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Re: Good set of DS 3 [#permalink]
03 Jan 2010, 02:14

Bunuel 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.

|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.

Answer: E.

This is the easiest way to solve this problem. In my previous solution I just tried to show what each statement means algebraically. Please tell which part needs more clarification.

Not able to get why do you say this (marked in red)... I know that |a| would always give a positive value, but why cant we consider a as 2 or 3...? Any reason for this? _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: Good set of DS 3 [#permalink]
03 Jan 2010, 08:23

Expert's post

jeeteshsingh wrote:

Bunuel 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.

|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.

Answer: E.

This is the easiest way to solve this problem. In my previous solution I just tried to show what each statement means algebraically. Please tell which part needs more clarification.

Not able to get why do you say this (marked in red)... I know that |a| would always give a positive value, but why cant we consider a as 2 or 3...? Any reason for this?

OK. For statement (2): First of all I'm not saying that a can ONLY be 1. I'm saying that it MAY be 1 and in this case b can take any value statement (2): "|a|^b is a non-zero integer " to hold true. So b can take positive as well as negative values, hence |a|b > 0 may or may not be true. That's why (2) is not sufficient.

To elaborate further: we are told that |a|^b is A. not zero and B. it's an integer.

A. Means that a is not zero, as the only way|a|^b to be zero, is a to be zero.

B.|a|^b is an integer (non=zero). What does that mean?

When a is any integer so that |a|>1 (eg 2, -2, 3, -3, ...), b can take any value but negative. Example: a=-3 --> |-3|=3, 3 in integer power (given a and b are integers) to be an integer, power (b), must be positive or zero, as when b is negative, let's say -2, we'll have: 3^{-2}=\frac{1}{3^2} which is not integer.

But when |a|=1 (1 or -1), then b can take ANY integer value as 1 in any power equals to 1 which is integer.

So from statement (2) we'll have:

a does not equals to 0. Plus, b can be ANY integer (positive or negative or zero) when |a|=1 and b must be positive or zero when |a|>1.

Re: Good set of DS 3 [#permalink]
18 Feb 2010, 12:38

jeeteshsingh,

To further clarify what Bunuel was saying. If b was negative and a was not 1 or -1, then |a|^b would be a fraction due to the properties of negative exponents. If |a|=1 (meaning a=1 or a=-1) then the denominator of the fraction would be some power of 1 which is also one. In this case the fraction would be 1 over 1 which is again 1 and satisfies the conditions.

Also, Bunuel, for the second portion I noticed that you didn't include in your listed possibilities that for |a|^b is a non zero integer, b=0 is possible as it would create an answer of 1 for any value of a including 0.

Re: Good set of DS 3 [#permalink]
18 Feb 2010, 13:07

Expert's post

theturk123 wrote:

jeeteshsingh,

To further clarify what Bunuel was saying. If b was negative and a was not 1 or -1, then |a|^b would be a fraction due to the properties of negative exponents. If |a|=1 (meaning a=1 or a=-1) then the denominator of the fraction would be some power of 1 which is also one. In this case the fraction would be 1 over 1 which is again 1 and satisfies the conditions.

Also, Bunuel, for the second portion I noticed that you didn't include in your listed possibilities that for |a|^b is a non zero integer, b=0 is possible as it would create an answer of 1 for any value of a including 0.

Welcome to the Gmat Club.

Your logic is correct. Though b=0 IS included as one of the possibilities. See the bolded part from the explanation:

Bunuel wrote:

So from statement (2) we'll have:

a does not equals to 0. Plus, b can be ANY integer (positive or negative or zero) when |a|=1 and b must be positive or zero when |a|>1.

