2018 AMC 10 A

Let be the number of cent coins that Joe has. Therefore, he must have cent coins and cent coins. Since the total value of his collection is cents, we can write .

Joe has cent coins, cent coins, and cent coins. Thus, our answer is .

Let be the number of cent coins Joe has, be the number of cent coins, and the number of cent coins. We are solving for .

We know that the value of the coins add up to cents. Thus, we have . Let this be ().

We know that there are coins. Thus, we have . Let this be ().

We know that there are more dimes than nickels, which also means that there are less nickels than dimes. Thus, we have .

Plugging into the other two equations for , () becomes and () becomes . () then becomes , and () then becomes .

Multiplying () by , we have (or ). Subtracting () from () gives us , which means .

Plugging into , .

Plugging and into the () we had at the beginning of this problem, .

Thus, the answer is .

So you set the number of cent coins as , the number of cent coins as , and the number of quarters .

You make the two equations:,

.

From there, you multiply the second equation by to get .

You subtract the first equation from the multiplied second equation to get . You can plug that value into one of the equations to get . So, the answer is .