Primary 6 PSLE Mathematics Fractions Quiz

<!-- TuitionGoWhere generation metadata: stage=5-1; model=moonshotai/kimi-k2.6:free; model_label=Kimi K2.6 Free; generated=2026-06-07; Sources: Stage 4-0 LLM templates, syllabus context, and Stage 2 evidence where available. -->

Primary 6 PSLE Mathematics Quiz - Fractions

Name: ___________________________ Class: __________ Date: __________

Duration: 45 minutes

Total Marks: 40 marks

Instructions:

• Answer all questions.
• Show your working clearly in the spaces provided.
• Write your answers in the simplest form.
• For questions requiring units, include them in your answer.

---

Section A: Direct Computation (Questions 1–8)

8 questions | 8 marks

---

1. Calculate: $\frac{3}{4} \div 6$

Answer: ________________

[1 mark]

---

2. Calculate: $5 \div \frac{2}{3}$

Answer: ________________

[1 mark]

---

3. Calculate: $\frac{5}{8} \div \frac{1}{4}$

Answer: ________________

[1 mark]

---

4. Simplify: $\frac{7}{9} \times \frac{3}{14} \div \frac{1}{6}$

Answer: ________________

[1 mark]

---

5. Find the value of: $\frac{2}{5} + \frac{3}{4} \times \frac{2}{9}$

Answer: ________________

[1 mark]

---

6. Arrange the following fractions in ascending order: $\frac{3}{5}$, $\frac{5}{8}$, $\frac{7}{12}$

Answer: ________________

[1 mark]

---

7. What is $\frac{4}{7}$ of $\frac{21}{16}$?

Answer: ________________

[1 mark]

---

8. Express $\frac{18}{24}$ as a fraction with denominator 32.

Answer: ________________

[1 mark]

---

Section B: Problem Solving (Questions 9–15)

7 questions | 14 marks

---

9. A baker had $\frac{3}{4}$ kg of flour. She used $\frac{2}{5}$ of it to make cakes. How much flour did she use? Give your answer in kilograms.

Working:

Answer: ________________ kg [2 marks]

---

10. Raj had 480 stickers. He gave $\frac{5}{8}$ of them to his sister and $\frac{1}{3}$ of the remainder to his cousin. How many stickers did he have left?

Working:

Answer: ________________ stickers [2 marks]

---

11. A tank was $\frac{5}{6}$ full of water. After 18 litres of water were poured out, it was $\frac{2}{3}$ full. How many litres of water could the tank hold when completely full?

Working:

Answer: ________________ litres [2 marks]

---

12. Mdm Tan spent $\frac{2}{5}$ of her money on groceries and $\frac{1}{4}$ of the remainder on transport. She had $72 left. How much money did she have at first?

Working:

Answer: $ ________________ [2 marks]

---

13. Peter and John had 360 marbles altogether. Peter had $\frac{3}{5}$ as many marbles as John. How many marbles did Peter have?

Working:

Answer: ________________ marbles [2 marks]

---

14. A ribbon was 2$\frac{1}{4}$ m long. It was cut into pieces, each measuring $\frac{3}{8}$ m. What is the maximum number of such pieces that can be obtained? How much ribbon is left over?

Working:

Answer: ________________ pieces, ________________ m left [2 marks]

---

15. The mass of a watermelon is $\frac{7}{4}$ times the mass of a honeydew. The honeydew is $\frac{2}{3}$ as heavy as a pumpkin. If the pumpkin has a mass of 4$\frac{1}{2}$ kg, find the mass of the watermelon.

Working:

Answer: ________________ kg [2 marks]

---

Section C: Multi-Step and Challenging Problems (Questions 16–20)

5 questions | 18 marks

---

16. (a) Calculate: $\left(\frac{3}{4} + \frac{2}{3}\right) \div \frac{5}{6}$

Working:

Answer for (a): ________________ [2 marks]

(b) A number is such that when it is divided by $\frac{3}{5}$, the result is $\frac{15}{16}$. What is the number?

Working:

Answer for (b): ________________ [2 marks]

---

17. Amy, Ben, and Cathy shared some money. Amy received $\frac{2}{5}$ of the total amount. Ben received $\frac{3}{8}$ of the remainder. Cathy received the rest, which was $105. How much money was shared altogether?

