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# Find the value of

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Intern
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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)
[Reveal] Spoiler: OA

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\sqrt{[square_root][square_root][square_root][square_root][square_root]}[/square_root][/square_root][/square_root][/square_root][/square_root]????????

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]

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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 $$\frac{|1+a|+|a-1|}{|1+a|-|a-1|}=\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{1}{a}$$.

Hope it's clear.
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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 $$\frac{|1+a|+|a-1|}{|1+a|-|a-1|}=\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{1}{a}$$.

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 $$\frac{|1+a|+|a-1|}{|1+a|-|a-1|}=\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{1}{a}$$.

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
Expert's post
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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 $$\frac{|1+a|+|a-1|}{|1+a|-|a-1|}=\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{1}{a}$$.

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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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]

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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:

$$\frac{(1+a) + (-a+1)}{(1+a) - (-a+1)}$$
$$\frac{2}{2a}$$
$$\frac{1}{a}$$

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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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27 May 2013, 11:38
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Expert's post
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 $$\sqrt{x^2}=x$$?

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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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Expert's post
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 $$\frac{|1+a|+|a-1|}{|1+a|-|a-1|}=\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{1}{a}$$.

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 $$\frac{|1+a|+|a-1|}{|1+a|-|a-1|}=\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{1}{a}$$.

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-1
2/2a = 1/a

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Re: Find the value of   [#permalink] 09 Jul 2013, 17:45

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