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Primary 6 · Fractions · Question type 2.3

P6 Fractions: Fraction of the Remainder Problems

The second fraction is taken from the remainder, which is a new and smaller whole, so the two fractions cannot be added. Identify what is left after the first change before finding a fraction of that amount.

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Primary 6, on the MOE Primary Mathematics syllabus.

In the Fractions chapter7 question types · see the chapter

The lesson · 2 minutes 29 seconds

How do you solve P6 fraction-of-the-remainder problems?

Watch: 1/3 of the stickers given away, then 2/5 of what is left

Another worked example

A worked example

Kumar spent 1/3 of his money on a badminton racket. He then spent 3/8 of the remainder on a pair of shoes. The racket cost $46 more than the shoes. How much money did Kumar have left in the end?

The working · Nine lines

The step-by-step working

Written out in full, the way it should appear on paper.

  1. the bar = 3 × 8 = 24 units
  2. the racket = 1/3 of 24 units, so 8 units
  3. fraction left after the racket = 1 - 1/3 = 2/3 of 24 units, so 16 units
  4. the shoes = 3/8 of 16 units, so 6 units
  5. the racket cost 8 - 6 = 2 units more than the shoes
  6. 2 units = $46
  7. 1 unit = $46 ÷ 2 = $23
  8. he had 16 - 6 = 10 units left
  9. 10 units = 10 × $23 = $230

The answer is $230.

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Find the notebooks at first

Draw your own model and find the amount at first.

Mrs Lim had some notebooks. She gave 1/5 of them to the library and 1/3 of the remainder to her class. She had 56 notebooks left. How many notebooks did she have at first?

Show the step-by-step solution
  1. take the denominators of the two fractions, 5 and 3
  2. the bar = 5 × 3 = 15 units
  3. fraction remaining after the library = 1 - 1/5 = 4/5
  4. 4/5 × 15 = 12, so 12 units are left after the library
  5. fraction of that remainder left = 1 - 1/3 = 2/3
  6. 2/3 × 12 = 8, so 8 units are left after her class
  7. 8 units = 56
  8. 1 unit = 56 ÷ 8 = 7
  9. 15 units = 15 × 7 = 105

Ans: 105

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