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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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Question Stats: 24% (01:43) correct 76% (01:57) wrong based on 73 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
Director  G
Joined: 09 Mar 2018
Posts: 994
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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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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Quote which i can relate to.
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Originally posted by KanishkM on 22 Jan 2019, 04:51.
Last edited by KanishkM on 23 Jan 2019, 08:57, edited 1 time in total.
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Schools: DeGroote "22 (S)
GMAT 1: 500 Q39 V21 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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@KanishkM:- Isn't A sufficient,

The question asks A/B>B/A

If you cross multiply the question is

Is A^2>B^2
Director  G
Joined: 09 Mar 2018
Posts: 994
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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Manat wrote:
@KanishkM:- Isn't A sufficient,

The question asks A/B>B/A

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 S
Joined: 07 Mar 2016
Posts: 49
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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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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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, 14:02.
Last edited by ocelot22 on 22 Jan 2019, 14:26, edited 3 times in total.
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Posts: 49
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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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

The answer is A

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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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

The answer is A

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

I just caught it bro, editing my post as we speak. Director  G
Joined: 09 Mar 2018
Posts: 994
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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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: 5031
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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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
Manager  G
Joined: 16 Oct 2011
Posts: 108
GMAT 1: 570 Q39 V41 GMAT 2: 640 Q38 V31 GMAT 3: 650 Q42 V38 GMAT 4: 650 Q44 V36 GMAT 5: 570 Q31 V38 GPA: 3.75
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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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?
Manager  G
Joined: 16 Oct 2011
Posts: 108
GMAT 1: 570 Q39 V41 GMAT 2: 640 Q38 V31 GMAT 3: 650 Q42 V38 GMAT 4: 650 Q44 V36 GMAT 5: 570 Q31 V38 GPA: 3.75
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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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, 08:56
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