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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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New post 21 Jan 2019, 21:15
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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
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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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New post 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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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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New post 22 Jan 2019, 09:25
@KanishkM:- Isn't A sufficient,

The question asks A/B>B/A

If you cross multiply the question is

Is A^2>B^2
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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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New post 22 Jan 2019, 09:28
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.
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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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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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New post 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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New post 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

Answer is E

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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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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New post 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

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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New post 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

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. ;)
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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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New post 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

Answer is E


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
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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.

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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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New post 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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New post 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

Answer is E


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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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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New post 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.
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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] 23 Jan 2019, 07:56
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