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Inequalities

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| ≠ 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: — Data Sufficiency |
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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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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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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.
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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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