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Primary 6 PSLE Mathematics Fractions Quiz

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Questions

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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: 34÷6\frac{3}{4} \div 6

Answer: ________________

[1 mark]

2. Calculate: 5÷235 \div \frac{2}{3}

Answer: ________________

[1 mark]

3. Calculate: 58÷14\frac{5}{8} \div \frac{1}{4}

Answer: ________________

[1 mark]

4. Simplify: 79×314÷16\frac{7}{9} \times \frac{3}{14} \div \frac{1}{6}

Answer: ________________

[1 mark]

5. Find the value of: 25+34×29\frac{2}{5} + \frac{3}{4} \times \frac{2}{9}

Answer: ________________

[1 mark]

6. Arrange the following fractions in ascending order: 35\frac{3}{5}, 58\frac{5}{8}, 712\frac{7}{12}

Answer: ________________

[1 mark]

7. What is 47\frac{4}{7} of 2116\frac{21}{16}?

Answer: ________________

[1 mark]

8. Express 1824\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 34\frac{3}{4} kg of flour. She used 25\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 58\frac{5}{8} of them to his sister and 13\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 56\frac{5}{6} full of water. After 18 litres of water were poured out, it was 23\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 25\frac{2}{5} of her money on groceries and 14\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 35\frac{3}{5} as many marbles as John. How many marbles did Peter have?

Working:

Answer: ________________ marbles [2 marks]

14. A ribbon was 214\frac{1}{4} m long. It was cut into pieces, each measuring 38\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 74\frac{7}{4} times the mass of a honeydew. The honeydew is 23\frac{2}{3} as heavy as a pumpkin. If the pumpkin has a mass of 412\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: (34+23)÷56\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 35\frac{3}{5}, the result is 1516\frac{15}{16}. What is the number?

Working:

Answer for (b): ________________ [2 marks]

17. Amy, Ben, and Cathy shared some money. Amy received 25\frac{2}{5} of the total amount. Ben received 38\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 35\frac{3}{5} full of water.

<image_placeholder> id: Q18-fig1 type: diagram linked_question: Q18 description: A rectangular tank with labelled dimensions showing water level labels: Tank length 60 cm, width 40 cm, height 30 cm; water level indicated at 3/5 height values: dimensions 60 cm × 40 cm × 30 cm, water fraction 3/5 must_show: Rectangular tank outline, dimension labels on three sides, horizontal water line at 3/5 height with "3/5 full" label, arrows showing length, width, height measurements </image_placeholder>

(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, 37\frac{3}{7} of the pupils are boys. 25\frac{2}{5} of the boys and 14\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. 58\frac{5}{8} of the fruits were apples. She gave away 25\frac{2}{5} of the apples and 34\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

Answers

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Primary 6 PSLE Mathematics Quiz - Fractions: Answer Key

Section A: Direct Computation

1. Calculate: 34÷6\frac{3}{4} \div 6

Answer: 18\frac{1}{8} [1 mark]

Explanation: Dividing by a whole number is the same as multiplying by its reciprocal. So 34÷6=34×16=324=18\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 16\frac{1}{6}. Simplify by dividing numerator and denominator by 3.

Common mistake: Forgetting to invert the whole number or incorrectly writing 6=616 = \frac{6}{1} and then inverting to 16\frac{1}{6}.

2. Calculate: 5÷235 \div \frac{2}{3}

Answer: 7127\frac{1}{2} or 152\frac{15}{2} [1 mark]

Explanation: Dividing by a fraction means multiplying by its reciprocal (flipping the fraction). The reciprocal of 23\frac{2}{3} is 32\frac{3}{2}. So 5÷23=5×32=152=7125 \div \frac{2}{3} = 5 \times \frac{3}{2} = \frac{15}{2} = 7\frac{1}{2}. Think of it as: "How many 23\frac{2}{3}s fit into 5?"

