2018/19 AHSMC Part 1 Problem 5

Problem 5 of 16

Pete has $\$x$ in his pocket. If he buys two apples using his $\$x$ he will have $\$3$ in change remaining. If he buys three bananas using his $\$x$ he will have $\$2$ in change remaining. Similarly, if he buys four apples and one banana using his $\$x$ he will also have $\$2$ in change remaining. What is the maximum number of bananas Pete can afford to buy?

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Solution

Let $A$ be the price of one apple and $B$ the price of one banana. We are given $x = 2A + 3$ , $x = 3B + 2$ , $x = 4A + B + 2$ . Thus, $2A = 3B - 1$ and $4A = 2B$ implying $6B - 2 = 2B$ or $B = 0.5$ . This gives $A = 0.25$ and $x = 3.5$ . Thus, Pete can buy a maximum of $7$ bananas. The answer is (A).

University of Alberta

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