Primary 6 PSLE Mathematics Fractions Quiz

Primary 6 PSLE Mathematics Quiz - Fractions — Answer Key

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Section A: Direct Calculation (5 marks)

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

Working: $\frac{3}{4} \div 6 = \frac{3}{4} \times \frac{1}{6} = \frac{3}{24} = \frac{1}{8}$

Key concept: Dividing by a whole number is the same as multiplying by its reciprocal ($\frac{1}{6}$). Always simplify the final answer.

Common mistake: Forgetting to flip the whole number or not simplifying.

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

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

Working: $\frac{5}{8} \div \frac{1}{4} = \frac{5}{8} \times 4 = \frac{20}{8} = \frac{5}{2} = 2\frac{1}{2}$

Key concept: Dividing by a fraction = multiplying by its reciprocal. $\frac{1}{4}$ flipped becomes $4$.

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

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3. Calculate $12 \div \frac{3}{5}$

Working: $12 \div \frac{3}{5} = 12 \times \frac{5}{3} = \frac{60}{3} = 20$

Key concept: Whole number ÷ fraction = whole number × reciprocal of fraction.

Answer: $20$ [1]

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4. Calculate $\frac{7}{9} \div \frac{2}{3}$

Working: $\frac{7}{9} \div \frac{2}{3} = \frac{7}{9} \times \frac{3}{2} = \frac{21}{18} = \frac{7}{6} = 1\frac{1}{6}$

Key concept: Multiply by reciprocal, then simplify by finding common factors. Here, $21$ and $18$ share factor $3$.

Answer: $\frac{7}{6}$ or $1\frac{1}{6}$ [1]

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5. Calculate $\frac{4}{5} \div 8 \times \frac{2}{3}$

Working: $\frac{4}{5} \div 8 \times \frac{2}{3} = \frac{4}{5} \times \frac{1}{8} \times \frac{2}{3} = \frac{8}{120} = \frac{1}{15}$

Key concept: For division and multiplication, work left to right. Simplify before multiplying: $\frac{4}{5} \times \frac{1}{8} = \frac{1}{10}$, then $\frac{1}{10} \times \frac{2}{3} = \frac{2}{30} = \frac{1}{15}$.

Answer: $\frac{1}{15}$ [1]

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Section B: Word Problems — Short Response (10 marks)

6. What fraction of original cookies went into each box?

Working:

• Given away: $\frac{1}{3} \times 48 = 16$ cookies
• Remaining: $48 - 16 = 32$ cookies
• Per box: $32 \div 4 = 8$ cookies
• Fraction of original: $\frac{8}{48} = \frac{1}{6}$

Alternative (fraction method):

• Remaining fraction: $1 - \frac{1}{3} = \frac{2}{3}$
• Fraction per box: $\frac{2}{3} \div 4 = \frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6}$

Key concept: "Of remainder" problems — find what remains first, then divide equally.

Answer: $\frac{1}{6}$ [2]

Marking: Method to find remainder (1), correct answer (1)

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7. Capacity of tank

Working:

• Difference: $\frac{5}{6} - \frac{2}{3} = \frac{5}{6} - \frac{4}{6} = \frac{1}{6}$
• So $\frac{1}{6}$ of tank = 9 litres
• Full tank: $9 \times 6 = 54$ litres

Key concept: The difference in fractions equals the actual amount used. This connects fraction to concrete measurement.

Answer: $54$ litres [2]

Marking: Find fraction difference (1), find whole (1)

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8. Leftover ribbon length

Working:

• Number of pieces: $\frac{7}{8} \div \frac{1}{4} = \frac{7}{8} \times 4 = \frac{28}{8} = 3\frac{1}{2}$
• So 3 whole pieces can be cut, with $\frac{1}{2}$ of a piece remaining
• Leftover: $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$ m

Alternative:

• Length used for 3 pieces: $3 \times \frac{1}{4} = \frac{3}{4}$ m
• Leftover: $\frac{7}{8} - \frac{3}{4} = \frac{7}{8} - \frac{6}{8} = \frac{1}{8}$ m

Key concept: Division gives how many divisors fit. The decimal/whole number part tells complete pieces; fractional remainder needs conversion back to actual length.

Answer: $\frac{1}{8}$ m [2]

Marking: Find number of pieces or equivalent (1), find actual leftover (1)

Common mistake: Stopping at "$3\frac{1}{2}$ pieces" without converting back to metres.

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9. Fraction of stock left after Tuesday

Working:

• After Monday: $1 - \frac{2}{5} = \frac{3}{5}$ remains
• Tuesday sold: $\frac{1}{2} \times \frac{3}{5} = \frac{3}{10}$
• Left: $\frac{3}{5} - \frac{3}{10} = \frac{6}{10} - \frac{3}{10} = \frac{3}{10}$

Alternative:

• After Tuesday, $\frac{1}{2}$ of remainder left: $\frac{1}{2} \times \frac{3}{5} = \frac{3}{10}$

Key concept: "Of remainder" — each fraction operates on what's left, not the original. Sequential multiplication works for finding what's left directly.

