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Question 544575: A man had dollar bills and 5 cent coins,he spent 1/3 of his money and had < $50 left,he now has 4 times as many dollars as he had 5 cent pieces and half as many 5 cent pieces as he had dollars.How much did he start with and spend ? Thanks.I am not sure how to asign values @ the start.How about 10 x + y ?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A man had dollar bills and 5 cent coins, he spent 1/3 of his money and had < $50 left, he now has 4 times as many dollars as he had 5 cent pieces and half as many 5 cent pieces as he had dollars.
How much did he start with and spend?
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Not alot of information here, we will have to depend on the fact that an integer solution will be unique.
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Let d = no. of dollar bills
Let n = no. 5 cents coins
1d + .05n = original amt
then
 (1d+.05n) = amt left after spending 1/3 of it
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"he now has 4 times as many dollars as he had 5 cent pieces and half as many 5 cent pieces as he had dollars."
(1d + .05n) = 4(n) + .5(.05d)
(1d + .05n) = 4n + .025d
multiply both sides by 3
2(1d + .05n) = 3(4.00n + .025d)
2d + .10n = 12n + .075d
2d - .075d = 12n - .10n
1.925d = 11.9n
d =  n
Put this equation into a graphing calc
y = x, the table show only one integer solution, keeping the <$50 rule in mind, for y which represent the original no. of dollars, d=68 and x is the no. of nickels, n=11
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the original amt then is: 68 + 11(.05) = $68.55
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Find the amt spent, he now has 4(11) = 44, .05(68/2) = 1.70. Total 45.70
therefore he spent, 68.55 - 45.70 = $22.85
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Check our solution by confirming the amt spent is 1/3 the original:
 =  ; confirms our solution
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Did this make sense to you?
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