Maths olympiad question · P6★★★★ Expert
Ratios: olympiad question 8
An expert question on ratios for P6, with a step-by-step solution and the answer. Ratio questions appear in the Hua Xia Cup's P4 to P6 first rounds, qualifying rounds and finals, such as linked ratios and ratios that change.
The question
A schools' 3000 m race is run on a 400 m track.
Ho-yin, Tsz-ching and Ka-lok start together from the same spot, each at a steady speed to the end.
When Ho-yin finishes, Tsz-ching still has 600 m to go; when Tsz-ching finishes, Ka-lok still has 750 m to go.
During the race, when Ho-yin first catches Ka-lok from behind (a lap ahead), how far has Ka-lok run?
Show the hints (3)
- Hint 1: In the same time, distance is in proportion to speed. When Ho-yin runs 3000 m, how far has Tsz-ching run? And Ka-lok when Tsz-ching runs 3000 m?
- Hint 2: Chain the two speed ratios to get Ka-lok : Ho-yin. "A lap ahead" means their distances differ by 400 m.
- Hint 3: Ka-lok : Ho-yin = 3 : 5. A gap of 2 parts is 400 m.
Worked solution
Have a go first, then read the solution step by step.
- 1.When Ho-yin runs 3000 m, Tsz-ching runs 3000 − 600 = 2400 m, so Tsz-ching : Ho-yin = 2400 : 3000. Likewise Ka-lok : Tsz-ching = 2250 : 3000.
- 2.Chained: Ka-lok : Ho-yin = (2400 × 2250) : (3000 × 3000) = 3 : 5.
- 3.In the same time Ka-lok runs 3 parts and Ho-yin 5, 2 parts apart. At the first catch Ho-yin is exactly one lap, 400 m, ahead, so 1 part = 400 ÷ 2 = 200 m.
- 4.Ka-lok has run 3 × 200 = 600 m (Ho-yin 1000 m, short of the finish, so it fits).
- 5.Check: 1000 − 600 = 400 ✓; 600 : 1000 = 3 : 5 ✓
- ⚠️Common mistake: thinking Ka-lok is 600 + 750 m short when Ho-yin finishes, adding the gaps. Tsz-ching's speed differs; chain the ratios, don't add.
Answer: 600 m
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