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Bookstore problem

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by akshatgupta87 » Sun Jun 26, 2011 10:33 am
Q.)A bookstore owner bought a shipment of n books at a total cost of E dollars, and plans to sell each of the books at the same price and make a 15 percent profit on the entire shipment,. If x of the books were damaged and impossible to sell, how many dollars more than she had originally planned must the owner charge for each of the remaining books in order to make the desired profit on the shipment?
A)1.15E/(n-x)
B)1.15Ex/(n(n-x))
C)1.15E/(n(n-x))
D)1.15Ex/(n-x)
E)1.15(n-x)/E
Someone explain..
Thanks,
~Akshat
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by winniethepooh » Sun Jun 26, 2011 10:48 am
Your question seems to be grammatically wrong: have you missed some words? Please check the question again.
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by winniethepooh » Sun Jun 26, 2011 11:05 am
From what I understand:
Let the number of books in the shipment be (n)1000 @ $100 each.
Therefore total cost = E = $100,000.

To earn 15 % the books should be sold at $100,000 * 15/100 = $115,000,
Or $115 each book,


Now, say (x)100 books were damaged. So total price to be charged on 900 remaining books to earn 15 % overall profit = $115000/900 = $127.77
Additional price to be charged = $127.77 - $115 = $12.77 each book.(Desired answer)

Now, plug in the values of n, x & E in each equation to get $12.77

In answer B:
1.15Ex/n(n-x) = 1.15 * 100000 * 100/ 1000(1000-100)
= 115000 * 100/ 1000 * 900
= 12.77
Hence, answer is B.
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by akshatgupta87 » Sun Jun 26, 2011 11:18 am
The question is from princeton.
i have copied the question correctly. ;)
OA is B
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by Gissinggy » Sun Jun 26, 2011 2:55 pm
First, without considering the X damaged books, in order to make 15% profit on all books, the unit selling price of each book should be 1.15E/n. Since X books were damaged, the unit selling price becomes 1.15E/(n-x). So the difference between these two selling prices is the answer:

1.15E/(n-x)-1.15E/n

The answer is 1.15Ex/n(n-x).

Hope my explanation is helpful.
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