SOLUTION: A newspaper carrier has $3.55 in change. He has three more quarters than dimes but four time as many nickels as quarters. How many coins of each type have?
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Question 1103893: A newspaper carrier has $3.55 in change. He has three more quarters than dimes but four time as many nickels as quarters. How many coins of each type have? Found 2 solutions by josgarithmetic, ikleyn:Answer by josgarithmetic(39630) (Show Source):
Simplify the system and solve for n, d, q.
Note that you should be able to write a single equation in just one variable, q, and solve this equation first; and then use the result to get the values for d and n.
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A newspaper carrier has $3.55 in change. He has three more quarters than dimes but four time as many nickels as quarters.
How many coins of each type have?
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Let x be the number of dimes.
Then the number of quarters is (x+3) ("three more quarters than dimes").
and the number of nickels is 4*(x+3) ("four time as many nickels as quarters")
Nickels contribute 5*(4*(x+3)) cents to the total.
Dimes contribute 10x cents to the total.
Quarters contribute 25*(x+3) cents to the total.
The equation is
5*(4*(x+3)) + 10x + 25*(x+3) = 355 cents (total).
Simplify and solve step by step
20*(x+3) + 10x + 25x + 75 = 355,
20x + 60 + 10x + 25x + 75 = 355,
55x + 135 = 355
55x = 355 - 135 = 220 ====> x = = 4.
Answer. There are 4 dimes, 4+3 = 7 quarters and 4*7 = 28 nickels.
Check. 4*10 + 7*25 + 28*5 = 355 cents. ! Correct !
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