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Find the value of
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Updated on: 27 May 2013, 11:12
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Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a1)^2}}{\sqrt{(1+a)^2}  \sqrt{(a1)^2}}\) for 0<a<1 A. a B. 1/a C. (a1)/(a+1) D. (a+1)/(a1) E. a/(a1)
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Originally posted by manimani on 28 Jun 2012, 02:10.
Last edited by Bunuel on 27 May 2013, 11:12, edited 2 times in total.
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28 Jun 2012, 02:19
manimani wrote: Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a1)^2}}{\sqrt{(1+a)^2}  \sqrt{(a1)^2}}\) for 0<a<1
A. a B. 1/a C. (a1)/(a+1) D. (a+1)/(a1) E. a/(a1) Note that \(\sqrt{x^2}=x\). \(\frac{\sqrt{(1+a)^2} + \sqrt{(a1)^2}}{\sqrt{(1+a)^2}  \sqrt{(a1)^2}}=\frac{1+a+a1}{1+aa1}\) Now, since \(0<a<1\), then: \(1+a=1+a\) and \(a1=(a1)=1a\). Hence \(\frac{1+a+a1}{1+aa1}=\frac{(1+a)+(1a)}{(1+a)(1a)}=\frac{1}{a}\). Answer: B. Hope it's clear.
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06 Dec 2012, 00:01
Remember these things regarding absolute values in the GMAT (1) \(\sqrt{x^2}=x\) (2) x = x ==> if x > 0 (3) x = x ==> if x < 0
Golden rules! Must memorize!
Solution: ***Transform the equation: \(\frac{1+a + a1}{1+a  a1}\)
***Since we know that a is a positive fraction as given: 0<a<1 ***This means 1+a is always positive. Using property#2 above, 1+a = 1+a ***For a1 = (fraction)  1, we know that a1 is negative. Using property #3, a1 = (a1)
Combine all that:
\(\frac{(1+a) + (a+1)}{(1+a)  (a+1)}\) \(\frac{2}{2a}\) \(\frac{1}{a}\)
Answer: B




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30 Jun 2012, 15:15
Hi Bunuel, Could you please explain me how a1 becomes 1a in the numerator and denominator in the second part of the problem? Thanks!



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Re: Find the value of
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30 Jun 2012, 20:39
Bunuel wrote: manimani wrote: Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a1)^2}}{\sqrt{(1+a)^2}  \sqrt{(a1)^2}}\) for 0<a<1
A. a B. 1/a C. (a1)/(a+1) D. (a+1)/(a1) E. a/(a1) Note that \(\sqrt{x^2}=x\). \(\frac{\sqrt{(1+a)^2} + \sqrt{(a1)^2}}{\sqrt{(1+a)^2}  \sqrt{(a1)^2}}=\frac{1+a+a1}{1+aa1}\) Now, since \(0<a<1\), then: \(1+a=1+a\) and \(a1=(a1)=1a\). Hence \(\frac{1+a+a1}{1+aa1}=\frac{(1+a)+(1a)}{(1+a)(1a)}=\frac{1}{a}\). Answer: B. Hope it's clear. As 0<a<1 Let's say a =0.5 then (1+0.5+0.51)/(1+0.50.5+1) = 0.5 ..the answer is A...I am not getting it ...Can you explain a bit more ?Thnaks