You also mention 0^0 issue, this not the case for statement 2 (as a can not be zero), but still as you brought this up:

0^0, in some sources equals to 1, some mathematicians say it's undefined. Anyway you won't need this for GMAT:

"During the past decade, mathematicians argued extensively about the value of 0^0. Some answer that 0^0 = 1, while others answer that 0^0 is undefined. In the unlikely event that this question appears in some format or is a required intermediary calculation, the correct answer is more likely that 0^0 = 1." http://www.platinumgmat.com/gmat_study_ ... ial_powers

Re: Good set of DS 3 [#permalink]
15 Sep 2010, 09:06

jax91 wrote:

Bunuel wrote:

5. 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

|a|b > 0?

|a| is always +ve. So we need to know if b is +ve or -ve.

1.) mod of any number is +ve. Insuff.

2.) |a|^b is an integer.

we know a and b are integers.

so |a| is a +ve integer.

any +ve integer raised to a -ve integer will give us a fraction.

e.g. 4 ^ -3 = 1/ (4^3)

which will never be an integer.

so for |a|^b to be an integer b has to be +ve.

So its suff.

So B.

Hi,Can anyone explain me how |a| is positive?We generally look at the sign of any number before deciding its abs value..Right?But how come that is not followed here? Generally we take |x|=-x (x <0) |x|=x (x>0) but y is |a| taken positive directly?In the question,it is just mentioned as an integer

Re: Good set of DS 3 [#permalink]
15 Sep 2010, 09:16

Expert's post

ravitejapandiri wrote:

Hi,Can anyone explain me how |a| is positive?We generally look at the sign of any number before deciding its abs value..Right?But how come that is not followed here? Generally we take |x|=-x (x <0) |x|=x (x>0) but y is |a| taken positive directly?In the question,it is just mentioned as an integer

But when x=negative<0 then |x|=-x=-(negative)=positive and when x=positive>0 then |x|=x=positive, so in any case when x is either positive or negative |x| is still positive. There is one more case though: when x=0 then |x|=0.

So generally we can say that absolute value of an expression is alway non-negative: |some \ expression|\geq{0} --> |x|\geq{0}.

As for this particular question: see my posts on it on pages 1, 2, and 3.

Re: Good set of DS 3 [#permalink]
19 Dec 2010, 16:08

yangsta8 wrote:

Bunuel wrote:

4. Is y – x positive? (1) y > 0 (2) x = 1 – y

Statement 1) y>0 Not suff, X could be anything larger or smaller than X. Statement 2) x=1-y x+y=1 Let x=3 and y=-2 then y-x < 0. But if x=1/4 and y=3/4 then y-x >0 Not suff.

1 and 2 together) From the example above we have: if x=1/4 and y=3/4 then y-x >0 but if we flip it around: if x=3/4 and y=1/4 then y-x <0 not suff.

ANS = E

I selected that both stmts were sufficient together because the examples I chose worked both ways so if x=-2 y=3 you get 3-(-2)=5 or if y=2 than x = (-1) so 2-(-1)=3. Can the insufficiency only be seen with fractional numbers? Thanks.

Re: Good set of DS 3 [#permalink]
19 Dec 2010, 16:33

Expert's post

gettinit wrote:

yangsta8 wrote:

Bunuel wrote:

4. Is y – x positive? (1) y > 0 (2) x = 1 – y

Statement 1) y>0 Not suff, X could be anything larger or smaller than X. Statement 2) x=1-y x+y=1 Let x=3 and y=-2 then y-x < 0. But if x=1/4 and y=3/4 then y-x >0 Not suff.

1 and 2 together) From the example above we have: if x=1/4 and y=3/4 then y-x >0 but if we flip it around: if x=3/4 and y=1/4 then y-x <0 not suff.

ANS = E

I selected that both stmts were sufficient together because the examples I chose worked both ways so if x=-2 y=3 you get 3-(-2)=5 or if y=2 than x = (-1) so 2-(-1)=3. Can the insufficiency only be seen with fractional numbers? Thanks.

Yes, in order to get NO answer you should choose values of x and y from (0, 1), note that this range can give you the YES answer as well.