3. Calculate: 58÷14\frac{5}{8} \div \frac{1}{4}

Answer: 2122\frac{1}{2} or 52\frac{5}{2} [1 mark]

Explanation: 58÷14=58×41=208=52=212\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 14\frac{1}{4} is 41=4\frac{4}{1} = 4. This shows that dividing by 14\frac{1}{4} is the same as multiplying by 4, which makes sense: there are four quarters in one whole.

4. Simplify: 79×314÷16\frac{7}{9} \times \frac{3}{14} \div \frac{1}{6}

Answer: 11 [1 mark]

Explanation: Work left to right. First: 79×314=21126=16\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: 16÷16=16×61=1\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: 79×314×61=126126=1\frac{7}{9} \times \frac{3}{14} \times \frac{6}{1} = \frac{126}{126} = 1.

5. Find the value of: 25+34×29\frac{2}{5} + \frac{3}{4} \times \frac{2}{9}

Answer: 1120\frac{11}{20} [1 mark]

Explanation: Follow order of operations (BODMAS/PEMDAS): multiplication before addition. First: 34×29=636=16\frac{3}{4} \times \frac{2}{9} = \frac{6}{36} = \frac{1}{6}. Then: 25+16=1230+530=1730\frac{2}{5} + \frac{1}{6} = \frac{12}{30} + \frac{5}{30} = \frac{17}{30}.

Correction: Rechecking: 25+16=12+530=1730\frac{2}{5} + \frac{1}{6} = \frac{12+5}{30} = \frac{17}{30}.

Wait — let me recheck: 34×29=636=16\frac{3}{4} \times \frac{2}{9} = \frac{6}{36} = \frac{1}{6}. Then 25+16=12+530=1730\frac{2}{5} + \frac{1}{6} = \frac{12+5}{30} = \frac{17}{30}.

Final Answer: 1730\frac{17}{30} [1 mark]

6. Arrange in ascending order: 35\frac{3}{5}, 58\frac{5}{8}, 712\frac{7}{12}

Answer: 712\frac{7}{12}, 58\frac{5}{8}, 35\frac{3}{5} [1 mark]

Explanation: Convert to common denominator or decimals. Using decimals: 35=0.6\frac{3}{5} = 0.6, 58=0.625\frac{5}{8} = 0.625, 712≈0.583\frac{7}{12} \approx 0.583. Ascending order: 0.583<0.6<0.6250.583 < 0.6 < 0.625. Using common denominator 120: 35=72120\frac{3}{5}=\frac{72}{120}, 58=75120\frac{5}{8}=\frac{75}{120}, 712=70120\frac{7}{12}=\frac{70}{120}. Order: 70120<72120<75120\frac{70}{120} < \frac{72}{120} < \frac{75}{120}.

7. What is 47\frac{4}{7} of 2116\frac{21}{16}?

Answer: 34\frac{3}{4} [1 mark]

Explanation: "Of" means multiply. 47×2116=84112=34\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 17×213164=34\frac{1}{\cancel{7}} \times \frac{\cancel{21}^{3}}{\cancel{16}_{4}} = \frac{3}{4} (after simplifying: 11×34=34\frac{1}{1} \times \frac{3}{4} = \frac{3}{4}... let me redo: 47×2116\frac{4}{7} \times \frac{21}{16}: 4 goes into 16 four times; 7 goes into 21 three times. So 11×34=34\frac{1}{1} \times \frac{3}{4} = \frac{3}{4}).

8. Express 1824\frac{18}{24} as a fraction with denominator 32.

Answer: 2432\frac{24}{32} [1 mark]

Explanation: First simplify 1824=34\frac{18}{24} = \frac{3}{4}. Then find equivalent: 34=?32\frac{3}{4} = \frac{?}{32}. Since 4×8=324 \times 8 = 32, multiply numerator by 8: 3×8=243 \times 8 = 24. So 2432\frac{24}{32}. Alternatively from original: 1824=x32\frac{18}{24} = \frac{x}{32}, so x=18×3224=57624=24x = \frac{18 \times 32}{24} = \frac{576}{24} = 24.