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

Marking: Correct operation on remainder (1), correct answer (1)

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10. Water level height

Working:

• Water poured in 8 minutes: $\frac{3}{4} \times 8 = 6$ litres = $6000 \text{ cm}^3$
• Volume = base area × height: $6000 = 240 \times h$
• $h = 6000 \div 240 = 25$ cm

Key concept: Connected to volume of cuboids (P6 syllabus). Unit conversion essential — litres to $\text{cm}^3$.

Answer: $25$ cm [2]

Marking: Volume calculation (1), height calculation (1)

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Section C: Word Problems — Long Response (20 marks)

11. Mrs Lim's money

(a) Fraction spent on shoes

Working:

• After handbag: $1 - \frac{2}{5} = \frac{3}{5}$ remains
• Shoes: $\frac{1}{3} \times \frac{3}{5} = \frac{1}{5}$

(b) Original amount

Working:

• Fraction left: $\frac{3}{5} - \frac{1}{5} = \frac{2}{5}$ (or $\frac{2}{3}$ of $\frac{3}{5} = \frac{2}{5}$)
• $\frac{2}{5}$ of total = $160
• Total: $160 \div \frac{2}{5} = 160 \times \frac{5}{2} = 400$

Key concept: Classic "fraction of remainder" — track changing base carefully. Each step's "whole" is different.

(a) Answer: $\frac{1}{5}$ [2]

(b) Answer: \400$ [3]

Marking (a): Find remainder fraction (1), find shoes fraction (1)

Marking (b): Find final remainder fraction (1), set up equation (1), solve (1)

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12. Ahmad and Ben's money

Let original amount for each = 1 unit

(a) Ahmad's remaining fraction

Working:

• After giving to Ben: $1 - \frac{1}{4} = \frac{3}{4}$
• After giving to sister: $\frac{3}{4} - \frac{1}{3} \times \frac{3}{4} = \frac{3}{4} \times \frac{2}{3} = \frac{1}{2}$

Or: $\frac{3}{4} \times \frac{2}{3} = \frac{1}{2}$ (keeps $\frac{2}{3}$ of remainder)

(b) Ben's final fraction of total

Working:

• Ben receives: $\frac{1}{4}$
• Total = 2 units (since equal at start)
• Ben's final: $1 + \frac{1}{4} = \frac{5}{4}$ of his original, but as fraction of total:
• Ahmad's final: $\frac{1}{2}$, Ben's final: $1 + \frac{1}{4} = \frac{5}{4}$. Check: Total relative = $1+1=2$
• Ben: $\frac{5}{4}$ out of "2 units" where 1 unit = original each... Let me use common denominator.

Clearer approach:

• Let each have $12$ units (LCM of 4 and 3)
• Ahmad gives Ben 3 units, keeps 9
• Ahmad gives sister $\frac{1}{3}$ of 9 = 3, keeps 6
• Ben has $12 + 3 = 15$
• Total: $6 + 15 = 21$... wait, sister has 3. Total should be 24.

Actually "total amount of money" means what Ahmad and Ben have together.

• Ahmad final: 6 units, Ben final: 15 units
• Total: 21 units... but original was 24.

Let's recalculate: Sister is external, so money leaves the pair.

• Original pair total: 24 units
• After giving to sister (not in pair): pair has $6 + 15 = 21$ units

Actually the question says "total amount of money" — typically means the original total or current total? Usually interpreted as original total.

Ben's fraction of original total: $\frac{15}{24} = \frac{5}{8}$

Or if "total" means what they have now: $\frac{15}{21} = \frac{5}{7}$

Given typical PSLE conventions, "in the end, what fraction of the total amount" = fraction of original total.

(a) Answer: $\frac{1}{2}$ [2]

(b) Answer: $\frac{5}{8}$ [3]

Marking (a): Find remainder after first gift (1), find final fraction (1)

Marking (b): Track Ben's amount (1), determine total reference (1), correct fraction (1)

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13. Baker's tarts

(a) Fraction packed

Working:

• Morning: sold $\frac{3}{8}$, so $\frac{5}{8}$ remains
• Afternoon: sold $\frac{2}{5}$ of $\frac{5}{8} = \frac{1}{4}$, so $\frac{3}{5}$ of $\frac{5}{8} = \frac{3}{8}$ remains
• Packed: $\frac{3}{8}$ of total

Check: $\frac{3}{8} + \frac{1}{4} + \frac{3}{8} = \frac{3+2+3}{8} = 1$ ✓

(b) Total tarts

Working:

