]
(b) the inter-quartile range,
LQ = 5
n
200
=
= 50th term
4
4
3
3
(n) =
(200) = 150 term
4
4
UQ = 7
IQR = UQ − LQ = 7 − 5 = 2
2
Answer: (b)
s [2]
(c) the number of students with a reaction time of more than 4 seconds.
Total number of student − mark on 4 seconds
200 − 20
180
180
Answer: (c)
[2]
May / June 2015 Paper 43
4The table shows the time, t minutes, taken by 200 students to complete an IGCSE paper.
Time (t minutes)
Frequency
40 < t ≤ 60 60 < t ≤ 70
10
70< t ≤ 75
75 < t ≤ 90
80
60
50
(a) By using mid interval values, calculate an estimate of the mean time.
mean = ∑fx ∑f
mean =
50 × 10 + 65 × 50 +72.5 × 80 + 82.5 × 60
200
mean = 72.5
Answer: (a)
72.5
min
[3]
409
Workbook of Variant 3
(b) On the grid, draw a histogram to show the information in the table.
20
18
16
14
12
Frequency
10
density
8
6
4
2
0
40
Frequency density =
50
60
Time (minutes)
70
80
90
t
frequency
class width
For 40 < t ≤ 60
fd =
1
10
=
2
20
For 60 < t ≤ 70
fd =
50
=5
10
For 70 < t ≤ 75
fd =
80
= 16
5
For 75 < t ≤ 90
fd =
60
=4
15
[4]
5
A
410
A
A
A
B
B
C
Statistics and Probability
(a) One of these 7 cards is chosen at random.
Write down the probability that the card
(i) shows the letter A,
Answer: (a) (i)
4
7
[1]
(ii) shows the letter A or B,
Answer: (a) (ii)
6
7
[1]
(iii) does not show the letter B.
Answer: (a) (iii)
5
7
[1]
(b) Two of the cards are chosen at random, without replacement.
Find the probability that
(i) both show the letter A,
A and A
4
3
12
2
×
=
=
7
6
42
3
Answer: (b) (i)
2
3
[2]
(ii) the two letters are different.
method 1
1 − P(same letter)
1 − [AA or BB]
1−[
2
2
1
+
×
]
3
7
6
2
28
=
3
42
method 2
A, not A, or B, not B, or C
4
3
2
5
1
2
28
×
+
×
+
=
=
7
6
7
6
7
3
42
Answer: (b) (ii)
2
3
[3]
(c) Three of the cards are chosen at random, without replacement
Find the probability that the cards do not show the letter C.
not C, not C , not C
4
120
6
5
4
×
×
=
= 7
210
7
6
5
411
Workbook of Variant 3
Answer: (c)
4
7
[2]
October / November 2015 Paper 23
4
The probability that it will rain on any day is
1
.
5
Calculate an estimate of the number of the days it will rain in a month with 30 days.
30 ×
1
=6
5
Answer:
6
[1]
18Samira takes part in two charity runs.
The probability that she finishes each run is 0.8.
First run
0.8
0.2
Second run
0.8
finishes
0.2
does not
finish
0.8
finishes
0.2
does not
finish
finishes
does not
finish
Find the probability that Samira finishes at least one run.
1 − P [does not finish x does not finish]
1 − 0.2 × 0.2
0.96
Answer:
412
0.96
[3]
Statistics and Probability
24
80
70
60
50
Median
Cumulative
frequency 40
30
30th percentile
20
10
0
1
2
3
4
5
6
7
8
9
10
Time (minutes)
The cumulative frequency diagram shows information about the times, in minutes, taken by 80
students to complete a short test.
Find
(a) the median,
n
80
=
= 40th term
2
2
40 term will give median
median = 6.2
6.2
Answer: (a)
[1]
(b) the 30th percentile,
30
× 80 = 24th term
100
24th term will give 30 percentile
Answer: (b)
5.8
min
[2]
(c) the number of students taking more than 5 minutes.
Total number of student − mark on 5 minutes
80 − 10
70
Answer: (c)
70
[2]
413
Workbook of Variant 3
October / November 2015 Paper 43
6
The table shows information about the masses, m grams, of 160 apples.
Mass (m grams)
30 < m ≤ 80
80 < m ≤ 100
Frequency
50
30
100 < m ≤ 120 120 < m ≤ 200
40
40
(a) Calculate an estimate of the mean.
mean = ∑fx
∑f
55 × 50 + 90 × 30 +110 × 40 + 160 × 40
160
mean =
mean = 101.5625
102
Answer: (a)
g [4]
(b) On the grid complete the histogram to show the information in the frequency table
2.5
2
1.5
Frequency
density
1
0.5
0
40
frequency
class width
For intervel 30 < m ≤ 80
Frequency density =
fd =
414
50
=1
50
80
120
Mass (grams)
160
200
m
Statistics and Probability
For intervel 80 <m ≤ 100
fd =
30
= 1.5
20
For intervel 100 < m ≤ 120
fd =
40
=2
20
[3]
(c) An apple is chosen at random from the 160 apples.
Find the probability that its mass is more than 120 g.
40 apple have mass of more than 120g
Probability =
1
40
=
4
160
1
4
Answer: (c)
[1]
(d) Two apples are chosen at random from the 160 apples, without replacement.
Find the probability that
(i) they both have a mass of more than 120 g,
40
39
1560
×
=
=
160
150
25440
13
212
Answer: (d) (i)
13
212
[2]
(ii) one has a mass of more than 120 g and one has a mass of 80 g or less.
(more than 120 g, 80 or less) or (80 or less, more than 120)
25
40
50
50
40
4000
×
+
×
=
=
159
160
159
160
159
25440
Answer: (d) (ii)
25
159
[3]
415