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If A and B are nonzero numbers such that AB >0, is A  B > A  B
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25 May 2020, 20:02
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Competition Mode Question If A and B are nonzero numbers such that \(AB >0\), is \(A  B > A  B\)? (1) \(\frac{1}{A} < \frac{1}{B}\) (2) \(2A + B < 0\) Are You Up For the Challenge: 700 Level Questions
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Re: If A and B are nonzero numbers such that AB >0, is A  B > A  B
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25 May 2020, 22:50
Basically our question stem is whether A<B
Statement 1
\(\frac{1}{A} < \frac{1}{B}\)
Since A and B has same sign, we can cross multiply
A>B
If A,B<0, B>A (Yes) If A,B>0, B<A (No)
Insufficient
Statement 2
B<2A Since A and B have same sign, B must be negative.
We have no idea about whether A is smaller or bigger than B
Insufficient
Combining both statements
A,B<0 (from statement 2)
Hence B>A (Yes) (from statement 1)
Sufficient
C



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Re: If A and B are nonzero numbers such that AB >0, is A  B > A  B
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25 May 2020, 23:03
Ans: C
AB>0, is A−B>A−B AB>0, means A & B both same sign(both ve or both +ve)
(1) 1/A<1/B,(1/A)(1/B)<0,(BA)/AB<0 as AB >0, so multiplying both sides by AB BA<0,B<A Say, A=2,B=1..putting in the equation: A−B>A−B,2−1>2−1,1>1 No. A=2,B=4..putting in the equation: A−B>A−B,2+4>2−4,2>2 Yes Both yes and no possible.Not Sufficient.
(2) 2A+B<0,A<(B/2) Say,B=4,A<2..so A must be ve.But as per the question both should have the same sign.So not possible. Say,B=4,A<2,we can take only ve values for A(1,2,3,4,5,6 etc) Taking: B=4 and A=4 putting in the equation: A−B>A−B,4+4>4−4,0>0 No. Taking: B=4 and A=2 putting in the equation: A−B>A−B,2+4>2−4,2>2 Yes Both yes and no possible.Not Sufficient.
Both 1+2 A>B & Both A&B is ve Taking both 1 and 2 we can say that the given equation is possible. C is the answer.



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Re: If A and B are nonzero numbers such that AB >0, is A  B > A  B
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25 May 2020, 23:03
If A and B are nonzero numbers such that AB>0, is A−B>A−B? A(+)/B(+) OR A()/B() both have same sign. Three things to take care while solving: 1. Sign of A and B. 2. Whether A>B OR A<B 3. Whether A and B are fractions OR one of them is a fraction (1) \(\frac{1}{A}<\frac{1}{B}\) Multiplying both sides by AB as AB>0 \(\frac{AB}{A}<\frac{AB}{B}\) B < A But nothing about the signs and fractions Case I: A = 6, B = 2; 6−2>6−2 i.e. 4 > 4 NO Case II: A = 2, B = 6; 2−(6)>2−6 i.e. 4 > 4 YES INSUFFICIENT. (2) 2A+B<0 Hence both A and B are negative. But whether A>B OR A<B and fractions ?Case I: A = 6, B = 2; 6−(2)>6−2 i.e. 4 > 4 NO Case II: A = 2, B = 6; 2−(6)>2−6 i.e. 4 > 4 YES INSUFFICIENT. Together 1 and 2 All the three conditions are known now. Sufficient then. Just to verify. Case II of statements applicable still. YES case. Case III: \(A = \frac{1}{3}, B = \frac{1}{2}; \frac{1}{3}−(\frac{1}{2})>\frac{1}{3}−\frac{1}{2} i.e. \frac{1}{6} > \frac{1}{6}\) YES SUFFICIENT. Answer C. PS: Case III would have been only necessary had the signs differed with some other conditions. Thus, it's unnecessary in this question to check fractions.
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If A and B are nonzero numbers such that AB >0, is A  B > A  B
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Updated on: 27 May 2020, 10:05
AB > 0 > A and B have same sign
Q. is A  B > A  B?
NOTE: If A>B and both are +ve, then A  B = A  B If A>B and both are ve, then A  B > A  B
If A<B and both are +ve, then A  B > A  B If A<B and both are ve, then A  B = A  B
From above observation, one have to know whether A>B and whether both A, B are +ve or ve
(1) 1/A < 1/B > A > B But we don't know the sign of A and B (both are +ve or ve) NOT SUFFICIENT
(2) 2A + B < 0 > B > 2A or B < 2A  A and B cannot be +ve, so A and B must be ve  B < 2A, Bu we are not sure whether A>B.
(1)+(2) A > B ; A, B are both ve So, as discussed above , A  B > A  B
FINAL ANSWER IS (C)
Posted from my mobile device
Originally posted by chondro48 on 25 May 2020, 23:05.
Last edited by chondro48 on 27 May 2020, 10:05, edited 2 times in total.



