SOLUTION: In a collection of coins made up of dimes and nickels, there are three more nickels than dimes. If there is a total of $1.95 in the collection, how many of each type of coin is the

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Question 1105145: In a collection of coins made up of dimes and nickels, there are three more nickels than dimes. If there is a total of $1.95 in the collection, how many of each type of coin is there?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
system%28n=d%2B3%2C5n%2B10d=195%29

Use elimination method to solve system%28n%2B2d=39%2Cn-d=3%29.

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Let n be the number of dimes; then the number of nickels is n+3.

The total value of the n dimes (10 cents each) and (n+3) nickels (5 cents each) is $1.95, or 195 cents:

10%28n%29%2B5%28n%2B3%29+=+195

Solve using basic algebra.

You can also solve this using logical reasoning; the calculations are the same as those used in the formal algebraic solution.

(1) Take away the "extra" 3 nickels (15 cents), leaving equal numbers of nickels and dimes, worth a total of 180 cents.
(2) Since 1 dime and 1 nickel makes 15 cents, the number of nickels and dimes is each 180/15 = 12.
(3) Now put back the 3 nickels you took away.

You end up with the answer, which is 12 dimes and 15 nickels.