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Secondary 3 Additional Mathematics Numbers Ratio Proportion Quiz

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Questions

Secondary 3 Additional Mathematics Quiz - Numbers Ratio Proportion

Name: ___________________________

Class: ___________________________

Date: ___________________________

Score: ________ / 40

Duration: 50 minutes

Total Marks: 40

Instructions

Answer ALL questions.
Show all working clearly. Answers without working may not be awarded full marks.
The number of marks for each question is shown in brackets [ ].
Non-programmable scientific calculators may be used.
Give non-exact answers correct to 3 significant figures unless otherwise stated.

Section A: Ratios and Proportions (Questions 1–10)

Questions 1–10 are short-answer questions. Each question carries 2–3 marks.

1. Express the ratio 450 g : 2 kg in its simplest form. Give your answer in the form a : b where a and b are integers with no common factor. [2]

2. The ratio of the number of boys to girls in a class is 5 : 4. There are 15 boys. How many students are in the class altogether? [2]

3. Divide $720 in the ratio 3 : 5. [2]

4. Given that a : b = 3 : 7 and b : c = 14 : 5, find a : b : c in its simplest form. [3]

5. A map has a scale of 1 : 25 000. The distance between two towns on the map is 6.8 cm. Calculate the actual distance between the two towns, giving your answer in kilometres. [2]

6. It takes 8 workers 15 days to build a wall. Assuming all workers work at the same rate, how many days would it take 12 workers to build the same wall? [3]

7. The price of a laptop is $1 200 before GST. If GST is 9%, calculate the total price of the laptop including GST. [2]

8. A recipe for 12 cupcakes requires 300 g of flour and 200 g of sugar.

(a) How much flour is needed for 30 cupcakes? [1]

(b) How much sugar is needed for 8 cupcakes? [2]

9. The length of a rectangle is increased in the ratio 5 : 4 and the width is decreased in the ratio 3 : 5. Find the ratio of the original area to the new area. [3]

10. Three friends, Amir, Bala, and Clara, share a sum of money in the ratio 2 : 5 : 3. If Bala receives $45 more than Clara, find the total sum of money shared. [3]

Section B: Direct and Inverse Proportion (Questions 11–16)

Questions 11–16 are structured questions. Each question carries 3–4 marks.

11. Given that y is directly proportional to x, and y = 24 when x = 6,

(a) express y in terms of x, [2]

(b) find the value of y when x = 10, [1]

12. The time taken, T hours, to complete a journey is inversely proportional to the average speed, v km/h.

(a) Write down a formula for T in terms of v. [1]

(b) If it takes 4 hours at an average speed of 60 km/h, find the constant of proportionality. [1]

13. The mass, M kg, of a solid metal sphere is directly proportional to the cube of its radius, r cm.

(a) Write down the relationship between M and r, introducing a constant k. [1]

(b) A sphere of radius 3 cm has a mass of 108 g. Find the value of k. [2]

14. The cost, $*C*, of printing a magazine is partly constant and partly varies as the number of copies, *n*, printed. When 500 copies are printed, the cost is$ 3 200. When 800 copies are printed, the cost is $4 400.

(a) Express C in terms of n. [3]

(b) Find the cost of printing 1 200 copies. [1]

15. The force of attraction, F, between two objects is inversely proportional to the square of the distance, d, between them. When d = 5 m, F = 80 N.

(a) Find the formula connecting F and d. [2]

(b) Find the value of F when d = 10 m. [1]

16. The volume, V cm³, of a gas is inversely proportional to the pressure, P Pa, exerted on it (at constant temperature). When the pressure is 200 Pa, the volume is 150 cm³.

(a) Find the formula connecting V and P. [2]

(b) Find the volume when the pressure is 375 Pa. [1]

Section C: Applications and Problem Solving (Questions 17–20)

Questions 17–20 are multi-step application questions. Each question carries 4–5 marks.

17. A car travels at a constant speed of 75 km/h for the first 120 km of a journey. It then increases its speed and travels the next 180 km at a constant speed of 90 km/h.

(a) Calculate the time taken for the first part of the journey. [2]

(b) Calculate the time taken for the second part of the journey. [1]

18. The surface area, A cm², of a sphere is directly proportional to the square of its radius, r cm. A sphere of radius 4 cm has a surface area of 201.1 cm² (correct to 1 decimal place).

(a) Find the formula connecting A and r, giving the constant of proportionality correct to 3 significant figures. [3]

(b) Use your formula to find the surface area of a sphere of radius 7 cm. [1]

19. A school is planning a fundraising event. The cost of hiring a hall is $500. The cost of food is$ 12 per person attending. The school charges $20 per ticket.

(a) Write down an expression for the total cost, $C, if n people attend. [1]

(b) Write down an expression for the total income, $I, if n tickets are sold. [1]

(d) If 200 people attend, find the profit made. [1]

20. Two alloys, Alloy P and Alloy Q, are made by mixing copper and zinc.

Alloy P contains copper and zinc in the ratio 5 : 3.

Alloy Q contains copper and zinc in the ratio 2 : 3.