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Re: If A and B are nonzero numbers such that AB >0, is A  B > A  B
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25 May 2020, 23:05
If A and B are nonzero numbers such that AB>0, is A−B>A−B? (1) 1/A<1/B (2) 2A+B<0 Given: A,B \(\ne0\), A,B can be (+,+) or (,). statement 1: 1/A < 1/B A \(\ne\) B possible cases: (+,+): A > B then A−B>A−B is false; consider (4,2) or (0.5,0.3) (,): B > A then A−B>A−B is true; consider (2,4) or (0.3,0.5) not sufficient statement 2: 2A + B < 0 A,B \(\ne\)(+,+) possible cases (,): A > B then A−B>A−B is false; consider (4,2) (,): B > A then A−B>A−B is true; consider (2,4) not sufficient combining both statements, only possible case : (,) & B > A. sufficient Ans: C



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Re: If A and B are nonzero numbers such that AB >0, is A  B > A  B
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26 May 2020, 03:42
Quote: If A and B are nonzero numbers such that AB>0, is A−B>A−B?
(1) 1/A<1/B (2) 2A+B<0 ab>0 = same signs ab>ab when a<b everything else, condition is false (1) sufic a b 1/a < 1/b 3 2 1/3 < 1/2 a>b 3 2 1/3 < 1/2 invalid 1/3 1/2 3 < 2 a>b 1/3 1/2 3 < 2 invalid (2) insufic 2a+b<0 a,b=1,1: 21<0 a=b a,b=2,1: 41<0 a<b a,b=1,2: 22<0 a>b ans (A)



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Re: If A and B are nonzero numbers such that AB >0, is A  B > A  B
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26 May 2020, 07:13
target is A−B>A−B given AB>0 possible when both ab are +ve or both are ve #1 1/A<1/B B/A<1 means A>B since given that both (a,b) are + or ve case 1 ; both a & b are +ve a = 2 and b = 1 A−B>A−B 21=21 ; no case 2 both a&b are ve a=1 and b = 2 l1+2l>l1ll2l 1>1 ; yes insufficient #2 2A+B<0 given that both (a,b) are + or ve a = b=1 A−B>A−B no and both are ve a=1 and b = 2 l1+2l>l1ll2l 1>1 ; yes ; insufficient from 1 &2 only possible case is that both a&b are ve integers and a>b OPTION C ; sufficient
If A and B are nonzero numbers such that AB>0, is A−B>A−B? (1) 1/A<1/B
(2) 2A+B<0



