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Awli
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Is x > 0 ? (1) x^3 < x^4 (2) x^3 < x^5 13 Apr 2015, 10:54
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

65% (hard)

Question Stats:

54% (02:02) correct 46% (01:32) wrong based on 89 sessions

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Date
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Not Attempted Yet
Is x > 0 ?

(1) x^3 < x^4

(2) x^3 < x^5
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E
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Is x > 0 ? (1) x^3 < x^4 (2) x^3 < x^5 13 Apr 2015, 12:44
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Hi Awli,

This DS question is a real test of one's "thoroughness"; if you're not considering all of the possibilities, then you'll likely get this question wrong. You can use Number Properties and/or TESTing VALUES to get to the correct answer.

We're asked if X > 0. This is a YES/NO question.

Fact 1: (X^3) < (X^4)

IF....
X = 2
8 < 16
The answer to the question is YES

IF....
X = -2
-8 < 16
The answer to the question is NO
Fact 1 is INSUFFICIENT

Fact 2: (X^3) < (X^5)

IF....
X = 2
8 < 32
The answer to the question is YES

IF...
X = -1/2
-1/8 < -1/32
The answer to the question is NO
Fact 2 is INSUFFICIENT

Combined, we know....
(X^3) < (X^4)
(X^3) < (X^5)

IF....
X = 2
8 < 16
8 < 32
The answer to the question is YES

IF...
X = -1/2
-1/8 < 1/16
-1/8 < -1/32
The answer to the question is NO
Combined, INSUFFICIENT

Final Answer:
Show Spoiler
E

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Is x > 0 ? (1) x^3 < x^4 (2) x^3 < x^5 14 Apr 2015, 04:01
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Awli
Is x > 0 ?

1. (x^3) < (x^4)

2. (x^3) < (x^5)
Show more

HINT:
We know that we cannot divide an inequality by a variable if we don't know its sign. But since x^3 < x^4 and x^3 < x^5 imply that x is not zero, then x to the even power (such as x^2, x^4, x^6, ...) will be positive, which means that we CAN divide both inequalities by x^2 to simplify the statements. Thus we'd get:
(1) x < x^2

(2) x < x^3

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Is x > 0 ? (1) x^3 < x^4 (2) x^3 < x^5 22 Apr 2015, 04:34
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We know that we cannot divide an inequality by a variable if we don't know its sign.[u] But since x^3 > x^4 and x^3 > x^5 imply that x is not zero, then x to the even power [/u](such as x^2, x^4, x^6, ...) will be positive, which means that we CAN divide both inequalities by x^2 to simplify the statements. Thus we'd get:
(1) x < x^2

(2) x < x^3

Bunuel can you please shed some more light on the underlined portion i am unable to understand that portion.

thanks
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Is x > 0 ? (1) x^3 < x^4 (2) x^3 < x^5 22 Apr 2015, 04:44
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anshul2014
We know that we cannot divide an inequality by a variable if we don't know its sign.[u] But since x^3 > x^4 and x^3 > x^5 imply that x is not zero, then x to the even power [/u](such as x^2, x^4, x^6, ...) will be positive, which means that we CAN divide both inequalities by x^2 to simplify the statements. Thus we'd get:
(1) x < x^2

(2) x < x^3

Bunuel can you please shed some more light on the underlined portion i am unable to understand that portion.

thanks
Show more


There is a little misprint in this part: "But since x^3 > x^4 and x^3 > x^5" should be "x^3 < x^4 and x^3 < x^5"
But this misprint change nothing. We know that \(x^3<x^4\) so we can infer that \(x\) not equal to \(0\)
Because if \(x = 0\) than \(x^3=x^4\) but we know from statements that this is not true.

and we know that any number in even power will be positive. As we know sign of \(x^2\) we can divide our equations on \(x^2\)
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Is x > 0 ? (1) x^3 < x^4 (2) x^3 < x^5 22 Apr 2015, 08:00
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Awli
Is x > 0 ?

1. (x^3) < (x^4)

2. (x^3) < (x^5)
Show more

An alternate solution to this question is by using The Wavy Line method :)

Analyzing Statement 1 first:

\(x^3\) < \(x^4\)

Subtracting both sides of an inequality with the same number doesn't affect the sign of inequality.

So, let's subtract both sides of the inequality with \(x^3\). We get:

0 < \(x^4\) - \(x^3\)

This inequality can also be written as:

\(x^4\) - \(x^3\) > 0

Factorizing the expression on the Left Hand Side:

\(x^3\)(x - 1) > 0

The value of the LHS will be zero for x = 0 and x = 1. So, 0 and 1 are known as the Zero Points of this expression.

Let's plot these two points on the number line. Then, starting from the top right corner, let's draw a wavy line that passes through these two points.



The given expression will be > 0 (that is, positive) in the regions where the Wavy Line is above the number line.

And, the given expression will be < 0 in the regions where the Wavy Line is below the number line.

So, we see that the inequality given in St. 1 (\(x^3\) < \(x^4\)) will hold for x < 0 and for x > 1.

Therefore, we cannot say for sure if x > 0 or not.

So, Statement 1 alone is not sufficient.


Let's now analyze Statement 2 alone:

\(x^3\) < \(x^5\)

Subtracting both sides of this inequality with \(x^3\), we get:

0 < \(x^5\) - \(x^3\)

This inequality can also be written as:

\(x^5\) - \(x^3\) > 0

Factorizing the expression on the Left Hand Side:

\(x^3\)\((x^2 - 1)\) > 0

This can be further factorized as:

\(x^3\) (x - 1)(x+1) > 0

The value of the LHS will be zero for x = -1, x = 0 and x = 1. So, -1, 0 and 1 are the Zero Points of this expression.

Let's plot these three points on the number line. Then, starting from the top right corner, we'll draw a wavy line that passes through these two points.



So, we see that the inequality given in St. 2 (\(x^3\) < \(x^5\)) will hold for -1 < x < 0 and for x > 1.

Therefore, we cannot say for sure if x > 0 or not.

So, Statement 2 alone is not sufficient.


Let's now combine the information from the two statements:

From Statement 1, we inferred that:

Either x < 0 Or x > 1

From Statement 2, we inferred that:

Either -1 < x < 0 Or x > 1

By combining these two statements, we get:

Either -1 < x < 0 Or x > 1

Therefore, we still have not been able to determine for sure if x is positive.

So, the correct answer is E.

Hope this helped! :)

- Japinder
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Is x > 0 ? (1) x^3 < x^4 (2) x^3 < x^5 22 Sep 2023, 19:50
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