Working:

Answer: $ ________________ [4 marks]

---

18. A rectangular tank measuring 60 cm by 40 cm by 30 cm was $\frac{3}{5}$ full of water.

Generated diagram for Q18.

(a) Find the volume of water in the tank.

Working:

Answer for (a): ________________ cm³ [2 marks]

(b) All the water was poured into another empty tank with a square base of side 30 cm. What would be the height of the water in the new tank?

Working:

Answer for (b): ________________ cm [2 marks]

---

19. In a school, $\frac{3}{7}$ of the pupils are boys. $\frac{2}{5}$ of the boys and $\frac{1}{4}$ of the girls wear glasses. What fraction of all the pupils wear glasses?

Working:

Answer: ________________ [4 marks]

---

20. Mrs Lim had some apples and oranges. $\frac{5}{8}$ of the fruits were apples. She gave away $\frac{2}{5}$ of the apples and $\frac{3}{4}$ of the oranges. In the end, she had 90 fruits left.

(a) What fraction of the fruits did she give away altogether?

Working:

Answer for (a): ________________ [2 marks]

(b) How many fruits did Mrs Lim have at first?

Working:

Answer for (b): ________________ fruits [2 marks]

---

END OF QUIZ

<!-- TuitionGoWhere generation metadata: stage=5-1; model=moonshotai/kimi-k2.6:free; model_label=Kimi K2.6 Free; generated=2026-06-07; Sources: Stage 4-0 LLM templates, syllabus context, and Stage 2 evidence where available. -->

Primary 6 PSLE Mathematics Quiz - Fractions: Answer Key

---

Section A: Direct Computation

---

1. Calculate: $\frac{3}{4} \div 6$

Answer: $\frac{1}{8}$ [1 mark]

Explanation: Dividing by a whole number is the same as multiplying by its reciprocal. So $\frac{3}{4} \div 6 = \frac{3}{4} \times \frac{1}{6} = \frac{3}{24} = \frac{1}{8}$. Remember: a whole number has a denominator of 1, so its reciprocal is $\frac{1}{6}$. Simplify by dividing numerator and denominator by 3.

Common mistake: Forgetting to invert the whole number or incorrectly writing $6 = \frac{6}{1}$ and then inverting to $\frac{1}{6}$.

---

2. Calculate: $5 \div \frac{2}{3}$

Answer: $7\frac{1}{2}$ or $\frac{15}{2}$ [1 mark]

Explanation: Dividing by a fraction means multiplying by its reciprocal (flipping the fraction). The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$. So $5 \div \frac{2}{3} = 5 \times \frac{3}{2} = \frac{15}{2} = 7\frac{1}{2}$. Think of it as: "How many $\frac{2}{3}$s fit into 5?"

---

3. Calculate: $\frac{5}{8} \div \frac{1}{4}$

Answer: $2\frac{1}{2}$ or $\frac{5}{2}$ [1 mark]

Explanation: $\frac{5}{8} \div \frac{1}{4} = \frac{5}{8} \times \frac{4}{1} = \frac{20}{8} = \frac{5}{2} = 2\frac{1}{2}$. The reciprocal of $\frac{1}{4}$ is $\frac{4}{1} = 4$. This shows that dividing by $\frac{1}{4}$ is the same as multiplying by 4, which makes sense: there are four quarters in one whole.

---

4. Simplify: $\frac{7}{9} \times \frac{3}{14} \div \frac{1}{6}$

Answer: $1$ [1 mark]

Explanation: Work left to right. First: $\frac{7}{9} \times \frac{3}{14} = \frac{21}{126} = \frac{1}{6}$ (cross-simplify: 7 and 14 share 7; 3 and 9 share 3). Then: $\frac{1}{6} \div \frac{1}{6} = \frac{1}{6} \times \frac{6}{1} = 1$. Alternatively, convert division to multiplication by reciprocal at the start: $\frac{7}{9} \times \frac{3}{14} \times \frac{6}{1} = \frac{126}{126} = 1$.

---

5. Find the value of: $\frac{2}{5} + \frac{3}{4} \times \frac{2}{9}$

Answer: $\frac{11}{20}$ [1 mark]

Explanation: Follow order of operations (BODMAS/PEMDAS): multiplication before addition. First: $\frac{3}{4} \times \frac{2}{9} = \frac{6}{36} = \frac{1}{6}$. Then: $\frac{2}{5} + \frac{1}{6} = \frac{12}{30} + \frac{5}{30} = \frac{17}{30}$.

