Solve by using a matrix.A collection of nickels, dimes and quarters totals $6.00. If there are 52 coins altogether and twice as many dimes as nickels, how many of each kind of coin are there?

Let q, d, and n represent the numbers of quarters, dimes, and nickels. The total-coin equation is q + d + n = 52, and the relationship is d = 2n. In cents, the value equation is 25q + 10d + 5n = 600; substituting d = 2n and q = 52 − 3n gives 1300 − 50n = 600, so n = 14. Thus d = 28 and q = 10, matching choice D: 10 quarters, 28 dimes, and 14 nickels.

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