Secondary 2 Mathematics Numbers Ratio Proportion Quiz

Secondary 2 Mathematics Quiz - Numbers Ratio Proportion (Answer Key)

Total Marks: 40

---

Section A: Short Answer Questions (Questions 1–10, 2 marks each)

1. Express the ratio $48 : 72$ in its simplest form.

Answer: $2 : 3$
Marks: 2
Working:

• Find HCF of 48 and 72: $48 = 2^4 \times 3$, $72 = 2^3 \times 3^2$, HCF $= 2^3 \times 3 = 24$
• Divide both parts by 24: $48 \div 24 = 2$, $72 \div 24 = 3$
• Simplest form: $2 : 3$

Common mistake: Stopping at $4 : 6$ or $8 : 12$ without fully simplifying.

---

2. A sum of money is divided between Ali and Bala in the ratio $3 : 5$. If Bala receives $40 more than Ali, find the total sum of money.

Answer: $160$
Marks: 2
Working:

• Difference in ratio parts $= 5 - 3 = 2$ parts
• $2$ parts = \40$
• $1$ part = \20$
• Total parts $= 3 + 5 = 8$ parts
• Total sum = 8 \times \20 = $160$

Alternative: Let total be $x$. Ali gets $\frac{3}{8}x$, Bala gets $\frac{5}{8}x$. Difference $= \frac{2}{8}x = 40 \Rightarrow x = 160$.

---

3. The scale of a map is $1 : 25\,000$. The distance between two towns on the map is $6.4$ cm. Find the actual distance between the two towns in kilometres.

Answer: $1.6$
Marks: 2
Working:

• Actual distance $= 6.4 \times 25\,000 = 160\,000$ cm
• Convert to km: $160\,000 \div 100\,000 = 1.6$ km

Key concept: Scale $1 : n$ means $1$ cm on map represents $n$ cm in reality. Remember unit conversion: $1$ km $= 100\,000$ cm.

---

4. $y$ is inversely proportional to the square of $x$. When $x = 3$, $y = 12$. Find the value of $y$ when $x = 6$.

Answer: $3$
Marks: 2
Working:

• $y = \frac{k}{x^2}$
• Substitute $x = 3$, $y = 12$: $12 = \frac{k}{9} \Rightarrow k = 108$
• Equation: $y = \frac{108}{x^2}$
• When $x = 6$: $y = \frac{108}{36} = 3$

Alternative (using proportion): Since $x$ doubles ($3 \to 6$), $x^2$ quadruples ($9 \to 36$), so $y$ becomes $\frac{1}{4}$ of original: $12 \div 4 = 3$.

---

5. A car travels $180$ km on $15$ litres of petrol. How many litres of petrol are needed to travel $300$ km at the same rate?

Answer: $25$
Marks: 2
Working:

• Rate $= \frac{180}{15} = 12$ km per litre
• Petrol needed $= \frac{300}{12} = 25$ litres

Alternative (proportion): $\frac{15}{180} = \frac{x}{300} \Rightarrow x = \frac{15 \times 300}{180} = 25$.

---

6. The ratio of the number of boys to the number of girls in a class is $4 : 5$. After $6$ boys join the class, the ratio becomes $5 : 5$. Find the original number of students in the class.

Answer: $54$
Marks: 2
Working:

• Let original boys $= 4u$, girls $= 5u$
• After 6 boys join: boys $= 4u + 6$, girls $= 5u$
• New ratio $5 : 5 = 1 : 1$, so $4u + 6 = 5u \Rightarrow u = 6$
• Original total $= 4u + 5u = 9u = 9 \times 6 = 54$

Check: Original: 24 boys, 30 girls. After: 30 boys, 30 girls. Ratio $30:30 = 1:1 = 5:5$. ✓

---

7. Simplify the ratio $0.75 : 1.5 : 2.25$.

Answer: $1 : 2 : 3$
Marks: 2
Working:

• Multiply all parts by 100 to remove decimals: $75 : 150 : 225$
• Divide by HCF (75): $75 \div 75 = 1$, $150 \div 75 = 2$, $225 \div 75 = 3$
• Simplest form: $1 : 2 : 3$

Alternative: Divide by smallest value $0.75$: $0.75 \div 0.75 = 1$, $1.5 \div 0.75 = 2$, $2.25 \div 0.75 = 3$.

