13. ncse 2019

NCSE Mathematics 2019 Paper II

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1

NCSE 2019 PAPER 2

SECTION I ANSWER ALL QUESTIONS IN THIS SECTION

Write your answers in the spaces provided. Remember to show all working.

1. (a) Calculate, showing all working, the value of

3 "#−&

"

1 (()

Numerator: 3 "#−&

"= (+

#− &

"= #,

(#− ()

(#= ,,

(# Denominator: ((

()

𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 =

,,(#((()

= 4415 ÷

1110 =

4415 ×

1011

=43×

21 =

83 = 2

23

(3 marks)

(b) (i) The first four figures in a sequence are shown below.

Draw Figure 6 of the sequence.

(2 marks)

(ii) How many dots would be in Figure 8? (1 mark)

Figure No. 1 2 3 4 5 6 7 8 No.of dots 1 3 6 10 15 21 28 36

+2 +3 +4 +5 +6 +7 +8

The number of dots in Figure 8 is 36

(Total 6 marks)

Figure 1 Figure 2 Figure 3 Figure 4 ... Figure 6



2

2. The Universal Set, U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

T and N are subsets of U.

T = {multiples of 3}

N = {odd numbers}

(a) Complete the Venn diagram to represent this information.

(2 marks)

(b) List the members of the set (T Ç N) /.

N ={1, 3, 5, 7, 9, 11} T= {3, 6, 9, 12}

T ∩ N = {3, 9} (T ∩ N)D = {1, 2, 4, 5, 6, 7, 8, 10, 11, 12}

(2 marks)

(c) What is the probability that a number, chosen at random, is BOTH odd and a multiple of three? 𝑛(T ∩ N) = 2. Hence, there are two numbers that are both odd and a multiple of three.

P(Number is odd and a multiple of 3) = E(F∩G)E(H)

= &(&= (

I

(2 marks)

(Total 6 marks)

U

T N

6

12

3

9

1

5

7

11

2 4 8

10



3

3. (a) Factorise completely.

(i) 𝑥2 + 𝑥𝑦 (1 mark)

= 𝑥(𝑥 + 𝑦)

(ii) 𝑥2 − 1

= (x)2 – (1)2 which is a difference of two squares (1 mark)

(𝑥 + 1)(𝑥 − 1)

(b) Solve for 𝒙 in

3(𝑥 − 3) = 15 (2 marks)

𝑥 − 3 = 5 (÷ 3)

𝑥 = 5 + 3

𝑥 = 8

(c) Given that 𝑎 = 2 and 𝑏 = −3, calculate the value of (𝑎 − b)2 (2 marks) (𝑎 − 𝑏)& = [2 − (−3)]&

= (2 + 3)&

= 5&

= 25

(Total 6 marks)



4

4. A science lab has two cylindrical containers A and B (not drawn to scale), as shown below.

(a) Calculate the volume of container A (use p = &&R

). (2 marks)

The radius of Container A= S+&cm = 49 cm

The volume = 𝜋𝑟&ℎ = &&R× 49 × 49 × 120𝑐𝑚" = 905520𝑐𝑚"

(b) State the number of litres of water that container A can hold. (1 000 cm3 = 1 litre). (1 mark)

905520𝑐𝑚" = S)##&)()))

𝑙 = 905.52𝑙

(c) If ALL of the water in container A is emptied into container B, what will be the height of water in container B?

(3 marks)

Let the height of water in container B be h Radius of Container B = (SI

&cm = 88 cm

Volume of water in container B = 𝜋𝑟&ℎ = &&R× 98 × 98 × ℎ𝑐𝑚" = 30184ℎ

Both volumes are the same, equating volumes 30184ℎ = 905520𝑐𝑚" ℎ = S)##&)

")(+,= 30𝑐𝑚

The height of water in Container B is 30 cm

(Total 6 marks)

A

B 98 cm

196 cm

120 cm



5

5. The diagram below shows the triangle PQR with coordinates P (-5, 5), Q (-2, 5) and R (-5, 1).

(a) The triangle PQR undergoes a reflection in the x-axis to form its image P′Q′R′. On the diagram above, draw and label the image P′Q′R′.