• Packed tarts: $6 \times 15 = 90$
• This is $\frac{3}{8}$ of total
• Total: $90 \div \frac{3}{8} = 90 \times \frac{8}{3} = 240$

(a) Answer: $\frac{3}{8}$ [2]

(b) Answer: $240$ [3]

Marking (a): Track remainder correctly (1), final fraction (1)

Marking (b): Find packed amount (1), set up equation (1), solve (1)

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14. Chen's marbles

(a) Fraction kept

Working:

• After losing: $1 - \frac{1}{5} = \frac{4}{5}$ remains
• After giving to brother: $\frac{1}{4}$ given away, so $\frac{3}{4}$ of $\frac{4}{5} = \frac{3}{5}$ kept

(b) Original marbles

Working:

• $\frac{3}{5}$ of original = 12
• Original: $12 \div \frac{3}{5} = 12 \times \frac{5}{3} = 20$

(a) Answer: $\frac{3}{5}$ [2]

(b) Answer: $20$ [3]

Marking (a): Apply sequential fractions (1), correct final fraction (1)

Marking (b): Link fraction to amount (1), division method (1), accuracy (1)

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15. Mei and Nina's stickers

Given: Ratio $5:3$. Let Mei = 5 units, Nina = 3 units.

(a) New ratio

Working:

• Mei gives: $\frac{1}{5} \times 5 = 1$ unit to Nina
• Mei now: 4 units, Nina now: 4 units
• Nina gives: $\frac{1}{4} \times 4 = 1$ unit back to Mei
• Mei now: 5 units, Nina now: 3 units

Wait — let me recheck: $\frac{1}{4}$ of Nina's new total.

• After first transfer: Mei = 4, Nina = 4
• Nina gives $\frac{1}{4}$ of 4 = 1 to Mei
• Final: Mei = 5, Nina = 3

Actually same ratio! Let me verify with different numbers or re-read.

Ah, "Nina then gave $\frac{1}{4}$ of her new total back to Mei."

So: Mei = 5, Nina = 3

• Mei gives 1 to Nina: Mei = 4, Nina = 4
• Nina gives $\frac{1}{4}$ of 4 = 1 to Mei: Mei = 5, Nina = 3

Ratio is $5:3$ again. This seems like a trick question or I need to check.

Actually, let me re-read: "Mei gave $\frac{1}{5}$ of her stickers to Nina"

If ratio is 5:3, Mei has 5 parts. $\frac{1}{5}$ of 5 = 1 part. Mei: 4, Nina: 4 (since she had 3, gets 1)

Then "Nina gave $\frac{1}{4}$ of her new total back to Mei" Nina has 4, gives 1 to Mei. Mei: 5, Nina: 3. Back to start.

Hmm, this seems trivial. Let me re-interpret: perhaps "Nina then gave $\frac{1}{4}$ of her original" or the problem is testing observation. Given PSLE style, maybe it's intentional — or I should change my interpretation.

Actually re-checking: if ratio is 5:3 and Mei gives $\frac{1}{5}$ of her stickers, then:

• Mei: 5u - 1u = 4u, Nina: 3u + 1u = 4u
• Nina gives $\frac{1}{4}$ of her total (4u) = 1u to Mei
• Mei: 5u, Nina: 3u

The ratio cycles back. For a more interesting problem, perhaps interpret as $\frac{1}{4}$ of what Nina received, or the problem is correct as stated to test careful reading.

For exam purposes, I'll state clearly:

(a) The new ratio is $5:3$ (same as original; the operations are inverses).

(b) Nina had 75 × $\frac{3}{5}$ = 45 at first, so 45 in end (or 3u = 45).

Wait: "If Mei had 75 stickers at first" — 5 units = 75, so 1 unit = 15. Nina at first: 3 × 15 = 45. In end: 3 × 15 = 45.

(a) Answer: $5:3$ [3]

(b) Answer: $45$ [2]

Marking (a): Correct transfers (2), simplified ratio (1)

Marking (b): Use ratio unit (1), correct answer (1)

Note to teacher: This question demonstrates that fraction operations can restore original states. Students should verify their answer makes sense.

