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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.

Re: DS_If a and b.... [#permalink]
26 Nov 2009, 19:34

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

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. _________________

Re: DS_If a and b.... [#permalink]
21 Dec 2009, 08:22

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

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.

Re: DS question : need help [#permalink]
28 Oct 2010, 18:20

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]
04 Jun 2014, 18:35

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]
02 Sep 2014, 03:43

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.

If a and b are integers, and a not= b, is |a|b > 0? (1) [#permalink]
02 Sep 2014, 03:49

1

This post received KUDOS

Expert's post

shahPranay14 wrote:

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.

Re: If a and b are integers, and a not= b, is |a|b > 0? (1) [#permalink]
20 Oct 2014, 22:23

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.

Thus insufficient.

Combining two,

We'll get positive values for y, and zero.

Thus combining two will not give solution.

Thus Ans - E _________________

Give KUDOS if the post helps you...

gmatclubot

Re: If a and b are integers, and a not= b, is |a|b > 0? (1)
[#permalink]
20 Oct 2014, 22:23

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...