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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
\(|x|=-x\), when \(x<0\) and \(|x|=x\) when \(x>0\) --> CORRECT.
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
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29 Jul 2012, 04:02
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