If abc 0, is a < b < c ? (1) 1/c < 1/b < 1/a (2) c > a

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"

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.

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..

Hi LightHouse,
Can you explain how you find the answer by plugging in number?

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.

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.

PLEASE SEARCH FOR A QUESTION BEFORE POSTING.

Hi, Please explain why/if this is wrong,
For statement 1,
\(1/c < 1/b < 1/a\)
Multiply throughout by abc

\(abc*1/c < abc* 1/b < abc*1/a\)
\(ab < ac < bc\)
\(So:: ab<ac ; ac<bc ; ab<bc\)
\(=> b<c ; a<b ; a<c\)
\(=> a<b<c\)

Ans (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

[quote="Boltions"]Hi, Please explain why/if this is wrong,
For statement 1,
[m]1/c < 1/b < 1/a[/m]
Multiply throughout by abc

[m]abc*1/c < abc* 1/b < abc*1/a[/m]
[m]ab < ac < bc[/m]
[m]So:: ab<ac ; ac<bc ; ab<bc[/m]
[m]=> b<c ; a<b ; a<c[/m]
[m]=> a<b<c[/m]

Ans (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.[/quote]


The operation would not be correct if abc<0.

When an ineauality is multiplied with a negetive number then the signs change.

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St1: 1/a > 1/c.
If both a and c are positive, then c>a (simply cross multiply).
If a is positive and c is negative, then a>c.
Insufficient.

St2: c>a
No information on b.
Insufficient.

St1+2:
As 1/a<1/c and c>a, we can conclude that the cross multiplication is applicable. That means both a and c and therefore 1/a and 1/c are positive.
As 1/b falls between 1/a and 1/c, we can deduce that 1/b is also positive and b lies between a and c.
So, inequality is c>b>a.
Sufficient.

Hence, Ans C

What I did was drawing a number line with a b and c, with S1 consider 4 situations:
1. all three variables are >0, in which case a<b<c
2. all three are <0, in which case a<b<c
3. 2 variables >0, in which case c<b<a
4. 1 variable >0, in which case b<c<a
and that's all the cases, clearly not sufficient to answer if a<b<c?

but S2 tells you c>a, which means only case 1 and 2 are possible, hence answer is C.

(1) NOT SUFFICIENT: If a = 1, b = 2, and c = 3, then the statement is satisfied (1/3 < 1/2 < 1), and the answer to the question is yes, a < b < c.

If a = 1, b = –2, and c = –1, then the statement is satisfied (–1 < –1/2 < 1), but the answer to the question is no, it’s not true that a < b < c.

(2) NOT SUFFICIENT: In this case a < c, yes, but the statement indicates nothing about b.

(1) AND (2) SUFFICIENT: First, consider only a and c. According to statement two, a < c. There are three possibilities: both negative, both positive, or a negative, c positive. According to statement one, though, 1/c < 1/a. If a were negative and c positive, then taking the reciprocals would leave 1/a less than 1/c. (For example, if a = -1, c = 1, then 1/a would have to be less than 1/c because 1/a is negative and 1/c is positive.) Therefore, a and c must have the same sign: either both positive or both negative.

From statement 1, because 1/b is between the other two, it must also have the same sign as the other two. Within numbers that all have the same sign, reciprocals are inversely proportional to the numbers themselves (i.e., the bigger the number, the smaller the reciprocal). For example, 3 is smaller than 100, but the reciprocal of 3 is larger than the reciprocal of 100.

Therefore, if all three variables have the same sign and 1/c < 1/b < 1/a, then it must be true that a < b < c.

The correct answer is (C).

Here's my take-

Firstly, abc != 0 => none of them = 0. They could be any other number, positive or negative.

Secondly, by taking a quick look the statements, I know the concept of inequalities and reciprocals and positives/negatives is being tested here.

I begin by statement 2 because it seems pretty straightforward: c > a. => nothing about b. NOT SUFFICIENT.

Then I come to statement 1: 1/c <1/b < 1/a . How can I deduce anything about the relationship among a, b and c? Well, there are a couple of rules:

Rule no. 1 - If x< y and x and y are both positive, then 1/x > 1/y .
Rule no. 2 - If x< y and x and y are both negative, then 1/x > 1/y .

The problem with statement 1 is that it doesn't tell me anything about the signs of a, b and c.

Case 1: What if a and b > 0 but c < 0? Then, the inequality in statement 1 can hold and the answer to our main question: a < b < c, will be NO.

Case 2: What if all three a, b and c > 0? Then the inequality in statement 1 can still hold and the answer to our main question: a < b < c, will be YES.

NOT SUFFICIENT.

On combining statements, we have: a < c and 1/c < 1/a, then there will be two cases as per the rules above:

1) a and c both positive. then b will also be positive definitely. b can't be negative because statement 1 doesn't allow it. And the answer to our main question will be YES.

2) a and c both negative. then b will also be negative definitely. b can't be positive because statement 1 doesn't allow it. And the answer to our main question will be YES.

SUFFICIENT. (C)

Any error in the logic? KarishmaB

Concept:
If variables have the same sign, flip the inequality sign when taking the reciprocal
If variables have different signs, keep inequality sign the same when taking the reciprocal

Question: Is a < b < c?

S1 provides the relationship between the reciprocals: 1/c < 1/b < 1/a

Case 1: signs can all be the same in which case the inequality sign would flip, thus generating a definitive YES to the question.
After sign flip: c > b > a

Case 2: signs can be different in which case the inequality sign would stay as is, thus generating a definitive NO to the question.
No sign flip: c < b < a

INSUFFICIENT

S2 provides the relationship between c and a only. No info on b

INSUFFICIENT

Combining both statements:
As per S2, the sign has clearly flipped as c > a, which means c and a have the same sign. Because b is between c and a, b must also have the same sign. Therefore, we can definitively answer YES to the question "is a < b < c?"

SUFFICIENT

You nailed it by inequality flip KarishmaB