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Are both x and y positive?
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09 Oct 2013, 23:36
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58% (01:34) correct 42% (01:14) wrong based on 389 sessions
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Are both x and y positive? (1) \(\sqrt{x^2}=x\) (2) \(y=\sqrt{2x}\)
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Re: Are both x and y positive?
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10 Oct 2013, 00:04
innocous wrote: Are both x and y positive?
1) \(\sqrt{(x^2)}=x\)
(2) \(y=\sqrt{(2x)}\) statement 1: \(\sqrt{(x^2)}=x\) this means x>0 or x= 0 nothing about y hence insufficient. statement 2: \(y=\sqrt{(2x)}\) this gives y> 0 or y = 0 but x<2 not sufficient. combining both y>0or =0 and x<2 again insufficient since if x=y=0 then x and y are not positive and in other cases they are positive. hence E
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Re: Are both x and y positive?
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10 Oct 2013, 00:55
Are both x and y positive?(1) \(\sqrt{x^2}=x\) > \(x=x\) > \(x\geq{0}\). Not sufficient. (2) \(y=\sqrt{2x}\) > y is equal to the square root of some number, thus \(y\geq{0}\). 2x is under the square root, thus \(2x\geq{0}\) > \(x\leq{2}\). Not sufficient. (1)+(2) \(0\leq{x}\leq{2}\) and \(y\geq{0}\). If \(x=y=1\), then the answer is YES but if \(x=2\) and \(y=0\), then the asnwer is NO. Not sufficient. Answer: E.
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Re: Are both x and y positive?
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27 Dec 2013, 06:32
Bunuel wrote: Are both x and y positive?
(1) \(\sqrt{x^2}=x\) > \(x=x\) > \(x\geq{0}\). Not sufficient.
(2) \(y=\sqrt{2x}\) > y is equal to the square root of some number, thus \(y\geq{0}\). 2x is under the square root, thus \(2x\geq{0}\) > \(x\leq{2}\). Not sufficient.
(1)+(2) \(0\leq{x}\leq{2}\) and \(y\geq{0}\). If \(x=y=1\), then the answer is YES but if \(x=2\) and \(y=0\), then the asnwer is NO. Not sufficient.
Answer: E. They don't mention they have to be integers either no?



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Re: Are both x and y positive?
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31 Oct 2014, 15:08
Bunuel wrote: Are both x and y positive?
(1) \(\sqrt{x^2}=x\) > \(x=x\) > \(x\geq{0}\). Not sufficient.
(2) \(y=\sqrt{2x}\) > y is equal to the square root of some number, thus \(y\geq{0}\). 2x is under the square root, thus \(2x\geq{0}\) > \(x\leq{2}\). Not sufficient.
(1)+(2) \(0\leq{x}\leq{2}\) and \(y\geq{0}\). If \(x=y=1\), then the answer is YES but if \(x=2\) and \(y=0\), then the asnwer is NO. Not sufficient.
Answer: E. I'm a bit unclear about this. Doesn't this simply mean x=x? I really didn't know what to do with this statement... if we had 4, then\(\sqrt{(4)^2}\) would still give a 4, wouldn't it?



