Secondary 1 Mathematics Practice Paper 5

TuitionGoWhere Practice Paper — Answer Key

Secondary 1 Mathematics — Numbers, Ratio & Proportion

Version 5

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Section A: Short Answer Questions (20 marks)

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1. Express 360 as a product of its prime factors. (2 marks)

Answer: $360 = 2^3 \times 3^2 \times 5$

Working: $360 \div 2 = 180$
$180 \div 2 = 90$
$90 \div 2 = 45$
$45 \div 3 = 15$
$15 \div 3 = 5$
$5 \div 5 = 1$

So $360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 2^3 \times 3^2 \times 5$

Marking: 1 mark for correct prime factorization process; 1 mark for correct answer in index notation.

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2. Find the HCF of 48 and 84. (2 marks)

Answer: HCF = 12

Working: $48 = 2^4 \times 3$
$84 = 2^2 \times 3 \times 7$

HCF = $2^2 \times 3 = 4 \times 3 = 12$

Marking: 1 mark for correct prime factorizations; 1 mark for correct HCF.

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3. Find the LCM of 18 and 30. (2 marks)

Answer: LCM = 90

Working: $18 = 2 \times 3^2$
$30 = 2 \times 3 \times 5$

LCM = $2 \times 3^2 \times 5 = 2 \times 9 \times 5 = 90$

Marking: 1 mark for correct prime factorizations; 1 mark for correct LCM.

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4. Evaluate $\dfrac{3}{4} - \dfrac{2}{5} + \dfrac{1}{10}$. (2 marks)

Answer: $\dfrac{9}{20}$

Working: LCM of 4, 5, and 10 = 20

$\dfrac{3}{4} = \dfrac{15}{20}$, $\dfrac{2}{5} = \dfrac{8}{20}$, $\dfrac{1}{10} = \dfrac{2}{20}$

$\dfrac{15}{20} - \dfrac{8}{20} + \dfrac{2}{20} = \dfrac{15 - 8 + 2}{20} = \dfrac{9}{20}$

Marking: 1 mark for correct common denominator conversion; 1 mark for correct final answer in lowest terms.

Common mistake: Students may incorrectly find LCM or make arithmetic errors when combining numerators.

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5. Arrange in ascending order: $\dfrac{2}{3},\; 0.65,\; 68\%,\; \dfrac{7}{10}$. (2 marks)

Answer: $0.65,\; \dfrac{2}{3},\; \dfrac{7}{10},\; 68\%$

Working: Convert all to decimals:

• $\dfrac{2}{3} = 0.666...$
• $0.65 = 0.65$
• $68\% = 0.68$
• $\dfrac{7}{10} = 0.7$

Ascending order: $0.65 < 0.666... < 0.68 < 0.7$

So: $0.65,\; \dfrac{2}{3},\; 68\%,\; \dfrac{7}{10}$

Marking: 1 mark for correct conversions; 1 mark for correct order.

Common mistake: Students may confuse $68\%$ with $0.068$ or miscompare recurring decimals.

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6. Write the inequality: $-7$ ____________ $-3$. (2 marks)

Answer: $-7 < -3$

Working: On the number line, $-7$ is to the left of $-3$, so $-7$ is less than $-3$.

Marking: 2 marks for correct symbol. Accept only $<$.

Common mistake: Students may write $>$ because $7 > 3$, forgetting that negative numbers reverse the inequality.

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7. Round 4.7385 to (a) 2 d.p. and (b) 3 s.f. (2 marks)

Answer: (a) $4.74$
(b) $4.74$

Working: (a) 4.7385 → Look at the 3rd decimal place (8). Since $8 \geq 5$, round up: $4.74$
(b) 4.7385 → First 3 significant figures are 4, 7, 3. The next digit is 8, so round up: $4.74$

Marking: 1 mark for each correct answer.

Note: In this case both answers happen to be the same, but students should understand the different rules.