(3 marks)

(b) P′′Q′′R′′ is the image of P′Q′R′ after it undergoes a translation T = 7 . State the co-ordinates of the image P′′Q′′R′′. 2 P′′ ( 2 , −3 )

Q′′( 5 , −3 )

R′′ ( 2 , 1 ) (3 marks)

(Total 6 marks)

𝑸

R

𝑷

𝑸′ 𝑷′

𝑹′

𝑷′′

𝑹′′

𝑸′′

^72`



6

6. The marks of 35 students in a Mathematics class are as follows:

22 11 16 15 21 14 20

18 20 21 18 24 16 18

21 21 23 14 25 20 25

15 25 20 10 18 21 23

13 18 15 22 16 23 22

(a) Complete the frequency table to represent the data above:

Exam Marks Tally Frequency

6 - 10 | 1

11 - 15 | | | | | | 7

16 - 20 | | | | | | | | | | 12

21 - 25 | | | | | | | | | | | | 15

(3 marks)

(b) How many students scored 15 marks or less in Mathematics? = 1 + 7 = 8 students

(1 mark)

(c) What is the probability that a student chosen at random scored a mark greater than 20? (2 marks)

Number of students who scored greater than 20 = 15 Total number of students = 1+7+12+15=35 P(student scored greater than 20)

=𝑁𝑢𝑚𝑏𝑒𝑟𝑤ℎ𝑜𝑠𝑐𝑜𝑟𝑒𝑑𝑔𝑟𝑒𝑎𝑡𝑒𝑟𝑡ℎ𝑎𝑛20

𝑇𝑜𝑡𝑎𝑙𝑛𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 =1535

=37

(Total 6 marks)



7

SECTION II

ANSWER TWO (2) QUESTIONS ONLY FROM THIS SECTION

7 (a)

PLAN A PLAN B

10% Discount for Cash Payment Down payment $500.

30 Monthly Instalments of $150.

(i) Calculate the discount given in PLAN A. (1 mark)

Discount given in Plan A = 10%𝑜𝑓$3450 = ()())

× 3450 = $345

(ii) Calculate the cost of the washing machine if bought using PLAN A. (2 marks) Cost of washing machine using Plan A = $3 450 −$345 = $3 105.

(iii) Calculate the cost of the washing machine if bought using PLAN B. (2 marks)

Down payment = $500.

30 installments @ $150 each = $150 × 30 = $4 500.

Total = $5 000.

(iv) Calculate the difference in the costs between PLAN A and PLAN B. (1 mark)

Difference = $5 000 − $3 105 = $1 895.

$3 450

Washing Machine



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(b) Using a pair of compasses, a ruler and a pencil ONLY, draw a line AB of length 5.5cm.

We draw a straight line longer that 5.5 cm and cut off AB = 5.5 cm

(1 mark)

Construct a 60o angle at A.

(2 marks)

(c) Construct the triangle PQR such that PQ = 6 cm, QR = 8 cm and angle PQR = 90o. (3 marks) We draw a straight line longer than 6 cm and cut off PQ = 6 cm 6 cm

A B 5.5 cm

A B

P Q



9

We construct an angle of 900 at Q

We locate R such that QR = 8 cm. Then, we join P to R to complete triangle PQR

(Total 12 marks)

6 cm

6 cm

8 cm



10

8. (a) The diagram shows a ladder, 15 metres long, resting against a vertical pole. The foot of the ladder is 12 metres away from the base of the pole.

(i) Calculate the height, h, of the pole.

Using Pythagoras’ Theorem ℎ& + 12& = 15& ℎ& = 15& − 12& ℎ& = 225 − 144 ℎ& = 81 ℎ = √81 = 9 m

(3 marks)

(ii) If the ladder slides 3 metres down the pole, calculate the new distance from the foot

of the ladder to the base of the pole, to 1 decimal place.

