If x and y are positive numbers, is x^3 y?

Question

If x and y are positive numbers, is  x^3 > y ?

(1)  \sqrt{x} > y

(2)  x > y

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yashikaaggarwal · Answer

(1) √x > y
Case 1: X and y are in fraction.
1/2 > 1/3
X = 1/4 and Y = 1/3
X^3 = 1/64 < 1/3 (not satisfying)

Case 2: x and y are integers.
4 > 3
X = 16 and Y = 3
X^3 = 4096 > 3 (satisfying)
(Insufficient)

(2) x > y
Case 1: X and y are in fraction.
1/2 > 1/3
X = 1/2 and Y = 1/3
X^3 = 1/8 < 1/3 (not satisfying)

Case 2: x and y are integers.
4 > 3
X = 4 and Y = 3
X^3 = 64 > 3 (satisfying)
(Insufficient)

Combining both: still no solution,
(Insufficient)

IMO E

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GMATinsight · Answer

Answer: Option E

Please check the video for the step-by-step solution.

GMATinsight's Solution

https://www.youtube.com/watch?v=IChKPw4D2_o

Archit3110 · Answer

target is x^3>y
and x&y are +ve numbers
#1
 \sqrt{x} > y 
x=4 , y = 1 ; we get yes
and x= 1/4 and y = 1/2 we get no
insufficient
#2
x>y
again for integers we get yes and fraction we get no
from 1 &2
integers values it would be yes always and for fractions ; x=1/2, y ; 1/3 it would be no
insufficient
OPTION E

KaramveerBakshi · Answer

Is x^3>y Statements: (1) Sqrt(x) > y Let x<1 x = 1/4 and y = 1/5 Sqrt(x) > y but x^31 x=9 and y = 2 Sqrt(x)> y and x^3>y (ANS-YES) Two different answers, Insufficient. (2) x > y Let x<1 x = 1/4 and y = 1/5 x > y but x^31 x=9 and y = 2 x> y and x^3>y (ANS-YES) Two different answers, Insufficient. Combining, we still can't say anything for sure. Insufficient. Hence, the answer is Option (E).

RaghavKhanna · Answer

Given: x,y are positive numbers
--> x,y > 0

Question: Is \(x^{3}\) > y?

Statement 1:
\(\sqrt{x}\) > y
=> x > \(y^{2}\)

(i) if x = 5, y = 2
Then, 5 > 4 and 125 > 2 --- Yes

(ii) if x = 1/4, y = 1/3
Then, 1/4 > 1/9 But 1/8 < 1/3 --- No

Statement 1 is insuffficient.

Statement 2:
x > y
(i) x = 5, y = 4
Then, 5 > 4 and 125 > 4 --- Yes

(ii) x = 1/2, y = 1/4
Then, 1/2 > 1/4 But 1/8 < 1/4 --- No

Statement 2 is insuffficient.

Statement 1 & Statement 2:
From 1: x > \(y^{2}\)
From 2: x > y

(i) 0 ----- 1 ----- y ----- \(y^{2}\) ----- x ----- \(x^{3} \)
=> \(x^{3}\) > y --- Yes

(ii) ------- 0 ------ \(y^{2}\) ----- y ----- \(x^{3} \) ----- x
=> \(x^{3}\) > y --- Yes

But,
(ii) ------- 0 ------ \(y^{2}\) ----- \(x^{3} \) ------ y ----- x
=> \(x^{3}\) < y --- No

=> Statement 1 and Statement 2 are together insuffficient.

Answer: E

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If x and y are positive numbers, is x^3 > y?

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