Olympiad topic · P4–P6
Fraction problems
Common in the Hua Xia Cup's P5 and P6 qualifying rounds and finals; harder ones work backwards through fractions, or need clever shortcuts with fractions.
Try one
★★ MediumIn Class 6C, a quarter of the pupils are boys, and there are 20 more girls than boys.
How many pupils are there in Class 6C?
- Hint 1: Think of the class as 4 equal parts, with the boys taking 1. How many parts are girls?
- Hint 2: How many more parts are girls than boys? Those parts are the 20 pupils.
- 1.Think of the class as 4 equal parts: boys 1, girls 4 − 1 = 3.
- 2.Line them up: the girls have 3 − 1 = 2 more parts, and that is the 20 more.
- 3.One part = 20 ÷ 2 = 10.
- 4.The class is 4 parts: 10 × 4 = 40 pupils.
- ⚠️Common mistake: treating 20 as the number of girls. It is only how many more girls there are than boys: 2 parts.
Answer: 40 pupils
Four levels, step by step
- ★EasyFree2 question types · unlimited questionsA part, or the whole from a part.
- ★★MediumFree2 question types · unlimited questionsA fraction of what's left.
- ★★★HardFree2 question types · unlimited questions"a sixth less 18 pages", two people spending the same.
- ★★★★Expert (competition level)VIP5 question types · unlimited questionsFractions of what's left again and again, whole amounts with a few extra sold, fraction sums that split and cancel (three factors, two factors, plus and minus by turns).
Fraction problems: what it is and how to solve it
Fractions as "this is so much of that": find a part, or the whole.
For example: A 54 cm ribbon; two ninths of it is cut off. How long is that piece? 54 ÷ 9 × 2 = 12 cm.
The key idea: First find which amount is "the whole", and whether it says "of all" or "of what's left". A part = whole × fraction; the whole = part ÷ fraction. Olympiad books call the whole "the unit 1"; the usual method draws it split into equal parts — the same idea.
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