# A change machine contains nickels, dimes, and quarters. There are 75 coins in the machine, and the value of the coins is \$7.25

#1

A change machine contains nickels, dimes, and quarters. There are 75 coins in the machine, and the value of the coins is \$7.25. There are 5 times as many nickels as dimes. ?

Let’s set x equal to the number of dimes in the change machine. So if we know that there are 5 times as many nickels as dimes, we can say that the number of nickels is equal to 5 * x, or 5x. With this established, we can determine the total number of nickels and dimes as: dimes + nickels, or 5x + x, which is 6x. The remaining coins must be quarters, so the total number of coins minus the number of nickels and dimes must be equal to the number of quarters, which we know is: 75 - 6x = number of quarters.

Now, we know that the total value of all the coins is equal to 7.25, so the sum of the quantity of coin * value of coin for each denomination will be equal to 7.25. Here’s what the equation looks like:

nickels*.05 + dimes*.1 + quarters*.25 = 7.25

…when we substitute the values:

5x*.05 + x*.1 + (75 - 6x)*.25 = 7.25

Now you just have to solve for x. Remember, x we said was equal to the number of dimes in the coin machine. Once we find x, we can plug it into the expressions we came up with before to find the quantity of each coin.

If we solve for x, we find that x = 10. Let’s plug that back into those expressions:

Nickels = 5x = 5(10) = 50 nickels
Dimes = x = 10 = 10 dimes
Quarters = 75-6x = 75-6(10) = 75-60 = 15 quarters

There must be 50 nickels, 10 dimes, and 15 quarters in the coin machine.