Question

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?

$0$

$1$

$2$

$3$

$4$

Solution:

Let $x$ be the number of $5$ -cent coins that Joe has. Then the number of $10$ -cent coins he has is $x + 3.$

Therefore, Joe has $23 - x - (x + 3) = 20 - 2x$ $25$ -cent coins.

The total value of all these coins is $5x + 10(x + 3) + 25(20 - 2x)$ $= 530 - 35x.$

We know that $530 - 35x = 320 \Rightarrow x = 6.$

This means that Joe has $20 - 2 \cdot 6 = 8$ $25$ -cent coins. Therefore, he has $8 - 6 = 2$ more $25$ -cent coins than $5$ -cent coins.

Thus, C is the correct answer.

2018 AMC 10A Problem 8

Solution:

Problem 8 in Other Years