Question

Sasha has 3.20 in U.S. coins. She has the same number of quarters and nickels. What is the greatest number of quarters she could have?

Answer 1

Solution 1:
Suppose Sasha has $q$ quarters. Then she also has $q$ nickels, and the total value of her quarters and nickels is $.25q + .05q = .30q$. Since $3.20/0.30 = 10\frac{2}{3}$, this means that she has at most 10 quarters

Answer 2

10 quarters = $2.50

10 nickels = $0.50

that leaves $0.20 for other coins (dimes / pennies)

Explanation:

First, suppose she has only quarters and nickels and no other coins. Then if C is the identical number of coins of each type, then 5C + 25C = 320, so 30C = 320 and 3C = 32, but there is no integer solution to this. So she must have at least one other type of coin.

Assume she has only quarters, nickels, and dimes. Then if D is the number of dimes, 5C + 25C + 10D = 320, which means 30C + 10D = 320, or 3C + D = 32. The smallest D can be is 2, leaving 3C = 30 and thus C = 10. So in this scenario she would have 10 quarters, 10 nickels, and two dimes to make $2.50 + $0.50 + $0.20 = $3.20.

This has to be the highest number, because if she had 11 quarters and 11 nickels, that alone would add up to 11(0.25) + 11(0.05) = $3.30, which would already be too much.