Section B: Problem Solving

9. Baker's flour problem

Answer: 310\frac{3}{10} kg [2 marks]

Working:

  • Flour used = 25\frac{2}{5} of 34\frac{3}{4} kg = 25×34\frac{2}{5} \times \frac{3}{4} [1 mark]
  • = 620=310\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 34\frac{3}{4} kg and used 25\frac{2}{5} of that amount, not 25\frac{2}{5} kg extra. Simplify 620\frac{6}{20} by dividing by 2.

10. Raj's stickers

Answer: 120 stickers [2 marks]

Working:

  • Given to sister: 58×480=300\frac{5}{8} \times 480 = 300 stickers
  • Remainder: 480−300=180480 - 300 = 180 stickers [1 mark]
  • Given to cousin: 13×180=60\frac{1}{3} \times 180 = 60 stickers
  • Left: 180−60=120180 - 60 = 120 stickers [1 mark]

Alternative: Remainder after sister = 38×480=180\frac{3}{8} \times 480 = 180. Then keep 23\frac{2}{3} of remainder: 23×180=120\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 58\frac{5}{8} away is 38\frac{3}{8} (or 480−300=180480-300=180). Then take 13\frac{1}{3} of this remainder, not of the original.

11. Tank water problem

Answer: 108 litres [2 marks]

Working:

  • Fraction poured out: 56−23=56−46=16\frac{5}{6} - \frac{2}{3} = \frac{5}{6} - \frac{4}{6} = \frac{1}{6} [1 mark]
  • 16\frac{1}{6} of tank = 18 litres
  • Full tank = 18×6=10818 \times 6 = 108 litres [1 mark]

Alternative with algebra: Let full capacity be CC. Then 56C−18=23C\frac{5}{6}C - 18 = \frac{2}{3}C, so 56C−23C=18\frac{5}{6}C - \frac{2}{3}C = 18, thus 16C=18\frac{1}{6}C = 18, C=108C = 108.

Explanation: The difference between two fractional amounts equals 18 litres. Convert to common denominator to subtract fractions. Then use the unitary method: if 16\frac{1}{6} = 18, then whole = 18×618 \times 6.

12. Mdm Tan's money

Answer: $160 [2 marks]

Working:

  • Spent on groceries: 25\frac{2}{5}
  • Remainder: 35\frac{3}{5}
  • Spent on transport: 14\frac{1}{4} of 35=320\frac{3}{5} = \frac{3}{20}
  • Total spent: 25+320=820+320=1120\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, 35\frac{3}{5} remains. This is split: 14\frac{1}{4} on transport, so 34\frac{3}{4} of remainder = 34×35=920\frac{3}{4} \times \frac{3}{5} = \frac{9}{20} left. If 920=72\frac{9}{20} = 72, then 120=8\frac{1}{20} = 8, so whole = 20×8=16020 \times 8 = 160.

Explanation: The "remainder" changes after each spending. Track carefully: first remainder is 35\frac{3}{5} of original. Transport is 14\frac{1}{4} of this remainder, not of original. The final amount left represents 920\frac{9}{20} of original.

13. Peter and John's marbles

Answer: 135 marbles [2 marks]

Working:

  • Peter : John = 3 : 5 (since Peter has 35\frac{3}{5} as many as John) [1 mark]
  • Total parts = 3+5=83 + 5 = 8
  • Peter's marbles = 38×360=135\frac{3}{8} \times 360 = 135 [1 mark]

Alternative: Let John have JJ. Then Peter has 35J\frac{3}{5}J. So J+35J=360J + \frac{3}{5}J = 360, giving 85J=360\frac{8}{5}J = 360, so J=225J = 225, Peter = 360−225=135360 - 225 = 135.