Correction: Rechecking: $\frac{2}{5} + \frac{1}{6} = \frac{12+5}{30} = \frac{17}{30}$.

Wait — let me recheck: $\frac{3}{4} \times \frac{2}{9} = \frac{6}{36} = \frac{1}{6}$. Then $\frac{2}{5} + \frac{1}{6} = \frac{12+5}{30} = \frac{17}{30}$.

Final Answer: $\frac{17}{30}$ [1 mark]

---

6. Arrange in ascending order: $\frac{3}{5}$, $\frac{5}{8}$, $\frac{7}{12}$

Answer: $\frac{7}{12}$, $\frac{5}{8}$, $\frac{3}{5}$ [1 mark]

Explanation: Convert to common denominator or decimals. Using decimals: $\frac{3}{5} = 0.6$, $\frac{5}{8} = 0.625$, $\frac{7}{12} \approx 0.583$. Ascending order: $0.583 < 0.6 < 0.625$. Using common denominator 120: $\frac{3}{5}=\frac{72}{120}$, $\frac{5}{8}=\frac{75}{120}$, $\frac{7}{12}=\frac{70}{120}$. Order: $\frac{70}{120} < \frac{72}{120} < \frac{75}{120}$.

---

7. What is $\frac{4}{7}$ of $\frac{21}{16}$?

Answer: $\frac{3}{4}$ [1 mark]

Explanation: "Of" means multiply. $\frac{4}{7} \times \frac{21}{16} = \frac{84}{112} = \frac{3}{4}$. Cross-simplify first: 4 and 16 share 4; 7 and 21 share 7. So $\frac{1}{\cancel{7}} \times \frac{\cancel{21}^{3}}{\cancel{16}_{4}} = \frac{3}{4}$ (after simplifying: $\frac{1}{1} \times \frac{3}{4} = \frac{3}{4}$... let me redo: $\frac{4}{7} \times \frac{21}{16}$: 4 goes into 16 four times; 7 goes into 21 three times. So $\frac{1}{1} \times \frac{3}{4} = \frac{3}{4}$).

---

8. Express $\frac{18}{24}$ as a fraction with denominator 32.

Answer: $\frac{24}{32}$ [1 mark]

Explanation: First simplify $\frac{18}{24} = \frac{3}{4}$. Then find equivalent: $\frac{3}{4} = \frac{?}{32}$. Since $4 \times 8 = 32$, multiply numerator by 8: $3 \times 8 = 24$. So $\frac{24}{32}$. Alternatively from original: $\frac{18}{24} = \frac{x}{32}$, so $x = \frac{18 \times 32}{24} = \frac{576}{24} = 24$.

---

Section B: Problem Solving

---

9. Baker's flour problem

Answer: $\frac{3}{10}$ kg [2 marks]

Working:

• Flour used = $\frac{2}{5}$ of $\frac{3}{4}$ kg = $\frac{2}{5} \times \frac{3}{4}$ [1 mark]
• = $\frac{6}{20} = \frac{3}{10}$ kg [1 mark]

Explanation: "Of" indicates multiplication. When finding a fraction of a quantity, we multiply. The baker started with $\frac{3}{4}$ kg and used $\frac{2}{5}$ of that amount, not $\frac{2}{5}$ kg extra. Simplify $\frac{6}{20}$ by dividing by 2.

---

10. Raj's stickers

Answer: 120 stickers [2 marks]

Working:

• Given to sister: $\frac{5}{8} \times 480 = 300$ stickers
• Remainder: $480 - 300 = 180$ stickers [1 mark]
• Given to cousin: $\frac{1}{3} \times 180 = 60$ stickers
• Left: $180 - 60 = 120$ stickers [1 mark]

Alternative: Remainder after sister = $\frac{3}{8} \times 480 = 180$. Then keep $\frac{2}{3}$ of remainder: $\frac{2}{3} \times 180 = 120$.

Explanation: "Of the remainder" is crucial — calculate sequentially. First find what remains after each step. The remainder after giving $\frac{5}{8}$ away is $\frac{3}{8}$ (or $480-300=180$). Then take $\frac{1}{3}$ of this remainder, not of the original.

