Question 57225: A box of coins contains dimes, quarters, and half dollars. The face value of the half dollars and the dimes is the same as the face value of the quarters. Together the face value of the dimes and quarters is $6.50. The total number of coins is 43. How many dimes, quarters, and half dollars are in the box?
PS- This problem came off a worksheet otherwise I would give the information about the book. Basically I want you to know I'm not a slacker.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A box of coins contains dimes, quarters, and half dollars. The face value of the half dollars and the dimes is the same as the face value of the quarters. Together the face value of the dimes and quarters is $6.50. The total number of coins is 43. How many dimes, quarters, and half dollars are in the box?
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Write an equation for each statement:
:
" The face value of the half dollars and the dimes is the same as the face value of the quarters"
1st Eq: .50h + .10d = .25q
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"Together the face value of the dimes and quarters is $6.50."
2nd eq: .10d + .25q = 6.50
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"The total number of coins is 43":
3rd eq: d + q + h = 43
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Try this: (.50h+.10d) = .25q, so substitute that for .25q in the 2nd equation:
:
.10d + (50h + .10d) = 6.50
.20d + .50h = 6.50
.50h = 6.50 - .2d
Mult eq by 2:
h = (13 - .4d)
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Substitute (13-.4d) for h in the 3rd equation:
d + q + (13-.4d) = 43
d -.4d + q = 43 - 13
.6d + q = 30
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pair this up with the 2nd equation:
.60d + 1.0q = 30
.10d + .25q = 6.50> mult by 4 and you have:
:
.60d + 1.0q = 30
.40d + 1.0q = 26
------------------- subtract
.20d + 0q = 4
d = 4/.2
d = 20 dimes
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Find the quarters using the 2nd equation, substitute 20 for d
.10(20) + .25q = 6.50
2 + .25q = 6.60
.25q = 6.50 - 2
.25q = 4.50
q = 4.50/.25
q = 18 quarters
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Using the 3rd equation: 43 - 20 - 18 = 5 half dollars
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Check our solution in the 1st equation:
.10(20) + .50(5) = .25(18)
2.00 + 2.50 = 4.50
:
Not a very neat process but we finally got a solution:
20 dimes, 18 quarters, 5 halfs
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