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Re: Are both x and y positive?
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01 Nov 2014, 05:18
usre123 wrote: Bunuel wrote: Are both x and y positive?
(1) \(\sqrt{x^2}=x\) > \(x=x\) > \(x\geq{0}\). Not sufficient.
(2) \(y=\sqrt{2x}\) > y is equal to the square root of some number, thus \(y\geq{0}\). 2x is under the square root, thus \(2x\geq{0}\) > \(x\leq{2}\). Not sufficient.
(1)+(2) \(0\leq{x}\leq{2}\) and \(y\geq{0}\). If \(x=y=1\), then the answer is YES but if \(x=2\) and \(y=0\), then the asnwer is NO. Not sufficient.
Answer: E. I'm a bit unclear about this. Doesn't this simply mean x=x? I really didn't know what to do with this statement... if we had 4, then\(\sqrt{(4)^2}\) would still give a 4, wouldn't it? You really need to brush up fundamentals on roots and absolute values. This is basic staff! First, of all, when the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. So, \(\sqrt{(4)^2}=\sqrt{16}=4\), not 4 and not +/4, ONLY 4! Next, about \(\sqrt{x^2}=x\). The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\). So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to? Let's consider following examples: If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\). So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=x\), if \(x<0\). What function does exactly the same thing? The absolute value function: \(x=x\), if \(x\geq{0}\) and \(x=x\), if \(x<0\). That is why \(\sqrt{x^2}=x\). Best GMAT Math Prep Books (Reviews & Recommendations): bestgmatmathprepbooksreviewsrecommendations77291.htmlHope it helps.
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Are both x and y positive?
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01 Nov 2014, 10:56
Bunuel wrote: usre123 wrote: Bunuel wrote: Are both x and y positive?
(1) \(\sqrt{x^2}=x\) > \(x=x\) > \(x\geq{0}\). Not sufficient.
(2) \(y=\sqrt{2x}\) > y is equal to the square root of some number, thus \(y\geq{0}\). 2x is under the square root, thus \(2x\geq{0}\) > \(x\leq{2}\). Not sufficient.
(1)+(2) \(0\leq{x}\leq{2}\) and \(y\geq{0}\). If \(x=y=1\), then the answer is YES but if \(x=2\) and \(y=0\), then the asnwer is NO. Not sufficient.
Answer: E. I'm a bit unclear about this. Doesn't this simply mean x=x? I really didn't know what to do with this statement... if we had 4, then\(\sqrt{(4)^2}\) would still give a 4, wouldn't it? You really need to brush up fundamentals on roots and absolute values. This is basic staff! First, of all, when the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. So, \(\sqrt{(4)^2}=\sqrt{16}=4\), not 4 and not +/4, ONLY 4! Next, about \(\sqrt{x^2}=x\). The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\). So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to? Let's consider following examples: If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\). So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=x\), if \(x<0\). What function does exactly the same thing? The absolute value function: \(x=x\), if \(x\geq{0}\) and \(x=x\), if \(x<0\). That is why \(\sqrt{x^2}=x\). Best GMAT Math Prep Books (Reviews & Recommendations): bestgmatmathprepbooksreviewsrecommendations77291.htmlHope it helps. thank you for taking the time out to explain... I'm studying math after a very very long time so I'm extremely rusty and most of my questions involve very very basic concepts. I just reversed the OG13 DS question one concept that when x^2=4, x can be + or2. then perhaps underoot 4 can be + or ve 2 as well. I read this post, and perhaps this mixed me up (or I suppose I fail to understand rule 3). Could someone please comment? http://www.manhattangmat.com/blog/2012/ ... thegmat/



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Re: Are both x and y positive?
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02 Nov 2014, 04:16
usre123 wrote: Bunuel wrote: usre123 wrote: I'm a bit unclear about this. Doesn't this simply mean x=x? I really didn't know what to do with this statement... if we had 4, then\(\sqrt{(4)^2}\) would still give a 4, wouldn't it?
You really need to brush up fundamentals on roots and absolute values. This is basic staff! First, of all, when the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. So, \(\sqrt{(4)^2}=\sqrt{16}=4\), not 4 and not +/4, ONLY 4! Next, about \(\sqrt{x^2}=x\). The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\). So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to? Let's consider following examples: If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\). So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=x\), if \(x<0\). What function does exactly the same thing? The absolute value function: \(x=x\), if \(x\geq{0}\) and \(x=x\), if \(x<0\). That is why \(\sqrt{x^2}=x\). Best GMAT Math Prep Books (Reviews & Recommendations): bestgmatmathprepbooksreviewsrecommendations77291.htmlHope it helps. thank you for taking the time out to explain... I'm studying math after a very very long time so I'm extremely rusty and most of my questions involve very very basic concepts. I just reversed the OG13 DS question one concept that when x^2=4, x can be + or2. then perhaps underoot 4 can be + or ve 2 as well. I read this post, and perhaps this mixed me up (or I suppose I fail to understand rule 3). Could someone please comment? http://www.manhattangmat.com/blog/2012/ ... thegmat/Rule 3 there says that \(\sqrt{x^2}=3\) means that \(x=3\) or \(x=3\). \(\sqrt{x^2}=3\) > square: \(x^2=9\) > \(x=3\) or \(x=3\). Or: \(\sqrt{x^2}=3\) > \(x=3\) > \(x=3\) or \(x=3\). Hope it's clear.
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Re: Are both x and y positive?
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04 Nov 2014, 10:08
I'm sorry, I still don't see it. if they had written option one =x (in the original question), then i could safely say x is negative, right? Since they say it equals x, then x must be positive. Is that the jist of what your saying? Also, is it reasonable to assume (as mgmat says) that when we have an actual number under the square root, take only the positive root. but if we have a variable, take positive and negative both?



