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If abc ≠ 0, is a < b < c ?
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Updated on: 12 Aug 2014, 06:05
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If abc ≠ 0, is a < b < c ? (1) 1/c < 1/b < 1/a (2) c > a Can some one explain why we cannot do the following 1/c<1/b<1/a = c>b>a by taking reciprocals ??
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Originally posted by ashutoshbarawkar on 04 Aug 2014, 17:40.
Last edited by Bunuel on 12 Aug 2014, 06:05, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: If abc ≠ 0, is a < b < c ?
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04 Aug 2014, 19:35
ashutoshbarawkar wrote: If abc ≠ 0, is a < b < c ? (1) 1/c < 1/b < 1/a (2) c > a  Can some one explain why we cannot do the following 1/c<1/b<1/a = c>b>a by taking reciprocals ?? You may have forgotten to take into account negative numbers? Here are two scenarios that show that statement 1 alone is not sufficient: a=1; b=2; c=3 > Statement 1 is still valid and your response to the question would be "True" a=1; b=2; c=3 > Statement 1 is still valid, but your response to the question would be "False"



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Re: If abc ≠ 0, is a < b < c ?
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05 Aug 2014, 09:00
ashutoshbarawkar wrote: If abc ≠ 0, is a < b < c ? (1) 1/c < 1/b < 1/a (2) c > a  Can some one explain why we cannot do the following 1/c<1/b<1/a = c>b>a by taking reciprocals ?? I agree with Lighthouseprep. You need to consider the situation that each number is negative or positive. Below is my solution for this problem: 1/c <1/ b >c> b if bc >0 (I) or c < b if bc <0 (II) ( bc > 0 means b and c are both positive or both negative numbers and bc < means one number is positive and the other is negative) 1/b < 1/ a > b>a if ab > 0 (III) or b < a if ab <0 (IV) => Statement (1) 1/c < 1/b < 1/a: not sufficient Statement (2) c>a: not sufficient because we don't know anything about b. Combine statement (1) & (2) 1/c <1/a > c> a if ac >0 or c < a if ac <0 From (2): c> a > ac>0 then a > 0 and c> 0 (V) or a < 0 and c< 0 (VI) From (I) (II) (III) (IV) (V) (VI), we have 4 situation. Situation 1: (IV) a > 0 and c> 0 & (II): c<b if bc < 0. This situation CANNOT happen because bc<0 and c >0 so b <0 which cannot make c< b. Situation 2: (IV) a>0 and c> 0 & (I) c> b if bc >0. In this situation c>b>0. Because b>0 and a> 0 then we have (II) b>a if ab > 0. We prove that c>b>aSituation 3: (V) a < 0 and c< 0 & (II): c<b if bc < 0. In this situation, b>0>c. When b>0>a, we have ab<0 which is the condition for (III) b < a. This situation CANNOT happen. Situation 4: (V) a<0 and c< 0 & (I) c> b if bc >0. In this situation, 0>c>b. When 0>b and 0> a we have ab >0 which is the condition for (III): b>a. We prove that c>b>aConclusion: Statement (1) and (2) together are sufficient. It takes me 7 minutes to answer and 20 minutes to think how I can explain my answer. If anyone can solve this problem differently, please let me know. I would appreciate your help and send you kudos.



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Re: If abc ≠ 0, is a < b < c ?
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05 Aug 2014, 09:03
linhntle wrote: ashutoshbarawkar wrote: If abc ≠ 0, is a < b < c ? (1) 1/c < 1/b < 1/a (2) c > a  Can some one explain why we cannot do the following 1/c<1/b<1/a = c>b>a by taking reciprocals ?? I agree with Lighthouseprep. You need to consider the situation that each number is negative or positive. Below is my solution for this problem: 1/c <1/ b >c> b if bc >0 (I) or c < b if bc <0 (II) ( bc > 0 means b and c are both positive or both negative numbers and bc < means one number is positive and the other is negative) 1/b < 1/ a > b>a if ab > 0 (III) or b < a if ab <0 (IV) => Statement (1) 1/c < 1/b < 1/a: not sufficient Statement (2) c>a: not sufficient because we don't know anything about b. Combine statement (1) & (2) 1/c <1/a > c> a if ac >0 or c < a if ac <0 From (2): c> a > ac>0 then a > 0 and c> 0 (V) or a < 0 and c< 0 (VI) From (I) (II) (III) (IV) (V) (VI), we have 4 situation. Situation 1: (IV) a > 0 and c> 0 & (II): c<b if bc < 0. This situation CANNOT happen because bc<0 and c >0 so b <0 which cannot make c< b. Situation 2: (IV) a>0 and c> 0 & (I) c> b if bc >0. In this situation c>b>0. Because b>0 and a> 0 then we have (II) b>a if ab > 0. We prove that c>b>aSituation 3: (V) a < 0 and c< 0 & (II): c<b if bc < 0. In this situation, b>0>c. When b>0>a, we have ab<0 which is the condition for (III) b < a. This situation CANNOT happen. Situation 4: (V) a<0 and c< 0 & (I) c> b if bc >0. In this situation, 0>c>b. When 0>b and 0> a we have ab >0 which is the condition for (III): b>a. We prove that c>b>aConclusion: Statement (1) and (2) together are sufficient. It takes me 7 minutes to answer and 20 minutes to think how I can explain my answer. If anyone can solve this problem differently, please let me know. I would appreciate your help and send you kudos. Hi linhntle, Did you try the plugging in numbers method that I used? With that, I found the answer much quicker.



