SOLUTION: Mike has $1.45 in dimes and nickels. If he has 8 more nickels than dimes, how many of each coin does he have?

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Question 1156188: Mike has $1.45 in dimes and nickels. If he has 8 more nickels than dimes, how many of each coin does he have?
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


With formal algebra... if it is required....

x = # dimes
x+8 = # nickels

10x%2B5%28x%2B8%29+=+145

Solve using basic algebra; I leave that to you.

Informally....

Count the "extra" 8 nickels first. That is 40 cents; what is left is equal numbers of dimes and nickels, with a value of 145-40 = 105 cents.

The value of one dime and one nickel is 15 cents; to make the remaining 105 cents, the number of dimes and nickels has to be 105/15 = 7.

ANSWER: 7 dimes and 7+8=15 nickels

CHECK: 7(10)+15(5) = 70+75 = 145