Multiple step coin puzzle

Question

In a collection of coins there were twelve fewer dimes than nickles and six times as many quarters as the sum of the nickels and dimes. If the coins are worth $43.80, how many of each kind of coin is there?

I am not able to figure out the step of the equation for the quarters. I have the dimes and nickles set up. Can you please help me.

Answer 1

Hi Shannon, let n represent the nickels, d represent the dimes, and q represent the quarters.

There are n nickels.

There are d dimes, but there are n - 12 dimes.

There are q quarters, but there are 6 * (n + (n - 12)) quarters.

Equation to solve:

.05 * n + .10 * d + .25 * q = 43.80

Substitute using nickels:

.05 * n + .10 * (n - 12) + .25 * (6 * (n + (n - 12))) = 43.80

Clear parenthesis:

.05 * n + .10 * n - 1.20 + 3.00 * n - 18 = 43.80

Combine terms:

3.15 * n - 19.20 = 43.80

Add 19.20 to both sides:

3.15 * n = 63.00

Divide both sides by 3.15:

n = 20

The number of nickels is 20.

The number of dimes is n - 12, or 20 - 12 = 8. There are 8 dimes.

The number of quarters is 6 * (n + n - 12), or 6 * (40 -12). There are 168 quarters.

20 nickels = $1.00

8 dimes = $.80

168 quarters = $42.00

Total = $43.80

Questions?