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Moderator  Joined: 18 Apr 2015
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a > 0 [#permalink]
Expert's post 00:00

Question Stats: 83% (00:36) correct 16% (00:34) wrong based on 18 sessions
$$a > 0$$

 Quantity A Quantity B $$(a + a^{-1})^2$$ $$a^2 + a^{-2}$$

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.

kudo for the right solution and explanation
[Reveal] Spoiler: OA

_________________ Manager Joined: 25 Nov 2017
Posts: 51
Followers: 0

Kudos [?]: 41  , given: 5

Re: a > 0 [#permalink]
1
KUDOS
QA (A + A^-1)^2
A^2 + 2(A)(1/A) + (1/A)^2
A^2 + 2 + 1/A^2
QB
A^2 + A^-2 = A^2 + 1/A^2

Lets substract both side 1/A^2

QA A^2 +2
QB A^2
since we know that is positive, we can substract each side by A^2
QA = +2
QB = 0
So A

If we want to plu values, if this case we need to test fraction and interger positive right? (E.g. 1/2 and 2)
If this question did not mention anything about a, we could not do what I did?
How could we deal if we only had the two quantities and nothing about a, it could either -4, -(1/4), 1/4 and 4? Director Joined: 20 Apr 2016
Posts: 824
WE: Engineering (Energy and Utilities)
Followers: 10

Kudos [?]: 601  , given: 121

Re: a > 0 [#permalink]
2
KUDOS
Carcass wrote:
$$a > 0$$

 Quantity A Quantity B $$(a + a^{-1})^2$$ $$a^2 + a^{-2}$$

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.

kudo for the right solution and explanation

From Statement 1 , we have

$$(a + a^{-1})^2$$ = $$(a + \frac{1}{a})^2$$ = $$(a^2 + \frac{1}{a^2}) +2$$

From Statement 2, we have

$$a^2 + a^{-2}$$ = $$a^2 + \frac{1}{a^2}$$

From the two statement we find out that Statement 1 has number 2 added to $$(a^2 + \frac{1}{a^2})$$

so for any values of a>0 , Statement 1 will always be greater than statement 2

Hence Option - A
_________________

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Rules for Posting https://greprepclub.com/forum/rules-for ... -1083.html Re: a > 0   [#permalink] 22 Dec 2017, 22:32
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