GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Jan 2019, 08:30

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### Key Strategies to Master GMAT SC

January 26, 2019

January 26, 2019

07:00 AM PST

09:00 AM PST

Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.
• ### Free GMAT Number Properties Webinar

January 27, 2019

January 27, 2019

07:00 AM PST

09:00 AM PST

Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes.

# If abc ≠ 0, is a < b < c ?

Author Message
TAGS:

### Hide Tags

Intern
Joined: 26 Jun 2014
Posts: 6
If abc ≠ 0, is a < b < c ?  [#permalink]

### Show Tags

Updated on: 12 Aug 2014, 06:05
2
14
00:00

Difficulty:

75% (hard)

Question Stats:

50% (02:04) correct 50% (01:38) wrong based on 361 sessions

### HideShow timer Statistics

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, 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.
Manager
Joined: 21 Jul 2014
Posts: 124
Re: If abc ≠ 0, is a < b < c ?  [#permalink]

### Show Tags

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"
Intern
Joined: 29 May 2014
Posts: 6
Re: If abc ≠ 0, is a < b < c ?  [#permalink]

### Show Tags

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>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.
Manager
Joined: 21 Jul 2014
Posts: 124
Re: If abc ≠ 0, is a < b < c ?  [#permalink]

### Show Tags

05 Aug 2014, 09:03
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.
Intern
Joined: 05 Jun 2012
Posts: 32
GMAT 1: 480 Q48 V9
Re: If abc ≠ 0, is a < b < c ?  [#permalink]

### Show Tags

05 Aug 2014, 10:01
2
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..
Intern
Joined: 29 May 2014
Posts: 6
Re: If abc ≠ 0, is a < b < c ?  [#permalink]

### Show Tags

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>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?
Manager
Joined: 21 Jul 2014
Posts: 124
Re: If abc ≠ 0, is a < b < c ?  [#permalink]

### Show Tags

06 Aug 2014, 05:36
1
2
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.
Senior Manager
Joined: 07 Apr 2012
Posts: 359
Re: If abc ≠ 0, is a < b < c ?  [#permalink]

### Show Tags

08 Aug 2014, 05:57
Guys do we have an OA for this?
Intern
Joined: 29 May 2014
Posts: 6
Re: If abc ≠ 0, is a < b < c ?  [#permalink]

### Show Tags

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.
Retired Moderator
Joined: 18 Sep 2014
Posts: 1112
Location: India
If abc ≠ 0, is a < b < c ?  [#permalink]

### Show Tags

Updated on: 26 Sep 2015, 23:37
3
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

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

Originally posted by Nevernevergiveup on 13 Aug 2015, 03:57.
Last edited by Nevernevergiveup on 26 Sep 2015, 23:37, edited 2 times in total.
CEO
Joined: 20 Mar 2014
Posts: 2636
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: If abc ≠ 0, is a < b < c ?  [#permalink]

### Show Tags

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.
Non-Human User
Joined: 09 Sep 2013
Posts: 9462
Re: If abc ≠ 0, is a < b < c ?  [#permalink]

### Show Tags

20 Mar 2018, 04:07
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: If abc ≠ 0, is a < b < c ? &nbs [#permalink] 20 Mar 2018, 04:07
Display posts from previous: Sort by