Vincen wrote:Is x^3 > y^2?
(1) y = x^2
(2) x > x^2
The OA is the option C.
How can I use both statements together to solve this DS question? I don't know what should I do. Experts, I'd appreciate your help.
We have to determine whether x^3 > y^2.
(1) y = x^2
Replaing the value of y = x^2 in x^3 > y^2, we get x^3 > y^2 => x^3 ? (x^2)^2 => x^3 ? x^4
Case 1: x is negative
Is x is negative, then x^3 is negative and x^4 is positive, thus x^3 < x^4. The answer is No.
Case 2: 0 < x < 1
Say x = 1/2, then x^3 = (1/2)^3 = 1/8 and x^4 = (1/2)^4 = 1/16. We see that x^3 > x^4. The answer is Yes. No unique answer.
For the sake of understanding, we discuss the third case too.
Case 3: x ≥ 1
Say x = 2, then x^3 = 2^3 = 8 and x^4 = 2^4 = 16. We see that x^3 < x^4. The answer is No. Insufficient.
(2) x > x^2
We do not have any information about y. Insufficient.
(1) and (2) combined
Among the three cases discussed above, only Case 2 is applicable for x > x^2. Thus, the answer is Yes. No unique answer.
The correct answer:
C
Hope this helps!
-Jay
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