CSEC Mathematics Paper 2 Solutions: June 2024, Exams of Mathematics

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CSEC Mathematics

June 202 4 – Paper 2

Solutions

SECTION I

Answer ALL questions.

All working must be clearly shown.

1. (a) (i) Calculate the value of √

2

2

, giving your answer correct to

(a) 2 significant figures [1]

2

2

= 7. 7 (to 2 significant figures)

(b) 2 decimal places [1]

2

2

= 7. 67 (to 2 decimal places)

(ii) Write the following quantities in ascending order. [1]

12

25

12

25

In ascending order, the numbers should be arranged as 0. 46 , 47 % ,

12

25

12

25

EC$

US$

(c) The present population of Portmouth is 550 000. It is expected that this

population will increase 42% by the year 2030.

(i) Write the number 550 000 in standard form. [1]

550 000 = 5. 5 × 10

5

(ii) Calculate the expected population in Portmouth in 2030. [1]

Final percentage =

Final percentage = 142 %

Expected population = 142% of 550 000

Expected population =

142

100

×

550 000

1

Expected population = 781 000

∴ The expected population in Portmouth in 2030 is 781 000.

(d) The graph below can be used to convert between United States dollars (US$)

and Eastern Caribbean dollars (EC$).

Using the graph,

(i) convert US$2 to EC$ [1]

From graph, US$2 = EC$5.40.

(ii) convert EC$70 to US$ [1]

From graph, EC$ 7 = US$2.6 0.

Now,

EC$70 = US$ 2. 60 × 10

EC$70 = US$ 26. 00

Total: 9 marks

We are told that altogether, she uses 60 𝑚 of mesh.

So, we have,

÷ 4

∴ A simplified expression for 𝑦 in terms of 𝑥 is 𝑦 = 15 − 2 𝑥.

(b) The area of the rectangular seedbed is equal to the area of the square

seedbed.

(i) Use this information and your answer in (a) to write down a quadratic

equation, in terms of 𝑥, and show it simplifies to

2

− 60 𝑥 + 225 = 0 [2]

Area of rectangular seedbed = length × width

Area of rectangular seedbed = 3 𝑥 × 𝑥

Area of rectangular seedbed = 3 𝑥

2

Area of square seedbed = side × side

Area of square seedbed = 𝑦 × 𝑦

Area of square seedbed = 𝑦

2

We are given that the area of the rectangular seedbed is equal to the

area of the square seedbed.

Q.E.D.

So, we have,

2

2

2

2

[from part (i)]

2

2

2

2

2

2

(ii) Solve the equation 𝑥

2

− 60 𝑥 + 225 = 0 using the quadratic

formula. [3]

2

− 60 𝑥 + 225 = 0 which is in the form 𝑎𝑥

2

where 𝑎 = 1 , 𝑏 = − 60 and 𝑐 = 225.

Using the quadratic formula,

−𝑏±

√ 𝑏

2

− 4 𝑎𝑐

2 𝑎

−

( − 60

) ±

√( − 60

)

2

− 4

( 1

)( 225

)

2

( 1

)

60 ±√ 3600 − 900

2

60 ±√ 2700

2

Either 𝑥 =

60 −√ 2700

2

or 𝑥 =

60 +√ 2700

2

𝑥 = 4. 019 (to 4 s.f.) 𝑥 = 55. 98 (to 4 s.f.)

3. (a) A vertical flagpole, 𝐹𝑃, stands on horizontal ground and is held by two ropes,

𝑃𝐺 and 𝑃𝑅, as shown in the diagram below.

𝑃𝐺 = 18 𝑚, 𝐹𝑅 = 9. 7 𝑚 and angle 𝐹𝐺𝑃 = 38°.

(i) Calculate the height of the flagpole, 𝐹𝑃. [2]

Consider the right-angled triangle 𝐹𝐺𝑃.

sin 𝜃 =

𝑜𝑝𝑝

ℎ𝑦𝑝

sin 38 ° =

𝐹𝑃

18

𝐹𝑃 = 18 sin 38 °

𝐹𝑃 = 11. 1 𝑚 (to 3 significant figures)

∴ The height of the flagpole, 𝐹𝑃, is 11. 1 𝑚.

(ii) Find 𝑃𝑅, the length of one of the pieces of rope used to hold the

flagpole. [2]

Consider the right-angled triangle 𝐹𝑃𝑅.

Using Pythagoras’ Theorem,

2

2

2

2

2

2

2

2

• ( 18 sin 38 °)

2

2

𝑃𝑅 = 14. 7 𝑚 (to 3 significant figures)

∴ The length of 𝑃𝑅 is 14. 7 𝑚.

