Cambridge Secondary checkpoint Mathematics April Paper 1 2019 Details Solution

CHECKPOINT MATHEMATICS/1112/01 2019

SOLUTION TO APRIL 2019 – PAPER 1
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Solution Marks Notes

S1 6 × 3 + = 23 1

18 + = 23

= 23 – 18

= 5

S2 Let the number be . Back to QUICK ACCESS GRID
1
3 × – 6 = 15
3 – 6 = 15
3 = 15 + 6
3 = 21

21
= 3
= 7

S3 Back to QUICK ACCESS GRID
1 1 bottle → 500 ml
3 × 500 ml = 1500 ml

1500 ml = 1500 ÷ 1000 = 1.5 3 bottles → 3 × 500 ml
1 litre = 1000 ml

To convert ml to → divide by 1000

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Solution Marks Notes

S4 12 biscuits → 125 g butter 1

125
1 biscuit → 12 g butter

125
36 biscuits → 12 × 36 g butter

125
12 × 36 = 125 × 3 = 375

36 biscuits → 375 g butter

S5 3 2 3 Back to QUICK ACCESS GRID
8 × 5 = 20 1

34
8×5

32 Back to QUICK ACCESS GRID
=8 ×5 ×2

3
= 20 × 2


=

S6

1 Test for divisibility by 9

If the sum of all the digits in a number is
exactly divisible by 9, then the number is
exactly divisible by 9.

54 → 5 + 4 = 9
297 → 2 + 9 + 7 = 18

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Solution Marks Notes

S7

1 Multiply by 4
× 4 = 4

then subtract 2
4 – 2

S8 12.7 × 0.3 Back to QUICK ACCESS GRID
1 127 × 3 = 381
127 3
= 10 × 10

381
= 100

= 3.81

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S9 Number of days it rains in November 2 ALTERNATIVE ANSWER

3 Number of days it does not rain in
= 5 × 30
=3×6 November
2

= 5 × 30

= 18 = 2 × 6

Number of days it does not rain in = 12
November = 30 – 18 = 12

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Solution Marks Notes

S10

Area of cross-section of prism 2 Area of a rectangle = length × width
=3×7×4
= 12 × 7 Volume of the prism
= 84 cm2 = area of cross-section × 10
Volume of the prism
= 84 × 10
= 840 cm3

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Solution Marks Notes

S11

Rice A Rice B ✔ 2

Free rice in bag A
= 25% of 500 g

25
= 100 × 500
= 25 × 5
= 125 g
Free rice in bag B

1
= 5 of 750 g

1
= 5 × 750
= 150 g
150 g > 125 g
∴ Bag B gives more free rice than bag A.

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Solution Marks Notes

S12 Lily ✔ 1

Mia

√121 = 11
√100 = 10
120 is closer to 121 than to 100.
∴ √120 is closer to √121 than to √100,
that is, √120 is closer to 11 than to 10.

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S13 Primes between 60 and 70: 1

61 and 67

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S14

1 For a fair coin, both head and tail are
equally likely outcomes.
Accordingly, if a fair coin is tossed 200
times, the probability of obtaining heads is
50%, that is, 100 times.
Hence, coin C is most likely to be a fair coin.

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Solution Marks Notes

S15 1. 14 ÷ 0.2 = 70 2 1 2
2. 16 × 1.25 = 20
3. 20 × 0.5 = 10 14 ÷ 0.2 16 × 1.25
4. 36 ÷ 0.75 = 48 125
2
= 14 ÷ 10 = 16 × 100
125
10
= 14 × 2 = 4 × 25
=4×5
= 7 × 10 = 20

= 70

3 4

20 × 0.5 36 ÷ 0.75
5 75

= 20 × 10 =36 ÷ 100
= 10 100

= 36 × 75
4

= 36 × 3
= 12 × 4
= 48

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S16 Size of exterior angle of a regular 2 Size of exterior angle of a regular n-sided
10-sided polygon
polygon
360
= 10
=

= 36°

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Solution Marks Notes

S17
(a)

1

13 24
– 11 06
18
02

Oliver’s journey will take 02 hours 18
minutes.