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8. Simplify the ratio $36 : 60$. (2 marks)

Answer: $3 : 5$

Working: HCF of 36 and 60 = 12

$36 \div 12 = 3$, $60 \div 12 = 5$

So $36 : 60 = 3 : 5$

Marking: 1 mark for finding HCF; 1 mark for correct simplified ratio.

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9. The ratio of boys to girls is $4 : 5$. There are 24 boys. How many girls? (2 marks)

Answer: 30 girls

Working: $4$ parts = 24 boys
$1$ part = $24 \div 4 = 6$
$5$ parts = $5 \times 6 = 30$ girls

Marking: 1 mark for finding the value of 1 part; 1 mark for correct answer.

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10. Express 240 as a percentage of 600. (2 marks)

Answer: $40\%$

Working: $\dfrac{240}{600} \times 100\% = \dfrac{2}{5} \times 100\% = 40\%$

Marking: 1 mark for correct fraction; 1 mark for correct percentage.

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Section B: Structured Questions (25 marks)

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11. (a) Find the HCF and LCM of 72 and 108 using prime factorization. (3 marks)

Answer: HCF = 36, LCM = 216

Working: $72 = 2^3 \times 3^2$
$108 = 2^2 \times 3^3$

HCF = $2^2 \times 3^2 = 4 \times 9 = 36$
LCM = $2^3 \times 3^3 = 8 \times 27 = 216$

Marking: 1 mark for each correct prime factorization; 1 mark for correct HCF and LCM.

(b) Two traffic lights change every 72 s and 108 s. If they change together at 8:00 a.m., when will they next change together? (2 marks)

Answer: 8:03:36 a.m. (or 8:03 a.m. and 36 seconds)

Working: They will next change together after LCM(72, 108) = 216 seconds.

$216$ seconds = $3$ minutes $36$ seconds

$8{:}00{:}00 + 3\text{ min }36\text{ s} = 8{:}03{:}36$ a.m.

Marking: 1 mark for using LCM = 216; 1 mark for correct time.

Common mistake: Students may use HCF instead of LCM, getting 36 seconds.

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12. A recipe for 6 servings requires $\dfrac{3}{4}$ cup of sugar and $\dfrac{1}{2}$ cup of flour.

(a) Sugar for 15 servings. (2 marks)

Answer: $1\dfrac{7}{8}$ cups (or $\dfrac{15}{8}$ cups)

Working: Sugar per serving = $\dfrac{3}{4} \div 6 = \dfrac{3}{4} \times \dfrac{1}{6} = \dfrac{3}{24} = \dfrac{1}{8}$ cup

For 15 servings: $15 \times \dfrac{1}{8} = \dfrac{15}{8} = 1\dfrac{7}{8}$ cups

Marking: 1 mark for correct method (finding per-serving amount or using proportion); 1 mark for correct answer.

(b) Flour for 10 servings. (2 marks)

Answer: $\dfrac{5}{6}$ cup

Working: Flour per serving = $\dfrac{1}{2} \div 6 = \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{12}$ cup

For 10 servings: $10 \times \dfrac{1}{12} = \dfrac{10}{12} = \dfrac{5}{6}$ cup

Marking: 1 mark for correct method; 1 mark for correct answer in lowest terms.

(c) Maximum servings with 3 cups of sugar. (1 mark)

Answer: 24 servings

Working: $3 \div \dfrac{1}{8} = 3 \times 8 = 24$ servings

Marking: 1 mark for correct answer.

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13. Maths : Science = $7 : 5$; Science : English = $3 : 4$.

(a) Find Maths : Science : English. (2 marks)

Answer: $21 : 15 : 20$

Working: Maths : Science = $7 : 5 = 21 : 15$ (multiply by 3)
Science : English = $3 : 4 = 15 : 20$ (multiply by 5)

So Maths : Science : English = $21 : 15 : 20$

Marking: 1 mark for correctly equating the Science parts; 1 mark for correct combined ratio.

Common mistake: Students may simply write $7 : 5 : 4$ without making the Science parts equal.

