Mod problem - anything wrong with this approach

Sris,
I think Pardeep did the same thing as u di but the oa seems to be E). I have included my thougths too. Look it over and provide your feedback.


Pradeep,
Whats the source and do they provide any explanation on why its E)?

Regards,
Cramya

cramya wrote:Sris,
I think Pardeep did the same thing as u di but the oa seems to be E). I have included my thougths too. Look it over and provide your feedback.


Pradeep,
Whats the source and do they provide any explanation on why its E)?

Regards,
Cramya

I think the OA is wrong...If not I badly need the explanation

Lets say we have to find the value of x.

equation 1 is (x-1)(x-4) = 0
Then x can be either 1 or 4

equation 2 is x-1 != 0
THen x can take any value except 1

But x= 4 is the only value which will satisfy both the equations.
here IMO C not E

If this approach is wrong then probably I have to redo all the problems in equations and inequations

You could be right also! I have PM'd Stuart and we will see how this turns out.

austin wrote:Tryingmybest,

Statement 1: |x+1| = 2|x-1|
Squaring, (x+1)^2 = 4(x-1)^2
=>x^2+1+2x = 4x^2+4-8x
=> 3x^2-10x+3 = 0
=> 3x^2-9x-x+3 = 0
=> 3x(x-3) -1(x-3) = 0
=> (3x-1) (x-3) = 0

How do you get img. roots?

This is a great approach to this question. I agree that the answer is (C), since statement (2) elminates one of the two possible values for x, leaving you with the certainty that x = 1/3.

One note on saving some time. Once you arrived at:

3x^2-10x+3 = 0

you knew that statement (1) gave you two possible values for x, which is enough to say that (1) is insufficient by itself.

When we get to the combination stage (since (2) is clearly insufficient by itself), all we really care about is if x=3 is one of the solutions for statement (1). So, if you're not sure how to factor it, just try plugging in x=3. In this case it works, so we know that statement (2) eliminates one of the two solutions and that combined the statements are sufficient.

Even if we have no clue what the second solution is, knowing that x has only one possible value is all we need to prove sufficiency.

Thanks again Stuart for the explanation/clarification!