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P6 Fractions & Decimals Questions (with Worked Solutions)

P6 prelim questions on fractions and decimals — the four operations, conversions, and word problems. Every question is from a real Singapore P6 preliminary exam, marked instantly, with step-by-step solutions.

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Example Fractions & Decimals questions

A few real fractions & decimals questions from P6 prelim papers. The full set — with marking and progress tracking — is inside SG Maths Exam.

Tao Nan 2025 · Paper 2 · Q8 · 3 marks
After Aini gave $\frac{1}{6}$ of her stickers away and Bala gave $\frac{1}{4}$ of his stickers away, both of them had the same number of stickers left. The total number of stickers Aini and Bala gave away was 2280. How many stickers did Bala have at first?
Worked solution
  1. Aini: gave $\frac{1}{6} = \frac{3}{18}$
  2. Bala: gave $\frac{1}{4} = \frac{5}{20}$. Left equal: Aini $\frac{15}{18}$, Bala $\frac{15}{20}$ -> same numerator. Gave: 3 units (Aini) + 5 units (Bala) where 3 units = $\frac{3}{18}$ of Aini total. Setting equal numerator method: Aini total 18 units, gave 3 units
  3. Bala total 20 units, gave 5 units. 3 + 5 = 8 units = 2280
  4. 1 unit = 285
  5. Bala = 20 units = 5700.
Nanyang 2025 · Paper 2 · Q8 · 3 marks
Mr Soh spent $\frac{1}{5}$ of his money on a pair of shoes and $\frac{3}{4}$ of his remaining money on a bag, a shirt and a hat. The bag cost 3 times as much as the hat. The shirt cost $\frac{1}{2}$ as much as the hat. The pair of shoes cost $\$48$ more than the shirt. How much money did Mr Soh have at first?
Worked solution
  1. Let total money $= 20$ units. Shoes $= \frac{1}{5} \times 20 = 4$ units. Remaining $= 16$ units. Spent on bag, shirt, hat $= \frac{3}{4} \times 16 = 12$ units. Let the hat $= 2$ parts
  2. bag $= 6$ parts
  3. shirt $= 1$ part
  4. total $= 9$ parts $= 12$ units. The shirt $= \frac{1}{9} \times 12 = \frac{4}{3}$ units. Shoes $-$ shirt $= 4 - \frac{4}{3} = \frac{8}{3}$ units $= \$48$, so $1$ unit $= \$18$. Total $= 20 \times \$18 = \$360$.
ACS Junior 2025 · Paper 1 Booklet A · Q13 · 2 marks
The table below shows the rates for renting a bicycle at a shop Daniel rented a bicycle at 8.20 a.m. What is the latest time Daniel must return his bicycle if he only has \$20 to spend on rental fees?
Manual crop
  • 1. 10.20 a.m.
  • 2. 10.50 a.m.
  • 3. 11.20 a.m.
  • 4. 11.50 a.m.
Worked solution
  1. \(\text{Budget} = \$20\)
  2. \(\text{First hour} = \$6\)
  3. \(\text{Remaining} = \$20 - \$6 = \$14\)
  4. \(\text{Additional 30-min blocks} = \$14 \div \$4 = 3 \text{ full blocks (= 90 min)}\)
  5. \(\text{Total time} = 1 \text{ h} + 90 \text{ min} = 2 \text{ h } 30 \text{ min}\)
  6. \(\text{Latest return} = 8.20 \text{ a.m.} + 2 \text{ h } 30 \text{ min} = 10.50 \text{ a.m.}\)
ACS Junior 2025 · Paper 1 Booklet A · Q15 · 2 marks
Mollie had a container full of flour. Nellie and Ollie each had $\frac{2}{7}$ of what Mollie had. Mollie gave away some flour to Nellie and Ollie so that all 3 of them have the same amount of flour in the end. What fraction of flour did Mollie give away?
  • 1. $\frac{5}{7}$
  • 2. $\frac{5}{14}$
  • 3. $\frac{5}{21}$
  • 4. $\frac{10}{21}$
Worked solution
  1. \(\text{Let Mollie's flour} = m\)
  2. \(\text{Nellie} = \tfrac{2}{7}m,\ \text{Ollie} = \tfrac{2}{7}m\)
  3. \(\text{Total} = m + \tfrac{2}{7}m + \tfrac{2}{7}m = \tfrac{11}{7}m\)
  4. \(\text{Each ends with} = \tfrac{1}{3} \times \tfrac{11}{7}m = \tfrac{11}{21}m\)
  5. \(\text{Mollie gave away} = m - \tfrac{11}{21}m = \tfrac{10}{21}m\)
  6. \(\text{Fraction} = \tfrac{10}{21}\)
Catholic High 2025 · Paper 1 Booklet A · Q12 · 2 marks
Alex and Ben had \$190 altogether at first. After Alex gave Ben \$30, Alex had \$40 more than Ben. How much did Ben have at first?
  • 1. $45
  • 2. $75
  • 3. $120
  • 4. $160
Worked solution
  1. \(\text{Let Ben have } \$b\text{; Alex had } \$(190 - b)\)
  2. \(\text{After Alex gives Ben } \$30: \text{Alex} = 160 - b,\ \text{Ben} = b + 30\)
  3. \(\text{Alex has } \$40 \text{ more}: (160 - b) - (b + 30) = 40\)
  4. \(130 - 2b = 40\)
  5. \(2b = 90\)
  6. \(b = 45\)
Catholic High 2025 · Paper 1 Booklet A · Q13 · 2 marks
Arrange these volumes from the smallest to the largest. 2.25 l $2\frac{2}{5}$ l 2 l 225 ml
Manual crop
  • 1. $2\frac{2}{5}$ l 2.25 l 2 l 225 ml
  • 2. $2\frac{2}{5}$ l 2 l 225 ml 2.25 l
  • 3. 2 l 225 ml 2.25 l $2\frac{2}{5}$ l
  • 4. 2.25 l 2 l 225 ml $2\frac{2}{5}$ l
Worked solution
  1. \(2.25 \text{ l} = 2.250 \text{ l}\)
  2. \(2\frac{2}{5} \text{ l} = 2.4 \text{ l}\)
  3. \(2 \text{ l } 225 \text{ ml} = 2.225 \text{ l}\)
  4. \(\text{Smallest to largest}: 2.225 < 2.250 < 2.4\)
  5. \(\text{So: } 2 \text{ l } 225 \text{ ml},\ 2.25 \text{ l},\ 2\frac{2}{5} \text{ l}\)

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