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DS Doubt

by gmatmillenium » Tue Jun 15, 2010 12:41 am
X range is -1 <X <1, Q Which of the following must be true?
Answers are:
1. X ^ 2> X ^ 3,
2. (X +1 / 2) (X-1 / 2) <0

Explanation given by one the experts was as under..

Obviously x^2 > x^3 does not hold for x = 0.
Taking the second statement we have that (X +1 / 2) (X-1 / 2) <0
So we have that either (a) (X +1 / 2)>0 and (X-1 / 2)< 0
Or
(b) (X +1 / 2)<0 and (X-1 / 2)>0. This would mean x<-1/2 and x>1/2 which is not possible and so (b) is automatically rejected.
So we have that (X +1 / 2) (X-1 / 2) <0 implies -1/2<x<1/2 which is possible because given range of x is -1 <x<1.

The second statement is true.

My query - for x= .6, statement 2 doesn't hold good so how is the answer b?
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by kvcpk » Tue Jun 15, 2010 1:02 am
HI gmatmillenium,

The question is specifying a rangle for X as -1 to 1

in the 2 premises given below, our task is to find if th range of x is equal to or subset of the range specified in the question.

Clearly statement1 x^2-x^3>0 doesnt hold for x=0. So eliminated.

in statement2, (x+1/2)(x-1/2)<0
whenever, (x+a)(x+b)<0, implies that b<x<a
so x has to be between -1/2 and 1/2 for the statement2 to be true.
This is clearly a subset of the range in question.

So Answer is B
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by gmatmillenium » Tue Jun 15, 2010 1:39 am
key word is "must" and hence for all values of x within -1 and 1, the inequality should hold good.....here it doesnot hold good at x=.6

your approach is finding a subset of the range from the given inequality whereas the ques asked for all situation validity of the equation after specifying a range....

May I request one of the quant experts to comment on this please?....




kvcpk wrote:HI gmatmillenium,

The question is specifying a rangle for X as -1 to 1

in the 2 premises given below, our task is to find if th range of x is equal to or subset of the range specified in the question.

Clearly statement1 x^2-x^3>0 doesnt hold for x=0. So eliminated.

in statement2, (x+1/2)(x-1/2)<0
whenever, (x+a)(x+b)<0, implies that b<x<a
so x has to be between -1/2 and 1/2 for the statement2 to be true.
This is clearly a subset of the range in question.

So Answer is B
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by sumanr84 » Tue Jun 15, 2010 5:51 am
It cannot be B

(X +1 / 2) (X-1 / 2) = X^2 - (1/2)^2 = (X ^2) - (1/4)

Let say we plug in X = 1/2 then it eqn becomes = 0, and hence does not satisfy 2.
I am on a break !!
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by Rich@VeritasPrep » Tue Jun 15, 2010 6:11 am
gmatmillenium: .6 does not work, because it is outside of the given range (-1/2<x<1/2). I'm guessing you just accidentally overlooked this. Happens to the best of us!

sumanr84: I agree that your method is the best way to approach Statement (2)...

We have (X +1 / 2) (X-1 / 2) <0, and one thought could be to consider each case where one of the parentheticals is positive, the other negative. The problem with this method is that (as sumanr84 pointed out), it fails to account for the possibility that the left side of the inequality could equal 0.

Instead, do what sumanr84 did: Realize that plugging in x=-1/2 or x=1/2 makes the left side 0, and thus the inequality does not have to be true.

You could also verify this with a little algebra, although you certainly don't need to:

(X +1 / 2) (X-1 / 2) <0

(X ^2) - (1/4) < 0

x^2 < 1/4

Now, it's easy to see that this doesn't hold for either x=-1/2 or x=1/2.

Make sense?
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by Rich@VeritasPrep » Tue Jun 15, 2010 6:21 am
Also, you are saying this is a DS question?

Because DS question prompts never have phrasings like "Which of following must be true?"

Either they'd ask something like "What is the value...?" or "Is this true...?"

What was the source?
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by gmatmillenium » Tue Jun 15, 2010 7:07 am
Hi Raz,

Source is beat the gmat for both the ques and the explanation (strikes me very odd now because I only follow expert commented ques) - I should have posted the query in the same thread..


well the range given was -1 to 1 as is even mentioned in my ques so no, not a case of oversight....i pretty much too arrived at "Not B", which conflicted with the answer I had seen on beat the gmat as I mentioned.

Thanks for chipping in.

Thanks Suman!


raz1024 wrote:gmatmillenium: .6 does not work, because it is outside of the given range (-1/2<x<1/2). I'm guessing you just accidentally overlooked this. Happens to the best of us!

sumanr84: I agree that your method is the best way to approach Statement (2)...

We have (X +1 / 2) (X-1 / 2) <0, and one thought could be to consider each case where one of the parentheticals is positive, the other negative. The problem with this method is that (as sumanr84 pointed out), it fails to account for the possibility that the left side of the inequality could equal 0.

Instead, do what sumanr84 did: Realize that plugging in x=-1/2 or x=1/2 makes the left side 0, and thus the inequality does not have to be true.

You could also verify this with a little algebra, although you certainly don't need to:

(X +1 / 2) (X-1 / 2) <0

(X ^2) - (1/4) < 0

x^2 < 1/4

Now, it's easy to see that this doesn't hold for either x=-1/2 or x=1/2.

Make sense?
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by kvcpk » Tue Jun 15, 2010 7:44 am
What is the final answer for this question?
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