(b) If 135 students take Science, how many take Mathematics? (2 marks)

Answer: 189 students

Working: Science = 15 parts = 135 students
1 part = $135 \div 15 = 9$
Maths = 21 parts = $21 \times 9 = 189$ students

Marking: 1 mark for finding 1 part = 9; 1 mark for correct answer.

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14. $4\,200 shared among Amir, Bella, Chris in ratio$3 : 4 : 7$.

(a) How much does each receive? (3 marks)

Answer: Amir = $900, Bella =$1,200, Chris = $2,100

Working: Total parts = $3 + 4 + 7 = 14$
1 part = $4\,200 \div 14 = 300$

Amir: $3 \times 300 = 900$
Bella: $4 \times 300 = 1\,200$
Chris: $7 \times 300 = 2\,100$

Marking: 1 mark for total parts and value of 1 part; 1 mark for Amir and Bella; 1 mark for Chris.

(b) Bella's amount as a percentage of the total. (1 mark)

Answer: $28\dfrac{4}{7}\%$ (or approximately $28.6\%$)

Working: $\dfrac{1\,200}{4\,200} \times 100\% = \dfrac{2}{7} \times 100\% = \dfrac{200}{7}\% = 28\dfrac{4}{7}\%$

Marking: 1 mark for correct answer. Accept $\dfrac{200}{7}\%$ or $28\dfrac{4}{7}\%$.

(c) Chris gives $\dfrac{1}{4}$ of his share to Amir. How much does Amir have now? (2 marks)

Answer: $1,425

Working: Chris gives: $\dfrac{1}{4} \times 2\,100 = 525$

Amir now has: $900 + 525 = 1\,425$

Marking: 1 mark for finding $\dfrac{1}{4}$ of Chris's share; 1 mark for correct final amount.

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15. Solve $-5x \leq 20$ and illustrate on the number line. (3 marks)

Answer: $x \geq -4$

Working: $-5x \leq 20$

Divide both sides by $-5$ (reverse the inequality sign):

$x \geq \dfrac{20}{-5}$

$x \geq -4$

Number line: Closed circle at $-4$, shading to the right.

←——|——|——|——|——●====|——|——|——|——|——|——|——|——→
   -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6

Marking: 1 mark for dividing by $-5$; 1 mark for reversing the inequality sign; 1 mark for correct number line (closed circle at $-4$, arrow/shading to the right).

Common mistake: Students forget to reverse the inequality sign when dividing by a negative number, getting $x \leq -4$.

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Section C: Problem-Solving Questions (15 marks)

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16. Price ratio of Brand P to Brand Q is $5 : 3$. Brand P costs $12 more than Brand Q.

(a) Find the price of each brand. (3 marks)

Answer: Brand P = $30, Brand Q =$18

Working: Difference in parts = $5 - 3 = 2$ parts
2 parts = $12 1 part =$6$

Brand P = $5 \times 6 = 30$
Brand Q = $3 \times 6 = 18$

Marking: 1 mark for finding the difference in parts; 1 mark for finding 1 part = $6; 1 mark for both correct prices.

(b) Brand P discounted by $20\%$, Brand Q by $10\%$. Find the new ratio. (3 marks)

Answer: $40 : 27$

Working: New price of Brand P: $30 \times (1 - 0.20) = 30 \times 0.80 = 24$
New price of Brand Q: $18 \times (1 - 0.10) = 18 \times 0.90 = 16.20$

New ratio = $24 : 16.20$

Multiply both by 100: $2\,400 : 1\,620$

Divide by 60: $40 : 27$

Marking: 1 mark for correct discounted price of Brand P; 1 mark for correct discounted price of Brand Q; 1 mark for correct simplified ratio.

Common mistake: Students may try to apply discounts directly to the ratio parts without first finding actual prices.

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17. Population of $12\,000$ increases by $15\%$ in Year 1, then decreases by $10\%$ in Year 2.