Explanation: "35\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, 08\frac{0}{8} or 0 m left... rechecking: [2 marks]

Working:

  • 214=942\frac{1}{4} = \frac{9}{4} m [0.5 mark]
  • Number of pieces: 94÷38=94×83=7212=6\frac{9}{4} \div \frac{3}{8} = \frac{9}{4} \times \frac{8}{3} = \frac{72}{12} = 6 [1 mark]
  • Leftover: 6×38=188=946 \times \frac{3}{8} = \frac{18}{8} = \frac{9}{4}. Since 6×38=188=228=2146 \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 6×38=946 \times \frac{3}{8} = \frac{9}{4}? Yes, exactly, so no remainder. This is a "measurement division" problem — how many 38\frac{3}{8}s in 94\frac{9}{4}?

15. Mass of fruits

Answer: 5145\frac{1}{4} kg or 214\frac{21}{4} kg [2 marks]

Working:

  • Pumpkin: 412=924\frac{1}{2} = \frac{9}{2} kg
  • Honeydew: 23×92=186=3\frac{2}{3} \times \frac{9}{2} = \frac{18}{6} = 3 kg [1 mark]
  • Watermelon: 74×3=214=514\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 23\frac{2}{3} as heavy as pumpkin" means honeydew = 23×\frac{2}{3} \times pumpkin.

Section C: Multi-Step and Challenging Problems

16(a). Calculate: (34+23)÷56\left(\frac{3}{4} + \frac{2}{3}\right) \div \frac{5}{6}

Answer: 1710\frac{17}{10} or 17101\frac{7}{10} [2 marks]

Working:

  • Brackets first: 34+23=912+812=1712\frac{3}{4} + \frac{2}{3} = \frac{9}{12} + \frac{8}{12} = \frac{17}{12} [1 mark]
  • Then divide: 1712÷56=1712×65=10260=1710\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: 916\frac{9}{16} [2 marks]

Working:

  • Let the number be xx
  • x÷35=1516x \div \frac{3}{5} = \frac{15}{16} [0.5 mark]
  • x=1516×35=4580=916x = \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 35\frac{3}{5} gives 1516\frac{15}{16}, then multiply 1516\frac{15}{16} by 35\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: 25\frac{2}{5} of total
  • Remainder after Amy: 35\frac{3}{5} of total [1 mark]
  • Ben: 38\frac{3}{8} of remainder = 38×35=940\frac{3}{8} \times \frac{3}{5} = \frac{9}{40} of total [1 mark]
  • Cathy's fraction: 1−25−940=40−16−940=1540=381 - \frac{2}{5} - \frac{9}{40} = \frac{40-16-9}{40} = \frac{15}{40} = \frac{3}{8} [1 mark]
  • If 38\frac{3}{8} = 105,thentotal=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: 38×24=9\frac{3}{8} \times 24 = 9 units
  • Cathy: 24−9=1524 - 9 = 15 units = $105
  • 1 unit = $7
  • Total: 7×40=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 38\frac{3}{8} as 38\frac{3}{8} of total instead of 38\frac{3}{8} of remainder.

18(a). Volume of water

Answer: 43,200 cm³ [2 marks]

Working:

  • Full volume: 60×40×30=7200060 \times 40 \times 30 = 72 000 cm³ [1 mark]
  • Water volume: 35×72000=43200\frac{3}{5} \times 72 000 = 43 200 cm³ [1 mark]

Or directly: 35×60×40×30=43200\frac{3}{5} \times 60 \times 40 \times 30 = 43 200 cm³.

