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Question 5378: The petty-cash drawer of a small office contained $16.25. If there were twice as many nickels as quarters and as many dimes as nickels and quarters combined; how many coins of each type were there?
Answer by Abbey(339)   (Show Source):
You can put this solution on YOUR website! Let x = the number of quarters
Let 2x = number of nickels (twice as many as quarters)
Let (2x+x) = the number of dimes (as many dimes as quarters and nickels combined)
Let's work in cents, rather than dollars for this equation:
Total value = 1,625 cents
The total value of each of these coins is determined by multiplying the value by the number of coins (If I have two nickels worth 5 cents, then 2*5=10 - I have 10 cents worth of coins):
let 5(2x) = the total value of nickels
let 10(2x+x) = the total value of dimes
let 25x = the total value of quarters
Total value = value of nickels + value of dimes + value of quarters
1625 = 5(2x) + 10(2x+x) + 25x
1625 = 10x + (20x+10x) + 25x
1625 = 65x
divide both sides by 65
25 = x
So the number of quarters = 25
The number of nickels is twice that, or 50
and the number of dimes is the same as the nickels and quarters, or 75
Check by looking at the values:
25 quarters = $6.25
50 nickels = $2.50
75 dimes = $7.50
6.25 + 2.5 +7.5 = 16.25
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