A bookstore owner bought a shipment of n books at a total cost of E do

Question

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. \(\frac{1.15E}{n-x}\)

B. \(\frac{1.15Ex}{n(n-x)}\)

C. \(\frac{1.15E}{n(n-x)}\)

D. \(\frac{1.15Ex}{n-x}\)

E. \(\frac{1.15(n-x)}{E}\)

Show Answer

Official Answer

B

A.  \frac{1.15E}{n-x}

B.  \frac{1.15Ex}{n(n-x)}

C.  \frac{1.15E}{n(n-x)}

D.  \frac{1.15Ex}{n-x}

E.  \frac{1.15(n-x)}{E}

sahilvermani · Answer

Selling price of n books 1.15E
Planned SP per book = 1.15E/n

x books were damaged.
New SP per book = 1.15E/(n-x)

So, increment = 1.15E/n - 1.15E/(n-x) = 1.15E   = 1.15Ex/n(n-x)

So, C.

stne · Answer

Hi, sahilvermani , 
Good Job, you did everything correct except writing the option. Small typo. 
Your answer is B , in error you have written C.

sahilvermani · Answer

Yep...that's right

Palak74 · Answer

Total Price that the retailer needs to recover from all N books (including 15% profit) = 1.15E
Price of one book planned =1.15E/n

since X books are damaged

Profit needed remains same but No. of books available to be sold decreases

{1.15E/n + A} (n-x)= 1.15E ------ Main equation to be solved

In the above equation 1.15E/n is the price planned per book
A is the addition needed per book to make up for the loss
n-x is the number of books available to be sold 
1.15E is the total amount that is needed.

Solving the equation we get option B

1.15Exn(n−x)1.15Exn(n−x)

salilbagai · Answer

HI,
A minor mistake in the solution,
So increment = 1.15E/(n-x) - 1.15E/n = 1.15E (1/(n-x) - 1/n) = 1.15E ((n - n +x)/ ((n-x)(n))) = 1.15Ex/(n(n-x))
Ans B

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A bookstore owner bought a shipment of n books at a total cost of E do

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