The total if all 236 coins were nickels would be $11.80, which is $3.95 short of the actual amount.
Replacing a nickel with a dime adds $0.05 to the total value, so there must have been $3.95/$0.05 = 79 such replacements.
There are 79 dimes.
There are 236 -79 = 157 nickels.
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Using the given variables, the problem statement gives rise to two equations. One is the based on the number of coins. The other is based on their value.
- n + d = 236
- .05n +.10d = 15.75
Solving the first for n, we get
... n = 236 - d
Substituting that into the second equation, we have
... .05(236 -d) +.10d = 15.75
... .05d = 15.75 -236·.05 . . . . . collect terms, subtract .05·236
... d = 3.95/.05 . . . . . . . . . . . . . divide by .05
... d = 79
... n = 236-79 = 157
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The solution should look familiar, as it matches the verbal description at the beginning.
