Question 951120: I am having a problem breaking down word problems.
This is the problem i am trying to solve:
I have a mixture of 742 coins which are a mixture of pennies, nickels, dimes and quarters. There are 10 times as many pennies as dimes and half as many quarters as dimes. If i have $24.60 how many dimes are there?
I know i have to pull out the following terms
742 total coins
10x pennies as dimes
1/2 quarters as dimes
and the total of 24.60
Answer by quada(2)   (Show Source):
You can put this solution on YOUR website! Let d = number of dimes.
Then 10d = number of pennies, and d/2 = number of quarters. Let n = number of nickels.
1(10d) + 10(d) + 25(d/2) + 5n = 2660
10d + 10d + 12.5d + 5n = 2660
32.5d + 5n = 2660
Also, all types of coins added up = 742, so
10d + n + d + d/2 = 742
11.5d + n = 742
n = 742 - 11.5d
32.5d + 5(742-11.5d) = 2660
32.5d + 3710 - 57.5d = 2660
-25d = -1050
d = 42
42 dimes, 420 pennies, 21 quarters, which leave
259 nickels.
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