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Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a-1)^2}}{\sqrt{(1+a)^2} - \sqrt{(a-1)^2}}\) for 0<a<1
A. a
B. 1/a
C. (a-1)/(a+1)
D. (a+1)/(a-1)
E. a/(a-1)
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Bunuel on 27 May 2013, 11:12, edited 2 times in total.
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Re: Find the value of [#permalink ]
28 Jun 2012, 02:19
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Re: Find the value of [#permalink ]
30 Jun 2012, 15:15
Hi Bunuel, Could you please explain me how a-1 becomes 1-a in the numerator and denominator in the second part of the problem? Thanks!
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Re: Find the value of [#permalink ]
30 Jun 2012, 20:39
Bunuel wrote:
manimani wrote:
Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a-1)^2}}{\sqrt{(1+a)^2} - \sqrt{(a-1)^2}}\) for 0<a<1 A. a B. 1/a C. (a-1)/(a+1) D. (a+1)/(a-1) E. a/(a-1)
Note that \(\sqrt{x^2}=|x|\).
\(\frac{\sqrt{(1+a)^2} + \sqrt{(a-1)^2}}{\sqrt{(1+a)^2} - \sqrt{(a-1)^2}}=\frac{|1+a|+|a-1|}{|1+a|-|a-1|}\)
Now, since \(0<a<1\), then: \(|1+a|=1+a\) and \(|a-1|=-(a-1)=1-a\).
Hence \(\frac{|1+a|+|a-1|}{|1+a|-|a-1|}=\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{1}{a}\).
Answer: B.
Hope it's clear.
As 0<a<1
Let's say a =0.5 then (1+0.5+0.5-1)/(1+0.5-0.5+1) = 0.5 ..the answer is A...I am not getting it ...Can you explain a bit more ?Thnaks
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Re: Find the value of [#permalink ]
01 Jul 2012, 02:02
Bunuel wrote:
manimani wrote:
Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a-1)^2}}{\sqrt{(1+a)^2} - \sqrt{(a-1)^2}}\) for 0<a<1 A. a B. 1/a C. (a-1)/(a+1) D. (a+1)/(a-1) E. a/(a-1)
Note that \(\sqrt{x^2}=|x|\).
\(\frac{\sqrt{(1+a)^2} + \sqrt{(a-1)^2}}{\sqrt{(1+a)^2} - \sqrt{(a-1)^2}}=\frac{|1+a|+|a-1|}{|1+a|-|a-1|}\)
Now, since \(0<a<1\), then: \(|1+a|=1+a\) and \(|a-1|=-(a-1)=1-a\).
Hence \(\frac{|1+a|+|a-1|}{|1+a|-|a-1|}=\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{1}{a}\).
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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Re: Find the value of [#permalink ]
01 Jul 2012, 02:06
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sanjoo wrote:
Bunuel wrote:
manimani wrote:
Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a-1)^2}}{\sqrt{(1+a)^2} - \sqrt{(a-1)^2}}\) for 0<a<1 A. a B. 1/a C. (a-1)/(a+1) D. (a+1)/(a-1) E. a/(a-1)
Note that \(\sqrt{x^2}=|x|\).
\(\frac{\sqrt{(1+a)^2} + \sqrt{(a-1)^2}}{\sqrt{(1+a)^2} - \sqrt{(a-1)^2}}=\frac{|1+a|+|a-1|}{|1+a|-|a-1|}\)
Now, since \(0<a<1\), then: \(|1+a|=1+a\) and \(|a-1|=-(a-1)=1-a\).
Hence \(\frac{|1+a|+|a-1|}{|1+a|-|a-1|}=\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{1}{a}\).
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?
\(\frac{|1+a|+|a-1|}{|1+a|-|a-1|}=\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{1+a+1-a}{1+a-1+a}=\frac{2}{2a}=\frac{1}{a}\).
Hope it's clear now.
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Re: Find the value of [#permalink ]
01 Jul 2012, 02:20
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Re: Find the value of [#permalink ]
06 Dec 2012, 00:01
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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:
\(\frac{|1+a| + |a-1|}{|1+a| - |a-1|}\)
***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 a-1 = (fraction) - 1, we know that a-1 is negative. Using property #3, |a-1| = -(a-1)
Combine all that:
\(\frac{(1+a) + (-a+1)}{(1+a) - (-a+1)}\)
\(\frac{2}{2a}\)
\(\frac{1}{a}\)
Answer: B
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Re: Find the value of [#permalink ]
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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Re: Find the value of [#permalink ]
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{3-2x}= \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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Re: Find the value of [#permalink ]
27 May 2013, 10:38
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Re: Find the value of [#permalink ]
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{3-2x}= \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 \(\sqrt{(3x-2)}\)^2 = \(3-2x\)
Probably should sleep now !!
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Re: Find the value of [#permalink ]
27 May 2013, 11:01
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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{3-2x}= \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 \(\sqrt{(3x-2)}\)^2 = \(3-2x\)
Probably should sleep now !!
Right.
In that question we have \((\sqrt{3-2x})^2\), which equals to \(3-2x\), the same way as \((\sqrt{x})^2=x\).
