If a b, is 1/(a-b) ab ? (1) |a| |b| (2) a

Question

If a ≠ b, is 1/(a - b) > ab ?

(1) |a| > |b|
(2) a < b

Bunuel · Accepted Answer

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.

CrackverbalGMAT · Answer

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

mau5 · Answer

From F.S 1, let's assume a = -3 and b = -2. Thus, 1/(a-b) = -1 and a*b = 6. Thus, as -1<6, the answer to the question stem is No. Again, pick a = -3 and b = 2, and 1/(a-b) = -0.2, and a*b = -6. In this case we see that -0.2>-6, thus the answer to the question stem is a YES. Insufficient. From F.S 2, lets again assume a = -3 and b = -2. Just as above we still get a NO. Again choosing the same set for a = -3 and b = 2, we get a YES to the question stem. Insufficient. Combining both, we know that b-a>0 and mod(a)-mod(b)>0. Thus lets choose a=-7 and b=-2. We get 1/(a-b) = -0.2 and a*b = 14. Thus a NO. Again, choosing b=3 and a=-5, we get a YES . Insufficient. Basically, the two fact statements given together mean that (a+b)<0. It's because from F.S 1, we get a^2-b^2>0 or (a-b)*(a+b)>0. We have from F.S 2 that a-b<0. Thus, (a+b) has to be negative. E.

unceldolan · Answer

(1) |a| > |b| Clearly IS. Look at this: a > b -a > b - a > -b a > -b Would give you various answers for the YES/NO Question. IS! (2) a a*b. IS. Answer E.

prsnt11 · Answer

Hey Karishma & Bunuel,
Is there a faster way to solve this problem? I tried picking numbers but it took me more than 2 mins to arrive at the answer.
Thanks,
-Prasoon

AdjustedEbitdad · Answer

Do we assume that a & b are integers? I think Non-integer values also create some other cases and eventually lead to E as the right choice.

Bunuel · Answer

No, a and b are not necessarily integers, nor are they assumed to be in the solutions above.

P.S. Also, please note that pure algebraic questions like the above are no longer included in the syllabus of the new GMAT Focus exam format.

LamboWalker · Answer

Woah, I am pleasantly suprised to hear this out. Is this change only applicable for the new DS syllabus?

Bunuel · Answer

Yes. Ppure algebraic questions are no longer a part of the  DS syllabus  of the GMAT.

DS questions in GMAT Focus encompass various types of word problems, such as:
 
 Word Problems 
 Work Problems 
 Distance Problems 
 Mixture Problems 
 Percent and Interest Problems 
 Overlapping Sets Problems 
 Statistics Problems 
 Combination and Probability Problems

While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won't encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."

Check  GMAT Syllabus for Focus Edition

Hope it helps.

Prakruti_Patil · Answer

Thank you for sharing the number line visualization of testing cases  Bunuel  ! Super helpful!

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If a b, is 1/(a-b) > ab ? (1) |a| > |b| (2) a < b

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