A new alloy, Alloy R, is made by melting 12 kg of Alloy P with m kg of Alloy Q. Alloy R contains copper and zinc in the ratio 3 : 2.

(a) Find the mass of copper and the mass of zinc in 12 kg of Alloy P. [2]

(b) Find the mass of copper and the mass of zinc in m kg of Alloy Q, in terms of m. [1]

END OF QUIZ

Answers

Secondary 3 Additional Mathematics Quiz - Numbers Ratio Proportion

Answer Key

Section A: Ratios and Proportions

1. [2]

450 g : 2 kg = 450 g : 2000 g

= 450 : 2000

Divide both by 50:

= 9 : 40

[Marking notes:]

M1: Convert to same units (kg to g or g to kg)
A1: Correct simplified ratio 9 : 40
Common mistake: Forgetting to convert units before simplifying

2. [2]

Ratio boys : girls = 5 : 4

5 parts = 15, so 1 part = 3

Number of girls = 4 × 3 = 12

Total students = 15 + 12 = 27

[Marking notes:]

M1: Find the value of 1 part (= 3)
A1: Correct answer 27

3. [2]

Total parts = 3 + 5 = 8

1 part = $720 ÷ 8 =$ 90

Share 1 = 3 × $90 = **$ 270**

Share 2 = 5 × $90 = **$ 450**

[Marking notes:]

M1: Find value of 1 part
A1: Both shares correct ( $270 and$ 450)

4. [3]

a : b = 3 : 7 = 6 : 14 (multiply by 2 to match b)

b : c = 14 : 5

Therefore a : b : c = 6 : 14 : 5

[Marking notes:]

M1: Identify that b must be made equal in both ratios
M1: Correctly scale the first ratio (or second ratio)
A1: Correct combined ratio 6 : 14 : 5
Common mistake: Simply writing 3 : 7 : 5 without equalising the common term

5. [2]

Actual distance = 6.8 × 25 000 = 170 000 cm

= 170 000 ÷ 100 000 = 1.7 km

[Marking notes:]

M1: Multiply map distance by scale factor
A1: Correct answer in km (1.7 km)
Common mistake: Giving answer in cm without converting to km

6. [3]

This is inverse proportion: more workers means fewer days.

Total work = 8 × 15 = 120 worker-days

Days for 12 workers = 120 ÷ 12 = 10 days

[Marking notes:]

M1: Recognise inverse proportion and calculate total work (120 worker-days)
M1: Divide total work by 12
A1: Correct answer 10 days
Common mistake: Treating as direct proportion

7. [2]

GST = 9% of $1 200 = 0.09 × 1200 =$ 108

Total price = $1 200 +$ 108 = $1 308

[Marking notes:]

M1: Calculate GST amount (or multiply by 1.09 directly)
A1: Correct answer $1 308

8. [3]

(a) [1]

Flour for 30 cupcakes = 300 × (30/12) = 300 × 2.5 = 750 g

(b) [2]

Sugar for 8 cupcakes = 200 × (8/12) = 200 × (2/3) = 133 g (3 s.f.)

[Marking notes:]

(a) A1: 750 g
(b) M1: Set up proportion 200 × 8/12; A1: 133 g (accept 133.3 g or 400/3 g)

9. [3]

Let original length = 5L, original width = 3W

Original area = 5L × 3W = 15LW

New length = 4L, new width = 5W

New area = 4L × 5W = 20LW

Ratio of original area : new area = 15LW : 20LW = 3 : 4

[Marking notes:]

M1: Assign variables based on given ratios
M1: Calculate original and new areas
A1: Correct ratio 3 : 4
Common mistake: Adding or subtracting ratios instead of multiplying

10. [3]

Ratio Amir : Bala : Clara = 2 : 5 : 3

Bala − Clara = 5 parts − 3 parts = 2 parts

2 parts = $45, so 1 part =$ 22.50

Total = 2 + 5 + 3 = 10 parts = 10 × $22.50 = **$ 225**

[Marking notes:]

M1: Find difference in parts between Bala and Clara (= 2)
M1: Find value of 1 part (= $22.50)
A1: Correct total $225

Section B: Direct and Inverse Proportion

11. [4]

(a) [2]

y ∝ x, so y = kx

24 = k(6), so k = 4

y = 4x

(b) [1]

When x = 10: y = 4 × 10 = 40

When y = 60: 60 = 4x, so x = 15

[Marking notes:]

(a) M1: Set up y = kx and substitute; A1: y = 4x
(b) A1: 40
(c) A1: 15

12. [4]

(a) [1]

T = k / v (where k is a constant)

(b) [1]

4 = k / 60, so k = 240

T = 240 / 80 = 3 hours = 3 hours 0 minutes

[Marking notes:]

(a) A1: T = k/v
(b) A1: k = 240
(c) M1: Substitute v = 80; A1: 3 hours (accept 3 h 0 min)

13. [4]

(a) [1]

M = kr³

(b) [2]

108 = k(3)³ = 27k

k = 108 ÷ 27 = 4

M = 4 × (5)³ = 4 × 125 = 500 g

[Marking notes:]

(a) A1: M = kr³
(b) M1: Substitute M = 108, r = 3; A1: k = 4
(c) A1: 500 g

14. [4]

(a) [3]

Let C = a + bn, where a is the constant part and bn varies with n.