(b) In the diagram below, 𝑃𝑄 is parallel to 𝑀𝑁, 𝐿𝑅𝑇 is an isosceles triangle and

𝑆𝐿𝑇 is a straight line.

Find the value of 𝑥. [2]

Consider the triangle 𝐿𝑅𝑇. The base angles in an isosceles triangle are equal.

180 °− 28 °

2

152 °

2

(i) Describe fully the single transformation that maps Shape 𝑇 onto

Shape 𝑄. [2]

The single transformation that maps Shape 𝑇 onto Shape 𝑄 is a

reflection in the line 𝑦 = 𝑥.

(ii) On the diagram above, draw the image of Shape 𝑇 after it undergoes a

translation by the vector (

). Label this image 𝑀. [1]

See diagram above.

Total: 9 marks

4. (a) A rectangle, 𝑃𝑄𝑅𝑆, has a diagonal, 𝑃𝑅, where 𝑃 is the point (− 3 , 10 ) and 𝑅 is

the point

(i) Calculate the length of the line 𝑃𝑅. [2]

Points are 𝑃(− 3 , 10 ) and 𝑅( 4 , − 4 ).

Length of 𝑃𝑅 = √

2

1

2

2

1

2

Length of 𝑃𝑅 =

2

2

Length of 𝑃𝑅 = √( 7 )

2

2

Length of 𝑃𝑅 = √

Length of 𝑃𝑅 = 7 √

Length of 𝑃𝑅 = 15. 7 units (to 3 significant figures)

∴ The length of the line 𝑃𝑅 is 15. 7 units.

(ii) Determine the equation of the line 𝑃𝑅. [3]

Points are 𝑃(− 3 , 10 ) and 𝑅( 4 , − 4 ).

Gradient, 𝑚 =

𝑦

2

−𝑦

1

𝑥

2

−𝑥

1

Gradient, 𝑚 =

− 4 − 10

4 −(− 3 )

Gradient, 𝑚 =

− 14

7

Gradient, 𝑚 = − 2

5. (a) Mr. Morgan administered a spelling test to his class. The table below shows

the number of words out of 10 that each students spelt correctly.

Number of Words 5 6 7 8 9 10

Frequency 8 4 2 2 3 4

(i) For the data set shown above, state the

(a) mode [1]

The mode is 5 words spelt correctly.

(b) median [1]

The median is 6 words spelt correctly.

(ii) Calculate the mean number of words spelt correctly. [2]

Mean =

∑ 𝑓𝑥

∑ 𝑓

Mean =

( 5 × 8

)

( 6 × 4

)

( 7 × 2

)

( 8 × 2

)

( 9 × 3

)

( 10 × 4

)

8 + 4 + 2 + 2 + 3 + 4

Mean =

40 + 24 + 14 + 16 + 27 + 40

23

Mean =

161

23

Mean = 7 words

∴ The mean number of words spelt correctly is 7 words.

(b) The attendance officer at a particular school recorded the time, 𝑡, in minutes,

taken by each student in a group to travel to school. The data collected is

shown on the cumulative frequency curve below.

Using the cumulative frequency curve, find an estimate of

(i) the number of students who took at MOST 32 minutes to travel to

school [1]

From graph, the number of students who took at MOST 32 minutes to

travel to school is 52 students.

P O S I T I V

I T

Y

(c) The letters in the word “POSITIVITY” are each written on separate cards and

placed in a bag. Dacia picks 2 of these cards, at random, with replacement.

Find the probability that she picks the letter “I” then the letter “V”. [2]

Probability she picks “I” =

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠

𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠

Probability she picks “I” =

3

10

Probability she picks “V” =

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠

𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠

Probability she picks “V” =

1

10

Now, the events are independent.

So, we have,

Probability that she picks “I” then “V” =

3

10

×

1

10

Probability that she picks “I” then “V” =

3

100

Total: 9 marks

6. [In this question, take 𝜋 =

22

7

. ]

(a) The diagram below shows a gold bar in the shape of a trapezoidal prism. Its

volume is 2 886 𝑐𝑚

3

. The length and height of the prism are indicated on the

diagram.

(i) Calculate the area of the shaded cross-section of the trapezoidal

prism. [1]

Volume of prism = Cross-sectional area × length

Cross-sectional area =

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑝𝑟𝑖𝑠𝑚

𝑙𝑒𝑛𝑔𝑡ℎ

Cross-sectional area =

2886

31. 2

Cross-sectional area = 92. 5 𝑐𝑚

∴ The area of the shaded cross-section is 92. 5 𝑐𝑚.