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Question 4912: A collection of nickels, dimes and quarters has a value of $7.50. If there are 3 fewer dimes than nickels and twice as many quarters as nickels, how many dimes are there?
Answer by Abbey(339)   (Show Source):
You can put this solution on YOUR website! Let x=number of nickels
let x-3=number of dimes (3 fewer than nickels)
let 2x=number of quarters (twice as many as nickels)
Let 5x=the value of all nickels (5 cents for each nickel)
Let 10(x-3)= the value of all dimes (10 cents for each dime)
Let 25(2x)= the value of all quarters (25 cents for each quarter)
Then adding all values together we know:
$7.50= total value of all coins
We are working in cents, so convert$7.50 to 750 cents and write the equation:
750=5x+10(x-3)+25(2x)
Clear the parentheses:
750=5x+10x-30+50x
Combine like terms:
750=65x-30
Add 30 to both sides:
780=65x
Divide both sides by 65:
12=x
Using our original statments we know their are 12 nickels, 9 dimes (3 fewer than the nickels) and 24 quarters (twice as many as the nickels).
Check your answer:
12 * $.05 = $.60
9 * $.10 = $.90
24 * $.25 =$6.00
$6.00+.90+.60=$7.50, So this is correct.
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