SOLUTION: The manager at Cream of the Crop bought a load of watermelons for $200. She priced the melons so that she would make $1.50 profit for each melon. When all but 30 had been sold, t
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-> SOLUTION: The manager at Cream of the Crop bought a load of watermelons for $200. She priced the melons so that she would make $1.50 profit for each melon. When all but 30 had been sold, t
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Question 153236: The manager at Cream of the Crop bought a load of watermelons for $200. She priced the melons so that she would make $1.50 profit for each melon. When all but 30 had been sold, the manager had recovered her initial investment. How many did she buy originally? Found 2 solutions by checkley77, stanbon:Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! 200=[(200/x)+1.50](x-30)
200=[(200+1.5x)/x](x-30)
200x=(200+1.5x)(x-30)
200x=200x+1.5x^2-6,000-45x
1.5x^2-45x-6,000=0
(1.5x-120)(x+50)=0
1.5x-120=0
1.5x=120
x=120/1.5
x=80 is the number of watermellons bought.
Proof:
200=[(200/80)+1.5](80-30)
200=[2.5+1.5]*50
200=4*50
200=200
You can put this solution on YOUR website! The manager at Cream of the Crop bought a load of watermelons for $200. She priced the melons so that she would make $1.50 profit for each melon. When all but 30 had been sold, the manager had recovered her initial investment. How many did she buy originally?
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Let the original number bought be "x".
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Average cost of each melon was $(200/x)
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She priced the melons so that she would make $1.50 profit for each melon.
Let her price per melon be "p".
EQUATION:
p = (200/x) + 1.5
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When all but 30 had been sold, the manager had recovered her initial investment.
EQUATION:
p(x-30) = 200
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Rearrange the two equations:
p = (200/x) + 1.5
p = 200/(x-30)
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Substitute to solve for "x".
200/(x-30) = (200/x) + 1.5
Multiply thru by x(x-30) to get:
200x = 200(x-30) + 1.5x(x-30)
6000 = 1.5x^2 - 45x
x^2 - 30x -4000 = 0
(x- 80)(x+50) = 0
Positive solution
x = 80 (original number of melons)
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Cheers,
Stan H.
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