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If abc ≠ 0, is a < b < c ?

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If abc ≠ 0, is a < b < c ?  [#permalink]

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New post Updated on: 12 Aug 2014, 07:05
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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 ??

Originally posted by ashutoshbarawkar on 04 Aug 2014, 18:40.
Last edited by Bunuel on 12 Aug 2014, 07:05, 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 ?  [#permalink]

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New post 04 Aug 2014, 20:35
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 ?  [#permalink]

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New post 05 Aug 2014, 10:00
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>a
Situation 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>a

Conclusion: 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 ?  [#permalink]

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New post 05 Aug 2014, 10:03
1
1
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>a
Situation 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>a

Conclusion: 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 ?  [#permalink]

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New post 05 Aug 2014, 11:01
1
1
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 ?  [#permalink]

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New post 05 Aug 2014, 13:22
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>a
Situation 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>a

Conclusion: 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 ?  [#permalink]

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New post 06 Aug 2014, 06:36
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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 ?  [#permalink]

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New post 08 Aug 2014, 06:57
Guys do we have an OA for this?
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Re: If abc ≠ 0, is a < b < c ?  [#permalink]

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New post 08 Aug 2014, 11:54
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 ?  [#permalink]

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New post Updated on: 27 Sep 2015, 00:37
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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.

Originally posted by Nevernevergiveup on 13 Aug 2015, 04:57.
Last edited by Nevernevergiveup on 27 Sep 2015, 00:37, edited 2 times in total.
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Re: If abc ≠ 0, is a < b < c ?  [#permalink]

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New post 13 Aug 2015, 06:38
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 ?  [#permalink]

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