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If a = -0.3, (OG160

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Source: — Problem Solving |

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by Brent@GMATPrepNow » Thu May 19, 2016 3:59 pm
I fixed answer choices C and D to reflect the intent of the question.
boomgoesthegmat wrote:If a = -0.3, which of the following is true?

A) a < a² < a³
B) a < a³ < a²
C) a² < a < a³
D) a² < a³ < a
E) a³ < a < a²

OA: B
If a = -0.3, then:
a = -0.3
a² = (-0.3)(-0.3) = 0.09
a³ = (-0.3)(-0.3)(-0.3) = -0.027

If we order these values, we get: -0.3 < -0.027 < 0.09
In other words, a < a³ < a²
Answer: B

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by [email protected] » Thu May 19, 2016 4:02 pm
Hi boomgoesthegmat,

For this question, it's important to remember WHY a number is bigger than another number. By definition, the farther to the RIGHT on a number line, the bigger the number is.

Since A = -.3, we can calculate the values of the other terms and then put them in order from least to greatest...

A^2 = +.09
A^3 = -.027

In this example, A^3 is GREATER than A (since it's closer to 0, and farther to the right on the number line), but since they're both negative, A^2 is greater than them both.

Final Answer: B

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by Jeff@TargetTestPrep » Tue Jan 02, 2018 9:49 am
boomgoesthegmat wrote:If a = -0.3, which of the following is true?

A) a < a^2 < a^3
B) a < a^3 < a^2
C) av2 < a < a^3
D) av2 < a^3 < a
E) a^3 < a < a^2
We should keep in mind that when a negative number is raised to an even power, the result is positive, and when a negative number is raised to an odd power, the result is negative. Thus, a^2 is the largest among a, a^2, and a^3. Furthermore, we should keep in mind that when we raise a negative proper fraction to the third power, the result is a larger negative number. Thus, since a = -0.3:

a < a^3 < a^2, or -0.3 < -0.027 < 0.09

Answer: B

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