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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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21 Jan 2019, 22:15
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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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Re: A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)
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22 Jan 2019, 14:12
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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A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)
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Updated on: 23 Jan 2019, 08:57
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, 04:51.
Last edited by KanishkM on 23 Jan 2019, 08:57, edited 1 time in total.



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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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Updated on: 22 Jan 2019, 14:26
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, 14:02.
Last edited by ocelot22 on 22 Jan 2019, 14:26, 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)
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22 Jan 2019, 10:25
@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)
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22 Jan 2019, 10:28
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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Re: A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)
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22 Jan 2019, 13:53
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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Re: A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)
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22 Jan 2019, 14:13
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)
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22 Jan 2019, 18:43
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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Re: A and B are non zero real numbers. Is A/B > B/A? (1) A^2 > B^2 (2)
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23 Jan 2019, 03:58
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)
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23 Jan 2019, 08:36
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 deviceI 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)
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23 Jan 2019, 08:56
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)
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23 Jan 2019, 08:56