(a) Population after Year 1. (2 marks)

Answer: $13\,800$

Working: Increase = $12\,000 \times 0.15 = 1\,800$
Population after Year 1 = $12\,000 + 1\,800 = 13\,800$

Marking: 1 mark for correct increase amount; 1 mark for correct population.

(b) Population after Year 2. (2 marks)

Answer: $12\,420$

Working: Decrease = $13\,800 \times 0.10 = 1\,380$
Population after Year 2 = $13\,800 - 1\,380 = 12\,420$

Marking: 1 mark for correct decrease amount; 1 mark for correct population.

(c) Overall percentage change. (2 marks)

Answer: $3.5\%$ increase

Working: Change = $12\,420 - 12\,000 = 420$ (increase)

Percentage change = $\dfrac{420}{12\,000} \times 100\% = 3.5\%$

Marking: 1 mark for correct change amount; 1 mark for correct percentage.

(d) Is the overall change an increase or decrease? Explain. (1 mark)

Answer: Overall increase. Although the population decreased in Year 2, the decrease was applied to a larger base ($13\,800), so the absolute decrease ($1,380) was smaller than the absolute increase in Year 1 ($1,800). The net effect is an increase.

Marking: 1 mark for correct identification (increase) with valid explanation.

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18. Three friends share money. Devi receives $\dfrac{2}{5}$ of total. Ethan receives $\dfrac{1}{3}$ of the remainder. Farah receives the rest.

(a) Fraction that Ethan receives. (2 marks)

Answer: $\dfrac{2}{15}$

Working: Devi receives $\dfrac{2}{5}$, so remainder = $1 - \dfrac{2}{5} = \dfrac{3}{5}$

Ethan receives $\dfrac{1}{3}$ of $\dfrac{3}{5} = \dfrac{1}{3} \times \dfrac{3}{5} = \dfrac{3}{15} = \dfrac{2}{15} \times \dfrac{3}{3} = \dfrac{3}{15}$

Wait — let me recalculate: $\dfrac{1}{3} \times \dfrac{3}{5} = \dfrac{3}{15} = \dfrac{1}{5}$

Answer: $\dfrac{1}{5}$

Working: Remainder after Devi = $\dfrac{3}{5}$
Ethan = $\dfrac{1}{3} \times \dfrac{3}{5} = \dfrac{1}{5}$

Marking: 1 mark for finding the remainder; 1 mark for correct fraction.

(b) Fraction that Farah receives. (2 marks)

Answer: $\dfrac{8}{15}$

Wait — let me recalculate:

Devi = $\dfrac{2}{5} = \dfrac{6}{15}$
Ethan = $\dfrac{1}{5} = \dfrac{3}{15}$
Farah = $1 - \dfrac{6}{15} - \dfrac{3}{15} = \dfrac{6}{15} = \dfrac{2}{5}$

Answer: $\dfrac{2}{5}$

Working: Total = 1
Devi = $\dfrac{2}{5}$
Ethan = $\dfrac{1}{5}$
Farah = $1 - \dfrac{2}{5} - \dfrac{1}{5} = \dfrac{2}{5}$

Marking: 1 mark for correct method; 1 mark for correct answer.

(c) Devi receives $48 more than Ethan. Find the total sum. (3 marks)

Answer: $240

Working: Devi's fraction = $\dfrac{2}{5}$, Ethan's fraction = $\dfrac{1}{5}$

Difference = $\dfrac{2}{5} - \dfrac{1}{5} = \dfrac{1}{5}$ of total

$\dfrac{1}{5}$ of total = $48 Total =$48 \times 5 = 240$

Marking: 1 mark for finding the difference in fractions; 1 mark for setting up the equation; 1 mark for correct answer.

Common mistake: Students may subtract the fractions incorrectly or use the wrong fraction for the difference.

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Mark Summary

Section	Marks
A: Questions 1–10	20
B: Questions 11–15	25
C: Questions 16–18	15
Total	60