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01 Jul 2012, 02:02
Bunuel wrote: manimani wrote: Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a1)^2}}{\sqrt{(1+a)^2}  \sqrt{(a1)^2}}\) for 0<a<1
A. a B. 1/a C. (a1)/(a+1) D. (a+1)/(a1) E. a/(a1) Note that \(\sqrt{x^2}=x\). \(\frac{\sqrt{(1+a)^2} + \sqrt{(a1)^2}}{\sqrt{(1+a)^2}  \sqrt{(a1)^2}}=\frac{1+a+a1}{1+aa1}\) Now, since \(0<a<1\), then: \(1+a=1+a\) and \(a1=(a1)=1a\). Hence \(\frac{1+a+a1}{1+aa1}=\frac{(1+a)+(1a)}{(1+a)(1a)}=\frac{1}{a}\). Answer: B. Hope it's clear. Bunuel ..i ddnt get it.. how did u get the 1/a in the end ?? can u plz explain bunuel..? m trying it but i cant get 1/a in the end?
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01 Jul 2012, 02:06
sanjoo wrote: Bunuel wrote: manimani wrote: Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a1)^2}}{\sqrt{(1+a)^2}  \sqrt{(a1)^2}}\) for 0<a<1
A. a B. 1/a C. (a1)/(a+1) D. (a+1)/(a1) E. a/(a1) Note that \(\sqrt{x^2}=x\). \(\frac{\sqrt{(1+a)^2} + \sqrt{(a1)^2}}{\sqrt{(1+a)^2}  \sqrt{(a1)^2}}=\frac{1+a+a1}{1+aa1}\) Now, since \(0<a<1\), then: \(1+a=1+a\) and \(a1=(a1)=1a\). Hence \(\frac{1+a+a1}{1+aa1}=\frac{(1+a)+(1a)}{(1+a)(1a)}=\frac{1}{a}\). Answer: B. Hope it's clear. Bunuel ..i ddnt get it.. how did u get the 1/a in the end ?? can u plz explain bunuel..? m trying it but i cant get 1/a in the end? \(\frac{1+a+a1}{1+aa1}=\frac{(1+a)+(1a)}{(1+a)(1a)}=\frac{1+a+1a}{1+a1+a}=\frac{2}{2a}=\frac{1}{a}\). Hope it's clear now.
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01 Jul 2012, 02:20
sharmila79 wrote: Hi Bunuel, Could you please explain me how a1 becomes 1a in the numerator and denominator in the second part of the problem? Thanks! Check this: mathabsolutevaluemodulus86462.htmlAbsolute value properties:When \(x\leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\); When \(x\geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\); So, for our case, since \(a<1\), then \(a1<0\) hence according to the above \(a1=(a1)=1a\). Hope it helps.
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Re: Find the value of
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27 Mar 2013, 02:37
since its mentioned that a is between 0 and 1,we can directly use any of the numbers,say a=1/2.Substitute this in the equation,in the end ul get 2 as the answer. Looking at ans choices,since we started out with a=1/2, 1/a gives us our answer that is 2



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27 May 2013, 10:28
Bunuel,
When to use 1.\sqrt{x} = Mod(X) or
2. \sqrt{X} = X
Here u have used \sqrt{x} = Mod(X) while in the problem below:
If \sqrt{32x}= \sqrt{2x} +1 then 4x2 =
(A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1
you have used \sqrt{X} = X
Why so?



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27 May 2013, 10:38
karjan07 wrote: Bunuel,
When to use 1.\sqrt{x} = Mod(X) or
2. \sqrt{X} = X
Here u have used \sqrt{x} = Mod(X) while in the problem below:
If \sqrt{32x}= \sqrt{2x} +1 then 4x2 =
(A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1
you have used \sqrt{X} = X
Why so? \(\sqrt{x^2}=x\). If \(x\leq{0}\). then \(\sqrt{x^2}=x=x\). For example if \(x=5\), then \(\sqrt{(5)^2}=\sqrt{25}=5=x=x\). If \(x\geq{0}\), then \(\sqrt{x^2}=x=x\). For example if \(x=5\), then \(\sqrt{5^2}=\sqrt{25}=5=x=x\). I guess you are talking about the following problem: ifroot32xroot2x1then4x135539.html Where did I write that \(\sqrt{x^2}=x\)? P.S. Please read this: rulesforpostingpleasereadthisbeforeposting133935.html#p1096628 Press m button when formatting. Thank you.
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27 May 2013, 10:52
Bunuel wrote: karjan07 wrote: Bunuel,
When to use 1.\sqrt{x} = Mod(X) or
2. \sqrt{X} = X
Here u have used \sqrt{x} = Mod(X) while in the problem below:
If \sqrt{32x}= \sqrt{2x} +1 then 4x2 =
(A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1
you have used \sqrt{X} = X
Why so? \(\sqrt{x^2}=x\). If \(x\leq{0}\). then \(\sqrt{x^2}=x=x\). For example if \(x=5\), then \(\sqrt{(5)^2}=\sqrt{25}=5=x=x\). If \(x\geq{0}\), then \(\sqrt{x^2}=x=x\). For example if \(x=5\), then \(\sqrt{5^2}=\sqrt{25}=5=x=x\). I guess you are talking about the following problem: Where did I write that \(\sqrt{x^2}=x\)? Got it... My mistake... was confused on \(\sqrt{(3x2)}\)^2 = \(32x\) Probably should sleep now !!



