SOLUTION: in a collection of coins there were twelve fewer dimes than nickels and six times as many quarters as the sum of the nickels and dimes. if the collection of coins was worth 43.80,

Question 1136814: in a collection of coins there were twelve fewer dimes than nickels and six times as many quarters as the sum of the nickels and dimes. if the collection of coins was worth 43.80, how many of each kind of coin are there?
Answer by ikleyn(52884) (Show Source): You can put this solution on YOUR website!
.

Let N = # of nickels.


Then the number of dimes = (N-12),
and the number of quarters = 6*(N + (N-12)) = 6*(2N-12) = 12N - 72.


The the total of money equation is


5N + 10*(N-12) + 25*(12N-72) = 4380  cents.


Simplify and solve for N.


Then evaluate the number of dimes and the number of quarters.

This problem is to be solved using one single equation for one unknown.

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