Question 1026988
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A newspaper carrier had $6.65 in change. He has ten more quarters than dimes but two times as many nickels as quarters. How many coins of each type does he have?
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Let q = # of quarters, d = # of dimes and n = # of nickels.

Then what you have from the condition is 

d = q - 10  and  n = 2q.     (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*(2q) + 10*(q-10) + 25q = 665.

Simplify and solve it for q:

10q + 10q - 100 + 25q = 665,

45q = 665 + 100,

45q = 765,

q = {{{765/45}}} = 17.

Now from (1)  d = q - 10  = 17 - 10 = 7  and  n = 2q = 2*17 = 34.

<U>Answer</U>.  34 nickels,  7 dimes  and  17 quarters.
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