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Average of P and its reciprocal [#permalink]
24 Jun 2019, 08:45

00:00

Question Stats:

77% (00:45) correct
22% (00:52) wrong based on 27 sessions

\(0<P<1\) where P is a fraction.

Quantity A

Quantity B

Average of P and its reciprocal

1

A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.

I already know by substituting the values we can get the answer. I want someone to solve this question using Algebra and not substitution.

Re: Sum of Fraction and its Reciprocal [#permalink]
25 Jun 2019, 05:43

4

This post received KUDOS

Expert's post

skjoshi267 wrote:

\(0<P<1\) where P is a fraction.

Quantity A

Quantity B

Average of P and its reciprocal

1

Average of P and its reciprocal = \(\frac{\frac{1}{P} + P}{2}\)

\(=\frac{\frac{1 + P^2}{P}}{\frac{2}{1}}\)

\(=\frac{1+P^2}{2P}\)

So, we have: QUANTITY A: \(\frac{1+P^2}{2P}\)

QUANTITY B: \(1\)

We can complete this solution using matching operations

Since \(2P\) is POSITIVE, we can safely multiply both quantities by \(2P\) to get: QUANTITY A: \(1+P^2\) QUANTITY B: \(2P\)

If you notice that we have a PERFECT SQUARE hiding among the two quantities, then we can first subtract \(2P\) from both quantities to get: QUANTITY A: \(1+P^2-2P\) QUANTITY B: \(0\)

Rearrange to get: QUANTITY A: \(P^2-2P+1\) QUANTITY B: \(0\)

Factor to get: QUANTITY A: \((P-1)^2\) QUANTITY B: \(0\)

Since we know that P-1 does not equal 0, we know that \((P-1)^2\) is POSITIVE

Answer: A

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Re: Sum of Fraction and its Reciprocal
[#permalink]
25 Jun 2019, 05:43