Question 245540
Brad has a collection of dimes and quarters totaling $53.50. The number of quarters is 10 more than three times the number of dimes. How many coins of each kind does he have?
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Equations:
Quantity Eq::: q = 3d + 10
Value Eq::::: 25q + 10d = 5350 cents
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Multiply thru the 1st Eq. by 25 and rearrange it.
25q - 75d = 250
25q + 10d = 5350
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Subtract 1st from 2nd and solve for "d":
85d = 5100
d = 60 (# of dimes)
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Since q = 3d+10, q = 3*60+10 ; q= 190 (# of quarters)
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Cheers,
Stan H.