Inequalities

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krishna kumar: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Fri Jun 11, 2010 10:02 pmFollowed by: 1 member

by krishna kumar » Tue Jun 15, 2010 4:42 am

Hi,

I have this problem from the Sets -(set 28 problem 5)

Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| â‰ 0

The OA is C.

I haven't been able to figure that one out.

I figured that both 1 and 2 are not sufficient.

Could any one help.

thanks

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Source: Beat The GMAT — Data Sufficiency | Tue Jun 15, 2010 4:42 am

kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Tue Jun 15, 2010 5:05 am

By solving for different cases using statement1, we get
x=1/3 or x=3

one value of x is less than1 and the other greater than 1

So we need to use the statement2 which says that x is not equal to 3

So x =1/3 which means that IxI<1

Hence C

Hope it Helps!!

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ssuarezo: Master | Next Rank: 500 Posts; Posts: 126; Joined: Wed Jun 24, 2009 1:12 pm; Location: Montreal; Thanked: 2 times; GMAT Score:510

by ssuarezo » Tue Jun 15, 2010 9:00 am

kvcpk wrote:By solving for different cases using statement1, we get
x=1/3 or x=3

one value of x is less than1 and the other greater than 1

So we need to use the statement2 which says that x is not equal to 3

So x =1/3 which means that IxI<1

Hence C

Hi kvcpk:
Could you give more detail of how u got x=1/3, 3 from stm1? I thought I knew abs values but your answer I will simplify my review.
Thanks
Silvia

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Testluv: GMAT Instructor; Posts: 1302; Joined: Mon Oct 19, 2009 2:13 pmFollowed by: 164 members; Location: Toronto; Thanked: 539 times; GMAT Score:800

by Testluv » Tue Jun 15, 2010 1:19 pm

Hi kvcpk:
Could you give more detail of how u got x=1/3, 3 from stm1? I thought I knew abs values but your answer I will simplify my review.
Thanks
Silvia

(1) |x+1| = 2|x-1|

case 1:

x + 1 = 2x - 2
x = 3

case 2:
-(x+1) = 2x - 2
x = 1/3

We always have 2 cases when dealing with absolute value. When we see the absolute value bars on both sides of the equation, we only have to take pos an neg of one side.

Kaplan Teacher in Toronto

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kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Tue Jun 15, 2010 8:35 pm

Testluv wrote:
Hi kvcpk:
Could you give more detail of how u got x=1/3, 3 from stm1? I thought I knew abs values but your answer I will simplify my review.
Thanks
Silvia

(1) |x+1| = 2|x-1|

case 1:

x + 1 = 2x - 2
x = 3

case 2:
-(x+1) = 2x - 2
x = 1/3

We always have 2 cases when dealing with absolute value. When we see the absolute value bars on both sides of the equation, we only have to take pos an neg of one side.

Thanks TestLuv.. I was late to respond..

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