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

[srikant]

Is x > 0?

(1) |x + 3| = 4x - 3

(2) |x - 3| = |2x - 3|

The CAT answer was wrong and I would like to find what others have to say.
Can anyone please tell the write answer with explanation.
I am more concerned about the second statement!

[srikant]

From first statement we can see that L.H.S. is always +ve i.e. 4x-3>=0; x>=3/4 sufficient.

From statement 2 there are 3 possible cases
x<3/2
3/2<x<3
3<x

for each case we see the following result
1) for x<3/2
x=0. satisfies
2) for 3/2<x<3
x=2 satisfies
3)for 3<x
x=0. not right

Since we get 2 values of x, one equal to 0 and another greater than 0, it can not be said whether x>0. Not sufficient

Therefore answer is A.

[StaceyKoprince]

Yes, the answer is A. You said in your first post that the CAT answer was wrong, but I just looked it up in the system and the answer is correctly listed as A. (Unless the second post is from some other person who also chose srikant as a username... ?)

So - I'm confused as to your question. Let me know!
:)

[sathishm.123]

Hi Stacey,

Can you please solve the problem, I am quite confused with the obsolute to eniquality conversion.

My approach was
1) x-3=2x-3 therefore x=0
2) x-3 = -(2x-3) therefore x=2 so sufficient, please point my mistake.

[mschwrtz]

I suspect that this is the same as the CAT explanation, but I think that I can explain it about as quickly as I can look it up.

Is x > 0?

(1) |x + 3| = 4x - 3

This admits, or seems to admit, two possibilities:

if x+3>= 0, then x+3=4x-3-->3x=6-->x=2
plug back in to verify. Yup, |2+3|=4*2-3.

if x+3<=0, then x+3=3-4x-->3x=0-->x=0
This is consistent with the assumption that x<=0, but plug back in to verify. Nope, |0-3| does not equal 4*0+3.
THIS CAN HAPPEN WHEN THERE IS A VARIABLE OUTSIDE THE ABSOLUTE VALUE AS WELL AS INSIDE.

So S1 means that x=2 SUFFICIENT

[sachin.w]

This is how I proved A is sufficient.

|x + 3| = 4x - 3
=>
since |x+3| is the absolute value, it is non negative and so

4x-3 >=0

=>

x>= 3/4

so x>0

[tim]

perfect!

[clarence.booth]

Hi,

As an FYI, the CAT Exam #4 still declares E as the correct answer as oppose to A. I was pulling my hair out trying justify how. I'm glad to see it is in fact A. The error was cited in 2007 and still persists. Please resolve.

Here's the question and explanation as listed on my CAT 4 exam:

Question 3

Is x > 0?

(1) |x + 3| < 4

(2) |x – 3| < 4


Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

(1) INSUFFICIENT: We can solve this absolute value inequality by considering both the positive and negative scenarios for the absolute value expression |x + 3|.
If x > -3, making (x + 3) positive, we can rewrite |x + 3| as x + 3:
x + 3 < 4
x < 1
If x < -3, making (x + 3) negative, we can rewrite |x + 3| as -(x + 3):
-(x + 3) < 4
x + 3 > -4
x > -7
If we combine these two solutions we get -7 < x < 1, which means we can’t tell whether x is positive.

(2) INSUFFICIENT: We can solve this absolute value inequality by considering both the positive and negative scenarios for the absolute value expression |x – 3|.
If x > 3, making (x – 3) positive, we can rewrite |x – 3| as x – 3:
x – 3 < 4
x < 7
If x < 3, making (x – 3) negative, we can rewrite |x – 3| as -(x – 3) OR 3 – x
3 – x < 4
x > -1
If we combine these two solutions we get -1 < x < 7, which means we can’t tell whether x is positive.

(1) AND (2) INSUFFICIENT: If we combine the solutions from statements (1) and (2) we get an overlapping range of -1 < x < 1. We still can’t tell whether x is positive.

The correct answer is E.

[clarence.booth]

As a follow up, I'm not sure I understand the method the explanation illustrates. I used the CLA (Criteria, List, Answer) method shown in the Data Sufficiency Lab. But this somehow seems quicker.

Can someone explain the manipulation of the expressions in statement 1 and 2 shown in the CAT Exam explanation?

[RonPurewal]

As a follow up, I'm not sure I understand the method the explanation illustrates. I used the CLA (Criteria, List, Answer) method shown in the Data Sufficiency Lab. But this somehow seems quicker.

Can someone explain the manipulation of the expressions in statement 1 and 2 shown in the CAT Exam explanation?

in those explanations, what specific things are giving you trouble?

[BrianW172]

I must be missing something. I am struggling with combining the inequalities from statements 1 & 2 on this question.

From statement 1: -7 < x < 1
From statement 1: -1 < x < 7

Therefore, shouldn't statements 1 & 2 combined be the following:
Stat 1 & 2: -8 < 2x < 8
which can be reduced to -4 < x < 4?

Not sure how we arrived at -1 < x < 1

As always, thanks for the help.

[tim]

Look at what you wrote in your own post! In one place you wrote x<1, and in another place you wrote -1<x. Combining these two things that you said generates the statement that -1<x<1. Of course it is also true that -4<x<4, but it is also true that -1000000<x<1000000. Your goal however should be to find the most precise description of x that you can, and -1<x<1 is definitely better than those other alternatives.

[BrianW172]

Thanks for the help, Tim!! This forum helps a great deal.

[RonPurewal]

excellent.

[ZadiaH581]

I know this is a old post, but can someone say which problem this is and where I can find it? Also can someone explain it in full detail?

Thank you!