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Re: If A and B are nonzero numbers such that AB >0, is A  B > A  B
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26 May 2020, 10:49
Tough one for me . Is the ans C ? Below is my thought process . If A and B are nonzero numbers such that AB>0, is A−B>A−B? So both A and B have same sign . Both positive or both negative. A−B>A−B? Distance of A to B > distance of A to zero  distance of B to zero ? This is only possible if A < B . So we need to find out A < B ? (1) 1/A<1/B Case 1 : If A ,B > 0 B < A B < A . Case 2 : If A , B < 0 . 1/A < 1/B lets put a value . 1/2 < 1/3 so A = 2 B = 3 so B < A but B > A.. That's why we cannot conclusively tell that A < B . SO insufficient . (2) 2A+B<0 = A < B/2 . Lets take B as 1 , A < 1/2 (Not possible since A B both same sign .) lets take B = 2 . A < 1 . A can be 1.9 ,2 or 3 . So for different value of A we can get different relation of A < B or A = B or A > B So not sufficient . Now combing 1 and 2 . A , B < 0 and B > A. so we can conclusively tell that A−B>A−B . since B = 2 and A = 1.2 0.8 > 1.2  2 => 0.8 > 0.8 . So option C could be my answer .
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Re: If A and B are nonzero numbers such that AB >0, is A  B > A  B
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26 May 2020, 11:28
Quote: If A and B are nonzero numbers such that AB>0, is A−B>A−B?
(1) \(\frac{1}{A}<\frac{1}{B}\)
(2) 2A+B<0 What an intriguing question! There's a lot going on here. Let's slow down and unpack it bit by bit. First, we have AB>0. That means A and B have the same sign: either they're both positive, or they're both negative. Now, what we want to know: is A−B>A−B?. I spent some time trying to simplify this, but the simplification took a long time, and was even harder to explain than it was to do! So, I think the right approach on test day would be to leave this how it is, write it down on your paper, and get ready to carefully and thoughtfully test cases. Let's do that now. Statement 1: 1/A < 1/B. I'd like to simplify this, but I'm wary, because I know that you usually can't multiply both sides of an inequality by a variable. That's because if you don't know whether the variable is positive or negative, you don't know whether you're supposed to flip the sign or not. But we have something we can actually use here! Because the problem says that AB > 0, we know that AB is positive. So, we can safely multiply both sides of the inequality by AB. 1/A * AB < 1/B * AB B < A This statement really says that B < A. Now let's try some cases where B < A. We're dealing with absolute values, so we should try both positive and negative cases. Case 1: B = 1, A = 10. Is A−B>A−B? A  B = 10  1 = 9 = 9 A  B = 101 = 10  1 = 9 Since they're equal, the answer is "no" Case 2: B = 10, A = 1. Is A−B>A−B? A  B = 1  (10) = 9 = 9 A  B = 1  10 = 1  10 = 9 The left side is bigger, so the answer is "yes." We got both types of answers, so this statement is insufficient. Statement 2: 2A+B<0 We know, from the original question, that A and B are either both positive or both negative. They definitely can't both be positive anymore, though, given this statement! They must both be negative. And, if they're both negative... this statement is definitely true! This will be true for any pair of negative values of A and B. So really, all this is telling us (in the context of this specific problem and the info we already have) is that A and B are both negative. Let's test some cases. Case 1: First, reuse 'case 2' from the previous statement. That case gave us a "yes" answer. Case 2: Now, I'm going to try to find a case that works the other way. It seemed to be important whether A is bigger or whether B is bigger, so this time I'm going to try a case where B is bigger than A. Let's say B = 3, and A = 8. Is A  B > A  B? A  B = 8  (3) = 5 = 5 A  B = 8  3 = 8  3 = 5 They're equal, so the answer is "no." Since we got both a "yes" and a "no," this statement is also insufficient. Statements 1 and 2 together: Let's evaluate what we know at this point. We know from Statement 2 that A and B are both negative. We also know from statement 1 that A > B. Is A  B > A  B? Since A > B, A  B will be a positive number. Therefore, A  B = A  B. Since A and B are both negative, A = A, and B = B. The question is really asking: Is A  B > A  (B)? Simplify: Is A  B > B  A? Is 2A > 2B? Is A > B? We already know that the answer to this question is "yes." So, the statements are sufficient together, and the answer to the problem is C. Nice one!
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Re: If A and B are nonzero numbers such that AB >0, is A  B > A  B
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26 May 2020, 14:29
If A and B are nonzero numbers such that AB>0AB>0, is A−B>A−BA−B>A−B?
(1) 1A<1B1A<1B
(2) 2A+B<0 Since, AB > 0, so both of them are of same sign. Another thing is that the left side will be positive. so if A and B are not both negative, the left side will always be greater than the right one. 1) A > B, when A = 2, B = 0.5. Both sides become 1.5. When A = 5, B =6, 1 > 1. Two different answers. Not sufficient 2) 2A + B < 0. B < 2A. B is negative, so is A. When A = 1, B = 3. Both sides are equal. when A = 1, B = 1, both sides become 0. as, A+B= BA =AB = AB and A  B= A B. Sufficient B is the answer



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Re: If A and B are nonzero numbers such that AB >0, is A  B > A  B
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26 May 2020, 22:48
AB>A  B only possible when A > 0 and B < 0 or A < 0 and B > 0
1. 1/A < 1/B A > B if A > 0 and B > 0 A < B if A < 0 and B < 0
no other information, not sufficient
2. 2A + B > 0 A should be greater than 0, B should be greater than zero, 2 is already greater than 0 Sufficient
Answer  B



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Re: If A and B are nonzero numbers such that AB >0, is A  B > A  B
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27 May 2020, 06:23
nick1816 wrote: Basically our question stem is whether A<BC Hello nick1816 how did you conclude that this is the question stem?



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If A and B are nonzero numbers such that AB >0, is A  B > A  B
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27 May 2020, 10:21
\(A  B > A  B\) Since AB>0, the magnitude of AB is equal to magnitude of AB. Hence only way LHS is greater than RHS, if RHS is negative. [Since LHS is always positive} A  B < 0 A < B If you still have doubt, you can ask. Ruchirkalra wrote: nick1816 wrote: Basically our question stem is whether A<BC Hello nick1816 how did you conclude that this is the question stem?




If A and B are nonzero numbers such that AB >0, is A  B > A  B
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