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27 Feb 2013, 13:17
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
51% (02:30) correct
49%
(02:31)
wrong
based on 742
sessions
History
Date
Time
Result
Not Attempted Yet
01 May 2014, 00:58
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prsnt11
For this problem I'd still advice to use number plugging at one point or another.
If a is not equal to b, is 1/(a-b) > ab ?
(1) |a| > |b|. This statement implies that a is further from 0 then b. We can have 4 cases:
--------0--b--a--
-----b--0-----a--
--a-----0--b-----
--a--b--0--------
For the second case the LHS is positive, while RHS is negative: 1/(a-b) > ab;
For the fourth case the LHS is negative, while RHS is positive: 1/(a-b) < ab.
Two different answers. Not sufficient.
(2) a < b --> a - b < 0. The LHS is negative:
If a=-2 and b=1, then (1/(a-b)=-1/3) > (ab=-2);
If a=-2 and b=-1, then (1/(a-b)=-1) < (ab=2).
Two different answers. Not sufficient.
(1)+(2) We can have only the third or fourth cases from (1):
--a-----0--b-----
--a--b--0--------
We can use the same example as for (2):
If a=-2 and b=1, then (1/(a-b)=-1/3) > (ab=-2);
If a=-2 and b=-1, then (1/(a-b)=-1) < (ab=2).
Two different answers. Not sufficient.
Answer: E.
Hope it helps.
30 May 2019, 23:39
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While solving DS inequality questions, the best approach is to always breakdown the question stem if possible. To breakdown the question stem, there are three hygiene factors, that if followed will simplify your analysis of the question.
1. Always keep the RHS of the inequality as 0
2. Simplify the LHS to a product or division of values (product and division of terms are easier to analyze)
3. Always try and maintain even powered terms (as the sign of them will always be 0 or positive)
The question here is, Is 1/(a - b) > ab?
1/(a - b) - ab > 0 ----> (1 - ab(a - b))/(a - b) > 0
For this equation to hold both the numerator and denominator need to have the same sign.
Statement 1 : |a| > |b|
Squaring both sides we get a^2 - b^2 > 0 ------> (a - b)(a + b) > 0.
This means that (a - b) and (a + b) can both be positive or both negative. This though answers the question with both a YES and a NO.
Statement 2 : a < b
This tells us that a - b < 0. This makes the denominator negative, but again we do not know the sign of ab in the numerator, ab. Insufficient.
Combining the statements we have (a - b)(a + b) > 0 and a - b < 0. This implies that (a + b) is also negative. But we still are unsure about the sign of ab, ab can either be positive or negative depending on the magnitudes we take for a and b. Insufficient.
E
Hope this helps!
Aditya
CrackVerbal Academics Team
1. Always keep the RHS of the inequality as 0
2. Simplify the LHS to a product or division of values (product and division of terms are easier to analyze)
3. Always try and maintain even powered terms (as the sign of them will always be 0 or positive)
The question here is, Is 1/(a - b) > ab?
1/(a - b) - ab > 0 ----> (1 - ab(a - b))/(a - b) > 0
For this equation to hold both the numerator and denominator need to have the same sign.
Statement 1 : |a| > |b|
Squaring both sides we get a^2 - b^2 > 0 ------> (a - b)(a + b) > 0.
This means that (a - b) and (a + b) can both be positive or both negative. This though answers the question with both a YES and a NO.
Statement 2 : a < b
This tells us that a - b < 0. This makes the denominator negative, but again we do not know the sign of ab in the numerator, ab. Insufficient.
Combining the statements we have (a - b)(a + b) > 0 and a - b < 0. This implies that (a + b) is also negative. But we still are unsure about the sign of ab, ab can either be positive or negative depending on the magnitudes we take for a and b. Insufficient.
E
Hope this helps!
Aditya
CrackVerbal Academics Team
