SOLUTION: A coin jar contains nickels, dimes, and quarters. There are 47 coins in all. There are 12 more nickels than quarters. The value of the dimes is $3.70 less than the value o

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: A coin jar contains nickels, dimes, and quarters. There are 47 coins in all. There are 12 more nickels than quarters. The value of the dimes is $3.70 less than the value o      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1107799: A coin jar contains nickels, dimes, and quarters.
There are 47 coins in all.
There are 12 more nickels than quarters.
The value of the dimes is $3.70 less than the value of the quarters.
How many coins of each type are in the jar?
Found 2 solutions by josgarithmetic, addingup:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
n, d, q
nickels dimes quarters

system%28n%2Bd%2Bq=47%2Cn-q=12%2C25q-10d=370%29
-
n=q%2B12
-
25q-370=10d
5q-74=2d
d=5q%2F2-37
-
n%2Bd%2Bq=47
%28q%2B12%29%2B%285q%2F2-37%29%2Bq=47
2q%2B5q%2F2%2B12-37=47
2q%2B5q%2F2=47-12%2B37
2q%2B5q%2F2=72
4q%2B5q=144
9q=144
highlight%28q=16%29
-
highlight%28n=28%29
-
highlight%28d=3%29

Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!

n + d + q = 47
25q - 10d = 370
n = q + 12
25q - 370 = 10d
5q - 74 = 2d
d = 5q/2 - 37
n + d + q = 47
(q + 12) + ((5q/2) - 37) + q = 17
2q + (5q/2) + 12 - 37 = 47
2q + (5q/2) = 47 - 12 + 37
2q + 5q/2 = 72
4q + 5q = 144
9q = 144
q = 16
n = 28
d = 3