---

8. It takes $8$ workers $12$ days to complete a job. How many days will it take $6$ workers to complete the same job, assuming they work at the same rate?

Answer: $16$
Marks: 2
Working:

• Total work $= 8 \times 12 = 96$ worker-days
• Days for 6 workers $= \frac{96}{6} = 16$ days

Key concept: Inverse proportion — more workers, fewer days. Worker-days constant.

---

9. The price of a watch increased from $200 to$250. Find the percentage increase.

Answer: $25$
Marks: 2
Working:

• Increase $= 250 - 200 = 50$
• Percentage increase $= \frac{50}{200} \times 100\% = 25\%$

Common mistake: Using new price as denominator: $\frac{50}{250} \times 100\% = 20\%$ (incorrect).

---

10. $p$ is directly proportional to the cube root of $q$. When $q = 27$, $p = 18$. Find an equation connecting $p$ and $q$.

Answer: $p = 6\sqrt[3]{q}$
Marks: 2
Working:

• $p = k\sqrt[3]{q}$
• Substitute $q = 27$, $p = 18$: $18 = k\sqrt[3]{27} = k \times 3 \Rightarrow k = 6$
• Equation: $p = 6\sqrt[3]{q}$

Check: When $q = 27$, $\sqrt[3]{27} = 3$, $p = 6 \times 3 = 18$. ✓

---

Section B: Structured Questions (Questions 11–15, 3 marks each)

11. A rectangle has length and breadth in the ratio $5 : 3$. The perimeter of the rectangle is $64$ cm.

(a) Find the length and breadth of the rectangle.
(b) Find the area of the rectangle.

Answer (a): Length = $20$ cm, Breadth = $12$ cm [2]
Answer (b): $240$ cm² [1]

Working:

• Let length $= 5x$, breadth $= 3x$
• Perimeter $= 2(5x + 3x) = 16x = 64 \Rightarrow x = 4$
• Length $= 5 \times 4 = 20$ cm, Breadth $= 3 \times 4 = 12$ cm
• Area $= 20 \times 12 = 240$ cm²

Mark breakdown: (a) 1 mark for $x = 4$, 1 mark for both dimensions; (b) 1 mark for correct area.

---

12. The map scale is $1 : 50\,000$. On the map, a rectangular plot of land measures $4$ cm by $3$ cm.

(a) Find the actual length and breadth of the plot in metres.
(b) Find the actual area of the plot in hectares. ($1$ hectare $= 10\,000$ m²)

Answer (a): Length = $2000$ m, Breadth = $1500$ m [2]
Answer (b): $300$ hectares [1]

Working:

• Actual length $= 4 \times 50\,000 = 200\,000$ cm $= 2000$ m
• Actual breadth $= 3 \times 50\,000 = 150\,000$ cm $= 1500$ m
• Actual area $= 2000 \times 1500 = 3\,000\,000$ m²
• Area in hectares $= \frac{3\,000\,000}{10\,000} = 300$ hectares

Mark breakdown: (a) 1 mark each for correct length and breadth in metres; (b) 1 mark for correct area in hectares.

---

13. $A$ is directly proportional to $B^2$. When $B = 4$, $A = 48$.

(a) Find an equation connecting $A$ and $B$.
(b) Find the value of $A$ when $B = 6$.
(c) Find the value of $B$ when $A = 108$.

Answer (a): $A = 3B^2$ [1]
Answer (b): $108$ [1]
Answer (c): $6$ [1]

Working:

• (a) $A = kB^2$. $48 = k(4^2) = 16k \Rightarrow k = 3$. Equation: $A = 3B^2$.
• (b) $A = 3(6^2) = 3 \times 36 = 108$.
• (c) $108 = 3B^2 \Rightarrow B^2 = 36 \Rightarrow B = 6$ (positive since $B$ typically represents a magnitude).

Mark breakdown: 1 mark each part.

---

14. A sum of $3600 is divided among three children, X, Y, and Z, in the ratio$2 : 3 : 4$.

(a) Find the amount each child receives.
(b) If X gives $100 to Y, find the new ratio of X's money to Y's money in its simplest form.