(b)

As per the train time-table on Saturday, 2
to be at Hankberg before 18:15, Yuri
needs to take the 15:30 train from
Flaghaven.

To be able to take the 15:30 train from
Flaghaven, he must be at the station
before 15:30.
The latest train from Dibside that can
reach him before 15:30 is the one at
12:36

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Solution Marks Notes

S18 539 11 × 7 × 7 1
(a) 847 = 7 × 11 × 11 =

(b) 55 11 × 5 1
539 = 11 × 7 × 7 =

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S19
(a)

2

(b) Rajiv is not correct ✔ 1 The median length for 40 birds
= (length of 20th bird + length of 21st bird) ÷
As per the given data, the lengths of 20th
as well as 21st birds lie in the interval 2
17 ≤ < 18.

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Solution Marks Notes

S20 2 + 8(40 – 5) 1 Alternative answer
= 2 + 8 × 40 – 8 × 5
= 2 + 320 – 40 2 + 8(35) = 2 + 280 = 282
= 2 + 280
= 282

S21 Back to QUICK ACCESS GRID Secondary
Primary

Ask the parents of his friends ✔

Look for survey results on the internet ✔

Go to the library to look up the results of the ✔
last election

1

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S22 20% of $30 = 40% of $ 15 1 20% of $30 =40% of $
20 40
100 × 30 = 100 ×
20
40 × 30 =
1
2 × 30 =
15 =

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Solution Marks Notes

S23 1 – 7 × – 5 = + 35
35

(a)

(b) 13 1 6 – (– 7) = 6 + 7 = 13

S24 ( – 1) × 5 = ( + 8) × 2 Back to QUICK ACCESS GRID
(a) 1 3 = 21
= 21 ÷ 3
5 – 5 = 2 + 16 = 7
5 – 2 = 16 + 5
3 = 21
OR
= 7

(b) 3 = 21 2

= 21 ÷ 3
= 7

Length of one line
= ( + 8) × 2
= (7 + 8) × 2
= 15 × 2
= 30 cm

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Solution Marks Notes

S25 A → (7, 2) 2

B → ( , )
M → (5, 6)

7 +
2 =5

7 + = 10
= 10 – 7
= 3
2 +

2 =6
2 + = 12
= 12 – 2
= 10
co-ordinates of B = (3, 10)

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Solution Marks Notes

S26

+124668
1235779
2 3 4 6 8 8 10

4 5 6 8 10 10 12
6 7 8 10 12 12 14
6 7 8 10 12 12 14

8 9 10 12 14 14 16

Total number of scores = 36 3

Number of total scores that are 10 or
more
= 17

probability that the total score is 10 or


more =

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Solution Marks Notes

S27

Positive gradient Zero gradient Negative gradient

= 4 + 1 = –1 = – 6

= 3 – 5 + = 11

2 Equation of a straight line: = m + c

m = gradient

for = –1, m = 0

+ = 11
= – + 11
m = –1

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Solution Marks Notes

S28
(a)

° + 62° + 62° = 180° 1 ° + 62° + 62° = 180° → sum of all angles
° + 124° = 180°
° = 180° – 124° of a triangle

↓ ° = 56° 1 ° + 132° + 90° + 118° = 360° → sum of all
angles of a quadrilateral
° + 132° + 90° + 118° = 360°
° + 340° = 360°
° = 360° – 340°
° = 40°

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Solution Marks Notes

S28 1 Alternative answer
(b) Not correct ✔
The diagonals of a Kite are
REASON: perpendicular to each other which is not
true for BCDE.
One pair of opposite angles of a Kite are
equal.
Both the pairs of opposite angles of
quadrilateral BCDE are unequal.

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S29
(a)

↓ 2

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Solution Marks Notes

S29
(b)

1
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Solution Marks Notes

S30

(1 + 1 1) ÷ 2

3 2

= (1 + 1 × 2 + 1) ÷ 2
2
3

= (31 + 32) ÷ 2

= (31××22 + 32××33) ÷ 2
= (26 + 69) ÷ 2

11
= 6 ÷2

11 1
= 6 ×2

11
= 12

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