SOLUTION: Gary has 42 coins in nickels, dimes, and quarters. If he has 8 more nickels than dimes and has $7.15 in all, how many of each does he have?

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Question 989611: Gary has 42 coins in nickels, dimes, and quarters. If he has 8 more nickels than dimes and has $7.15 in all, how many of each does he have?
Found 3 solutions by CubeyThePenguin, josgarithmetic, ikleyn:
Answer by CubeyThePenguin(3113)   (Show Source): You can put this solution on YOUR website!
n = number of nickels
d = number of dimes
q = number of quarters

n + d + q = 42
n = d + 8
5n + 10d + 25q = 715

Substitute the second equation into the first and third equations.

2d + q = 34
15d + 25q = 675

Multiply the first equation by 25.

50d + 25q = 850
-(15d + 25q = 675)
----------------------
35d = 175
d = 5

Gary has 5 dimes, 24 quarters, and 13 nickels.
Answer by josgarithmetic(39620)   (Show Source): You can put this solution on YOUR website!


Simplify and solve the system.

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Answer by ikleyn(52814)   (Show Source): You can put this solution on YOUR website!
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Gary has 42 coins in nickels, dimes, and quarters.
If he has 8 more nickels than dimes and has $7.15 in all, how many of each does he have?
~~~~~~~~~~~


            It is a typical problem to be solved using one single unknown and one single equation.

            See below on how to do it . . . 


Let x be the number of nickels;

then the number od dimes is (x-8) and the number of quarters is  (42 - x - (x-8)) = (42+8 - 2x) = (50-2x).


At this point, you can write the total money equation


    5x + 10*(x-8) + 25*(50-2x) = 715  cents.


Simplify and solve


    5x + 10x - 80 + 1250 - 50x = 715

           -35x                = 715 + 80 - 1250

           -35x                = -455

              x                =  = 13.


ANSWER.  13 nickels;  13-8 = 5 dimes  and  the rest 42-13-5 = 24 quarters.


CHECK.   13*5 + 5*10 + 25*24 = 715 cents, in total.    ! Correct !

Solved.



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