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## secondary checkpoint Mathematics April Paper 2 2019 Details Solution

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# Cambridge Secondary Checkpoint Mathematics April Paper 2 2019 Details Solution

### Keywords: Cambridge Secondary Checkpoint Mathematics April Paper 2 2019 Details Solution

CHECKPOINT MATHEMATICS/1112/02 2019

SOLUTION TO APRIL 2019 – PAPER 2

The solution to a particular question can be accessed instantly by
clicking on the desired question number in the QUICK ACCESS GRID.

12345
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

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EDUCATE. EXCELERATE.

DISCLAIMER
The solution to the said examination paper has been drafted by teachers who are duly

accredited and experienced with the CHECKPOINT MATHEMATICS curriculum.
It has been developed independently of the CAIE and is in no way endorsed by the CAIE.

CHECKPOINT MATHEMATICS/1112/02 2019

Solution Marks Notes

S1 25 × 1.8 + 32 1
(a)
Back to QUICK ACCESS GRID
= 45 + 32
= 77 °F

(b)

S2 142 + 29 1
3×22 + 7 Back to QUICK ACCESS GRID
256 + 29
= 3×4 + 7 1 Note:
285
= 12 + 7 The answer can be worked out in a single
285 step using the calculator.
= 19
= 15 Back to QUICK ACCESS GRID

S3

1 P (Mike arrives late to school)
15 3

= 20 = 4

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CHECKPOINT MATHEMATICS/1112/02 2019

Solution Marks Notes

S4 248 675 1
(a) = 249 000 to the nearest thousand

(b) 52.747 square kilometres 1 number in the second decimal place (4) < 5
= 52.7 square kilometres
∴ number in the first decimal place remains
S5 unchanged.

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2
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CHECKPOINT MATHEMATICS/1112/02 2019

Solution Marks Notes

S6 1
(a)

(b)

Line = 2 shown on the grid in red 1

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CHECKPOINT MATHEMATICS/1112/02 2019

Solution Marks Notes

S7 Diameter of the pond in the scale 1 1 cm → 2 m
(a) drawing = 3.1 cm
3.1 cm → 3.1 × 2 m = 6.2 m
Diameter of the pond in real life
= 6.2 metres

(b)

Length of the shed in the scale drawing 1 2 m → 1 cm
= 1.5 cm 3 m → 3 ÷ 2 cm = 1.5 cm
4 m → 4 ÷ 2 cm = 2 cm
Width of the shed in the scale drawing
= 2 cm

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CHECKPOINT MATHEMATICS/1112/02 2019

Solution Marks Notes

S8 Total distance Carlos swims 2 1000 m = 1 km
(a) To convert m to km → divide by 1000

= 90 × 25 m

= 2250 m 2250 m = 2250 ÷ 1000 km = 2.250 km

= 2.250 km

(b) number of lengths Carlos swims on his 2
front

4
= 4 + 1 × 90

4
= 5 × 90
= 72

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S9
(a)

1

(b)

= –2 1 solution to the simultaneous equations
= –3 = co-ordinates of point of intersection of
the two lines

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CHECKPOINT MATHEMATICS/1112/02 2019

Solution Marks Notes

S10 S = a × b × ÷ 2 2
(a)

S =

(b) × 7.6 × 9.22 2 2
V = 8 = 252.6 cm3 V= 8

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S11
(a)

(b) = 5.7 1

1 56 – 55.29 = 0.71
56.84 – 56 = 0.84
∴ 55.29 is closer to 56 than 56.84

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CHECKPOINT MATHEMATICS/1112/02 2019

Solution Marks Notes

S12
(a)

(b) 1 Blessy is older than Manjit.
< ∴ Blessy’s age > Manjit’s age

1 Anastasia is less than half the age of Blessy.

Anastasia’s age < ½ of Blessy’s age

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S13 Number of runners that take less than 1 2 1 hour = 60 minutes
hour = 20 + 15 + 10 = 45

Percentage of runners that take less than
1 hour

45
= 85 × 100

= 52.9%

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CHECKPOINT MATHEMATICS/1112/02 2019

Solution Marks Notes

S14

2
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S15
(a)