---

11. Tank water problem

Answer: 108 litres [2 marks]

Working:

• Fraction poured out: $\frac{5}{6} - \frac{2}{3} = \frac{5}{6} - \frac{4}{6} = \frac{1}{6}$ [1 mark]
• $\frac{1}{6}$ of tank = 18 litres
• Full tank = $18 \times 6 = 108$ litres [1 mark]

Alternative with algebra: Let full capacity be $C$. Then $\frac{5}{6}C - 18 = \frac{2}{3}C$, so $\frac{5}{6}C - \frac{2}{3}C = 18$, thus $\frac{1}{6}C = 18$, $C = 108$.

Explanation: The difference between two fractional amounts equals 18 litres. Convert to common denominator to subtract fractions. Then use the unitary method: if $\frac{1}{6}$ = 18, then whole = $18 \times 6$.

---

12. Mdm Tan's money

Answer: $160 [2 marks]

Working:

• Spent on groceries: $\frac{2}{5}$
• Remainder: $\frac{3}{5}$
• Spent on transport: $\frac{1}{4}$ of $\frac{3}{5} = \frac{3}{20}$
• Total spent: $\frac{2}{5} + \frac{3}{20} = \frac{8}{20} + \frac{3}{20} = \frac{11}{20}$ [1 mark]
• Left: \frac{9}{20} = \72$
• Total: 72 \div \frac{9}{20} = 72 \times \frac{20}{9} = 8 \times 20 = \160$ [1 mark]

Alternative (model method): After groceries, $\frac{3}{5}$ remains. This is split: $\frac{1}{4}$ on transport, so $\frac{3}{4}$ of remainder = $\frac{3}{4} \times \frac{3}{5} = \frac{9}{20}$ left. If $\frac{9}{20} = 72$, then $\frac{1}{20} = 8$, so whole = $20 \times 8 = 160$.

Explanation: The "remainder" changes after each spending. Track carefully: first remainder is $\frac{3}{5}$ of original. Transport is $\frac{1}{4}$ of this remainder, not of original. The final amount left represents $\frac{9}{20}$ of original.

---

13. Peter and John's marbles

Answer: 135 marbles [2 marks]

Working:

• Peter : John = 3 : 5 (since Peter has $\frac{3}{5}$ as many as John) [1 mark]
• Total parts = $3 + 5 = 8$
• Peter's marbles = $\frac{3}{8} \times 360 = 135$ [1 mark]

Alternative: Let John have $J$. Then Peter has $\frac{3}{5}J$. So $J + \frac{3}{5}J = 360$, giving $\frac{8}{5}J = 360$, so $J = 225$, Peter = $360 - 225 = 135$.

Explanation: "$\frac{3}{5}$ as many as" creates a ratio. Peter : John = 3 : 5. The total represents 8 equal parts. This is a "fraction as ratio" concept linking fractions to ratio work.

---

14. Ribbon pieces

Answer: 6 pieces, $\frac{0}{8}$ or 0 m left... rechecking: [2 marks]

Working:

• $2\frac{1}{4} = \frac{9}{4}$ m [0.5 mark]
• Number of pieces: $\frac{9}{4} \div \frac{3}{8} = \frac{9}{4} \times \frac{8}{3} = \frac{72}{12} = 6$ [1 mark]
• Leftover: $6 \times \frac{3}{8} = \frac{18}{8} = \frac{9}{4}$. Since $6 \times \frac{3}{8} = \frac{18}{8} = 2\frac{2}{8} = 2\frac{1}{4}$, there is 0 m left. [0.5 mark]

Answer: 6 pieces, 0 m left (or no remainder)

Explanation: Division determines how many pieces fit. Convert mixed number to improper fraction first. Check: does $6 \times \frac{3}{8} = \frac{9}{4}$? Yes, exactly, so no remainder. This is a "measurement division" problem — how many $\frac{3}{8}$s in $\frac{9}{4}$?

---

15. Mass of fruits

Answer: $5\frac{1}{4}$ kg or $\frac{21}{4}$ kg [2 marks]

Working:

• Pumpkin: $4\frac{1}{2} = \frac{9}{2}$ kg
• Honeydew: $\frac{2}{3} \times \frac{9}{2} = \frac{18}{6} = 3$ kg [1 mark]
• Watermelon: $\frac{7}{4} \times 3 = \frac{21}{4} = 5\frac{1}{4}$ kg [1 mark]

Explanation: Chain of "of" relationships requires sequential multiplication. First find honeydew from pumpkin, then watermelon from honeydew. Each "A is [fraction] of B" translates to multiplication, but read carefully: "honeydew is $\frac{2}{3}$ as heavy as pumpkin" means honeydew = $\frac{2}{3} \times$ pumpkin.

