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Find the value of [#permalink ]

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28 Jun 2012, 03:10
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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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Re: Find the value of [#permalink ]

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28 Jun 2012, 03:19
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Re: Find the value of [#permalink ]

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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 ]

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30 Jun 2012, 21:39

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 ]

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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 ]

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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 ]

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Re: Find the value of [#permalink ]

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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 ]

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27 Mar 2013, 03:37

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 ]

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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 ]

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Re: Find the value of [#permalink ]

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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 ]

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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 ]

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27 May 2013, 12:38

Thanks... Its crystal clear now...

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Re: Find the value of [#permalink ]

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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 ]

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29 May 2013, 13:01

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 ]

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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 ]

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29 May 2013, 13:11

Got it! Thanks a lot! Feels great to finally get it.

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Re: Find the value of [#permalink ]

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03 Jul 2013, 07:21

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 ]

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09 Jul 2013, 17:45

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

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