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# Is it true that ? |b-2|+|b+8|=10 (1) b less than or equal to

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Manager
Joined: 04 May 2007
Posts: 110
Is it true that ? |b-2|+|b+8|=10 (1) b less than or equal to [#permalink]

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20 Aug 2007, 16:14
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Is it true that ? |b-2|+|b+8|=10

(1) b less than or equal to 2
(2) b is greater than or equal to negative 8

Last edited by gmatiscoming on 21 Aug 2007, 08:26, edited 2 times in total.
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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21 Aug 2007, 09:13
gmatiscoming wrote:
Is it true that ? |b-2|+|b+8|=10

(1) b less than or equal to 2
(2) b is greater than or equal to negative 8

OK I get E...

here is how...

1) lets first see the break points they are 2 and -8

lets set a boundry condition

-8<b<2>0 then yes works out...10=10
if b<0 then (pick -6)

-6+2=-4 -6+8=2 -4+2=-2 which is not equal to 10 Insuff

2) b < or equal to -8

then b-2+b+8 is equal to 10.
if b <-8 then its not!

insuff
Director
Joined: 26 Jul 2007
Posts: 535
Schools: Stern, McCombs, Marshall, Wharton

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21 Aug 2007, 11:38
fresinha12 wrote:
gmatiscoming wrote:
Is it true that ? |b-2|+|b+8|=10

(1) b less than or equal to 2
(2) b is greater than or equal to negative 8

OK I get E...

here is how...

1) lets first see the break points they are 2 and -8

lets set a boundry condition

-8<b<2>0 then yes works out...10=10
if b<0 then (pick -6)

-6+2=-4 -6+8=2 -4+2=-2 which is not equal to 10 Insuff

2) b < or equal to -8

then b-2+b+8 is equal to 10.
if b <-8 then its not!

insuff

The break points are -8 and 2
Stmt 1 gives you the upper point 2
Stmt 2 gives you the lower point -8

Ans is C
VP
Joined: 10 Jun 2007
Posts: 1438

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21 Aug 2007, 11:48
gmatiscoming wrote:
Is it true that ? |b-2|+|b+8|=10

(1) b less than or equal to 2
(2) b is greater than or equal to negative 8

C.

These type of questions, it is much faster if you realize the critical points of the equation. In this case, critical points are b=2 and b=-8.
From this, you know that there are three intervals that the equation must satisfy.

A interval: b<-8
B interval: -8<b<2
C interval: b>2

If you find that b fall into one of these interval, then it must be right. (1) doesn't since it falls into intervals A and B, so INSUFFICIENT. (2) doesn't either because it falls into interval B and C, INSUFFICIENT. Together, it falls into interval B. SUFFICIENT.
Manager
Joined: 18 Jun 2007
Posts: 87

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21 Aug 2007, 12:09
I am not sure how C could be correct..

There are only 2 possible values of B that could satisfy the equation

|b-2|+|b+8|=10 , B is either2 or -8

Combining C

b<2>=-8

Combining both

B could have a value between -8......2 ?

Why are we not considering the values between ?
Thanks for clarrifying..
VP
Joined: 10 Jun 2007
Posts: 1438

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21 Aug 2007, 12:12
|b-2|+|b+8|=10

Plug in any number between -8 and 2 will satisfy the equation.
VP
Joined: 10 Jun 2007
Posts: 1438

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21 Aug 2007, 12:21
gmatiscoming wrote:
Is it true that ? |b-2|+|b+8|=10

(1) b less than or equal to 2
(2) b is greater than or equal to negative 8

C.

These type of questions, it is much faster if you realize the critical points of the equation. In this case, critical points are b=2 and b=-8.
From this, you know that there are three intervals that the equation must satisfy.

A interval: b<-8
B interval: -8<b<2
C interval: b>2

If you find that b fall into one of these interval, then it must be right. (1) doesn't since it falls into intervals A and B, so INSUFFICIENT. (2) doesn't either because it falls into interval B and C, INSUFFICIENT. Together, it falls into interval B. SUFFICIENT.

*Additional comment: for this method to be useful, you should at least plug in a value from those interval to see if which one equals 10.
SVP
Joined: 01 May 2006
Posts: 1796

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21 Aug 2007, 13:02
(C) for me too

|b-2|+|b+8|=10

I suggest to analyse quickly what is going on by replacing the absolutes.

o If b < -8 then
|b-2|+|b+8|
= -(b-2) - (b+8)
= -2*b - 6 >>> Not 10 and could not be

o If -8 =< b =< 2 then
|b-2|+|b+8|
= -(b-2) + (b+8)
= 10 >>> Sounds good

o If b > 2 then
|b-2|+|b+8|
= +(b-2) + (b+8)
= 2*b + 6 >>> Not 10 and could not be

From 1
b <= 2 >>>> So, we cannot conclude as it could be that -8 =< b =< 2 or b < -8

INSUFF.

From 2
b >= -8 >>>> So, we cannot conclude again as it could be that -8 =< b =< 2 or b > 2.

INSUFF.

Both (1) & (2)
Bongo.... It's -8 =< b =< 2 and we have seen that |b-2|+|b+8| = 10.

SUFF.
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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21 Aug 2007, 13:10
If I can only freakin add on D day..arrrgggggggh...

guys any suggestions on how to avoid such mistakes????
VP
Joined: 10 Jun 2007
Posts: 1438

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21 Aug 2007, 13:29
fresinha12 wrote:
If I can only freakin add on D day..arrrgggggggh...

guys any suggestions on how to avoid such mistakes????

It happens to me too when I try to do it too fast or do it at work. I suggest writing it out step by step or get a good sleep. Or maybe it is just rainy today :D
Manager
Joined: 03 Sep 2006
Posts: 233

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21 Aug 2007, 19:19
bkk145 wrote:
These type of questions, it is much faster if you realize the critical points of the equation. In this case, critical points are b=2 and b=-8.
From this, you know that there are three intervals that the equation must satisfy.

A interval: b<-8
B interval: -8<b<2>2

If you find that b fall into one of these interval, then it must be right. (1) doesn't since it falls into intervals A and B, so INSUFFICIENT. (2) doesn't either because it falls into interval B and C, INSUFFICIENT. Together, it falls into interval B. SUFFICIENT.

Many thanks!

I also got C on this question =( But your explanation is great, it'll be definitely a shortcut for me!
Re: DS absolute value   [#permalink] 21 Aug 2007, 19:19
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