Author
Message
TAGS:

Hide Tags
Intern

Joined: 22 Jun 2012

Posts: 13

Kudos [? ]:
65 [7 ], given: 2

+1 Kudos

Location: India

GPA: 3.09

Find the value of [#permalink ]

Show Tags
28 Jun 2012, 03:10
7

This post received KUDOS

47

This post was BOOKMARKED

Question Stats:

46% (01:24) correct

54% (01:18) wrong

based on 930 sessions

Hide Show timer Statistics
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)

Official Answer and Stats are available only to registered users.

Register /

Login .

_________________

\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.

Edited the question

Kudos [? ]:
65 [7 ], given: 2

+1 Kudos

Math Expert

Joined: 02 Sep 2009

Posts: 41682

Kudos [? ]:
124394 [18 ], given: 12078

+1 Kudos

Re: Find the value of [#permalink ]

Show Tags
28 Jun 2012, 03:19
18

This post received KUDOS

Expert's post

13

This post was BOOKMARKED

Kudos [? ]:
124394 [18 ], given: 12078

+1 Kudos

Intern

Joined: 15 May 2012

Posts: 41

Kudos [? ]:
5 [0 ], given: 94

+1 Kudos

Re: Find the value of [#permalink ]

Show Tags
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!

Kudos [? ]:
5 [0 ], given: 94

+1 Kudos

Manager

Joined: 15 Apr 2012

Posts: 94

Kudos [? ]:
58 [0 ], given: 134

+1 Kudos

Location: Bangladesh

Concentration: Technology, Entrepreneurship

GPA: 3.56

Re: Find the value of [#permalink ]

Show Tags
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

Kudos [? ]:
58 [0 ], given: 134

+1 Kudos

Senior Manager

Joined: 06 Aug 2011

Posts: 393

Kudos [? ]:
229 [0 ], given: 82

+1 Kudos

Re: Find the value of [#permalink ]

Show Tags
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?

_________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

Kudos [? ]:
229 [0 ], given: 82

+1 Kudos

Math Expert

Joined: 02 Sep 2009

Posts: 41682

Kudos [? ]:
124394 [0 ], given: 12078

+1 Kudos

Re: Find the value of [#permalink ]

Show Tags
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.

_________________

New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders ; 8. Overlapping Sets | PDF of Math Book ; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years. 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 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. , 11 New DS set. What are GMAT Club Tests ? Extra-hard Quant Tests with Brilliant Analytics

Kudos [? ]:
124394 [0 ], given: 12078

+1 Kudos

Math Expert

Joined: 02 Sep 2009

Posts: 41682

Kudos [? ]:
124394 [6 ], given: 12078

+1 Kudos

Re: Find the value of [#permalink ]

Show Tags
01 Jul 2012, 03:20
6

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

Kudos [? ]:
124394 [6 ], given: 12078

+1 Kudos

Senior Manager

Joined: 13 Aug 2012

Posts: 462

Kudos [? ]:
526 [2 ], given: 11

+1 Kudos

Concentration: Marketing, Finance

GPA: 3.23

Re: Find the value of [#permalink ]

Show Tags
06 Dec 2012, 01:01
2

This post received KUDOS

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

_________________

Impossible is nothing to God.

Kudos [? ]:
526 [2 ], given: 11

+1 Kudos

Manager

Joined: 06 Jun 2010

Posts: 159

Kudos [? ]:
19 [0 ], given: 151

+1 Kudos

Re: Find the value of [#permalink ]

Show Tags
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

Kudos [? ]:
19 [0 ], given: 151

+1 Kudos

Intern

Joined: 28 Jan 2013

Posts: 9

Kudos [? ]:
5 [0 ], given: 27

+1 Kudos

Location: India

Concentration: Marketing, International Business

GPA: 3

WE: Marketing (Manufacturing)

Re: Find the value of [#permalink ]

Show Tags
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?

