Maths olympiad question · P6★★★★ Expert
Percentages and mixtures: olympiad question 9
An expert question on percentages and mixtures for P6, with a step-by-step solution and the answer. Common in the Hua Xia Cup's P6 first round and finals: discounts, a rise followed by a fall, and mixtures.
The question
A bottle holds 90 g of sugar water at 40%.
First some grams are poured out and the same weight of plain water added, then stirred; then the same weight is poured out again and replaced with water again.
The sugar water is now 10%.
How many grams were poured out each time?
Show the hints (3)
- Hint 1: Pouring some out, what fraction of the sugar goes? Does adding water change the sugar?
- Hint 2: The bottle always holds 90 g, so the concentration follows the sugar. Pouring the same amount each time keeps the same fraction of the sugar each time.
- Hint 3: After two pours the sugar is 10/40 = 1/4 of the start. Which fraction times itself gives 1/4?
Worked solution
Have a go first, then read the solution step by step.
- 1.Each pour takes the sugar out in proportion; the water added changes no sugar, and the bottle is always 90 g. So the concentration follows the sugar.
- 2.Say each pour keeps r of the sugar (r = what stays ÷ 90). After two pours it is r × r of the start, and so is the concentration: r × r = 10 ÷ 40 = 1/4.
- 3.1/2 × 1/2 = 1/4, so r = 1/2: each time 90 × 1/2 = 45 g stays.
- 4.Each pour takes 90 − 45 = 45 g.
- 5.Check: the sugar starts at 90 × 40% = 36 g; after one pour 36 × 1/2 = 18 g, after two 9 g, and 9 ÷ 90 = 10% ✓
- ⚠️Common mistake: thinking the concentration drops 40 − 10 = 30 points in two equal steps. Each pour removes less sugar than the last: the concentration is multiplied by the same fraction, not reduced by the same amount.
Answer: 45 g
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