Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 2.9 Junior Find the answer

Joe has a collection of $23$ coins, consisting of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins. He has $3$ more $10$ -cent coins than $5$ -cent coins, and the total value of his collection is $320$ cents. How many more $25$ -cent coins does Joe have than $5$ -cent coins?

Pick one

Solution

Let $x$ be the number of $5$ -cent coins that Joe has. Therefore, he must have $(x+3) \ 10$ -cent coins and $(23-(x+3)-x) \ 25$ -cent coins. Since the total value of his collection is $320$ cents, we can write
$\begin{align*} 5x + 10(x+3) + 25(23-(x+3)-x) &= 320 \\ 5x + 10x + 30 + 500 - 50x &= 320 \\ 35x &= 210 \\ x &= 6. \end{align*}$
Joe has six $5$ -cent coins, nine $10$ -cent coins, and eight $25$ -cent coins. Thus, our answer is
$8-6 = \boxed{\textbf{(C) } 2}.$
~Nivek

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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.