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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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Bunuel on 27 May 2013, 11:12, edited 2 times in total.

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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 ]
27 May 2013, 10:38
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Re: Find the value of [#permalink ]
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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Re: Find the value of [#permalink ]
29 May 2013, 12:03
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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 ]
29 Nov 2013, 08:59
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mumbijoh wrote:

Hi Bunuel, I was also thinking along the same lines as you in this question.However, I got stuck deciding which rule to apply to which equation. How did you decide this. Couldn't you have swapped the rules then? |1+a|=-1-a And |a-1|=a-1

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 .

Similarly as

0<a<1 , then

1+a>0 , hence

|1+a|=1+a .

Hope it helps.

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Re: Find the value of [#permalink ]
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 ]
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 ]
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 ]
01 Jul 2012, 02:06
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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Re: Find the value of [#permalink ]
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 ]
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 ]
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 ]
27 May 2013, 11:38

Thanks... Its crystal clear now...

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Re: Find the value of [#permalink ]
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 ]
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 ]
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 ]
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

Re: Find the value of
[#permalink ]
03 Jul 2013, 06:21