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If x^3 < x^2, which of the following must be negative?

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by BTGModeratorVI » Fri Jul 03, 2020 7:15 am
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If x^3 < x^2, which of the following must be negative?

A. x
B. −x
C. x^5
D. x − 1
E. x^(−1)

Answer: D
Source: Veritas Prep
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Source: — Problem Solving |

BTGModeratorVI wrote: ↑
Fri Jul 03, 2020 7:15 am
If x^3 < x^2, which of the following must be negative?

A. x
B. −x
C. x^5
D. x − 1
E. x^(−1)

Answer: D
Source: Veritas Prep
Given: x³ < x²
This tells us that x ≠ 0. So, we can be certain that x² is POSITIVE.
Since x² is POSITIVE, we can safely divide both sides of the inequality by x²
When we do this, we get: (x³)/(x²) < 1
Simplify: x < 1
Subtract 1 from both sides to get: x - 1 < 0
In other words, x - 1 is NEGATIVE

Answer: D
Brent Hanneson - Creator of GMATPrepNow.com
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BTGModeratorVI wrote: ↑
Fri Jul 03, 2020 7:15 am
If x^3 < x^2, which of the following must be negative?

A. x
B. −x
C. x^5
D. x − 1
E. x^(−1)

Answer: D
Source: Veritas Prep
Alternatively, we can TEST some values.

We're told that x^3 < x^2
So, one possible value is x = -1

Now let's test the answer choices. We get:
A. -1, which is negative. KEEP
B. −(-1) =1, which is positive. ELIMINATE
C. (-1)^5 = -1, which is negative. KEEP
D. (-1) − 1 = -2, which is negative. KEEP
E. (-1)^(−1) = 1/(-1) = -1, which is negative. KEEP

Okay, we're still left with A, C, D and E

We need to test another value.
Another possible value is x = 1/2
Test the remaining answer choices...
A. 1/2, which is positive. ELIMINATE
C. (1/2)^5 = 1/32, which is positive. ELIMINATE
D. (1/2) − 1 = -1/2, which is negative. KEEP
E. (1/2)^(−1) = 2/1 = 2, which is positive. ELIMINATE

By the process of elimination, the correct answer is D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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BTGModeratorVI wrote: ↑
Fri Jul 03, 2020 7:15 am
If x^3 < x^2, which of the following must be negative?

A. x
B. −x
C. x^5
D. x − 1
E. x^(−1)

Answer: D
Solution:

Since x^3 is less than x^2, there are two cases for x: 1) x could be a negative number, or 2) x could be a (positive) number whose value is between 0 and 1.

Thus, x - 1 will always be negative.

Answer: D

Scott Woodbury-Stewart
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