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Re: Are both x and y positive?
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04 Nov 2014, 10:36
usre123 wrote: I'm sorry, I still don't see it. if they had written option one =x (in the original question), then i could safely say x is negative, right? Since they say it equals x, then x must be positive. Is that the jist of what your saying? Also, is it reasonable to assume (as mgmat says) that when we have an actual number under the square root, take only the positive root. but if we have a variable, take positive and negative both? If it were \(\sqrt{x^2}=x\), then it would mean that x = x, thus \(x\leq{0}\). Try to play with numbers there and it might become clearer.
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Re: Are both x and y positive?
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04 Nov 2014, 10:47
yup, I got it! and can't believe I wasted your time asking such a basic question! I feel so silly. Thanks !



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Re: Are both x and y positive?
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03 Mar 2015, 23:22
Bunuel wrote: Are both x and y positive?
(1) \(\sqrt{x^2}=x\) > \(x=x\) > \(x\geq{0}\). Not sufficient.
(2) \(y=\sqrt{2x}\) > y is equal to the square root of some number, thus \(y\geq{0}\). 2x is under the square root, thus \(2x\geq{0}\) > \(x\leq{2}\). Not sufficient.
(1)+(2) \(0\leq{x}\leq{2}\) and \(y\geq{0}\). If \(x=y=1\), then the answer is YES but if \(x=2\) and \(y=0\), then the asnwer is NO. Not sufficient.
Answer: E. Can't y be the root of a negative number? Let's say x = 5 the y = \(\sqrt{25}\) = \(\sqrt{3}\) I know this concept is tested on GMAT.



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Re: Are both x and y positive?
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04 Mar 2015, 03:36
b2bt wrote: Bunuel wrote: Are both x and y positive?
(1) \(\sqrt{x^2}=x\) > \(x=x\) > \(x\geq{0}\). Not sufficient.
(2) \(y=\sqrt{2x}\) > y is equal to the square root of some number, thus \(y\geq{0}\). 2x is under the square root, thus \(2x\geq{0}\) > \(x\leq{2}\). Not sufficient.
(1)+(2) \(0\leq{x}\leq{2}\) and \(y\geq{0}\). If \(x=y=1\), then the answer is YES but if \(x=2\) and \(y=0\), then the asnwer is NO. Not sufficient.
Answer: E. Can't y be the root of a negative number? Let's say x = 5 the y = \(\sqrt{25}\) = \(\sqrt{3}\) I know this concept is tested on GMAT. The GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers so even roots from negative number are undefined on the GMAT: \(\sqrt[{even}]{negative}=undefined\), for example \(\sqrt{25}=undefined\).
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Re: Are both x and y positive?
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23 Mar 2016, 17:49
Bunuel wrote: b2bt wrote: Bunuel wrote: Are both x and y positive?
(2) \(y=\sqrt{2x}\) > y is equal to the square root of some number, thus \(y\geq{0}\). 2x is under the square root, thus \(2x\geq{0}\) > \(x\leq{2}\). Not sufficient.
How does this lead you to believe that y must be positive? Couldn't it be undefined as you said in another comment?



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Are both x and y positive?
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23 Mar 2016, 23:23
mcolbert wrote: How does this lead you to believe that y must be positive? Couldn't it be undefined as you said in another comment?
Hi, As you say y can be +ive or undefined number.. BUT GMAT uses only real numbers and undefined numbers are not tested in GMAT.. so we don't cater for any unreal numbers and assume/accept only the REAL values .. if you take x=3.. y=\(\sqrt{23}\)=\(\sqrt{1}\).. But what happens to y, y becomes undefined number.. however if a variable is given, it has to be real so this vakue is not correct..
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Re: Are both x and y positive?
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06 Oct 2016, 06:54
classic GMAT trap... not considering 0 as a valid value.
good question!



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