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Tricky Inequalities - please help me quant gurus

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Tricky Inequalities - please help me quant gurus

by Natashangmat » Tue Nov 04, 2008 3:10 pm
Need some help from you guys....
Can anyone please give me a "detailed step by step method with ALL different possibilities" for solving these 2 questions?

plzzzz no shortcuts etc. or IMO's!

I am finding a lot of problems in solving equaltions like "(x-1)^2<=1" and "x^2 - 1>0"

Question1 - Is |x - 1| less than 1 ?
1). (x - 1)^2 less than and equal to 1
2). x^2 - 1 greater than 0


Question2 - Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|


I know the answer and I have some explanation also BUT i m just not able to understand the "actual logic" of doing such questions

OA's after some posts!
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Re: Tricky Inequalities - please help me quant gurus

by logitech » Tue Nov 04, 2008 3:58 pm
Natashangmat wrote:

OA's after some posts!
.

:D Come on!!!!!! Please give us our OAs....lol
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by cramya » Tue Nov 04, 2008 4:34 pm
Question1 - Is |x - 1| less than 1 ?
1). (x - 1)^2 less than and equal to 1
2). x^2 - 1 greater than 0


I am getting B) Will post soln if its right

Please post OA's using SPOILER FUNCTION if possible.
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by cramya » Tue Nov 04, 2008 4:49 pm
Question2 - Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|



I am getting A) OA please?
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by logitech » Tue Nov 04, 2008 5:20 pm
One has the OA...the other has the solution...let's see what is going to happen...

Next Episode of BTG DS! :lol:
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by logitech » Tue Nov 04, 2008 5:23 pm
cramya wrote:Question2 - Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|



I am getting A) OA please?
Cramya,

Both equations have two solutions: X = (0,2)

So don't you think the OA should be E ?
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by logitech » Tue Nov 04, 2008 5:27 pm
cramya wrote:Question1 - Is |x - 1| less than 1 ?
1). (x - 1)^2 less than and equal to 1
2). x^2 - 1 greater than 0


I am getting B) Will post soln if its right

Please post OA's using SPOILER FUNCTION if possible.
II) x^2 - 1 > 0

x^2 > 1

so X > 1 and X < -1

Is |x - 1| less than 1 ?

what if x=1.1 ? |x - 1| will be less than 1 but if X is 3 |x - 1| will be more than 1 ?
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by cramya » Tue Nov 04, 2008 5:35 pm
fell in to considering only intgers trap.

U r right. It should be E) then

x^2>1

x^2 < = 2x

If we take x=1.1 then |x-1| < 1
If we take x=2 then |x-1|=1

INSUFF
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by Natashangmat » Wed Nov 05, 2008 10:36 am
Thanks for your replies....

I am not sure if the below answers r correct or not but this is what i have as "OA's" (source - got it from some other forum):-

Answer to 1st question: E
Answer to 2nd question: A
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by logitech » Wed Nov 05, 2008 11:02 am
I will post this seperately

Question2 - Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|
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by 4meonly » Thu Nov 06, 2008 12:49 am
Guys, absolute value questions are my weak point.
Please look through my reasoning for this Q.


Question2 - Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|

(1) |x + 3| = 4x – 3
a. x + 3 = 4x – 3, and x=2, or x >0
b. -(x + 3) = 4x – 3, and x=0, or x not >0. BUT x=0 doesn't satisfy |x + 3| = 4x – 3
x=2, or x >0
SUFF

(2) |x – 3| = |2x – 3|
a. x – 3 = 2x – 3, x=0, or x not >0
b. -(x – 3) = 2x – 3, -x + 3 = 2x – 3, x=2, or x >0
INSUFF

A
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by jimmiejaz » Thu Nov 06, 2008 4:18 am
4meonly wrote:Guys, absolute value questions are my weak point.
Please look through my reasoning for this Q.


Question2 - Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|

(1) |x + 3| = 4x – 3
a. x + 3 = 4x – 3, and x=2, or x >0
b. -(x + 3) = 4x – 3, and x=0, or x not >0. BUT x=0 doesn't satisfy |x + 3| = 4x – 3
x=2, or x >0
SUFF

(2) |x – 3| = |2x – 3|
a. x – 3 = 2x – 3, x=0, or x not >0
b. -(x – 3) = 2x – 3, -x + 3 = 2x – 3, x=2, or x >0
INSUFF

A

Hi 4meonly,natashangmat

Will suggest you a critical point approach to these questions.
For questions/choices with only one abs value as in choice 1.
(1) |x + 3| = 4x – 3 , for these type of questions,
check both positive or negative values
when x>0, x+3 = 4x-3
which gives x=2, which is correct since we supposed x>0. hence x=2 is a solution for this equation.
Now, x<0, -(x+3) = 4x-3
which gives x=0 which is not possible as we have assumed x<0.
hence, statement 1 gives us a solution x=2. its suff.
So far it was relatively easy.
Problem comes when we have 2 abs values in an equation.
To do these type of questions, follow this approach.
Initially if we go by the previous approach, we should come across 4 possible equations as 2 for each side as when each of them x>0 and one when x<0.
Instead, check the critical points ie.. the points at which the abs value becomes 0.
So , in 2.)
critical points are 3 and 3/2.
Draw a number line and mark these 2 points. there will be 3 regions.
x< 3/2, 3/2<x<3, x>3
when x< 3/2
Both |x-3| and |2x-3| will be negative
so, we get, -(x-3) = -(2x-3) or x=0
which is a valid solution since we took x<3/2.
Now for 3/2<x<3
|x-3| will be positive and |2x-3| will be negative
x-3=-(2x-3)
x=2 which is also correct since 3/2<2<3
Now for the last region, x>3
Both |x-3| and |2x-3| will be positive
x-3=2x-3
x=0 which is not possible since we have took x>3.
hence, 2.) gives two values of x. 0 and 2
hence 2 is insuff.
Ans is A.
It looks a bit lengthy but i am sure if you go by this approach, you will not face any prob and will get it under 2 mins.
hope it helps...
keep rocking.
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by 4meonly » Thu Nov 06, 2008 7:04 am
Thank you! This will really make my life easier!

But how you decided that |2x-3| will be negative? Why not |x-3|?
jimmiejaz wrote: Now for 3/2<x<3
|x-3| will be positive and |2x-3| will be negative
x-3=-(2x-3)
x=2 which is also correct since 3/2<2<3
Thank you for reply!
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by jimmiejaz » Thu Nov 06, 2008 7:18 am
if u draw this on a number line you will have three regions. Check by going from negative to positive from left to right for |x-3| and positive to negative from right to left for |2x-3|
or the easiest approach is to plug values and see in both.
while plugging values, ignore abs sign.
eg. the first region..x<3/2
take x=1
x-3 = 1-3 = -2 which is -ve.
2x-3 = 2-3 = -1 which is -ve

second region 3/2<x<3
x=2
x-3 = 2-3 = -1 which is -ve
2x-3 = 4-3 = 1 which is +ve

third region x>3
x=4
x-3 = 4-3 = 1 which is +ve.
2x-3 = 8-3 = 5 which is +ve

Hope this clears the confusion if at all any.. :)
cheers!!!
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