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Find the value of [#permalink] New post 28 Jun 2012, 03:10
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Difficulty:

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Question Stats:

28% (02:31) correct 71% (01:03) wrong based on 17 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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\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 28 Jun 2012, 03:12, edited 1 time 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 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 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] 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
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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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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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|\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.
_________________

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

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. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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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 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 :)
Re: Find the value of   [#permalink] 27 Mar 2013, 03:37
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