SOLUTION: A newspaper carrier has $4.60 in change. He has two more quarters than dimes but five times as many nickels as quarters. How many coins of each type does he have?
Question 1065120: A newspaper carrier has $4.60 in change. He has two more quarters than dimes but five times as many nickels as quarters. How many coins of each type does he have? Found 2 solutions by ikleyn, josgarithmetic:Answer by ikleyn(52835) (Show Source):
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A newspaper carrier has $4.60 in change. He has two more quarters than dimes but five times as many nickels as quarters.
How many coins of each type does he have?
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Let "q" be the number of quarters, "d" be the number of dimes and "n" be the number of nickels.
Then what you have from the condition is
d = q - 2 and n = 5q. (1)
Next, you have the "value" equation
5n + 10d + 25q = 665. (2)
Substitute (1) into the "value" equation. You will get a single equation for q:
5*(5q) + 10*(q-2) + 25q = 460.
Simplify and solve it for q:
25q + 10q - 20 + 25q = 460,
60q = 460 + 20,
60q = 480,
q = = 8.
Now from (1) d = q - 2 = 8 - 2 = 6 and n = 5q = 5*8 = 40.
Answer. 40 nickels, 6 dimes and 8 quarters.
The referred lessons are the part of this online textbook under the topic "Coin problems".
Your problem is very similar to the problem 2 of the lesson marked (*) in the list.
For this response I simply copied and pasted that solution and updated the input data in it.
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