---

Section C: Multi-Step and Challenging Problems

---

16(a). Calculate: $\left(\frac{3}{4} + \frac{2}{3}\right) \div \frac{5}{6}$

Answer: $\frac{17}{10}$ or $1\frac{7}{10}$ [2 marks]

Working:

• Brackets first: $\frac{3}{4} + \frac{2}{3} = \frac{9}{12} + \frac{8}{12} = \frac{17}{12}$ [1 mark]
• Then divide: $\frac{17}{12} \div \frac{5}{6} = \frac{17}{12} \times \frac{6}{5} = \frac{102}{60} = \frac{17}{10}$ [1 mark]

Explanation: BODMAS demands brackets before division. Common denominator for addition is 12. Division becomes multiplication by reciprocal; cross-simplify: 6 and 12 share 6.

---

16(b). Missing number

Answer: $\frac{9}{16}$ [2 marks]

Working:

• Let the number be $x$
• $x \div \frac{3}{5} = \frac{15}{16}$ [0.5 mark]
• $x = \frac{15}{16} \times \frac{3}{5} = \frac{45}{80} = \frac{9}{16}$ [1.5 marks]

Explanation: To find the original number, reverse the operation. If dividing by $\frac{3}{5}$ gives $\frac{15}{16}$, then multiply $\frac{15}{16}$ by $\frac{3}{5}$. This uses the "inverse operation" concept — multiplication undoes division.

---

17. Amy, Ben, Cathy money

Answer: $280 [4 marks]

Working: Method 1: Fraction tracking

• Amy: $\frac{2}{5}$ of total
• Remainder after Amy: $\frac{3}{5}$ of total [1 mark]
• Ben: $\frac{3}{8}$ of remainder = $\frac{3}{8} \times \frac{3}{5} = \frac{9}{40}$ of total [1 mark]
• Cathy's fraction: $1 - \frac{2}{5} - \frac{9}{40} = \frac{40-16-9}{40} = \frac{15}{40} = \frac{3}{8}$ [1 mark]
• If $\frac{3}{8}$ = $105, then total =$105 \times \frac{8}{3} = 35 \times 8 = $280 [1 mark]

Method 2: Model drawing (units)

• Total: 40 units
• Amy: 16 units, remaining 24 units
• Ben: $\frac{3}{8} \times 24 = 9$ units
• Cathy: $24 - 9 = 15$ units = $105
• 1 unit = $7
• Total: $7 \times 40 =$280

Explanation: Multiple remainders require careful tracking. Cathy's amount comes from the remainder after BOTH Amy and Ben have taken their shares. Always express as fraction of total for easier comparison. The unit method offers a visual alternative.

Common mistake: Treating Ben's $\frac{3}{8}$ as $\frac{3}{8}$ of total instead of $\frac{3}{8}$ of remainder.

---

18(a). Volume of water

Answer: 43,200 cm³ [2 marks]

Working:

• Full volume: $60 \times 40 \times 30 = 72 000$ cm³ [1 mark]
• Water volume: $\frac{3}{5} \times 72 000 = 43 200$ cm³ [1 mark]

Or directly: $\frac{3}{5} \times 60 \times 40 \times 30 = 43 200$ cm³.

Explanation: Volume of cuboid = length × width × height. The fraction $\frac{3}{5}$ applies to the volume (equivalently, to the height if base is constant).

---

18(b). Height in new tank

Answer: 48 cm [2 marks]

Working:

• New base area: $30 \times 30 = 900$ cm² [0.5 mark]
• Volume = base area × height, so height = $\frac{43 200}{900}$ [1 mark]
• = 48 cm [0.5 mark]

Explanation: Conservation of volume — water volume stays constant. The formula $V = \text{base} \times \text{height}$ rearranges to find height. The diagram (when rendered) should show water fills to height h in a tank with 30 cm square base.

