Ajar contains n nickels and d dimes. there is a total of 236 coins in the jar. the value of the coins is $15.75. how many nickels and how many dimes are in the jar

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.

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.

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.

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step-by-step explanation:

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