Is y-x>0? --> is y>x?

(1) y > 0, not sufficient as no info about x. (2) x = 1 - y --> x+y=1 --> the sum of 2 numbers equal to 1 --> we can not say which one is greater. Not sufficient.

(1)+(2) x+y=1 and y>0, still can not determine which one is greater: if y=0.1>0 and x=0.9 then y<x but if y=0.9>0 and x=0.1 then y>x. Not sufficient.

Re: Good set of DS 3 [#permalink]
31 Jan 2011, 05:48

Expert's post

subhashghosh wrote:

Hi Bunuel

For question # 8, please explain how this is true :

either 4y=32 y=8 x=-3, xy=-24 OR -4y=32

Regards, Subhash

If x and y are non-zero integers and |x| + |y| = 32, what is xy?

(1) -4x-12y=0 --> x=-3y --> x and y have opposite signs.

So either: |x|=x and |y|=-y --> in this case |x|+|y|=x-y=-3y-y=-4y=32: y=-8, x=24, xy=-24*8;

OR:|x|=-x and |y|=y --> |x|+|y|=-x+y=3y+y=4y=32 --> y=8 and x=-24 --> xy=-24*8, the same answer.

Sufficient.

(2) |x| - |y| = 16. Sum this one with th equations given in the stem --> 2|x|=48 --> |x|=24, |y|=8. xy=-24*8 (x and y have opposite sign) or xy=24*8 (x and y have the same sign). Multiple choices. Not sufficient.

Re: Good set of DS 3 [#permalink]
23 Feb 2011, 00:13

For question number 3, consider this:

1st number = 10w+x and 2nd number = 10c+x so three digit number = 100*(wc)+10*(w+c)+x^2

In 1, We are given, wc=x*(w+c)=x^2

or x*(w+c-x) = 0 implying either x=0 or w+c-x = 0 Since it is given that x is non-zero, we have w+c-x = 0 so 1 is sufficient w+c and x being odd just says that w+c-x is even number, so insufficient Answer A

Re: The sum of n consecutive positive integers is 45 [#permalink]
20 Apr 2012, 20:16

ENAFEX wrote:

Bunuel wrote:

Please find below new set of DS problems:

TIP: many of these problems act in GMAT zone, so beware of ZIP trap.

1. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n < 9

Bunuel, Is this approach correct?

sum of n consecutive +ve integers = \frac{n(n+1)}{2}

so, \frac{n(n+1)}{2} = 45 n(n+1) = 90 n^{2}+ n - 90 = 0 solving this equation gives n= -10 or 9

Both A and B doesn't help. Hence E.

Bunuel, Ignore my request. I found out the mistake. The question does not say sum of first consecutive integers, so, I cant use this formula. _________________

Re: Good set of DS 3 [#permalink]
02 May 2012, 07:11

Hi Bunuel/Karishma, Request you to clarify the below colored part. What difference does this makes in the question? Thanks H

Bunuel wrote:

ANSWERS:

7. If b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value of c? (1) d = 3 (2) b = 6

Note this part: "for all values of x" So, it must be true for x=0 --> c=d^2 --> b=2d (1) d = 3 --> c=9 Sufficient (2) b = 6 --> b=2d, d=3 --> c=9 Sufficient

Re: The sum of n consecutive positive integers is 45 [#permalink]
29 Jul 2012, 04:02

dvinoth86 wrote:

1. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n < 9

Cant we use the formula for sum of n consecutive +ve integers (n)*(n+1)/2 = 45.

[/color] n^2 +n -90 = 0 n=9 or -10

Cant we use the formula for sum of n consecutive +ve integers (n)*(n+1)/2 = 45.

This formula gives the sum of the first n consecutive positive integers: 1,2 3, ..., n. Nowhere is stated that our sequence starts with 1. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

gmatclubot

Re: The sum of n consecutive positive integers is 45
[#permalink]
29 Jul 2012, 04:02

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