|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 04 May 2007
Posts: 111
Followers: 1
Kudos [?]:
0
[0], given: 0
|
Is it true that ? |b-2|+|b+8|=10 (1) b less than or equal to [#permalink]
 20 Aug 2007, 16:14
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
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: 3437
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 11
Kudos [?]:
135
[0], given: 2
|
Re: DS absolute value [#permalink]
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: 541
Schools: Stern, McCombs, Marshall, Wharton
Followers: 4
Kudos [?]:
61
[0], given: 0
|
Re: DS absolute value [#permalink]
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 -6-2=-8 -6+8=-2 You added instead of subtracted
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: 1478
Followers: 5
Kudos [?]:
73
[0], given: 0
|
Re: DS absolute value [#permalink]
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: 89
Followers: 2
Kudos [?]:
0
[0], given: 0
|
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: 1478
Followers: 5
Kudos [?]:
73
[0], given: 0
|
|b-2|+|b+8|=10
Plug in any number between -8 and 2 will satisfy the equation.
|
|
|
|
|
|
VP
Joined: 10 Jun 2007
Posts: 1478
Followers: 5
Kudos [?]:
73
[0], given: 0
|
Re: DS absolute value [#permalink]
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: 1837
Followers: 8
Kudos [?]:
73
[0], given: 0
|
(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: 3437
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 11
Kudos [?]:
135
[0], given: 2
|
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: 1478
Followers: 5
Kudos [?]:
73
[0], given: 0
|
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: 237
Followers: 1
Kudos [?]:
5
[0], given: 0
|
Re: DS absolute value [#permalink]
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
|
|
|
|
|
|
|
|
|
|
|
close
