SOLUTION: A payphone is hold its maximum number of 500 coins consisting of nickels, dimes, and quarters. The number of quarters is twice the number of dimes. If the value of all coins is $88
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Question 1126289: A payphone is hold its maximum number of 500 coins consisting of nickels, dimes, and quarters. The number of quarters is twice the number of dimes. If the value of all coins is $88 how many nickels were in the payphone?
I have no idea how to approach is. Please help me write the equation?
Thank You!! Found 3 solutions by rothauserc, addingup, Alan3354:Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! let d be the number of dimes, then
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number of quarters is 2d
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let n be the number of nickels, then
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d +n +2d = 500
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1) n +3d = 500
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2) 0.05n + 0.10d +0.25(2d) = 88.00
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solve equation 1 for d
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3d = (500 - n)
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d = (500 - n)/3
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substitute for d in equation 2
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0.05n +0.10((500 - n)/3) +0.50((500 - n)/3) = 88.00
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multiply both sides of = by 3
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0.15n +50 -0.10n +250 -0.50n = 264.00
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-0.45n = -36
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n = 80
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There are 80 nickels
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check the answer
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500 - 80 = 420
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then there are 140 dimes and 280 quarters
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0.05(80) + 0.10(140) + 0.25(280) = 88.00
:
4 +14 +70 = 88.00
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88.00 = 88.00
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answer checks
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You can put this solution on YOUR website! n + d + q = 500
The number of quarters is twice the number of dimes:
n + d + q = 500
n + 3d = 500
n = 500 - 3d
so now we have that:
0.10d + 0.25(2d) + 0.05(500 - 3d) = 88
0.10d + 0.50d + 25 - 0.15d = 88
0.45d = 63
d = 140
There are 140 dimes and the quarters are twice as many: 140 x 2 = 280 there are 280 quarters and the nickels are 500-3d = 500 - (3 x 140) = 80
So we have:
n = 80
d = 140
q = 280
check:
0.05(80) + 0.10(140) + 0.25(280) = 88 Correct