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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Can anybody chime in? Although I agree with the answer, I don't think the bold statement below is correct. B does not have to be 0 or greater. It can be -4 if A is 1.

If b= -4 and a=1. this satisfiies the conditions: 1) a,b are integers 2) a is not equal to b 3) statement 2; 1^(-4)=1. that is a non-zero integer. So you can't say B is definitely positive or 0. Combining statement 1 and 2. You can't determine if A is a 1 or not so E.

Please correct me if I am wrong. Thanks.

Fig wrote:

(E) for me.

|a|*b > 0 ? <=> b > 0 ?

From (1) |a^b| > 0

o If b=-1 and a=1, 1 > 0 and b < 0 o If b=1 and a=-1, 1 > 0 and b > 0

INSUFF.

[b]From (2) |a|^b is a non zero integer implies that b must be positive or equal to 0 in order to not create a real number such as 2^-1.

So, we remains with the cases of b = 0 and b > 0.[/b] INSUFF.

If a and b are integers, and a not= b, is |a|b > 0?

(1) |a^b| > 0

(2) |a|^b is a non-zero integer

Clearly E. |a^b| and |a|^b are always >0, no matter what b is, because absolute value is always greater than 0. Thus, we can't know if b>0 or not.
_________________

If a and b are integers, and a not= b, is |a|b > 0?

(1) |a^b| > 0

(2) |a|^b is a non-zero integer

Question is |a|b>0 this can be proved if we can prove that a not=0 and b>0. 1. |a^b| > 0 implies that a not = 0. a can be +ve or -ve, and be can be 0, +ve or -ve....but we are sure that a not=0 else |a^b| = 0. Statement 1 itself is insuff.

2. |a|^b is a non zero integer. we already know a,b both are integers....|a| is > 0, so b >= 0. No clear value of b...

Given: a , b are ints. and a is different from b asking: |a| * b > 0

what the question is really asking if b > 0 [ |a| is always >0 ]

(1) |a^b| >0 ---------------- says nothing, cuz |x| is always > 0 statement 1 is insufficient

(2) |a|^b is not zero ------------------------- also says nothing .. we know |a|^b > 0 b could be -ve or +ve statement 2 is insufficient

(1) and (2) together ------------------------ both statements really say nothing about b

final answer is E

>>Please tell me how can |a| be taken as positive in the above steps without knowing its sign..I mean if a is negative,then |a| wil be negative..Right?Am i missing anythin badly?

>>Please tell me how can |a| be taken as positive in the above steps without knowing its sign..I mean if a is negative,then |a| wil be negative..Right?Am i missing anythin badly?

Absolute value of of an expression is alway non-negative: \(|some \ expression|\geq{0}\). Please check Walker's post on Absolute Value at: math-absolute-value-modulus-86462.html

As for the question:

If a and b are integers, and a does not equal to b, is |a|*b > 0? (1) |a^b| > 0 (2) |a|^b is a non-zero integer.

\(|a|*b>0\) is true when \(b>0\) and \(a\) does not equal to zero.

(1) \(|a^b| > 0\) --> \(a\) does not equal to zero, but we don't know about \(b\), it can be any value, positive or negative. Not sufficient.

(2) \(|a|^b\) is a non-zero integer --> \(a\) can be 1 and \(b\) any integer, positive or negative. Not sufficient.

(1)+(2) If \(a=1\) and \(b=2\), then \(|a|*b>0\), but if \(a=1\) and \(b=-2\), then \(|a|*b<0\). Not sufficient.

Basically just need to find out if b is positive or negative, since a will always be positive as it is inside of the | |.

1) Doesn't give you anything because everything is inside of the | |, so you can't tell if b is positive or negative, so insufficient.

2) Tells you that b is not negative since that would result in a non-integer. However, b could be 0 since a number raised to the 0 is 1, which is a non-integer number. Also insufficient.

Using both statements still doesn't provide anything because b can still be equal to 0, in which case |a|b > 0 is false. However, b can also be any positive number which would make |a|b > 0 true. Thus E.

Re: If a and b are integers, and a not= b, is |a|b > 0? (1) [#permalink]

Show Tags

04 Jun 2014, 19:35

The reason why s2 alone or taken together with s1 is not sufficient bcos we need info on the signs that is, a is +ve or -ve and whether b is +ve or -ve & not wether they are zero or non-zero integers.

Re: If a and b are integers, and a not= b, is |a|b > 0? (1) [#permalink]

Show Tags

02 Sep 2014, 04:43

Bunuel wrote:

ravitejapandiri wrote:

>>Please tell me how can |a| be taken as positive in the above steps without knowing its sign..I mean if a is negative,then |a| wil be negative..Right?Am i missing anythin badly?

Absolute value of of an expression is alway non-negative: \(|some \ expression|\geq{0}\). Please check Walker's post on Absolute Value at: math-absolute-value-modulus-86462.html

As for the question:

If a and b are integers, and a does not equal to b, is |a|*b > 0? (1) |a^b| > 0 (2) |a|^b is a non-zero integer.

\(|a|*b>0\) is true when \(b>0\) and \(a\) does not equal to zero.

(1) \(|a^b| > 0\) --> \(a\) does not equal to zero, but we don't know about \(b\), it can be any value, positive or negative. Not sufficient.

