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Find the value of [#permalink] New post 28 Jun 2012, 02:10
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Question Stats:

41% (02:20) correct 59% (01:09) wrong based on 381 sessions
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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Last edited by Bunuel on 27 May 2013, 11:12, edited 2 times in total.
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Re: Find the value of [#permalink] New post 28 Jun 2012, 02:19
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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 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] New post 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] New post 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] New post 01 Jul 2012, 02:06
Expert's post
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, 02:20
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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 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] New post 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] New post 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] New post 27 May 2013, 10:38
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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, 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] New post 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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Find the value of [#permalink] New post 27 May 2013, 11:38
Thanks... Its crystal clear now...
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Re: Find the value of [#permalink] New post 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] New post 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] New post 29 May 2013, 12:03
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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 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] New post 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] New post 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-1
2/2a = 1/a
Re: Find the value of   [#permalink] 09 Jul 2013, 16:45
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