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If −1 < x < 0 and 0 < y < 1, which of the foll

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by BTGmoderatorLU » Mon Nov 27, 2017 7:26 am
$$If\ \ −1 < x < 0\ \ and\ \ 0 < y < 1,$$ which of the following has the least value?

$$A.\ \ \frac{x}{y}$$
$$B.\ (x/y)^2$$
$$C.\ \ \frac{x^2}{y}$$
$$D.\ \ \frac{x^3}{y}$$
$$E.\ \frac{x^2}{y^3}$$

The OA is A.

I know the answer. I just have one confusion as to how a negative sign to the power 3 could be greater than the negative sign to the power 1. Just as in this question it is implied that -x <-x^3. If you do it simply isn't -3 > (-3)^3.

Kindly can any expert help clarify my confusion. Thanks!
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by DavidG@VeritasPrep » Mon Nov 27, 2017 7:34 am
LUANDATO wrote:$$If\ \ −1 < x < 0\ \ and\ \ 0 < y < 1,$$ which of the following has the least value?

$$A.\ \ \frac{x}{y}$$
$$B.\ (x/y)^2$$
$$C.\ \ \frac{x^2}{y}$$
$$D.\ \ \frac{x^3}{y}$$
$$E.\ \frac{x^2}{y^3}$$

The OA is A.

I know the answer. I just have one confusion as to how a negative sign to the power 3 could be greater than the negative sign to the power 1. Just as in this question it is implied that -x <-x^3. If you do it simply isn't -3 > (-3)^3.

Kindly can any expert help clarify my confusion. Thanks!
Well, yes, -3 > (-3)^3, but given the constraints of the question, x can't be -3!

If x is between 0 and -1, then when we cube it, the result will be less negative. If x = -1/2, for example, then (-1/2)^3 = -1/8, and (-1/2) < (-1/8)

To summarize, if we cube a number less than negative one, the result will be more negative.

If we cube a number between -1 and 0, the result will be less negative.
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by [email protected] » Mon Nov 27, 2017 5:51 pm
Hi LUANDATO,

We're told that -1 < X < 0 and 0 < Y < 1. We're asked for the answer that has the LEAST value. This question can be solved with Number Property rules or by TESTing VALUES.

IF...
X = -1/2
Y = +1/2

Answer A: (-1/2)/(1/2) = -1
Answer B: [(-1/2)(1/2)]^2 = (-1)^2 = +1
Answer C: [(-1/2)^2]/(1/2) = +1
Answer D: [(-1/2)^3]/(1/2) = (-1/8)/(1/2) = -1/16
Answer E: [(-1/2)^2]/[(1/2)^3] = (1/4)/(1/8) = +2

Final Answer: A

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