---

19. Pupils wearing glasses

Answer: $\frac{17}{70}$ [4 marks]

Working:

• Let total pupils = 1 (or 70 for concrete working)
• Boys: $\frac{3}{7}$, Girls: $\frac{4}{7}$ [1 mark]
• Boys with glasses: $\frac{2}{5} \times \frac{3}{7} = \frac{6}{35}$ [1 mark]
• Girls with glasses: $\frac{1}{4} \times \frac{4}{7} = \frac{4}{28} = \frac{1}{7} = \frac{5}{35}$... recheck: $\frac{1}{4} \times \frac{4}{7} = \frac{4}{28} = \frac{1}{7} = \frac{5}{35}$ [1 mark]

Let me use common denominator 70:

• Boys with glasses: $\frac{6}{35} = \frac{12}{70}$
• Girls with glasses: $\frac{1}{7} = \frac{10}{70}$
• Total with glasses: $\frac{12+10}{70} = \frac{22}{70} = \frac{11}{35}$

Rechecking girls: $\frac{1}{4} \times \frac{4}{7} = \frac{1}{7}$. Correct.

Total: $\frac{6}{35} + \frac{1}{7} = \frac{6}{35} + \frac{5}{35} = \frac{11}{35}$

Answer: $\frac{11}{35}$ [4 marks]

Working (clean):

• Boys: $\frac{3}{7}$, Girls: $\frac{4}{7}$ [1 mark]
• Boys with glasses: $\frac{2}{5} \times \frac{3}{7} = \frac{6}{35}$ [1 mark]
• Girls with glasses: $\frac{1}{4} \times \frac{4}{7} = \frac{1}{7} = \frac{5}{35}$ [1 mark]
• Total with glasses: $\frac{6}{35} + \frac{5}{35} = \frac{11}{35}$ [1 mark]

Explanation: "Of" means multiply for each subgroup. The key insight is finding the girls' fraction first ($\frac{4}{7}$), then applying $\frac{1}{4}$ to that subgroup. Common denominitors needed for final addition.

---

20(a). Fraction given away

Answer: $\frac{37}{80}$ [2 marks]

Working:

• Apples given away: $\frac{2}{5} \times \frac{5}{8} = \frac{10}{40} = \frac{1}{4}$ of total [0.5 mark]
• Oranges: $\frac{3}{8}$ of total (since $1 - \frac{5}{8} = \frac{3}{8}$)
• Oranges given away: $\frac{3}{4} \times \frac{3}{8} = \frac{9}{32}$ of total [0.5 mark]
• Total given away: $\frac{1}{4} + \frac{9}{32} = \frac{8}{32} + \frac{9}{32} = \frac{17}{32}$ [1 mark]

Rechecking with common fraction: Let me re-examine. Apples = $\frac{5}{8}$, oranges = $\frac{3}{8}$.

• Apples given: $\frac{2}{5} \times \frac{5}{8} = \frac{2}{8} = \frac{1}{4}$
• Oranges given: $\frac{3}{4} \times \frac{3}{8} = \frac{9}{32}$
• Total given: $\frac{8}{32} + \frac{9}{32} = \frac{17}{32}$

Answer: $\frac{17}{32}$ [2 marks]

---

20(b). Original number of fruits

Answer: 288 fruits [2 marks]

Working:

• Fraction left = $1 - \frac{17}{32} = \frac{15}{32}$ [0.5 mark]
• Or: Apples left = $\frac{3}{5} \times \frac{5}{8} = \frac{3}{8}$; Oranges left = $\frac{1}{4} \times \frac{3}{8} = \frac{3}{32}$. Total left = $\frac{12}{32} + \frac{3}{32} = \frac{15}{32}$ [0.5 mark]
• If $\frac{15}{32}$ = 90, then total = $90 \times \frac{32}{15} = 6 \times 32 = 192$... recheck: $90 \div 15 = 6$, $6 \times 32 = 192$.

Wait: $192 \times \frac{15}{32} = 192 \div 32 \times 15 = 6 \times 15 = 90$. ✓

Answer: 192 fruits [2 marks]

Working (clean):

• Fraction left: $\frac{15}{32}$ [0.5 mark]
• Total = $90 \div \frac{15}{32} = 90 \times \frac{32}{15} = 192$ fruits [1.5 marks]

Explanation: Two approaches: find fraction left directly, or calculate fruits left by type. The "fraction left" method is cleaner. Note: $\frac{15}{32}$ comes from careful tracking of both apple and orange remainders. Unitary method: if 15 parts = 90, then 1 part = 6, so 32 parts = 192.

Common mistake: Assuming the $\frac{2}{5}$ and $\frac{3}{4}$ apply to the same base fraction of total.

---

END OF ANSWER KEY