Question

Sandy held a garage sale, during which she charged a dime for everything but accepted a nickel if the buyer bargained well. At the end of the day she realized she sold a total of 12 nickels and dimes. She had raked in a grand total of $0.95. She only had dimes and nickels. How many of each coin did she have?

Answer 1

• 5 nickels
• 7 dimes

Explanation:

If all 12 of Sandy's coins were nickels, she would have $0.60. She actually has $0.35 more than that. Each dime that replaces a nickel adds $0.05 to the value, so she must have $0.35/$0.05 = 7 dimes. The remaining 5 coins are nickels.

Sandy has 5 nickels and 7 dimes.

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Alternate solution

If you like, you can write one equation. Let d represent the number of dimes. Then 12-d is the number of nickels, and Sandy's change is ...

0.05(12-d) +0.10d = 0.95

0.60 +0.05d = 0.95

0.05d = 0.35 . . . . . . . . subtract 60¢

d = 0.35/0.05 = 7 . . . . . divide by 0.05. (number of dimes)

12-d = 5 . . . . . . . . . . . . . (number of nickels)

The sequence of steps in this problem should look a lot like the verbal description above.

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You can also write two equations:

• n + d = 12
• .05n + .10d = .95

If you solve this set by substitution for n, you will have the equation above. The same is true if you multiply the first equation by 0.05 and subtract that result from the second equation.

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Comment on solution approach

In the above solutions, we have solved for the number of dimes (the higher-value coin). If you work the problem by solving for the number of nickels, you will find you end up doing arithmetic with negative numbers. Some folks find it easier to work with positive numbers, so that's how we've done it here.