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27 May 2013, 11:01
karjan07 wrote: Bunuel wrote: karjan07 wrote: Bunuel,
When to use 1.\sqrt{x} = Mod(X) or
2. \sqrt{X} = X
Here u have used \sqrt{x} = Mod(X) while in the problem below:
If \sqrt{32x}= \sqrt{2x} +1 then 4x2 =
(A) 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1
you have used \sqrt{X} = X
Why so? \(\sqrt{x^2}=x\). If \(x\leq{0}\). then \(\sqrt{x^2}=x=x\). For example if \(x=5\), then \(\sqrt{(5)^2}=\sqrt{25}=5=x=x\). If \(x\geq{0}\), then \(\sqrt{x^2}=x=x\). For example if \(x=5\), then \(\sqrt{5^2}=\sqrt{25}=5=x=x\). I guess you are talking about the following problem: Where did I write that \(\sqrt{x^2}=x\)? Got it... My mistake... was confused on \(\sqrt{(3x2)}\)^2 = \(32x\) Probably should sleep now !! Right. In that question we have \((\sqrt{32x})^2\), which equals to \(32x\), the same way as \((\sqrt{x})^2=x\). If it were \(\sqrt{(32x)^2}\), then it would equal to \(32x\), the same way as \(\sqrt{x^2}=x\). Check here: ifrot32xroot2x1then4x107925.html#p1223681Hope it's clear.
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27 May 2013, 11:38
Thanks... Its crystal clear now...



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29 May 2013, 11:45
Hello I am having a tough time figuring out why a1 = (a1) I know that for some absolute value problems, you test the positive and negative cases of a number. For example, 4=2x+3 I would solve for: I. 4=2x+3 II. 4=(2x+3) But I am not sure why you are deriving a1 = (a1) here. Does it have something to do with a being positive? (i.e. 0<a<1) Also, in another problem I just solved, I ended up with 5x = x5 which ended up being 5x = 5x Why wouldn't a+1 = a1 Thanks as always! Bunuel wrote: manimani wrote: Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a1)^2}}{\sqrt{(1+a)^2}  \sqrt{(a1)^2}}\) for 0<a<1
A. a B. 1/a C. (a1)/(a+1) D. (a+1)/(a1) E. a/(a1) Note that \(\sqrt{x^2}=x\). \(\frac{\sqrt{(1+a)^2} + \sqrt{(a1)^2}}{\sqrt{(1+a)^2}  \sqrt{(a1)^2}}=\frac{1+a+a1}{1+aa1}\) Now, since \(0<a<1\), then: \(1+a=1+a\) and \(a1=(a1)=1a\). Hence \(\frac{1+a+a1}{1+aa1}=\frac{(1+a)+(1a)}{(1+a)(1a)}=\frac{1}{a}\). Answer: B. Hope it's clear.



