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p + 1/p = p^2 + 1/p

8p = p^2 - 2p + 1
p^2 - 2p + 1 - 8p = 0
p^2 - 10p + 1 = 0

p^2+1 = 10p

p^2+1/p = 10p/p = 10. Ans - B.
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Interesting question, easy but could be missed if overthought. I think the trick is not to solve for p first but try to isolate the whole p+1/p. Cross multiplication gives p^2 - 10p + 1 = 0 ; p^2 + 1 = 10p ; p+1/p=10

In summary, collect like times and try to get rid of the exponent on p. Also, a familiarity with the form p+1/p as quadratic will help see the connection.
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aveek86
If 2p/(p^2-2p+1) = 1/4, then the value of p+1/p is

A) 8
B) 10
C) 12
D) 2√6
E) None of these

The question asks:
\(p+\frac{1}{p}=\frac{p^2+1}{p}=?\)

Given:
\(\frac{2p}{(p^2-2p+1)} = \frac{1}{4}\)

\(4*2p = p^2-2p+1\)

\(p^2 + 1 = 10p\)

Thus:
\(p+\frac{1}{p}=\frac{p^2+1}{p}=\frac{10p}{p}=10\).

Answer: B.

Hope it helps.

We are not informed that p is a nonzero number, right? So how can we divide the equation by p?
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aveek86
If 2p/(p^2-2p+1) = 1/4, then the value of p+1/p is

A) 8
B) 10
C) 12
D) 2√6
E) None of these

The question asks:
\(p+\frac{1}{p}=\frac{p^2+1}{p}=?\)

Given:
\(\frac{2p}{(p^2-2p+1)} = \frac{1}{4}\)

\(4*2p = p^2-2p+1\)

\(p^2 + 1 = 10p\)

Thus:
\(p+\frac{1}{p}=\frac{p^2+1}{p}=\frac{10p}{p}=10\).

Answer: B.

Hope it helps.

We are not informed that p is a nonzero number, right? So how can we divide the equation by p?

p cannot be 0 because p =0 does not satisfy 2p/(p^2-2p+1) = 1/4.
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I don't understand whether it is my mistake or author's mistake to read it (p+1)/p which is actually asked.
Without that bracket, I think people like me could understand it p+ (1/p).

I am sure actual GMAT won't make such ambiguous question.
Thanks!
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gvij2017
I don't understand whether it is my mistake or author's mistake to read it (p+1)/p which is actually asked.
Without that bracket, I think people like me could understand it p+ (1/p).

I am sure actual GMAT won't make such ambiguous question.
Thanks!

The question ask to find the value of \(p + \frac{1}{p}\), not \(\frac{p+1}{p}\). Also, p + 1/p can only mean \(p + \frac{1}{p}\). If it were \(\frac{p+1}{p}\), it would have been written as (p+1)/p.
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