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Section D: Challenging Problems (25 marks)

16. Raj's money

(a) Fraction spent on gift

Working:

• After food: $1 - \frac{1}{3} = \frac{2}{3}$ remains
• After book: $\frac{1}{2}$ of $\frac{2}{3} = \frac{1}{3}$ spent, so $\frac{1}{3}$ remains; or $\frac{2}{3} \times \frac{1}{2} = \frac{1}{3}$ spent
• Gift: $\frac{3}{4}$ of $\frac{1}{3} = \frac{1}{4}$

(b) Original amount

Working:

• After gift: $\frac{1}{4}$ of $\frac{1}{3} = \frac{1}{12}$ remains
• $\frac{1}{12}$ of original = $15
• Original: $15 × 12 =$180

(a) Answer: $\frac{1}{4}$ [2]

(b) Answer: \180$ [3]

Marking (a): Sequential tracking (1), answer (1)

Marking (b): Find final remainder fraction (1), set up equation (1), solve (1)

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17. Fraction wearing glasses

Working:

• Let total pupils = 1 (or LCM of 7, 5, 3 = 105)
• Boys: $\frac{3}{7}$, Girls: $\frac{4}{7}$
• Boys with glasses: $\frac{2}{5} \times \frac{3}{7} = \frac{6}{35}$
• Girls with glasses: $\frac{1}{3} \times \frac{4}{7} = \frac{4}{21}$
• Total glasses: $\frac{6}{35} + \frac{4}{21} = \frac{18+20}{105} = \frac{38}{105}$

With 105 pupils:

• Boys: 45, Girls: 60
• Boys with glasses: 18, Girls with glasses: 20
• Total: 38 out of 105 = $\frac{38}{105}$

Key concept: Different fractions have different bases (of boys vs of girls). Cannot add directly.

Answer: $\frac{38}{105}$ [5]

Marking: Find girls fraction (1), boys with glasses (1), girls with glasses (1), common denominator (1), correct sum (1)

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18. Oil container

(a) Oil at first

Working:

• Difference: $\frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6}$ of container
• Oil removed: $8 \times \frac{1}{4} = 2$ litres
• So $\frac{1}{6}$ of container = 2 litres
• At first: $\frac{2}{3}$ of container = $2 \times 4 = 8$ litres

(b) Bottles needed to fill

Working:

• Full container: $2 \times 6 = 12$ litres
• Currently: $\frac{1}{2} \times 12 = 6$ litres (or from half full)
• Need to add: $12 - 6 = 6$ litres
• Bottles: $6 \div \frac{1}{4} = 6 \times 4 = 24$

(a) Answer: $8$ litres [3]

(b) Answer: $24$ [2]

Marking (a): Fraction difference (1), link to actual amount (1), calculate original (1)

Marking (b): Find current/full amount (1), bottles calculation (1)

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19. Alice, Ben, Claire sharing

(a) Claire's fraction

Working:

• Alice: $\frac{2}{5}$
• Remainder: $\frac{3}{5}$
• Ben: $\frac{3}{4} \times \frac{3}{5} = \frac{9}{20}$
• Claire: $1 - \frac{2}{5} - \frac{9}{20} = \frac{20-8-9}{20} = \frac{3}{20}$

(b) Total amount

Working:

• $\frac{3}{20}$ of total = $45
• Total: $45 \div \frac{3}{20} = 45 \times \frac{20}{3} = 300$

(c) Alice's new fraction

Working:

• Ben's share: $\frac{9}{20} \times 300 =$135
• Ben gives to Alice: $\frac{1}{3} \times 135 =$45
• Alice's new: $\frac{2}{5} \times 300 + 45 = 120 + 45 = 165$
• As fraction: $\frac{165}{300} = \frac{11}{20}$

(a) Answer: $\frac{3}{20}$ [2]

(b) Answer: \300$ [2]

(c) Answer: $\frac{11}{20}$ [2]

Marking (a): Find Ben's fraction (1), Claire's fraction (1)

Marking (b): Set up equation (1), solve (1)

Marking (c): Calculate Alice's new amount (1), express as fraction (1)

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20. David's stamps

(a) Fraction given to sister

Working:

• After brother: $1 - \frac{1}{5} = \frac{4}{5}$
• To sister: $\frac{1}{4} \times \frac{4}{5} = \frac{1}{5}$ of original

(b) Original stamps

Working:

• After both gifts: $\frac{4}{5} - \frac{1}{5} = \frac{3}{5}$ remains (or $\frac{3}{4} \times \frac{4}{5} = \frac{3}{5}$)
• He bought 36, ends with $\frac{3}{2}$ of original
• Let original = $x$
• Equation: $\frac{3}{5}x + 36 = \frac{3}{2}x$
• $36 = \frac{3}{2}x - \frac{3}{5}x = \frac{15-6}{10}x = \frac{9}{10}x$
• $x = 36 \times \frac{10}{9} = 40$

Verification:

• Original: 40
• After brother: 32, after sister: 24
• Bought 36: $24 + 36 = 60$
• $\frac{3}{2} \times 40 = 60$ ✓

(a) Answer: $\frac{1}{5}$ [2]

(b) Answer: $40$ [4]

Marking (a): Sequential fraction (1), simplified answer (1)

Marking (b): Set up remaining fraction (1), create equation (1), solve equation (1), verification/reasonableness (1)

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TOTAL: 60 marks