Answer (a): X = $800$, Y = $1200$, Z = $1600$ [2]
Answer (b): $7 : 13$ [1]

Working:

• (a) Total parts $= 2 + 3 + 4 = 9$. 1 part $= 3600 \div 9 = 400$.
  X $= 2 \times 400 = 800$, Y $= 3 \times 400 = 1200$, Z $= 4 \times 400 = 1600$.
• (b) After transfer: X has $800 - 100 = 700$, Y has $1200 + 100 = 1300$.
  Ratio $= 700 : 1300 = 7 : 13$.

Mark breakdown: (a) 1 mark for 1 part value, 1 mark for all three amounts; (b) 1 mark for correct simplified ratio.

---

15. The time taken to complete a task is inversely proportional to the number of workers. It takes $10$ workers $15$ hours to complete the task.

(a) Find an equation connecting the time $T$ (in hours) and the number of workers $W$.
(b) How many workers are needed to complete the task in $6$ hours?
(c) If $25$ workers are employed, how long will it take to complete the task?

Answer (a): $T = \frac{150}{W}$ [1]
Answer (b): $25$ workers [1]
Answer (c): $6$ hours [1]

Working:

• (a) $T = \frac{k}{W}$. $15 = \frac{k}{10} \Rightarrow k = 150$. Equation: $T = \frac{150}{W}$.
• (b) $6 = \frac{150}{W} \Rightarrow W = \frac{150}{6} = 25$.
• (c) $T = \frac{150}{25} = 6$.

Mark breakdown: 1 mark each part.

---

Section C: Application Questions (Questions 16–20, 4 marks each)

16. A paint mixture is made by mixing red, blue, and yellow paint in the ratio $3 : 2 : 1$ by volume.

(a) How many litres of each colour are needed to make $36$ litres of the mixture?
(b) If only $10$ litres of red paint is available, what is the maximum volume of the mixture that can be made?
(c) For the maximum volume in (b), how many litres of blue and yellow paint are needed?

Answer (a): Red = $18$ L, Blue = $12$ L, Yellow = $6$ L [2]
Answer (b): $20$ litres [1]
Answer (c): Blue = $6.67$ L (or $6\frac{2}{3}$ L), Yellow = $3.33$ L (or $3\frac{1}{3}$ L) [1]

Working:

• (a) Total parts $= 3 + 2 + 1 = 6$. 1 part $= 36 \div 6 = 6$ L.
  Red $= 3 \times 6 = 18$ L, Blue $= 2 \times 6 = 12$ L, Yellow $= 1 \times 6 = 6$ L.
• (b) Red is limiting. 3 parts red $= 10$ L $\Rightarrow$ 1 part $= \frac{10}{3}$ L.
  Total mixture $= 6 \times \frac{10}{3} = 20$ L.
• (c) Blue $= 2 \times \frac{10}{3} = \frac{20}{3} = 6\frac{2}{3}$ L.
  Yellow $= 1 \times \frac{10}{3} = \frac{10}{3} = 3\frac{1}{3}$ L.

Mark breakdown: (a) 1 mark for 1 part value, 1 mark for all three volumes; (b) 1 mark; (c) 1 mark for both correct.

---

17. The scale of a floor plan is $1 : 100$. On the plan, a rectangular room measures $5.2$ cm by $3.8$ cm.

(a) Find the actual dimensions of the room in metres.
(b) Find the actual area of the room in square metres.
(c) Tiles measuring $40$ cm by $40$ cm are used to cover the floor. How many tiles are needed? (Assume no wastage.)

Answer (a): Length = $5.2$ m, Breadth = $3.8$ m [1]
Answer (b): $19.76$ m² [1]
Answer (c): $124$ tiles [2]

Working:

• (a) Actual length $= 5.2 \times 100 = 520$ cm $= 5.2$ m.
  Actual breadth $= 3.8 \times 100 = 380$ cm $= 3.8$ m.
• (b) Area $= 5.2 \times 3.8 = 19.76$ m².
• (c) Tile area $= 0.4 \times 0.4 = 0.16$ m².
  Number of tiles $= \frac{19.76}{0.16} = 123.5 \Rightarrow 124$ tiles (round up since partial tile needed).