Explanation: Volume of cuboid = length × width × height. The fraction 35\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: 30×30=90030 \times 30 = 900 cm² [0.5 mark]
  • Volume = base area × height, so height = 43200900\frac{43 200}{900} [1 mark]
  • = 48 cm [0.5 mark]

Explanation: Conservation of volume — water volume stays constant. The formula V=base×heightV = \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: 1770\frac{17}{70} [4 marks]

Working:

  • Let total pupils = 1 (or 70 for concrete working)
  • Boys: 37\frac{3}{7}, Girls: 47\frac{4}{7} [1 mark]
  • Boys with glasses: 25×37=635\frac{2}{5} \times \frac{3}{7} = \frac{6}{35} [1 mark]
  • Girls with glasses: 14×47=428=17=535\frac{1}{4} \times \frac{4}{7} = \frac{4}{28} = \frac{1}{7} = \frac{5}{35}... recheck: 14×47=428=17=535\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: 635=1270\frac{6}{35} = \frac{12}{70}
  • Girls with glasses: 17=1070\frac{1}{7} = \frac{10}{70}
  • Total with glasses: 12+1070=2270=1135\frac{12+10}{70} = \frac{22}{70} = \frac{11}{35}

Rechecking girls: 14×47=17\frac{1}{4} \times \frac{4}{7} = \frac{1}{7}. Correct.

Total: 635+17=635+535=1135\frac{6}{35} + \frac{1}{7} = \frac{6}{35} + \frac{5}{35} = \frac{11}{35}

Answer: 1135\frac{11}{35} [4 marks]

Working (clean):

  • Boys: 37\frac{3}{7}, Girls: 47\frac{4}{7} [1 mark]
  • Boys with glasses: 25×37=635\frac{2}{5} \times \frac{3}{7} = \frac{6}{35} [1 mark]
  • Girls with glasses: 14×47=17=535\frac{1}{4} \times \frac{4}{7} = \frac{1}{7} = \frac{5}{35} [1 mark]
  • Total with glasses: 635+535=1135\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 (47\frac{4}{7}), then applying 14\frac{1}{4} to that subgroup. Common denominitors needed for final addition.

20(a). Fraction given away

Answer: 3780\frac{37}{80} [2 marks]

Working:

  • Apples given away: 25×58=1040=14\frac{2}{5} \times \frac{5}{8} = \frac{10}{40} = \frac{1}{4} of total [0.5 mark]
  • Oranges: 38\frac{3}{8} of total (since 1−58=381 - \frac{5}{8} = \frac{3}{8})
  • Oranges given away: 34×38=932\frac{3}{4} \times \frac{3}{8} = \frac{9}{32} of total [0.5 mark]
  • Total given away: 14+932=832+932=1732\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 = 58\frac{5}{8}, oranges = 38\frac{3}{8}.

  • Apples given: 25×58=28=14\frac{2}{5} \times \frac{5}{8} = \frac{2}{8} = \frac{1}{4}
  • Oranges given: 34×38=932\frac{3}{4} \times \frac{3}{8} = \frac{9}{32}
  • Total given: 832+932=1732\frac{8}{32} + \frac{9}{32} = \frac{17}{32}

Answer: 1732\frac{17}{32} [2 marks]

20(b). Original number of fruits

Answer: 288 fruits [2 marks]

Working:

  • Fraction left = 1−1732=15321 - \frac{17}{32} = \frac{15}{32} [0.5 mark]
  • Or: Apples left = 35×58=38\frac{3}{5} \times \frac{5}{8} = \frac{3}{8}; Oranges left = 14×38=332\frac{1}{4} \times \frac{3}{8} = \frac{3}{32}. Total left = 1232+332=1532\frac{12}{32} + \frac{3}{32} = \frac{15}{32} [0.5 mark]
  • If 1532\frac{15}{32} = 90, then total = 90×3215=6×32=19290 \times \frac{32}{15} = 6 \times 32 = 192... recheck: 90÷15=690 \div 15 = 6, 6×32=1926 \times 32 = 192.

Wait: 192×1532=192÷32×15=6×15=90192 \times \frac{15}{32} = 192 \div 32 \times 15 = 6 \times 15 = 90. ✓

Answer: 192 fruits [2 marks]

Working (clean):

  • Fraction left: 1532\frac{15}{32} [0.5 mark]
  • Total = 90÷1532=90×3215=19290 \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: 1532\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 25\frac{2}{5} and 34\frac{3}{4} apply to the same base fraction of total.

END OF ANSWER KEY