Question 1044099: There were 38 nickels, dimes and quarters whose total value came to $4.00. How many coins of each kind were there if there were twice as many dimes as nickels?
Answer by jorel555(1290) (Show Source):
You can put this solution on YOUR website! Let d be dimes, n be nickels, q be quarters. Then
d+n+q=38
d=2n, so 3n+q=38
.05n+.1(2n)+.25q=4
.25n+.25q=4
3n+q=38
n+q=16
2n=22
n=11
d=22
q=5
There are 11 nickels, 22 dimes, and 5 quarters, totaling $4.00 . ☺☺☺☺
|
|
|