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Is a^3 > 20?

This topic has 1 expert reply and 0 member replies

Is a^3 > 20?

Post Tue Oct 31, 2017 9:16 am
$$Is\ \ a^3>20?$$

$$(1)\ \ a^4>80$$
$$(2)\ \ a^5>200$$

The OA is B.

I don't understand it. Please, can any expert help me with this DS question? Thanks.

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Post Fri Nov 10, 2017 12:01 am
swerve wrote:
$$Is\ \ a^3>20?$$
$$(1)\ \ a^4>80$$
$$(2)\ \ a^5>200$$
The OA is B.

I don't understand it. Please, can any expert help me with this DS question? Thanks.
(1) a^4 > 80

We can safely assume that a^4 ≥ 81 => a^4 ≥ 3^4 => -3 ≥ a or 3 ≤ a.

Case 1: If -3 ≥ a, then say at a = -3, we have a^3 > 20 => (-3)^3 > 20 => -27 < 20. The answer is No.
Case 2: If 3 ≤ a, then say at a = 3, we have a^3 > 20 => (3)^3 > 20 => 27 > 20. The answer is Yes. No unique answer.

(2) a^5 > 200

This is a difficult one. We cannot approximate 200 to 243 (= 3^5); This much of approximation is way too much.

However, since the exponent of a is an odd number, a cannot have a negative value.

@ a = 3, we have a^5 = 3^5 = 243 > 200. At a = 3, we have a^3 = 3^3 = 27 > 20. The answer is yes. However, if a is closer to 3 but less than 3, it would be interesting to know whether at that value of a, is a^3 > 20?

Let's devise another tactic.

Dividing a^5 > 200 by a^3 > 20, we get a^5/a^3 > 200/20 => a^2 > 10 => a > 3. Mind that we already deduced that a is positive. Since at a = 3, we have a^3 = 3^3 = 27 > 20. The answer is yes. We can safely conclude that Statement 2 is sufficient.

The correct answer: B

Hope this helps!

-Jay
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