Question

A box contains $8.05 in nickels, dimes, and quarters. There are 48 coins in all, and the sum of the numbers of nickels and dimes is 2 less than the number of quarters. How many coins of each kind are there?

Answer 1

Answer:

25 quarters, 23 dimes, and 0 nickels.

Explanation:

Let n = number of nickels, q = number of quarters, and d = number of dimes

n + d = q - 2

n + d + q = 48

8.05 = .25q + .1d + .05n

so

(q-2) + q = 48

2q = 50

q = 25

Using that q value:

n + d = q - 2 --> n + d = 25 - 2 --> n + d = 23 --> n = 23 - d

Using that n value and q value:

8.05 = .25(23) + .1d + .05(23 - d)

8.05 = 5.75 + .1d + 1.15 - .05d

2.30 = 1.15 + .05d

1.15 = .05d

23 = d

Using that d value and q value:

n + d + q = 48 --> n + 23 + 25 = 48

n + 48 = 48

n = 0

Therefore there are 25 quarters, 23 dimes, and 0 nickels.

Checking that with:

8.05 = .25q + .1d + .05n

8.05 = .25(23) + .1(23) + .05(0)

8.05 = 5.75 + 2.30 + 0

8.05 = 8.05 ✓