It is currently 10 Dec 2018, 12:18

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# a > 0

Author Message
TAGS:
Moderator
Joined: 18 Apr 2015
Posts: 5126
Followers: 76

Kudos [?]: 1028 [0], given: 4631

a > 0 [#permalink]  21 Dec 2017, 16:53
Expert's post
00:00

Question Stats:

84% (00:35) correct 15% (00:11) wrong based on 13 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 [1] , given: 5

Re: a > 0 [#permalink]  22 Dec 2017, 00:26
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: 756
Followers: 6

Kudos [?]: 511 [2] , given: 94

Re: a > 0 [#permalink]  22 Dec 2017, 22:32
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
_________________

If you found this post useful, please let me know by pressing the Kudos Button

Re: a > 0   [#permalink] 22 Dec 2017, 22:32
Display posts from previous: Sort by