(2) \(|a|^b\) is a non-zero integer --> \(a\) can be 1 and \(b\) any integer, positive or negative. Not sufficient.

(1)+(2) If \(a=1\) and \(b=2\), then \(|a|*b>0\), but if \(a=1\) and \(b=-2\), then \(|a|*b<0\). Not sufficient.

>>Please tell me how can |a| be taken as positive in the above steps without knowing its sign..I mean if a is negative,then |a| wil be negative..Right?Am i missing anythin badly?

Absolute value of of an expression is alway non-negative: \(|some \ expression|\geq{0}\). Please check Walker's post on Absolute Value at: math-absolute-value-modulus-86462.html

As for the question:

If a and b are integers, and a does not equal to b, is |a|*b > 0? (1) |a^b| > 0 (2) |a|^b is a non-zero integer.

\(|a|*b>0\) is true when \(b>0\) and \(a\) does not equal to zero.

(1) \(|a^b| > 0\) --> \(a\) does not equal to zero, but we don't know about \(b\), it can be any value, positive or negative. Not sufficient.

(2) \(|a|^b\) is a non-zero integer --> \(a\) can be 1 and \(b\) any integer, positive or negative. Not sufficient.

(1)+(2) If \(a=1\) and \(b=2\), then \(|a|*b>0\), but if \(a=1\) and \(b=-2\), then \(|a|*b<0\). Not sufficient.

Re: If a and b are integers, and a not= b, is |a|b > 0? (1) [#permalink]

Show Tags

20 Oct 2014, 23:23

If a and b are integers, and a does not equal to b, is |a|*b > 0? (1) |a^b| > 0 (2) |a|^b is a non-zero integer. to check whether |a|*b > 0, we need to identify whether b>0

1 - This statement has to be positive, irrespective of the value of a and b. This is insufficient

2. |a|^b is non-zero integer -

Possibilities - a = -.5 or .5 and b = -1 The value of expression would be 2 a = any number and b = 0. The value will be 1 a = any positive / negative number and b = any positive number Thus the result will be a positive number.

Re: If a and b are integers, and a not= b, is |a|b > 0? (1) [#permalink]

Show Tags

15 Apr 2016, 07:36

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.
_________________

If a and b are integers, and a not= b, is |a|b > 0? (1) [#permalink]

Show Tags

20 Mar 2017, 20:51

mm007 wrote:

If a and b are integers, and a not= b, is \(|a|b > 0\)?

(1) \(|a^b| > 0\)

(2) \(|a|^b\) is a non-zero integer

OFFICIAL SOLUTION

Let us start be examining the conditions necessary for \(|a|b > 0\). Since |a| cannot be negative, both |a| and b must be positive. However, since |a| is positive whether a is negative or positive, the only condition for a is that it must be non-zero.

Hence, the question can be restated in terms of the necessary conditions for it to be answered "yes":

“Do both of the following conditions exist: a is non-zero AND b is positive?”

(1) INSUFFICIENT: In order for a = 0, \(|a^b|\) would have to equal 0 since 0 raised to any power is always 0. Therefore (1) implies that a is non-zero. However, given that a is non-zero, b can be anything for \(|ab| > 0\) so we cannot determine the sign of b.

(2) INSUFFICIENT: If a = 0, |a| = 0, and \(|a|^b = 0\) for any b. Hence, a must be non-zero and the first condition (a is not equal to 0) of the restated question is met. We now need to test whether the second condition is met. (Note: If a had been zero, we would have been able to conclude right away that (2) is sufficient because we would answer "no" to the question is |a|b > 0?) Given that a is non-zero, |a| must be positive integer. At first glance, it seems that b must be positive because a positive integer raised to a negative integer is typically fractional (e.g., \(a^{-2} = \frac{1}{{a^2}}\). Hence, it appears that b cannot be negative. However, there is a special case where this is not so:

If |a| = 1, then b could be anything (positive, negative, or zero) since \(|1|^b\) is always equal to 1, which is a non-zero integer . In addition, there is also the possibility that b = 0. If |b| = 0, then \(|a|^0\) is always 1, which is a non-zero integer.

Hence, based on (2) alone, we cannot determine whether b is positive and we cannot answer the question.

An alternative way to analyze this (or to confirm the above) is to create a chart using simple numbers as follows:

a b Is \(|a|^b\) a non-zero integer? Is \(|a|b > 0\)? 1 2 Yes Yes 1 -2 Yes No 2 1 Yes Yes 2 0 Yes No

We can quickly confirm that (2) alone does not provide enough information to answer the question.

(1) AND (2) INSUFFICIENT: The analysis for (2) shows that (2) is insufficient because, while we can conclude that a is non-zero, we cannot determine whether b is positive. (1) also implies that a is non-zero, but does not provide any information about b other than that it could be anything. Consequently, (1) does not add any information to (2) regarding b to help answer the question and (1) and (2) together are still insufficient. (Note: As a quick check, the above chart can also be used to analyze (1) and (2) together since all of the values in column 1 are also consistent with (1)).

The correct answer is E.

Attachments

Untitled.png [ 14.55 KiB | Viewed 502 times ]

_________________

Be challenged at EVERY MOMENT.

Each stage of the journey is crucial to attaining new heights of knowledge.

gmatclubot

If a and b are integers, and a not= b, is |a|b > 0? (1)
[#permalink]
20 Mar 2017, 20:51

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...