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Find the value of

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manimani
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Find the value of [#permalink] New post   Updated on: 27 May 2013, 12:12
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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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B

Originally posted by manimani on 28 Jun 2012, 03:10.
Last edited by Bunuel on 27 May 2013, 12:12, edited 2 times in total.
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Re: Find the value of [#permalink] New post   28 Jun 2012, 03:19
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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] New post   06 Dec 2012, 01:01
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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] New post   30 Jun 2012, 16:15
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] New post   01 Jul 2012, 03:02
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] New post   01 Jul 2012, 03:06
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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] New post   01 Jul 2012, 03:20
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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 \(|some \ expression|={-(some \ expression)}\). 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 \(a-1<0\) hence according to the above \(|a-1|=-(a-1)=1-a\).

Hope it helps.
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Re: Find the value of [#permalink] New post   27 May 2013, 11:28
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] New post   27 May 2013, 11:38
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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: if-root-3-2x-root-2x-1-then-4x-135539.html Where did I write that \(\sqrt{x^2}=x\)?

P.S. Please read this: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 Press m button when formatting. Thank you.
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Re: Find the value of [#permalink] New post   27 May 2013, 11:52
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] New post   27 May 2013, 12:01
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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] New post   29 May 2013, 12:45
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] New post   29 May 2013, 13:03
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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] New post   28 Nov 2013, 08:14
Hi Bunuel,
I was also thinking along the same lines as you in this question.However, I got stuck deciding which rule to apply to which equation.
How did you decide this. Couldn't you have swapped the rules then?
|1+a|=-1-a
And |a-1|=a-1
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Re: Find the value of [#permalink] New post   29 Nov 2013, 09:59
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mumbijoh wrote:
Hi Bunuel,
I was also thinking along the same lines as you in this question.However, I got stuck deciding which rule to apply to which equation.
How did you decide this. Couldn't you have swapped the rules then?
|1+a|=-1-a
And |a-1|=a-1


Absolute 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 \(a-1<0\) hence according to the above \(|a-1|=-(a-1)=1-a\).

Similarly as \(0<a<1\), then \(1+a>0\), hence \(|1+a|=1+a\).

Hope it helps.
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Re: Find the value of [#permalink] New post   15 Apr 2023, 04:46
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