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

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Bunuel
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If x and y are positive numbers, is x^3 > y? [#permalink] New post   10 Aug 2020, 02:10
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E

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53% (02:01) correct 47% (01:28) wrong based on 143 sessions
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If x and y are positive numbers, is \(x^3 > y\)?


(1) \(\sqrt{x} > y\)

(2) \(x > y\)


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E
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Re: If x and y are positive numbers, is x^3 > y? [#permalink] New post   10 Aug 2020, 02:46
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(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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Re: If x and y are positive numbers, is x^3 > y? [#permalink] New post   10 Aug 2020, 04:14
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Bunuel wrote:
If x and y are positive numbers, is \(x^3 > y\)?


(1) \(\sqrt{x} > y\)

(2) \(x > y\)



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Answer: Option E

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Re: If x and y are positive numbers, is x^3 > y? [#permalink] New post   10 Aug 2020, 23:13
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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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Re: If x and y are positive numbers, is x^3 > y? [#permalink] New post   10 Aug 2020, 04:50
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


Bunuel wrote:
If x and y are positive numbers, is \(x^3 > y\)?


(1) \(\sqrt{x} > y\)

(2) \(x > y\)



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Re: If x and y are positive numbers, is x^3 > y? [#permalink] New post   10 Aug 2020, 06:10
Bunuel wrote:
If x and y are positive numbers, is \(x^3 > y\)?


(1) \(\sqrt{x} > y\)

(2) \(x > y\)



Is x^3>y

Statements:

(1) Sqrt(x) > y

Let x<1
x = 1/4 and y = 1/5
Sqrt(x) > y but x^3<y (ANS - NO)

Let x>1
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^3<y (ANS - NO)

Let x>1
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).
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Re: If x and y are positive numbers, is x^3 > y? [#permalink] New post   10 Jan 2024, 10:42
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Re: If x and y are positive numbers, is x^3 > y? [#permalink]
10 Jan 2024, 10:42
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