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If r,s and t are integers, is r+s =square root of t [#permalink]
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22 May 2018, 03:47
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If \(r,s\) and \(t\) are integers, is \(r+s =\sqrt{t}\) ? [1] \(r<s\) [2] \((r+s)^2=t\)
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If r,s and t are integers, is r+s =square root of t [#permalink]
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Updated on: 31 May 2018, 21:58
Princ wrote: If \(r,s\) and \(t\) are integers, is \(r+s =\sqrt{t}\) ?
[1] \(r<s\)
[2] \((r+s)^2=t\) (1) \(r<s\) \(r+s<0\). Since, \(\sqrt{t}\) is nonnegative positive, answer is NO. Sufficient. (2) \(r+s=\sqrt{t}\). Not sufficient. Answer: A
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Originally posted by Tulkin987 on 22 May 2018, 05:04.
Last edited by Tulkin987 on 31 May 2018, 21:58, edited 1 time in total.



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Re: If r,s and t are integers, is r+s =square root of t [#permalink]
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22 May 2018, 05:34
OA: A Statement 1 : \(r<s\) \(r+s<0\) \(\sqrt{t}\) can not be ve. So answer for is \(r+s=\sqrt{t}\) :no. Statement 1 is alone sufficient. Statement 2 : \((r+s)^2=t\) \(r+s=\sqrt{t}\) Case a :When \(r+s <0\) Then statement 2 becomes \((r+s)=\sqrt{t}\) Answer for is \(r+s=\sqrt{t}\) :no Case b : When \({r+s}\geq{0}\), then statement 2 becomes \(r+s=\sqrt{t}\). answer for is \(r+s=\sqrt{t}\) :yes. So statement 2 alone is insufficient.
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Re: If r,s and t are integers, is r+s =square root of t [#permalink]
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22 May 2018, 06:17
(1) \(r<s\) \(r+s<0\). So, r+s is negative. Since, \(\sqrt{t}\) is always NON NEGATIVE, answer is NO. Sufficient.(2) \(r+s=\sqrt{t}\). Now \((r+s)\) can be equal to \(\sqrt{t}\) or \(\sqrt{t}.\) Not sufficient.Answer: A
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Re: If r,s and t are integers, is r+s =square root of t [#permalink]
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23 May 2018, 01:14
gmatbustersLET t=36 ROOT36=+ 6 ROOT(t) can be either +ve or ve



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Re: If r,s and t are integers, is r+s =square root of t [#permalink]
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23 May 2018, 01:26
GYANENDRA88 wrote: gmatbustersLET t=36 ROOT36=+ 6 ROOT(t) can be either +ve or ve Let me clarify it with two specific cases: • If \(a^2\) = 16, then a = +4 or 4 It happens because from the equation \(a^2\) = 16, we can derive (a + 4)(a – 4) = 0, which gives us two values of a = +4 and 4 This can also be written as a = +\(\sqrt{16}\) or \(\sqrt{16}\) Now, \(\sqrt{16}\) = 4, Hence, the values of a = +4 or 4 • If a = 16, then \(\sqrt{a}\) = \(\sqrt{16}\) = 4 always. Therefore \(\sqrt{a}\) is always positive. Hope this clarifies your doubt.
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Re: If r,s and t are integers, is r+s =square root of t [#permalink]
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23 May 2018, 01:58
EgmatQuantExpertThanks for your reply. One can approach it in a different way let a=6 t=a^2 root(t)=root(a^2)=6



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Re: If r,s and t are integers, is r+s =square root of t [#permalink]
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23 May 2018, 02:15
Hii I am afraid your understanding is not correct. If x² = 36, x = +/6 But if x = \(\sqrt{36}\) x = only 6. The square root function gives only the positive value. This is as per the definition of sqaure root function. Please remember this concept, this is very important for GMAT. If you have any further queries, feel free to tag me. Happy Learning... GYANENDRA88 wrote: gmatbustersLET t=36 ROOT36=+ 6 ROOT(t) can be either +ve or ve
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Re: If r,s and t are integers, is r+s =square root of t [#permalink]
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23 May 2018, 02:34
gmatbustersThanks for the reply can you prove it mathematically that root36 is not equal to 6.



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Re: If r,s and t are integers, is r+s =square root of t [#permalink]
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23 May 2018, 03:03
Hii A function is a special relationship where each input has a single output.So to define square root function, root function must give a unique answer. it has been defined as to be positive.. There is no reason for this. See it like this, we say that the charge of electron is negative, actually it is defined to be negative. Scientists might have defined charge of electron to be positive and proton charge as negative. But who knows, God might be in favour of Proton that day. Similarly we say that numbers on the number line to left of zero is negative and to right of zero is positive. We could have defined the convention otherwise also. But it is defined like this. There is no explanation for this. Hope it is clear now. GYANENDRA88 wrote: gmatbustersThanks for the reply can you prove it mathematically that root36 is not equal to 6.
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Re: If r,s and t are integers, is r+s =square root of t [#permalink]
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23 May 2018, 05:20
The symbol √ does not mean square root. It means "principal square root", which is the positive square root. From GMAT prespective , check out bunuel's comment Quote: 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. That is:
\(\sqrt{9} = 3\), NOT +3 or 3; \(\sqrt[4]{16} = 2\), NOT +2 or 2;
Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and 3. Because \(x^2 = 9\) means that \(x =\sqrt{9}=3\) or \(x=\sqrt{9}=3\). Same matter has been discussed in below link https://gmatclub.com/forum/squarerootalwayspositive114114.htmlYou can also check below link from Manhattanprep :What’s the Deal with Square Roots on the GMAT? https://www.manhattanprep.com/gmat/blog ... thegmat/
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Re: If r,s and t are integers, is r+s =square root of t [#permalink]
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31 May 2018, 08:33
Hi Tulkin987We can not say that \(\sqrt{t}\) is positive, In fact \(\sqrt{t}\) is NON NEGATIVE.Tulkin987 wrote: Princ wrote: If \(r,s\) and \(t\) are integers, is \(r+s =\sqrt{t}\) ?
[1] \(r<s\)
[2] \((r+s)^2=t\) (1) \(r<s\) \(r+s<0\). Since, \(\sqrt{t}\) is positive, answer is NO. Sufficient. (2) \(r+s=\sqrt{t}\). Not sufficient. Answer: A
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If r,s and t are integers, is r+s =square root of t [#permalink]
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31 May 2018, 22:00
gmatbusters wrote: Hi Tulkin987We can not say that \(\sqrt{t}\) is positive, In fact \(\sqrt{t}\) is NON NEGATIVE.Tulkin987 wrote: Princ wrote: If \(r,s\) and \(t\) are integers, is \(r+s =\sqrt{t}\) ?
[1] \(r<s\)
[2] \((r+s)^2=t\) (1) \(r<s\) \(r+s<0\). Since, \(\sqrt{t}\) is positive, answer is NO. Sufficient. (2) \(r+s=\sqrt{t}\). Not sufficient. Answer: AHi, gmatbusters! Corrected the mistake. Thanks
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