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# A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)

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A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)  [#permalink]

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21 Jan 2019, 21:15
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Difficulty:

85% (hard)

Question Stats:

21% (01:23) correct 79% (01:42) wrong based on 62 sessions

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A and B are non zero real numbers. Is A/B > B/A?

(1) A^2 > B^2

(2) A^3 > B^3
VP
Joined: 09 Mar 2018
Posts: 1001
Location: India
A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)  [#permalink]

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Updated on: 23 Jan 2019, 07:57
1
amanvermagmat wrote:
A and B are non zero real numbers. Is A/B > B/A?

(1) A^2 > B^2

(2) A^3 > B^3

non zero real numbers, can be anything from +ive integers , -ive integers to fraction, rational numbers

Question Is A/B > B/A ??

(1) A^2 > B^2

Will take some examples here
A = -2, B = - 1, 4 > 1, Question will be 2/1 > 1/2, Yes
A = -2, B = 1, 4 > 1, Question will be -2 > - 0.5, No

(2) A^3 > B^3

Will take some examples here
A = -1, B = - 2, -1 > -8, Question will be 1/2 > 2/1, No
A = 1/2, B = 1/4, 1/8 > 1/64, Question will be 2 > 1/2 , Yes

Combine both the statements
A^2 > B^2 and A^3 > B^3

We have to choose a value which will satisfy both of them( i thought x will always be greater than y, for both the inequalities to be true and they cannot be both -ive as it will not satisfy the first statement)

Now A and B, both will be +ive
A = 2 and B = 1, 4> 1 and 8 > 1, Question will be 2/1 > 1/2, Yes

A = 1/2, B = 1/4, 1/4 > 1/16 and 1/8 > 1/64, Question will be 2 > 1/2 , Yes

IMO C
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If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

Originally posted by KanishkM on 22 Jan 2019, 03:51.
Last edited by KanishkM on 23 Jan 2019, 07:57, edited 1 time in total.
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Re: A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)  [#permalink]

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22 Jan 2019, 09:25
@KanishkM:- Isn't A sufficient,

If you cross multiply the question is

Is A^2>B^2
VP
Joined: 09 Mar 2018
Posts: 1001
Location: India
Re: A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)  [#permalink]

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22 Jan 2019, 09:28
Manat wrote:
@KanishkM:- Isn't A sufficient,

If you cross multiply the question is

Is A^2>B^2

Hi Manat

But we dont know the magnitude of A and B, if they are not having same signs then ??

(1) A^2 > B^2

Will take some examples here
A = -2, B = - 1, 4 > 1, Question will be 2/1 > 1/2, Yes
A = -2, B = 1, 4 > 1, Question will be -2 > - 0.5, No

Kindly let me know if those cases are apt.
_________________

If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

Chat Moderator
Joined: 07 Mar 2016
Posts: 51
Re: A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)  [#permalink]

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22 Jan 2019, 12:53
KanishkM wrote:
amanvermagmat wrote:
A and B are non zero real numbers. Is A/B > B/A?

(1) A^2 > B^2

(2) A^3 > B^3

non zero real numbers, can be anything from +ive integers , -ive integers to fraction, rational numbers

Question Is A/B > B/A ??

(1) A^2 > B^2

Will take some examples here
A = -2, B = - 1, 4 > 1, Question will be 2/1 > 1/2, Yes
A = -2, B = 1, 4 > 1, Question will be -2 > - 0.5, No

(2) A^3 > B^3

Will take some examples here
A = -1, B = - 2, -1 > -8, Question will be 1/2 > 2/1, No
A = 1/2, B = 1/4, 1/8 > 1/64, Question will be 2 > 1/2 , Yes

Combine both the statements
A^2 > B^2 and A^3 > B^3

Now A and B, both will be +ive
A = 2 and B = 1, 4> 1 and 8 > 1, Question will be 2/1 > 1/2, Yes

A = 1/2, B = 1/4, 1/4 > 1/16 and 1/8 > 1/64, Question will be 2 > 1/2 , Yes

IMO C

What about if B=2 and A=1? Then A/B < B/A!

Since we only know the +'ve/-'ve nature of the variables and do not know the value of either A or B - I think the answer is E!
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A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)  [#permalink]

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Updated on: 22 Jan 2019, 13:26
1
We are not told if A and B are positive real numbers, so avoid the temptation to cross multiply

(1) A^2 > B^2

Let A =-2 B =-1 Then -2/-1 >? -1/-2 --? 2 > 1/2 Yes

Let A =2 B =1 Then 2/1 > 1/2 No

NS

(2) A^3 > B^3

Let A =2 B =1 Then 2/1 > 1/2 Yes

Let A =-1 B =-2 Then -1/-2 >? -2/-1 --? 1/2 > 2 ? NO

NS

(1) and (2)

Dividing 2 by 1 give A> B

Let A = -1 B = -2 Then -1/-2 >? -2/-1 1/2 > 2? No

Let A =2 B =1 Then 2/1 >? 1/2 2 > 1/2? Yes

NS

Originally posted by ocelot22 on 22 Jan 2019, 13:02.
Last edited by ocelot22 on 22 Jan 2019, 13:26, edited 3 times in total.
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Joined: 07 Mar 2016
Posts: 51
Re: A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)  [#permalink]

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22 Jan 2019, 13:12
2
ocelot22 wrote:
amanvermagmat wrote:
A and B are non zero real numbers. Is A/B > B/A?

