Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Find the answer

1. (7 points) 45 candies cost as many rubles as can be bought for 20 rubles. How many candies can be bought for 50 rubles?

Answer: 75 candies.

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Solution. Let $x$ be the cost of one candy in rubles. Then $45 x=\frac{20}{x}$ , from which $x=\frac{2}{3}$ . Then with 50 rubles, one can buy $\frac{50}{x}=75$ candies.

Criteria. Any correct solution: 7 points.

The equation $45 x=\frac{20}{x}$ is correctly set up, but an arithmetic error is made in solving it or in subsequent steps: 5 points.

In the solution, it is claimed that the price of one candy is $\frac{2}{3}$ , it is verified that this price fits the condition of the problem, and the correct answer is obtained: 4 points.

Only the correct answer is provided: 1 point.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.