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DS Question help!

by dqure040 » Tue Dec 01, 2015 7:55 pm
If x is negative, is x <-3?
1) x^2 > 9
2) x^3 < -9

When I solve 1) I get -3 < x & 3 < x, which means x is great than -3, i.e. Anwser A. Is my thinking correct here?
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by GMATinsight » Tue Dec 01, 2015 10:10 pm
dqure040 wrote:If x is negative, is x <-3?
1) x^2 > 9
2) x^3 < -9

When I solve 1) I get -3 < x & 3 < x, which means x is great than -3, i.e. Anwser A. Is my thinking correct here?
Question : Is x < -3 ?

Statement 1: x^2 > 9
i.e. x > 3 (NO) or x < -3 (YES) Hence
NOT SUFFICIENT

Statement 2: x^3 < -9

because (-8)^(1/3) = -2
so (-9)^(1/3) = -2.1 (approximately)

i.e. x < -2.1 (Approximately)
so x may or may not be less than -3 hence
NOT SUFFICIENT

Combining the two statements
we get, x < -3 and x can't be positive, hence
SUFFICIENT

Answer: Option C
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by GMATinsight » Tue Dec 01, 2015 10:17 pm
dqure040 wrote:If x is negative, is x <-3?
1) x^2 > 9
2) x^3 < -9

When I solve 1) I get -3 < x & 3 < x, which means x is great than -3, i.e. Anwser A. Is my thinking correct here?
Hi @dqure040,

The RED highlighted part above is a mistake because x can't be greater than -3 when x is negative e.g. x can't be -2 because (-2)^2 = 4 which is not greater than 9. Mathematically you can solve this expression as mentioned below

x^2 > 9
i.e. x^2 - 9 >0
i.e. x^2 - 3^2 >0
i.e. (x-3)*(x+3) > 0
which is possible when either both (x-3)&(x+3) are positive or both negative

if both are positive then x must be greater than 3 and greater than -3 i.e. x > 3 for both to be true

if both are Negative then x must be less than 3 and also less than -3 i.e. x < -3 for both to be true

Conclusively we get that x < -3 OR x > 3 for x^2 > 9

I hope this helps!
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by [email protected] » Wed Dec 02, 2015 5:44 pm
Hi dqure040,

When dealing with any type of math 'step', if you're unsure about whether you're doing the work correctly or not, you can always check your thinking by TESTING values.

For Fact 1, we have X^2 > 9...

The prompt tells us that X has to be NEGATIVE. You clearly understand that X COULD be greater than 3, but we're told NOT to consider those possibilities. So what types of negative values would 'fit' that inequality?

Would -2 'fit'?
Would -3 'fit'?
Would -4 'fit'?

So what does that tell you about X? You should deduce that X < -3. Thus, when X is negative, it MUST be less than -3, so the answer to the question is ALWAYS YES. Fact 1 is SUFFICIENT.

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Last edited by [email protected] on Thu Dec 03, 2015 10:19 am, edited 1 time in total.
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by manik11 » Thu Dec 03, 2015 12:26 am
[email protected] wrote:Hi dqure040,

When dealing with any type of math 'step', if you're unsure about whether you're doing the work correctly or not, you can always check your thinking by TESTING values.

For Fact 1, we have X^2 > 9...

You clearly understand that X COULD be greater than 3 and that there are some other solutions to consider. So what types of negative values would 'fit' that inequality?

Would -2 'fit'?
Would -4 'fit'?

So what does that tell you about X? You should deduce that X < -3 is the other set of values that you have to consider when dealing with Fact 1. Thus, X might be less than -3, but it might not. Fact 1 is INSUFFICIENT.

GMAT assassins aren't born, they're made,
Rich
Rich,
I have a doubt.
The prompt says "If X is negative". Doesn't this mean that we should not consider positive values of X for our test cases?
Accordingly if X has to be negative (according to the question) then one of the range of values of X from statement 1 cannot be considered in which case we can't consider X > 3 leaving us with only X < -3.

Please let me know if I'm missing something here.

Thanks!
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by GMATGuruNY » Thu Dec 03, 2015 3:36 am
dqure040 wrote:If x is negative, is x <-3?
1) x² > 9
2) x³ < -9
Statement 1: x² > 9
No negative value for x such that -3 ≤ x < 0 will satisfy the constraint that x² > 9.
To illustrate:
If x = -3, then x² = 9.
If x = -2, then x² = 4.
If x = -1, then x² = 1.
If x = -1/2, then x² = 1/4.
Implication:
To satisfy the constraint that x² > 9, it must be true that x < -3.
SUFFICIENT.