Alternative for (c): Room $= 520 \times 380$ cm. Tiles along length $= 520 \div 40 = 13$. Tiles along breadth $= 380 \div 40 = 9.5 \Rightarrow 10$. Total $= 13 \times 10 = 130$. But "no wastage" implies exact division, so area method preferred: $123.5 \to 124$.

Mark breakdown: (a) 1 mark for both dimensions; (b) 1 mark; (c) 1 mark for tile area, 1 mark for correct rounding up.

---

18. The cost $C$ of producing $n$ items is given by $C = k\sqrt{n} + 50$, where $k$ is a constant. When $100$ items are produced, the cost is $350$.

(a) Find the value of $k$.
(b) Find the cost of producing $225$ items.
(c) How many items can be produced for a cost of $500$?

Answer (a): $30$ [1]
Answer (b): $500$ [1]
Answer (c): $225$ items [2]

Working:

• (a) $350 = k\sqrt{100} + 50 = 10k + 50 \Rightarrow 10k = 300 \Rightarrow k = 30$.
• (b) $C = 30\sqrt{225} + 50 = 30 \times 15 + 50 = 450 + 50 = 500$.
• (c) $500 = 30\sqrt{n} + 50 \Rightarrow 30\sqrt{n} = 450 \Rightarrow \sqrt{n} = 15 \Rightarrow n = 225$.

Mark breakdown: (a) 1 mark; (b) 1 mark; (c) 1 mark for equation setup, 1 mark for solving.

---

19. A car uses petrol at a rate of $1$ litre per $12$ km. Petrol costs $2.80$ per litre.

(a) Find the cost of petrol for a journey of $360$ km.
(b) If the car's fuel efficiency improves by $20\%$, find the new cost for the same journey.
(c) Express the savings as a percentage of the original cost.

Answer (a): $84$ [1]
Answer (b): $70$ [2]
Answer (c): $16.67\%$ (or $16\frac{2}{3}\%$) [1]

Working:

• (a) Petrol used $= \frac{360}{12} = 30$ litres. Cost = 30 \times 2.80 = \84$.
• (b) Improved efficiency: $12 \times 1.2 = 14.4$ km per litre.
  Petrol used $= \frac{360}{14.4} = 25$ litres. Cost = 25 \times 2.80 = \70$.
• (c) Savings $= 84 - 70 = 14$. Percentage $= \frac{14}{84} \times 100\% = \frac{1}{6} \times 100\% = 16\frac{2}{3}\% \approx 16.67\%$.

Mark breakdown: (a) 1 mark; (b) 1 mark for new efficiency, 1 mark for new cost; (c) 1 mark.

---

20. Three gears A, B, and C are connected in series. Gear A has $24$ teeth, Gear B has $36$ teeth, and Gear C has $48$ teeth.

(a) Find the ratio of the number of teeth of Gear A : Gear B : Gear C in its simplest form.
(b) If Gear A makes $120$ revolutions, how many revolutions does Gear C make?
(c) If Gear B makes $90$ revolutions per minute, find the speed of Gear A in revolutions per minute.

Answer (a): $2 : 3 : 4$ [1]
Answer (b): $60$ revolutions [2]
Answer (c): $135$ rpm [1]

Working:

• (a) $24 : 36 : 48$. Divide by HCF (12): $2 : 3 : 4$.
• (b) For gears in series, revolutions are inversely proportional to teeth.
  $\frac{\text{Rev}_A}{\text{Rev}_C} = \frac{\text{Teeth}_C}{\text{Teeth}_A} = \frac{48}{24} = 2$.
  $\text{Rev}_C = \frac{120}{2} = 60$.
  Alternative: A→B: $\text{Rev}_B = 120 \times \frac{24}{36} = 80$. B→C: $\text{Rev}_C = 80 \times \frac{36}{48} = 60$.
• (c) $\frac{\text{Rev}_A}{\text{Rev}_B} = \frac{\text{Teeth}_B}{\text{Teeth}_A} = \frac{36}{24} = \frac{3}{2}$.
  $\text{Rev}_A = 90 \times \frac{3}{2} = 135$ rpm.

Mark breakdown: (a) 1 mark; (b) 1 mark for method/ratio, 1 mark for answer; (c) 1 mark.

---

End of Answer Key