Data shown by red cross on graph 1

(b) 8.75 seconds (approximately) 1 time deduced from the line of best fit;
values in the range of 8 – 9 seconds will be
acceptable

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CHECKPOINT MATHEMATICS/1112/02 2019

Solution Marks Notes

S16

0.075 km 7.6 m 77 cm 780 mm
75 m 7.6 m 0.77 m 0.78 m

0.77 m 0.78 m 7.6 m 75 m

77 cm 780 mm 7.6 m 0.075 km
smallest largest

2 To compare the lengths, it is essential to
express them using the same units.

To convert:

km to m → multiply by 1000
cm to m → divide by 100
mm to m → divide by 1000

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S17 Chen earned a bonus ✔ 2 total distance
Average speed = total time

Total distance covered by Chen on his Total distance = average speed × total time
last journey

= 61 × 4.5

= 274.5 km

Distance covered per litre of fuel
274.5

= 96

= 2.859375 km/

2.859375 km/ > 2.8 km/

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CHECKPOINT MATHEMATICS/1112/02 2019

Solution Marks Notes

S18

1 2.3 kilograms = 2.3 × 1000 g = 2300 g

0.3 litres = 0.3 × 1000 ml = 300 ml
150 ml < 300 ml
150 ml < 0.3

5 metres = 5 × 1000 mm = 5000 mm
5000 millimetres = 5 metres

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S19 1 2 Area of a trapezium
Area of the trapezium = 2 × (a + b) × 6 1
1
2 × (a + b) × 6 = 30 = 2 × (sum of parallel sides) × height

Other possible values of a and b in cm:

(a + b) × 3 = 30 ab
30 19
28
(a + b) = 3 37
a + b = 10 cm

a = 4 cm
b = 6 cm

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CHECKPOINT MATHEMATICS/1112/02 2019

S20 Solution Marks Notes

% of children in film A audience 2 To convert a fraction to percentage
11 ↓

= 11+19 × 100 Multiply the fraction by 100

11
= 30 × 100

= 36.7

% of children in film B audience
5

= 5+7 × 100
5

= 12 × 100

= 41.7

∴ Film B has the greater proportion of
children in the audience.

Film B ✔

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CHECKPOINT MATHEMATICS/1112/02 2019

Solution Marks Notes

S21 3
Area of the square = 82 = 64 cm2

Area of the square covered by the star
= 40% of 64 cm2

40
= 100 × 64
= 25.6 cm2

Width of the rectangle = 8 cm
60

60% of 8 cm = 100 × 8 = 4.8 cm
Length of the rectangle
= 8 + 4.8 = 12.8 cm

Area of the rectangle
= 12.8 × 8
= 102.4 cm2

% of the rectangle that the star covers
25.6

= 102.4 × 100
= 25%

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CHECKPOINT MATHEMATICS/1112/02 2019

Solution Marks Notes

S22 4200 cm3 54 000 mm3 45 litres 52 000 m
4200 cm3 54 cm3 45 000 cm3 52 000 cm3

52 000 cm3 → largest measurement

1 To compare the measurements, it is
essential to express them using the same
units.

To convert:
mm3 to cm3 → divide by 1000
litres to cm3 → multiply by 1000

1 cm3 = 1m

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CHECKPOINT MATHEMATICS/1112/02 2019

Solution Marks Notes

S23

Age Number of Can Cannot % of people
remember that can
11 to 20 people tested remember remember
21 to 30 4 80
31 to 40 20 16 6 70
20 14 14 65
40 26

Yes No ✔ 2

Reason:

As age increases, the percentage of
people that can remember decreases.
This indicates that younger people can
remember better than older people.

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CHECKPOINT MATHEMATICS/1112/02 2019

Solution Marks Notes

S24 Volume of the cube = 33 = 27 cm3 3 Volume of a cube = side3

Volume of the cuboid Volume of a cuboid
= 1.2 × 100 × 30 × 50 = length × width × height
= 180 000 cm3
Number of cubes that fit inside the box Length of the cuboid = 1.2 m = 1.2 × 100 cm

180 000 180 000
= 27 27 = 6666.7
= 6666
∴ number of cubes that fit completely
inside the box = 6666

S25 Back to QUICK ACCESS GRID
√ . = +3.5 and –3.5 1

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