The ladder slides 3 m down the pole ℎ = (9 − 3)𝑚 = 6𝑚 Let the new distance from the foot of the ladder to the base of the pole be x. By Pythagoras’ Theorem 𝑥& + 6& = 15& 𝑥& = 15& − 6& 𝑥& = 225 − 36 𝑥& = 189 𝑥 = √189 𝑥 = 13.74𝑚

𝑥 = 13.7𝑚, (correct to one decimal places) (3 marks)

h

ladder pole

15 m h m

12 m

15 m 6 m

x m



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(b) (i) A flight from England to Dubai took 5(& hours.

If the average speed of the flight was 885 kmh-1, determine the distance, flown in

kilometres.

Distance = Average Speed × Time

= 885 × 5.5

= 4867.5𝑘𝑚

(3 marks)

(ii) A ship sailed the same distance from England to Greece at an average speed of

150 kmh-1. How long was the journey in hours?

𝑇𝑖𝑚𝑒𝑇𝑎𝑘𝑒𝑛 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝐴𝑣𝑒𝑟𝑎𝑔𝑒𝑆𝑝𝑒𝑒𝑑

=4867.5150 ℎ𝑟

= 32.45ℎ𝑜𝑢𝑟

(3 marks)

(Total 12 marks)



12

9. (a) Josh purchased 6 bottles of orange juice and 4 packs of biscuits for $32.

Kevin purchased 4 bottles of orange juice and 2 packs of biscuits for $20.

(i) Using x to represent the cost in dollars of one (1) bottle of orange juice and y to represent

the cost in dollars of one (1) pack of biscuit, write two (2) equations in x and y to

represent the information above.

6 bottles of orange juice @ x dollars each will cost 6x dollars 4 packs of biscuits @ y dollars each will cost 4y dollars Josh’s purchase: 6𝑥 + 4𝑦 = 32 (1) 4 bottles of orange juice @ x dollars each will cost 4x dollars 2 packs of biscuits @ y dollars each will cost 2y dollars Kevin’s purchase: 4𝑥 + 2𝑦 = 20 (2) (2 marks)

(ii) Using your equations from (i) above, determine the cost of one bottle of orange juice and

the cost of one pack of biscuit.

6𝑥 + 4𝑦 = 32 (1) 4𝑥 + 2𝑦 = 20 (2)

Equation (1) divided by 2: 3𝑥 + 2𝑦 = 16 (3) Eq (2) – Eq (3) 𝑥 = 4 Substituting 𝑥 = 4 in Equation (2)

4(4) + 2𝑦 = 20 16 + 2𝑦 = 20 2𝑦 = 20 − 16

2𝑦 = 4 𝑦 = 2

The cost of a bottle of orange juice is 4 dollars The cost of one pack of biscuits is 2 dollars (4 marks)



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(b) The equation 𝑦 = 3𝑥 + 3 represents the relation between two variables x and y.

(i) Use the equation given to complete the table below.

𝑥 -1 0 1 2 𝑦 0 3 6 9

(2 marks)

When 𝑥 = 0, 𝑦 = 3(0) + 3 = 0 + 3 = 3 When 𝑥 = 2, 𝑦 = 3(2) + 3 = 6 + 3 = 9

(ii) Using a scale of 2 cm to represent 1 unit on both axes, on the answer sheet given, draw the graph of𝑦 = 3𝑥 + 3.

(2 marks) (iii) Using the same answer sheet, in (b) above, draw the line y = 7.

(1 mark)

(iv) Label the point P, on the graph, where the lines y = 7 and 𝑦 = 3𝑥 + 3intersect. (1 mark)

(Total 12 marks)



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END OF TEST

-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

-1

1

2

3

4

5

6

7

8

9

10

11

12

x

x-axis

Answer Sheet for Question 9b. Scale x axis: 2 cm = 1 unit y axis: 2 cm = 1 unit

x-axis

y-axis

P

13. ncse 2019

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