Maths olympiad question · P6★★★★ Expert
Percentages and mixtures: olympiad question 10
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
Cups A, B and C of sugar water are 25%, 20% and 26%, and cup B holds 100 g.
Mixing all of A and B would give 24%; mixing all of B and C would give 24%.
Now all three are mixed and some plain water is added, making it 8%.
How many grams of water were added?
Show the hints (3)
- Hint 1: A and B make 24%: how far is each cup from 24%? What one has over, the other is short — they balance.
- Hint 2: That gives cup A's weight; the B-and-C mixture at 24% gives cup C's the same way.
- Hint 3: With cups A's and C's weights, work out how much sugar the three hold. Adding water doesn't change that: the weight after = the sugar ÷ 8%, less the three cups' weight before.
Worked solution
Have a go first, then read the solution step by step.
- 1.A and B make 24%: A is 1 point above and B 4 below, and they balance: A × 1 = 100 × 4, so A = 400 ÷ 1 = 400 g.
- 2.B and C make 24%: C × 2 = 100 × 4, so C = 400 ÷ 2 = 200 g.
- 3.The sugar in all three: 400 × 25% + 100 × 20% + 200 × 26% = 100 + 20 + 52 = 172 g, in 700 g.
- 4.Water adds no sugar; at 8% the whole is 172 ÷ 8% = 2150 g, so 2150 − 700 = 1450 g of water went in.
- 5.Check: A and B, (100 + 20) ÷ 500 = 24% ✓; B and C, (20 + 52) ÷ 300 = 24% ✓
- ⚠️Common mistake: treating the cups as equal and averaging the three, or averaging 24% and 24%. Mixing different weights, divide the total sugar by the total weight.
Answer: 1450 g
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