Question

Jacy has a collection of 140 coins worth $19.00. She has only nickels, dimes, and quarters. The difference in the number of dimes and the number of quarters is 10. Using matrices, how many nickels are in Jacy’s collection?
38
50
95
130

Answer 1

Let n, d, q be the numbers of nickels, dimes, and quarters (cents). Then
n + d + q = 140
5n + 10d + 25q = 1900
and the difference |d − q| = 10.

Case A: d − q = 10. Substitute d = q + 10 into the first two equations:
n + 2q + 10 = 140 → n = 130 − 2q.
5(130 − 2q) + 10(q + 10) + 25q = 1900 → 750 + 25q = 1900 → q = 46.
Then d = 56 and n = 38.

Case B: q − d = 10. Substitute d = q − 10:
n + 2q − 10 = 140 → n = 150 − 2q.
5(150 − 2q) + 10(q − 10) + 25q = 1900 → 650 + 25q = 1900 → q = 50.
Then d = 40 and n = 50.

Both (n,d,q) = (38,56,46) and (50,40,50) satisfy the conditions, so the problem as stated is ambiguous. If the intended meaning is “there are 10 more dimes than quarters” (d − q = 10), the number of nickels is 38.