mgmat inequalities

resilient · Posted: Fri May 02, 2008 11:53 pm Post subject: mgmat inequalities

Is |x| <1> 0

qa is c

Stuart Kovinsky · Posted: Sat May 03, 2008 11:48 am Post subject: Re: mgmat inequalities

resilient · Posted: Sat May 03, 2008 2:33 pm Post subject: pc problem mgmat inequalities

Is |x| < 1 ?

(1) |x + 1| = 2|x – 1|

(2) |x – 3| > 0

qa is c

zacharyz · Posted: Wed May 07, 2008 2:55 pm Post subject:

Here's what I have so far:

Question is asking - is -1 < x < 1

Let's look at statement 2:
|x - 3| > 0
This doesn't tell you anything except that x DOES NOT EQUAL 3. For any other value, you can subtract three, take the absolute value and it is positive. 3 does not work because it says 3 - 3 = 0 > 0 not true.
This is insufficient (can be between -1 and 1 or can be higher)

Statement 1:
Let's look at values of x greater than 1 (because the second part of the statement is x-1, so I want to keep this expression positive).
You get:
x + 1 = 2x - 2 rearrange
3 = x This is one answer. And now you know that |x| is not less than 1

You need to find another value of x

This is where I actually failed. I could not figure out how to translate the equations to find a second value of x. (just by trial and error, I know that -3 is NOT the other solution, so I know that at least)

How you would set it up is evaluate each option:
x > 1 (all absolute value sections are positive)
1 > x > -1 (first statement is positive and second statement is negative)
x < -1 (all absolute value sections are negative)

Answer:
At this point, I know the answer is either going to be A or C. If A had given a second answer that is not -1 < x < 1, then that is the answer. If A provides an answer that is in that range, then you need statement B to prove that 3 is not the answer, and the second solution proves that -1 < x < 1.

In the interest of time (and already ruling out -3 as an answer), I guessed C, which is the official answer.

Incidentally - the second solution is 1/3. I figured this out a lot later, but could still use someone's help to set up the absolute value equations properly.

netigen · Posted: Wed May 07, 2008 4:47 pm Post subject:

Two solutions can be found by equating the abs equation on the RHS by first considering it + and then -

so I soln

x+1 = 2 (+(x-1))
you get x = 3

II soln

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

so ans is C

Hope this helps

lunarpower · Posted: Thu May 08, 2008 12:49 am Post subject:

moneyman · Posted: Thu May 08, 2008 5:06 am Post subject:

Hey netigen..great explanation but I have just one doubt..did u try changing the expression of stat(1) on both the sides ?

what I mean is, since its absolution on both sides of stat(1) did you try the following:

-(x+1)=2(-(x-1)) This gives the answer for x=-1. Though subsitituting the value of x=-1 in the expression does not hold true for stat(1). But I am just wondering how to approach the problem.

lunarpower · Posted: Thu May 08, 2008 11:17 am Post subject: