Cambridge International Examinations
Cambridge Secondary 1 Checkpoint
1112/02
MATHEMATICS
Paper 2
October 2018
MARK SCHEME
Maximum Mark: 50
Published
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. It shows the basis on which Markers were instructed to award marks. It does not indicate the
details of the discussions that took place at an Markers’ meeting before marking began, which would have
considered the acceptability of alternative answers.
Mark schemes should be read in conjunction with the question paper and the End of Series Report.
Cambridge will not enter into discussions about these mark schemes.
Mark scheme annotations and abbreviations
M1
A1
B1
FT
dep
oe
cao
isw
soi
method mark
accuracy mark
independent mark
follow through after error
dependent
or equivalent
correct answer only
ignore subsequent working
seen or implied
This document consists of 9 printed pages.
IB18 10_1112_02/2RP
© UCLES 2018
[Turn over
Page 2 of 9
1 In either order.
1 Ignore any units given.
11 and 13
20
6
7
B1
2 Allow any unambiguous indication of the correct
answer.
Millilitres or ml
Kilograms or kg
Kilometres or km
2 correct
1
10
4(b)
5
1
B1
2
2
For either 18 or 20 seen
or
For a mark on the graph at 17:30
2 (°C)
Further Information
October 2018
4(a)
3
M1
2
($)74
2
103.60
5
× 40 or 1.85 or 0.71(...) or
56
7
1
Marks
32.01
Answer
Checkpoint Mathematics - Mark Scheme
1
Question
1112/02
or
3
1
correct method for comparison, e.g.
31
or 31 ÷ 80
80
or
0.41 × 80 oe
Page 3 of 9
M1
implied by 32.8 or 33
or 38(.75%) or 0.38(…) or 39(%) or 0.39
Note other correct methods are acceptable
or 0.38(..) or 39(%) or 0.39
or 33
2 Note other correct methods are acceptable.
Ticks Women
and
correct figure for comparison, e.g.
• (31 out of 80) = 38(.75%)
• (41% of 80) = 32(.8)
10
Further Information
1 Allow any unambiguous indication of the correct
answer.
1
1
Marks
October 2018
Point E marked on the grid at (2, 1)
3 9
,
or 0.375 or 37.5%
8 24
or 0.33(…) or 33(.3…)%
or an equivalent fraction e.g.
120
45
6
2
Answer
Checkpoint Mathematics - Mark Scheme
9
8(b)
8(a)
Question
1112/02
13
12
600
744
or
600
or 144 or 0.24
or 1.24 or 124(%)
24 (%)
C
Page 4 of 9
M1
2
1
Allow their expression in (a) = 62
6x + 10 + 2x + 4 = 62
8x + 14 = 62, 2(4x + 7) = 62
M1
Forming a correct equation e.g. 2(3x + 5) + 2(x + 2) = 62 oe
or
3 × their 6 + 5
Q
Implied by x = 6
M2
Correctly solving their linear equation from (a)
3
23
11(b)
Further Information
1 Allow any equivalents e.g.
8x + 14 , 6x + 10 + 2x + 4 , 2(4x + 7)
Marks
October 2018
2(3x + 5) + 2(x + 2) isw
Answer
Checkpoint Mathematics - Mark Scheme
11(a)
Question
1112/02
16
15
14
Question
1112/02

seen
It has the largest mean/ average value.
Page 5 of 9
Saturday and a correct reason relating to the mean or
average, e.g.
63.6
× 100 oe
45.4
or 1.4…. or 0.454 or 0.71… or 2.2…
140(.088…) (pounds)
2 correct answers.



Answer
M1
B1
Further Information
October 2018
1 Do not accept
Saturday has the smallest range, by itself.
2 Award 2 marks for answers rounding to 140
correct to the nearest whole number
2 Allow any unambiguous indication of the correct
answer.
Marks
Checkpoint Mathematics - Mark Scheme
17
Question
1112/02
making the coefficients of x or y equal followed by an
appropriate consistent subtraction or addition across all
3 terms
re-arranging one of the equations to make one variable
the subject and then substituting their arrangement into
the other equation,
e.g. x + 2(24 – 3x) = 13 or 3(13 – 2y) + y = 24
•
•
Page 6 of 9
correct substitution and evaluation from incorrect first
value implied by two values satisfying one of the
original equations.
M1
An attempt at eliminating either x or y
e.g.
•
M2
Further Information
October 2018
With no errors.
With no more than one arithmetic error.
Can be implied by 5y = 15 or 5x = 35
3 Do not accept trial and improvement as a
method.
Marks
An algebraic method leading to either
x = 7 or y = 3
Algebraic method seen leading to
(x =) 7
(y =) 3
Answer
Checkpoint Mathematics - Mark Scheme
19(b)
19(a)
18
Question
1112/02
or
Any one term correct.
1 2 3
, ,
3 5 7
(±) 3n ± k seen
or correct expression seen then spoilt
3n + 11
Page 7 of 9
B1
B1
October 2018
Regardless of order
2 In correct order.
Accept equivalent fractions or decimal
equivalents for 2 marks or for B1:
= 0.33(33…),
= 0.4, = 0.42(857…) or 0.43
Allow for just 3n
2 Mark the final answer for 2 marks.
Allow equivalent unsimplified, e.g.
14 + 3n – 3, 14 + 3(n – 1)
Do not accept: n = 3n + 11
Do not allow x1 − x2
or
2
e.g.
2
M1
(6, k) or (k, 6) for C
or
for sight of x1 + x2 or with correct numbers
,
Further Information
B2
3
Marks
(8, k) or (k, 10) for D
or
for finding coordinates of C, i.e. (6, 6)
(8, 10)
Answer
Checkpoint Mathematics - Mark Scheme
22
21
20
Question
1112/02
8
+
8
x +1
B1
8.6 in answer space.
Page 8 of 9
M1
A correct trial of x where 8.6 < x ≤ 8.65
72 (cm)
and
53 (°)
M1
M1
e.g. M1 for
32
8x
+
32
4( x + 1)
,
32
12x + 4
M1 implied by correct unsimplified answer
Further Information
October 2018
1 Both correct for the mark.
For both M1 marks to be awarded, one
appropriate trial to at least 1 decimal place and
one appropriate trial to at least 2 decimal places
must be seen, e.g. trial at 8.6 and trial at 8.65
3 Ignore the final column in the table when
marking.
2
Marks
Any correct trial of a number between 8 and 9
Must include all three marking points below.
A complete trial and improvement method leading to the
answer x = 8.6
e.g.
2x
Correct fractions with a common denominator
8
3x + 1
Answer
Checkpoint Mathematics - Mark Scheme
M1
2
1500 (kg)
26
Page 9 of 9
150 000 (m2) seen
or for a correct method
oe
e.g. 15 × 10 000 ×
or
for follow through of incorrect area conversion multiplied by
0.01 correctly.
1
–2
3
2
25
or
13
7
1
1.5
1.8(57…) or
B1
2
Marks
8 (hours)
•
•
Any of
gives correct supporting ratios, e.g.
1.8(57…) : 1 and 1.5 : 1
and
Shirt A
Answer
Checkpoint Mathematics - Mark Scheme
24
23
Question
1112/02
e.g. 15 × 10k ×
oe
Allow rounded e.g. 1.9
Allow fractions : 1 and : 1
Allow rounded e.g. 1.9:1 and 1.5:1
Further Information
October 2018