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## Manhattan Inequality and Absolute problem

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### Manhattan Inequality and Absolute problem

by Kaustubhk » Mon Jun 12, 2017 5:55 am
Is |x| < 1 ?

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

(2) |x - 3| > 0

The OA is C

Hi Mates,

My approach

Is |x| < 1?

We need to find -1<x<1, whether x falls in this range.

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

First i considered positive sign,

x+1 = 2x-2
x=3.. This doesn't fall in the range -1<x<1,

Now consider negative sign,

-(x+1) = -2(x-1)
-x-1 =-2x +2
x= 3 .. This doesn't fall in the range -1<x<1,

A is insufficient.

2) |x-3| >0

First consider positive sign,

x-3>0
x>3..This doesn't fall in the range -1<x<1,

Now consider negative sign

-(X-3) >0
-X+3>0
-X>-3

X<3..This doesn't fall in the range -1<x<1,

If i go by my approach the answer would be E.

When i checked the solution,

The step where i take negative sign for 1st case, therein the opposite sign is used.

Opposite signs

(x+1) = -2(x-1)
x+1 =-2x +2
x= 1/3 .. This falls in the range -1<x<1,

I'm confused as to why opposite signs are taken..

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by GMATGuruNY » Mon Jun 12, 2017 6:32 am
Kaustubhk wrote:1) |x+1| = 2|x-1|

First i considered positive sign,

x+1 = 2x-2
x=3.. This doesn't fall in the range -1<x<1,

Now consider negative sign,

-(x+1) = -2(x-1)
-x-1 =-2x +2
x= 3 .. This doesn't fall in the range -1<x<1,
If both sides of the blue equation are multiplied by -1, the result is the red equation.
Since each side of the blue equation is multiplied by the SAME VALUE, the resulting equation -- the equation in red -- is the SAME as the original equation.
Since both equations are essentially the same, the red equation will yield the same solution as the blue equation.

When both sides of an equation have absolute value, solve two cases:
1. The equation with signs unchanged
2. The equation with signs changed on ONLY ONE SIDE.

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by [email protected] » Mon Jun 12, 2017 6:53 am
Kaustubhk wrote:Is |x| < 1 ?

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

(2) |x - 3| > 0

The OA is C

Hi Mates,

My approach

Is |x| < 1?

We need to find -1<x<1, whether x falls in this range.

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

First i considered positive sign,

x+1 = 2x-2
x=3.. This doesn't fall in the range -1<x<1,

Now consider negative sign,

-(x+1) = -2(x-1)
-x-1 =-2x +2
x= 3 .. This doesn't fall in the range -1<x<1,

A is insufficient.

2) |x-3| >0

First consider positive sign,

x-3>0
x>3..This doesn't fall in the range -1<x<1,

Now consider negative sign

-(X-3) >0
-X+3>0
-X>-3

X<3..This doesn't fall in the range -1<x<1,

If i go by my approach the answer would be E.

When i checked the solution,

The step where i take negative sign for 1st case, therein the opposite sign is used.

Opposite signs

(x+1) = -2(x-1)
x+1 =-2x +2
x= 1/3 .. This falls in the range -1<x<1,

I'm confused as to why opposite signs are taken..
Hi Kaustubhk,

I think you need someone to teach the concepts of Data Sufficiency. Data Sufficiency questions test whether the information given in a statement/statements is sufficient or not.

Even if one is sure that the given information is sufficient to prove that we cannot reach a unique result, you are still in a position to answer the question.

In Statement 1, you twice write "This doesn't fall in the range." Since there is no contradiction, you can conclude or answer the question. (However, the red part is wrong!).

Secondly, you need to sharpen your fundamental about the concept of absolute numbers.

I correct the red part of your working.

Statement 1:

Now consider negative sign,

-(x+1) = -2(x-1)
-x-1 =-2x +2
x= 3 .. This doesn't fall in the range -1<x<1,

IT SHOULD BE:

Now consider negative sign,

(x+1) = -2(x-1); Take negative sign only on one side!
x+1 =-2x +2
3x= 1
x = 1/3 .. This falls in the range -1<x<1,

Since there is a contradiction in results: a. DOES NOT FALL for case 1 and b. FALLS for case 2, there is no unique answer!

Regarding Statement 2, you did all fine except that you concluded incorrectly that x < 3 means that This doesn't falls in the range -1<x<1.

What if x = 1/2? x = 1/2 satisfies both x < 3 as well as -1<x<1. So Statement 2 is also a case of contradictory results.

Hope now you can deal with both the statements together.

Hope this helps!