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# Average of P and its reciprocal

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Intern
Joined: 08 Aug 2018
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Average of P and its reciprocal [#permalink]  24 Jun 2019, 08:45
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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.
[Reveal] Spoiler: OA

Last edited by Carcass on 25 Jun 2019, 16:34, edited 3 times in total.
Edited by Carcass
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Re: Sum of Fraction and its Reciprocal [#permalink]  24 Jun 2019, 09:17
Expert's post
Please follow the rule how to format a question in the proper manner

https://greprepclub.com/forum/how-to-po ... 12752.html
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Founder
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Kudos [?]: 2044 [0], given: 8095

Re: Sum of Fraction and its Reciprocal [#permalink]  24 Jun 2019, 09:31
Expert's post
$$\frac{\frac{1}{P} + P}{2}$$

$$\frac{\frac{1 + P^2}{P}}{\frac{2}{1}}$$

$$\frac{1+P^2}{2P}$$

Clearly, 1 + a certain quantity $$P^2$$, even though is a fraction as a number is > of 1

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

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

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