Question 260695: Barbara has 20 coins consisting of nickels and dimes. If the nickels were dimes
and the dimes were nickels, she would have 30¢ more than she has now. How
many dimes did she have to begin with?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Barbara has 20 coins consisting of nickels and dimes. If the nickels were dimes
and the dimes were nickels, she would have 30¢ more than she has now. How
many dimes did she have to begin with?
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N = # of nickels
D = # of dimes
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5N + 10D = A (some amount)
10N + 5D = A + 30
N + D = 20
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Sub for N or D, I'll do N
N = 20 - D
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5(20-D) + 10D = A
10(20-D) + 5D = A + 30
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+5D + 100 = A
-5D + 170 = A
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Since the equal A, the equal each other
5D + 100 = -5D + 170
D = 7
N = 13
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You can check it.
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Do it this way:
A change in the total of 30 means a change of 6 coins from nickels to dimes.
That makes the problem:
N + D = 20
N - D = 6
2N = 26
N = 13
D = 7
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