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DS: #97, 12th Edition

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DS: #97, 12th Edition

by boy141 » Mon Oct 26, 2009 7:30 pm
If x is negative, is x<-3?

(1) x^2 > 9
(2) x^3 < -9

A

The book's answer is crazy for both 1 and 2 - esp 2. Can someone explain it?
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Re: DS: #97, 12th Edition

by connit » Mon Oct 26, 2009 8:36 pm
boy141 wrote:If x is negative, is x<-3?

(1) x^2 > 9
(2) x^3 < -9

A

The book's answer is crazy for both 1 and 2 - esp 2. Can someone explain it?
(1) x^2 > 9
=> x>3 or x<-3
Since x is negative, we have x<-3 => Sufficient to answer the question. So we can eliminate B,C and E.

Now we check (2)
x^3 < -9 => x < -9^1/3 and the value -9^1/3 > -3 so x can be < -3 but x can also > -3 => Not sufficient => Eliminate D

So the ans is A.
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by boy141 » Mon Oct 26, 2009 8:48 pm
connit, I cannot follow your reasoning. Can you elaborate?


I get 1 but 2. eh.

(2) x^3 < -9

If I use -3 for X, I will get -27 which satisfy (2) and will answer NO for "is x<-3?". If I choose -4, then I will get -64 which will satisfy (2) but will make the answer NO for "is x<-3?".


My answers still say NO but using -3, it's equal while anything less than -3 will be greater and still the answer is NO.

Am I right or crazy?
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by connit » Mon Oct 26, 2009 9:01 pm
You are not crazy :) ... If you picked -3, x^3 = -27 which satisfies (2) but -3 is not < -3 => No

If you pick -4, x^3 = -64 < -9 which satisfies (2) and -4 < -3 => Ans Yes

Therefore, it's not sufficient. It has to be always No, or always Yes.
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Re: DS: #97, 12th Edition

by palvarez » Mon Oct 26, 2009 10:35 pm
Whenever you see x^3 < a^3, it implies x < a, no matter what.

x^3 - a^3 = (x-a) (x^2 +xa + a^2)

(x^2 + xa + a^2) > 0 for all x and a.

In this case, x^3 < -9

x^3 + 9 < 0

Therefore, x < -cubicroot(9)

note that -cubicroot(9) > -3.
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