Kudos [? ]:
5 [0 ], given: 27

+1 Kudos

Math Expert

Joined: 02 Sep 2009

Posts: 41682

Kudos [? ]:
124394 [1 ], given: 12078

+1 Kudos

Re: Find the value of [#permalink ]

Show Tags
27 May 2013, 11:38
Kudos [? ]:
124394 [1 ], given: 12078

+1 Kudos

Intern

Joined: 28 Jan 2013

Posts: 9

Kudos [? ]:
5 [0 ], given: 27

+1 Kudos

Location: India

Concentration: Marketing, International Business

GPA: 3

WE: Marketing (Manufacturing)

Re: Find the value of [#permalink ]

Show Tags
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 !!

Kudos [? ]:
5 [0 ], given: 27

+1 Kudos

Math Expert

Joined: 02 Sep 2009

Posts: 41682

Kudos [? ]:
124394 [1 ], given: 12078

+1 Kudos

Re: Find the value of [#permalink ]

Show Tags
27 May 2013, 12:01
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.

_________________

New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders ; 8. Overlapping Sets | PDF of Math Book ; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years. 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 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. , 11 New DS set. What are GMAT Club Tests ? Extra-hard Quant Tests with Brilliant Analytics

Kudos [? ]:
124394 [1 ], given: 12078

+1 Kudos

Intern

Joined: 28 Jan 2013

Posts: 9

Kudos [? ]:
5 [0 ], given: 27

+1 Kudos

Location: India

Concentration: Marketing, International Business

GPA: 3

WE: Marketing (Manufacturing)

Re: Find the value of [#permalink ]

Show Tags
27 May 2013, 12:38

Thanks... Its crystal clear now...

Kudos [? ]:
5 [0 ], given: 27

+1 Kudos

Senior Manager

Joined: 13 May 2013

Posts: 466

Kudos [? ]:
193 [0 ], given: 134

+1 Kudos

Re: Find the value of [#permalink ]

Show Tags
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.

Kudos [? ]:
193 [0 ], given: 134

+1 Kudos

Senior Manager

Joined: 13 May 2013

Posts: 466

Kudos [? ]:
193 [0 ], given: 134

+1 Kudos

Re: Find the value of [#permalink ]

Show Tags
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.

Kudos [? ]:
193 [0 ], given: 134

+1 Kudos

Math Expert

Joined: 02 Sep 2009

Posts: 41682

Kudos [? ]:
124394 [1 ], given: 12078

+1 Kudos

Re: Find the value of [#permalink ]

Show Tags
29 May 2013, 13:03
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.

_________________

New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders ; 8. Overlapping Sets | PDF of Math Book ; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years. 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 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. , 11 New DS set. What are GMAT Club Tests ? Extra-hard Quant Tests with Brilliant Analytics

Kudos [? ]:
124394 [1 ], given: 12078

+1 Kudos

Senior Manager

Joined: 13 May 2013

Posts: 466

Kudos [? ]:
193 [0 ], given: 134

+1 Kudos

Re: Find the value of [#permalink ]

Show Tags
29 May 2013, 13:11

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

Kudos [? ]:
193 [0 ], given: 134

+1 Kudos

Manager

Joined: 04 Mar 2013

Posts: 87

Kudos [? ]:
10 [0 ], given: 6

+1 Kudos

Location: India

Concentration: General Management, Marketing

GPA: 3.49

WE: Web Development (Computer Software)

Re: Find the value of [#permalink ]

Show Tags
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

Kudos [? ]:
10 [0 ], given: 6

+1 Kudos

Senior Manager

Joined: 13 May 2013

Posts: 466

Kudos [? ]:
193 [0 ], given: 134

+1 Kudos

Re: Find the value of [#permalink ]

Show Tags
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

Kudos [? ]:
193 [0 ], given: 134

+1 Kudos

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

Go to page
1 2
Next
[ 27 posts ]