Secondary 1 Mathematics Practice Paper 5

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TuitionGoWhere Practice Paper - Mathematics Secondary 1

TuitionGoWhere Practice Paper (AI)

Subject: Mathematics
Level: Secondary 1 (G3)
Paper: Practice Paper — Numbers, Ratio & Proportion
Duration: 1 hour 30 minutes
Total Marks: 60
Name: ________________________
Class: ________________________
Date: ________________________

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Instructions

1. Write your name, class, and date in the spaces provided above.
2. Answer all questions in the spaces provided.
3. Show your working clearly. Marks are awarded for correct steps, not only for the final answer.
4. Do not use correction fluid or tape.
5. The use of a calculator is allowed unless otherwise stated.
6. The total marks for this paper is 60.
7. This paper consists of Section A, Section B, and Section C.

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Section A: Short Answer Questions (20 marks)

Answer all questions. Each question carries 2 marks unless otherwise stated.

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1. Express 360 as a product of its prime factors. Give your answer in index notation.

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2. Find the Highest Common Factor (HCF) of 48 and 84.

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3. Find the Lowest Common Multiple (LCM) of 18 and 30.

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4. Evaluate the following, giving your answer as a fraction in its lowest terms:

    $\dfrac{3}{4} - \dfrac{2}{5} + \dfrac{1}{10}$

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5. Arrange the following numbers in ascending order:

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

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6. Write the following inequality using the symbol $<$ or $>$:

    $-7$ ____________ $-3$

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7. Round 4.7385 to (a) 2 decimal places, and (b) 3 significant figures.

    (a) _______________________________________________________________

    (b) _______________________________________________________________

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8. Simplify the ratio $36 : 60$ to its simplest form.

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9. The ratio of boys to girls in a class is $4 : 5$. If there are 24 boys, how many girls are there?

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10. Express $240$ as a percentage of $600$.

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Section B: Structured Questions (25 marks)

Answer all questions. Show your working clearly.

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

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    (b) Two traffic lights at a junction change every 72 seconds and 108 seconds respectively. If they change together at 8:00 a.m., at what time will they next change together? (2 marks)

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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) How much sugar is needed for 15 servings? (2 marks)

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    (b) How much flour is needed for 10 servings? (2 marks)

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    (c) If you have exactly 3 cups of sugar, what is the maximum number of servings you can make? (1 mark)

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13. In a school, the ratio of students who take Mathematics to those who take Science is $7 : 5$. The ratio of students who take Science to those who take English is $3 : 4$.

    (a) Find the ratio of students taking Mathematics : Science : English. (2 marks)

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

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14. A sum of $4\,200 is divided among three siblings, Amir, Bella, and Chris, in the ratio$3 : 4 : 7$.

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

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    (b) Express the amount Bella receives as a percentage of the total sum. (1 mark)

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

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15. Solve the following inequality and illustrate the solution on the number line provided. (3 marks)

    $-5x \leq 20$

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    Number line:

    ←——|——|——|——|——|——|——|——|——|——|——|——|——|——|——→

        -6  -5  -4  -3  -2  -1  0  1  2  3  4  5  6  7

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Section C: Problem-Solving Questions (15 marks)

Answer all questions. Show your working clearly. Full marks will be awarded for complete solutions with clear reasoning.

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16. A shop sells two brands of rice, Brand P and Brand Q. The ratio of the price of Brand P to Brand Q is $5 : 3$. Brand P costs $12 more than Brand Q.

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

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    (b) During a sale, Brand P is discounted by $20\%$ and Brand Q is discounted by $10\%$. Find the new ratio of the price of Brand P to Brand Q. Give your answer in its simplest form. (3 marks)

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17. The population of a town increases by $15\%$ in the first year and then decreases by $10\%$ in the second year. The original population was $12\,000$.

    (a) Find the population after the first year. (2 marks)

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    (b) Find the population after the second year. (2 marks)

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    (c) Express the overall percentage change in population over the two years. (2 marks)

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

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

    (a) What fraction of the total does Ethan receive? (2 marks)

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    (b) What fraction of the total does Farah receive? (2 marks)

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    (c) If Devi receives $48 more than Ethan, find the total sum of money. (3 marks)

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End of Paper

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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