When n = 500: a + 500b = 3200 … (i)

When n = 800: a + 800b = 4400 … (ii)

Subtract (i) from (ii): 300b = 1200, so b = 4

Substitute into (i): a + 500(4) = 3200, so a = 3200 − 2000 = 1200

C = 1200 + 4n

(b) [1]

C = 1200 + 4(1200) = 1200 + 4800 = $6 000

[Marking notes:]

(a) M1: Set up C = a + bn; M1: Form two simultaneous equations and solve; A1: C = 1200 + 4n
(b) A1: $6 000

15. [4]

(a) [2]

F ∝ 1/d², so F = k/d²

80 = k/(5)² = k/25

k = 80 × 25 = 2000

F = 2000 / d²

(b) [1]

F = 2000/(10)² = 2000/100 = 20 N

20 = 2000/d², so d² = 100, d = 10 m

[Marking notes:]

(a) M1: Set up F = k/d² and substitute; A1: F = 2000/d²
(b) A1: 20 N
(c) A1: 10 m

16. [4]

(a) [2]

V ∝ 1/P, so V = k/P

150 = k/200, so k = 150 × 200 = 30 000

V = 30 000 / P

(b) [1]

V = 30 000/375 = 80 cm³

100 = 30 000/P, so P = 30 000/100 = 300 Pa

[Marking notes:]

(a) M1: Set up and find k; A1: V = 30 000/P
(b) A1: 80 cm³
(c) A1: 300 Pa

Section C: Applications and Problem Solving

17. [5]

(a) [2]

Time = Distance ÷ Speed = 120 ÷ 75 = 1.6 hours (or 1 h 36 min)

(b) [1]

Time = 180 ÷ 90 = 2 hours

Total distance = 120 + 180 = 300 km

Total time = 1.6 + 2 = 3.6 hours

Average speed = 300 ÷ 3.6 = 83.3 km/h (3 s.f.)

[Marking notes:]

(a) M1: Use time = distance/speed; A1: 1.6 hours
(b) A1: 2 hours
(c) M1: Total distance ÷ total time; A1: 83.3 km/h (3 s.f.)

18. [5]

(a) [3]

A ∝ r², so A = kr²

201.1 = k(4)² = 16k

k = 201.1 ÷ 16 = 12.56875 ≈ 12.6 (3 s.f.)

A = 12.6r²

(b) [1]

A = 12.6 × (7)² = 12.6 × 49 = 617 cm² (3 s.f.)

804.2 = 12.6 × r²

r² = 804.2 ÷ 12.6 = 63.825...

r = √63.825... = 7.99 cm (3 s.f.)

[Marking notes:]

(a) M1: Set up A = kr²; M1: Substitute and solve for k; A1: k = 12.6 (3 s.f.)
(b) A1: 617 cm² (accept 617.4 cm²)
(c) A1: 7.99 cm (3 s.f.)

19. [5]

(a) [1]

C = 500 + 12n

(b) [1]

I = 20n

Break even: C = I

500 + 12n = 20n

500 = 8n

n = 500 ÷ 8 = 62.5

Since number of people must be a whole number, the school needs 63 people to break even (or accept 62.5 as exact break-even point).

Accept: 63 people (rounding up) or 62.5 people.

(d) [1]

Profit = I − C = 20(200) − (500 + 12(200)) = 4000 − (500 + 2400) = 4000 − 2900 = $1 100

[Marking notes:]

(a) A1: C = 500 + 12n
(b) A1: I = 20n
(c) M1: Set C = I and solve; A1: 62.5 or 63 people
(d) A1: $1 100

20. [5]

(a) [2]

Alloy P: copper : zinc = 5 : 3, total parts = 8

Copper in 12 kg = (5/8) × 12 = 7.5 kg

Zinc in 12 kg = (3/8) × 12 = 4.5 kg

(b) [1]

Alloy Q: copper : zinc = 2 : 3, total parts = 5

Copper in m kg = (2/5)m kg

Zinc in m kg = (3/5)m kg

Alloy R: copper : zinc = 3 : 2

Total copper = 7.5 + (2/5)m

Total zinc = 4.5 + (3/5)m

$\frac{7.5 + \frac{2}{5}m}{4.5 + \frac{3}{5}m} = \frac{3}{2}$

Cross multiply:

$2\left(7.5 + \frac{2}{5}m\right) = 3\left(4.5 + \frac{3}{5}m\right)$

$15 + \frac{4}{5}m = 13.5 + \frac{9}{5}m$

$15 - 13.5 = \frac{9}{5}m - \frac{4}{5}m$

$1.5 = \frac{5}{5}m$

m = 1.5 kg

[Marking notes:]

(a) M1: Calculate fraction of copper or zinc; A1: Both 7.5 kg and 4.5 kg
(b) A1: Both expressions correct
(c) M1: Set up ratio equation correctly; A1: m = 1.5 kg
Common mistake: Incorrectly setting up the ratio equation or arithmetic errors in solving

END OF ANSWER KEY