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The total price of n (n > 1) equally priced copies of a

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The total price of n (n > 1) equally priced copies of a [#permalink] New post 30 Nov 2008, 18:17
The total price of n (n > 1) equally priced copies of a certain book is $50. In terms of n, which of the following gives the total price of n - 1 of these copies?

A) 50(n-1)

B) 50/(n-1)

C) {50(n-1)} / n

D) 50n/(n-1)

E) 50/{n(n-1)}

I'm confused by the wording of this problem. My answer was B but its wrong. Any help is appreciated! Also, is it better to pick numbers or solve with variable for such problems?

Thanks in advance!
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Re: Total price of n-1 copies [#permalink] New post 30 Nov 2008, 18:58
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I believe the answer is C

The wording can be a bit confusing. It took me a few seconds to realize (n>1) is just telling us that n is greater than 1 and is not part of the actual problem.

To break it down we need to know how much 1 book costs, and then multiply it by the new number of books. The new number of books is represented by n-1.

so if the total of n books is $50, then 1 book is the total divded by the number of books summed to get that total, or 50 /n

Now we take the price of 1 (represented by 50/n) and multiply that by the new quantity of books n-1.

\frac{50}{n}* n-1 This is the same as \frac{50(n-1)}{n} or Answer C

nikki25 wrote:
The total price of n (n > 1) equally priced copies of a certain book is $50. In terms of n, which of the following gives the total price of n - 1 of these copies?

A) 50(n-1)

B) 50/(n-1)

C) {50(n-1)} / n

D) 50n/(n-1)

E) 50/{n(n-1)}

I'm confused by the wording of this problem. My answer was B but its wrong. Any help is appreciated! Also, is it better to pick numbers or solve with variable for such problems?

Thanks in advance!

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: Total price of n-1 copies [#permalink] New post 30 Nov 2008, 19:13
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C it is.

n books = 50$, 1 book = 50/n $ => n-1 books will cost= 50*(n-1)/n
Re: Total price of n-1 copies   [#permalink] 30 Nov 2008, 19:13
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