Statement 2: x³ < -9
It's possible that x = -3, since (-3)³ = -27, which is less than -9.
In this case, x = -3, so the answer to the question stem is NO.

It's possible that x = -4, since (-4)³ < -64, which is less than -9.
In this case, x < -3, so the answer to the question stem is YES.

Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.

The correct answer is A.
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by GMATGuruNY » Thu Dec 03, 2015 3:52 am
dqure040 wrote:If x is negative, is x <-3?
1) x^2 > 9
2) x^3 < -9

When I solve 1) I get -3 < x & 3 < x, which means x is great than -3, i.e. Anwser A. Is my thinking correct here?
To evaluate statement 1 algebraically, use the CRITICAL POINT approach.

Step 1: Move all of the variables to one side of the inequality and factor.
x² > 9
x² - 9 > 0
(x+3)(x-3) > 0.

Step 2: Identify the CRITICAL POINTS.
The CRITICAL POINTS are where the lefthand side is EQUAL to 0:
x=-3 and x=3.

Step 3: To determine the valid ranges of x, test one value to the left and right of each critical point.
Here, since x must be negative, we need to test only two cases:
A negative value to the LEFT of -3.
A negative value to the RIGHT of -3.

x<-3:
Test x=-10 in x² > 9:
(-10)² > 9
100 > 9.
This works: x<-3 is a valid range.

-3 < x < 0:
Test x=-1 in x² > 9:
(-1)² > 9
1 > 9.
Doesn't work: -3<x<0 is NOT a valid range.

Thus, the only valid negative range is x < -3.
SUFFICIENT.

If you type "critical points" and "gmatguruny" into the search bar, you'll find other problems that I've solved with the CRITICAL POINT approach.
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by [email protected] » Thu Dec 03, 2015 10:21 am
Hi manik11,

Very nice catch. I ignored the restriction that X was negative and went straight to Fact 1 - there's a lesson there about paying attention to all of the information that you're given. I've updated my explanation accordingly.

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by Matt@VeritasPrep » Fri Dec 11, 2015 2:02 pm
manik11 wrote: I have a doubt.
The prompt says "If X is negative". Doesn't this mean that we should not consider positive values of X for our test cases?
Yup, you're right: only negative values can be considered at that point.
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by Mo2men » Wed Oct 26, 2016 2:26 pm
GMATinsight wrote:
dqure040 wrote:If x is negative, is x <-3?
1) x^2 > 9
2) x^3 < -9

When I solve 1) I get -3 < x & 3 < x, which means x is great than -3, i.e. Anwser A. Is my thinking correct here?
Question : Is x < -3 ?

Statement 1: x^2 > 9
i.e. x > 3 (NO) or x < -3 (YES) Hence
NOT SUFFICIENT

Statement 2: x^3 < -9

because (-8)^(1/3) = -2
so (-9)^(1/3) = -2.1 (approximately)

i.e. x < -2.1 (Approximately)
so x may or may not be less than -3 hence
NOT SUFFICIENT

Combining the two statements
we get, x < -3 and x can't be positive, hence
SUFFICIENT

Answer: Option C
Hi GMATInsight,

statement 1 is sufficient as it is stated clearly in the prompt that X is Negative. So X>3 is not viable.
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by Scott@TargetTestPrep » Mon Oct 31, 2016 10:34 am
dqure040 wrote:If x is negative, is x <-3?
1) x^2 > 9
2) x^3 < -9
We are given that x is negative, and we must determine whether x < -3.

Statement One Alone:

x^2 > 9

Taking the square root of both sides of the inequality in statement one we have:

√x^2 > √9

|x| > 3

x > 3 OR -x > 3

x > 3 OR x < -3

Since we are given that x is negative, we see that x must be less than -3. Statement one alone is sufficient to answer the question. We can eliminate answer choices B, C, and E.

Statement Two Alone:

x^3 < -9

Using the information in statement two, we see that x can be less than -3 or not be less than -3.

For example, if x = -4, (-4)^3 = -64, (which fulfills the statement) and -4 is less than -3.

However, if x = -3, (-3)^3 = -27, (which fulfills the statement) but -3 is not less than -3.

Statement two alone is not sufficient to answer the question.

The answer is A

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