(1) A^2 > B^2

(2) A^3 > B^3

We can rewrite the question by cross multiplying as A^2>B^2? Yes/NO

(1) A^2>B^2 Suff

(2) A^3 >B>^3 Let A =-1, B=-2. Then A^3 >B^3, but A^2 not >B^2 --- NO

Let A=2 B=1 Then A^3>B^3, and A^2 >B^2 --- Yes

Mate, what if A and/or B have a negative value?
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Re: A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)  [#permalink]

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22 Jan 2019, 13:13
1
emockus wrote:
ocelot22 wrote:
amanvermagmat wrote:
A and B are non zero real numbers. Is A/B > B/A?

(1) A^2 > B^2

(2) A^3 > B^3

We can rewrite the question by cross multiplying as A^2>B^2? Yes/NO

(1) A^2>B^2 Suff

(2) A^3 >B>^3 Let A =-1, B=-2. Then A^3 >B^3, but A^2 not >B^2 --- NO

Let A=2 B=1 Then A^3>B^3, and A^2 >B^2 --- Yes

Mate, what if A and/or B have a negative value?

I just caught it bro, editing my post as we speak.
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Joined: 09 Mar 2018
Posts: 1001
Location: India
A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)  [#permalink]

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22 Jan 2019, 17:43
ocelot22 wrote:
We are not told if A and B are positive real numbers, so avoid the temptation to cross multiply

(1) A^2 > B^2

Let A =-2 B =-1 Then -2/-1 >? -1/-2 --? 2 > 1/2 Yes

Let A =2 B =1 Then 2/1 > 1/2 No

NS

(2) A^3 > B^3

Let A =2 B =1 Then 2/1 > 1/2 Yes

Let A =-1 B =-2 Then -1/-2 >? -2/-1 --? 1/2 > 2 ? NO

NS

(1) and (2)

Dividing 2 by 1 give A> B

Let A = -1 B = -2 Then -1/-2 >? -2/-1 1/2 > 2? No

Let A =2 B =1 Then 2/1 >? 1/2 2 > 1/2? Yes

NS

Why is it that when i use the values
A=-1 & B =-2, i am unable to satisfy both the statements

A^2 > B^2 and A^3 > B^3

We have to choose a value which will satisfy both of them( i thought x will always be greater than y, for both the inequalities to be true and they cannot be both -ive as it will not satisfy the first statement)

(-1)^2 > (-2)^2 and -1 > -8
1 > 4 and -1 > -8

1 > 4 No and -1 > -8 Yes
This value cannot be used.

Can you please share your thoughts on this, i believe that value is not apt.

Posted from my mobile device
_________________

If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

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Joined: 18 Aug 2017
Posts: 1880
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)  [#permalink]

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23 Jan 2019, 02:58
amanvermagmat wrote:
A and B are non zero real numbers. Is A/B > B/A?

(1) A^2 > B^2

(2) A^3 > B^3

given
A/B>B/A
#1
A^>B^2

check with a=-2 & b=-1 and at a=-1 & b = -1/2

not sufficient
#2:
a^3>b^3

at a= 3/2 & b =1/2
and a= -1 & b= -2
not sufficeint

from 1 & 2
divide 2 by1
we get
a>b

so at a=2 and b=1
sufficient

IMO C
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Re: A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)  [#permalink]

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23 Jan 2019, 07:36
KanishkM wrote:
ocelot22 wrote:
We are not told if A and B are positive real numbers, so avoid the temptation to cross multiply

(1) A^2 > B^2

Let A =-2 B =-1 Then -2/-1 >? -1/-2 --? 2 > 1/2 Yes

Let A =2 B =1 Then 2/1 > 1/2 No

NS

(2) A^3 > B^3

Let A =2 B =1 Then 2/1 > 1/2 Yes

Let A =-1 B =-2 Then -1/-2 >? -2/-1 --? 1/2 > 2 ? NO

NS

(1) and (2)

Dividing 2 by 1 give A> B

Let A = -1 B = -2 Then -1/-2 >? -2/-1 1/2 > 2? No

Let A =2 B =1 Then 2/1 >? 1/2 2 > 1/2? Yes

NS

Why is it that when i use the values
A=-1 & B =-2, i am unable to satisfy both the statements

A^2 > B^2 and A^3 > B^3

We have to choose a value which will satisfy both of them( i thought x will always be greater than y, for both the inequalities to be true and they cannot be both -ive as it will not satisfy the first statement)

(-1)^2 > (-2)^2 and -1 > -8
1 > 4 and -1 > -8

1 > 4 No and -1 > -8 Yes
This value cannot be used.

Can you please share your thoughts on this, i believe that value is not apt.

Posted from my mobile device

I believe that what happened, is that by dividing A^3>B^3 by A^2>B^2 to get A>B, I buried one of the constraints of the problem. I am gonna make a note not to do this. Can anybody weigh in if this is the case?
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Joined: 16 Oct 2011
Posts: 61
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GMAT 5: 570 Q31 V38
Re: A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)  [#permalink]

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23 Jan 2019, 07:56
amanvermagmat wrote:
A and B are non zero real numbers. Is A/B > B/A?

(1) A^2 > B^2

(2) A^3 > B^3

Very high quality question - It is a great example of questions where you have to use attention to detail to not miss a case.
Re: A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)   [#permalink] 23 Jan 2019, 07:56
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