Question 1053052: JIMMY HAS 32 COINS : SOME NICLES , DIMES AND QUARTERS . THE TOTAL NUMBER OF DIMES , D , AND QUARTERS , Q , IS EQAL TO THE NUMBER OF NICKLES , N .THE VALUE OF JIMMY'S COINS IS $ 3.90 . ? HOW CAN I SOLVE THIS PROBLEM .
Answer by jorel555(1290)   (Show Source):
You can put this solution on YOUR website! (self-edited)
Let n be nickels, d be dimes, and q be quarters. Then:
n+d+q=32 and (total amount of coins)
5n+10d+25q=390 (total value of coins
Since n=d+q, 2n=32 thus n=16, and d+q=16. So we have:
d+q=16
n=16
10d+25q=390-5(16)=310
d+q=16
So, Solving for q, we get:
(10d-10d)+(25q-10q)=310-160
15q-150
q=10
d+q=16 so d=6; and we know n=16; so we have
10 quarters, 6 dimes, and 16 nickels. ☺☺☺☺
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