If it were \(\sqrt{(3-2x)^2}\), then it would equal to \(|3-2x|\), the same way as \(\sqrt{x^2}=|x|\).
Check here:
if-rot-3-2x-root-2x-1-then-4x-107925.html#p1223681 Hope it's clear.
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Re: Find the value of [#permalink ]
27 May 2013, 11:38
Thanks... Its crystal clear now...
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Re: Find the value of [#permalink ]
29 May 2013, 11:45
Hello
I am having a tough time figuring out why |a-1| = -(a-1)
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 |a-1| = -(a-1) 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 |5-x| = |x-5| which ended up being |5-x| = |5-x| Why wouldn't |a+1| = |a-1|
Thanks as always!
Bunuel wrote:
manimani wrote:
Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a-1)^2}}{\sqrt{(1+a)^2} - \sqrt{(a-1)^2}}\) for 0<a<1 A. a B. 1/a C. (a-1)/(a+1) D. (a+1)/(a-1) E. a/(a-1)
Note that \(\sqrt{x^2}=|x|\).
\(\frac{\sqrt{(1+a)^2} + \sqrt{(a-1)^2}}{\sqrt{(1+a)^2} - \sqrt{(a-1)^2}}=\frac{|1+a|+|a-1|}{|1+a|-|a-1|}\)
Now, since \(0<a<1\), then: \(|1+a|=1+a\) and \(|a-1|=-(a-1)=1-a\).
Hence \(\frac{|1+a|+|a-1|}{|1+a|-|a-1|}=\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{1}{a}\).
Answer: B.
Hope it's clear.
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Re: Find the value of [#permalink ]
29 May 2013, 12:01
So...0<a<1 which means that |a+1| will always be positive. However |a-1| = |some negative value| so, |some negative value| = -(some negative value) therefore, in this case:
|a-1| = -(a-1) ==> =1-a.
Is this correct?
Bunuel wrote:
sharmila79 wrote:
Hi Bunuel, Could you please explain me how a-1 becomes 1-a in the numerator and denominator in the second part of the problem? Thanks!
Check this:
math-absolute-value-modulus-86462.html Absolute 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 \(a-1<0\) hence according to the above \(|a-1|=-(a-1)=1-a\).
Hope it helps.
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Re: Find the value of [#permalink ]
29 May 2013, 12:03
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WholeLottaLove wrote:
Hello
I am having a tough time figuring out why |a-1| = -(a-1)
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 |a-1| = -(a-1) 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 |5-x| = |x-5| which ended up being |5-x| = |5-x| Why wouldn't |a+1| = |a-1|
Thanks as always!
Bunuel wrote:
manimani wrote:
Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a-1)^2}}{\sqrt{(1+a)^2} - \sqrt{(a-1)^2}}\) for 0<a<1 A. a B. 1/a C. (a-1)/(a+1) D. (a+1)/(a-1) E. a/(a-1)
Note that \(\sqrt{x^2}=|x|\).
\(\frac{\sqrt{(1+a)^2} + \sqrt{(a-1)^2}}{\sqrt{(1+a)^2} - \sqrt{(a-1)^2}}=\frac{|1+a|+|a-1|}{|1+a|-|a-1|}\)
Now, since \(0<a<1\), then: \(|1+a|=1+a\) and \(|a-1|=-(a-1)=1-a\).
Hence \(\frac{|1+a|+|a-1|}{|1+a|-|a-1|}=\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{1}{a}\).
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 \(a-1<0\), so \(|a-1|=|negative|=-(a-1)=1-a\).
Hope it's clear.
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Re: Find the value of [#permalink ]
29 May 2013, 12:11
Got it! Thanks a lot! Feels great to finally get it.
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Re: Find the value of [#permalink ]
03 Jul 2013, 06:21
manimani wrote:
Find the value of \(\frac{\sqrt{(1+a)^2} + \sqrt{(a-1)^2}}{\sqrt{(1+a)^2} - \sqrt{(a-1)^2}}\) for 0<a<1 A. a B. 1/a C. (a-1)/(a+1) D. (a+1)/(a-1) E. a/(a-1)
remember BODMAS,
so now breakup we get 2/2a
done
logic + Basic = Magic in Gmat
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Re: Find the value of [#permalink ]
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 |+ |a-1| / |1+a| - |a-1| for 0<a<1 lets try plugging in a value for a: a=1/2 |1+a |+ |a-1| / |1+a| - |a-1| |1+1/2| + |1/2-1| / |1+1/2| - |1/2-1| |1.5| + |.5| / |1.5| - |.5| 2\1 = 2 A. a .5 B. 1/a 1/.5 = 2 C. (a-1)/(a+1) D. (a+1)/(a-1) E. a/(a-1) (B) Also, another way we could solve: |1+a |+ |a-1| / |1+a| - |a-1| (1+a) + -(a-1) / (1+a) - -(a-1) 1+a -a+1 / 1+a - (-a+1) 1+a -a+1 / 1+a +a-12/2a = 1/a
Re: Find the value of
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