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Re: If abc ≠ 0, is a < b < c ?
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05 Aug 2014, 10:01
ok guys.. so here i considered a=1 b= 2 n c =3 ..also a=1 ,b=2 n c=3 ..n i found out out C is correct...so i hope numbers will help in this case...please let me know if am wrong..



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Re: If abc ≠ 0, is a < b < c ?
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05 Aug 2014, 12:22
LighthousePrep wrote: linhntle wrote: ashutoshbarawkar wrote: If abc ≠ 0, is a < b < c ? (1) 1/c < 1/b < 1/a (2) c > a  Can some one explain why we cannot do the following 1/c<1/b<1/a = c>b>a by taking reciprocals ?? I agree with Lighthouseprep. You need to consider the situation that each number is negative or positive. Below is my solution for this problem: 1/c <1/ b >c> b if bc >0 (I) or c < b if bc <0 (II) ( bc > 0 means b and c are both positive or both negative numbers and bc < means one number is positive and the other is negative) 1/b < 1/ a > b>a if ab > 0 (III) or b < a if ab <0 (IV) => Statement (1) 1/c < 1/b < 1/a: not sufficient Statement (2) c>a: not sufficient because we don't know anything about b. Combine statement (1) & (2) 1/c <1/a > c> a if ac >0 or c < a if ac <0 From (2): c> a > ac>0 then a > 0 and c> 0 (V) or a < 0 and c< 0 (VI) From (I) (II) (III) (IV) (V) (VI), we have 4 situation. Situation 1: (IV) a > 0 and c> 0 & (II): c<b if bc < 0. This situation CANNOT happen because bc<0 and c >0 so b <0 which cannot make c< b. Situation 2: (IV) a>0 and c> 0 & (I) c> b if bc >0. In this situation c>b>0. Because b>0 and a> 0 then we have (II) b>a if ab > 0. We prove that c>b>aSituation 3: (V) a < 0 and c< 0 & (II): c<b if bc < 0. In this situation, b>0>c. When b>0>a, we have ab<0 which is the condition for (III) b < a. This situation CANNOT happen. Situation 4: (V) a<0 and c< 0 & (I) c> b if bc >0. In this situation, 0>c>b. When 0>b and 0> a we have ab >0 which is the condition for (III): b>a. We prove that c>b>aConclusion: Statement (1) and (2) together are sufficient. It takes me 7 minutes to answer and 20 minutes to think how I can explain my answer. If anyone can solve this problem differently, please let me know. I would appreciate your help and send you kudos. Hi linhntle, Did you try the plugging in numbers method that I used? With that, I found the answer much quicker. Hi LightHouse, Can you explain how you find the answer by plugging in number?



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Re: If abc ≠ 0, is a < b < c ?
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06 Aug 2014, 05:36
linhntle,
Correct me if I'm wrong, but I think you are asking how I chose which numbers I should plug in?
I think for this question to be done quickly, it requires a conceptual understanding of how positive and negative numbers and the magnitude of those numbers impact fractions. Unfortunately, I don't know a hard and fast rule about which numbers you should plug in to questions, but I think using the question to give you clues on what to plug in typically works. In this particular case, I used statement 2 to guide the numbers I plugged in for statement 1. Knowing that statement 2 specified that c > a, I thought to check statement 1 by plugging in a number where c < a.
One other tip I would offer is that people often try to do the statements in order (first do statement 1, and then do statement 2). I think a more effective strategy is to go after the simpler statement first. In this case, statement 2 was much simpler to analyze than statement 1. It only took me a couple seconds to figure out that without mentioning anything about b, I would have no way of finding the solution.



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Re: If abc ≠ 0, is a < b < c ?
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08 Aug 2014, 05:57
Guys do we have an OA for this?



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Re: If abc ≠ 0, is a < b < c ?
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08 Aug 2014, 10:54
ronr34 wrote: Guys do we have an OA for this? OA stands for Official Answer, right? If yes, OA is C. Let me know if my response doesn't answer your question.



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If abc ≠ 0, is a < b < c ?
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Updated on: 26 Sep 2015, 23:37
If abc ≠ 0, is a < b < c ? (1) 1/c < 1/b < 1/a
take two cases of number sets satisfying the statement 1
since \(\frac{1}{3}<\frac{1}{2}<\frac{1}{1}\) case I: a=1 b=2 c=3 1<2<3 so a<b<c these values give an answer YES to above YES/NO question.
Also \(\frac{1}{3}<\frac{1}{2}<\frac{1}{1}\) so lets take case II: a=1 b=2 c=3 1<2 but not <3 so a<b<c is not true in this case Answer:NO
So statement I is not sufficient. so B, C and E are left.
(2) c > a
3>1 or 1>3 but there is no value or constraint for B. case I a=1 b=2 c=3 satisfies c>a but a<b<c is not true in this case case II a=1 b=5 c=3 satisfies c>a but a<b<c is not true in this case case III a=1 b=2 c=3 satisfies c>a and a<b<c is also true in this case case IV a=3 b=2 c=1 satisfies c>a and a<b<c is also true in this case
therefore Statement II is also not sufficient.
So Either C or E.
(1) and (2) together
c>a and \(\frac{1}{c}<\frac{1}{a}\) so both C and A should have same sign i.r., either positive or negative and b should be in between a and c as in below cases a=1 b=2 c=3 a=1 b=2 c=3
So the answer is C.



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Re: If abc ≠ 0, is a < b < c ?
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13 Aug 2015, 05:38
Mechmeera wrote: If abc ≠ 0, is a < b < c ? (1) 1/c < 1/b < 1/a (2) c > a PLEASE SEARCH FOR A QUESTION BEFORE POSTING.



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Re: If abc ≠ 0, is a < b < c ?
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20 Mar 2018, 04:07
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