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29 May 2013, 12:01
So...0<a<1 which means that a+1 will always be positive. However a1 = some negative value so, some negative value = (some negative value) therefore, in this case: a1 = (a1) ==> =1a. Is this correct? Bunuel wrote: sharmila79 wrote: Hi Bunuel, Could you please explain me how a1 becomes 1a in the numerator and denominator in the second part of the problem? Thanks! Check this: mathabsolutevaluemodulus86462.htmlAbsolute value properties:When \(x\leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression\leq{(some \ expression)}\). For example: \(5=5=(5)\); When \(x\geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression\leq{some \ expression}\). For example: \(5=5\); So, for our case, since \(a<1\), then \(a1<0\) hence according to the above \(a1=(a1)=1a\). Hope it helps.



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Re: Find the value of
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29 May 2013, 12:03
WholeLottaLove wrote: Hello I am having a tough time figuring out why a1 = (a1) I know that for some absolute value problems, you test the positive and negative cases of a number. For example, 4=2x+3 I would solve for: I. 4=2x+3 II. 4=(2x+3) But I am not sure why you are deriving a1 = (a1) here. Does it have something to do with a being positive? (i.e. 0<a<1) Also, in another problem I just solved, I ended up with 5x = x5 which ended up being 5x = 5x Why wouldn't a+1 = a1 Thanks as always! Bunuel wrote: manimani wrote: Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a1)^2}}{\sqrt{(1+a)^2}  \sqrt{(a1)^2}}\) for 0<a<1
A. a B. 1/a C. (a1)/(a+1) D. (a+1)/(a1) E. a/(a1) Note that \(\sqrt{x^2}=x\). \(\frac{\sqrt{(1+a)^2} + \sqrt{(a1)^2}}{\sqrt{(1+a)^2}  \sqrt{(a1)^2}}=\frac{1+a+a1}{1+aa1}\) Now, since \(0<a<1\), then: \(1+a=1+a\) and \(a1=(a1)=1a\). Hence \(\frac{1+a+a1}{1+aa1}=\frac{(1+a)+(1a)}{(1+a)(1a)}=\frac{1}{a}\). Answer: B. Hope it's clear. If \(x\leq{0}\). then \(\sqrt{x^2}=x=x\). For example if \(x=5\), then \(\sqrt{(5)^2}=\sqrt{25}=5=x=x\). If \(x\geq{0}\), then \(\sqrt{x^2}=x=x\). For example if \(x=5\), then \(\sqrt{5^2}=\sqrt{25}=5=x=x\). ___________________________________ We are given that \(0<a<1\). For this range \(1+a>0\), so \(1+a=positive=1+a\) For this range \(a1<0\), so \(a1=negative=(a1)=1a\). Hope it's clear.
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29 May 2013, 12:11
Got it! Thanks a lot! Feels great to finally get it.



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03 Jul 2013, 06:21
manimani wrote: Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a1)^2}}{\sqrt{(1+a)^2}  \sqrt{(a1)^2}}\) for 0<a<1
A. a B. 1/a C. (a1)/(a+1) D. (a+1)/(a1) E. a/(a1) remember BODMAS, so now breakup we get 2/2a done logic + Basic = Magic in Gmat



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09 Jul 2013, 16:45
Luckily for us, every value here is the square root of a square so we can take the absolute value of each one:
What is the value of: 1+a + a1 / 1+a  a1 for 0<a<1
lets try plugging in a value for a: a=1/2
1+a + a1 / 1+a  a1 1+1/2 + 1/21 / 1+1/2  1/21 1.5 + .5 / 1.5  .5 2\1 = 2
A. a .5 B. 1/a 1/.5 = 2 C. (a1)/(a+1) D. (a+1)/(a1) E. a/(a1)
(B)
Also, another way we could solve:
1+a + a1 / 1+a  a1 (1+a) + (a1) / (1+a)  (a1) 1+a a+1 / 1+a  (a+1) 1+a a+1 / 1+a +a1 2/2a = 1/a




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