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

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Joined: 02 Sep 2009
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If x and y are positive numbers, is x > y?  [#permalink]

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14 Jun 2016, 02:00
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40% (02:22) correct 60% (01:40) wrong based on 102 sessions

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

(1) $$\frac{x^3}{y} < 1$$

(2) $$\frac{\sqrt[3]{x}}{y} < 1$$

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Joined: 18 Jan 2010
Posts: 254
Re: If x and y are positive numbers, is x > y?  [#permalink]

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14 Jun 2016, 03:04
Bunuel wrote:
If x and y are positive numbers, is x > y?

(1) $$\frac{x^3}{y} < 1$$

(2) $$\frac{\sqrt[3]{x}}{y} < 1$$

Statement 1

$$\frac{x^3}{y} < 1$$

x and y are positive numbers so we can cross multiply.

$$x^3 < y$$

x = 2, y = 9
$$x^3$$ = 8. This is less than y. So statement (1) is satisfied.
Since x=2, and y =9, x is not more than y.

x = (1/2), y = (1/7)
$$x^3$$ = (1/8). This is less than y. So statement (1) is satisfied.
Since x=(1/2), and y =(1/7), x is more than y.

Statement (1) is not sufficient

Statement 2

$$\frac{\sqrt[3]{x}}{y} < 1$$

$${\sqrt[3]{x}} < y$$

x and y are positive numbers so we can raise each of them to cubic powers

so x < $$y^3$$

x = 7, y = 2

$$y^3$$ = 8. x =7. x is less than $$y^3$$. So statement (2) is satisfied.
Since x=7, and y =9, x is more than y.

x = 1, y = 3

$$y^3$$ = 27. x =1. x is less than $$y^3$$. So statement (2) is satisfied.
Since x=1, and y =3, x is not more than y.

Statement (2) is not sufficient

Combining Statement 1 and 2

$$x^3 < y$$

x < $$y^3$$
Since x is positive we can cube it. so $$x^3 < y^6$$

Now $$x^3 < y$$ and $$x^3 < y^6$$

This means that we can take $$x^3 < y$$

This is nothing but Statement (1) only. We know it was insufficient.

So even after combining - insufficient

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Joined: 13 Aug 2013
Posts: 44
If x and y are positive numbers, is x > y?  [#permalink]

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14 Jun 2016, 20:49
Bunuel wrote:
If x and y are positive numbers, is x > y?

Combining Statement 1 and 2

$$x^3 < y$$

x < $$y^3$$
Since x is positive we can cube it. so $$x^3 < y^6$$

Now $$x^3 < y$$ and $$x^3 < y^6$$

This means that we can take $$x^3$$ < $$y$$

This is nothing but Statement (1) only. We know it was insufficient.

So even after combining - insufficient

I am not completely sure about the above. May be I am wrong, but can't we simply multiply the inequalities since we know that they are positive numbers?

$$x^3 < y$$

x < $$y^3$$

we get: $$x^4$$ < $$y^4$$

Hence, x < y and Hence answer should be C and not E

Is there something wrong with this logic?
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Kudos plz if it helped

Manager
Joined: 05 Dec 2015
Posts: 122
If x and y are positive numbers, is x > y?  [#permalink]

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26 Jun 2016, 21:33
rockstar23 wrote:
Bunuel wrote:
If x and y are positive numbers, is x > y?

Combining Statement 1 and 2

$$x^3 < y$$

x < $$y^3$$
Since x is positive we can cube it. so $$x^3 < y^6$$

Now $$x^3 < y$$ and $$x^3 < y^6$$

This means that we can take $$x^3$$ < $$y$$

This is nothing but Statement (1) only. We know it was insufficient.

So even after combining - insufficient

I am not completely sure about the above. May be I am wrong, but can't we simply multiply the inequalities since we know that they are positive numbers?

$$x^3 < y$$

x < $$y^3$$

we get: $$x^4$$ < $$y^4$$

Hence, x < y and Hence answer should be C and not E

Is there something wrong with this logic?

Hi - yes, your logic is not correct. If you have an even exponent it masks the sign of the variable, for example:

if x = -1 and y = -2 then x>y
if x = -2 and y = 3 then y>x

(edit b/c I had it backwards ;D)
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Joined: 16 Oct 2010
Posts: 8281
Location: Pune, India
Re: If x and y are positive numbers, is x > y?  [#permalink]

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26 Jun 2016, 22:55
1
2
Bunuel wrote:
If x and y are positive numbers, is x > y?

(1) $$\frac{x^3}{y} < 1$$

(2) $$\frac{\sqrt[3]{x}}{y} < 1$$

x and y are positive. So we can multiply the inequality by x or y.

(1) $$\frac{x^3}{y} < 1$$
$$x^3 < y$$

How do cubes behave on the right of 0? If x lies between 0 and 1, the cube will be less than x. If x is greater than 1, x^3 will be more than x. We don't know whether x lies in between 0 and 1 or is greater than 1. Not sufficient.

(2) $$\frac{\sqrt[3]{x}}{y} < 1$$
$$\sqrt[3]{x} < y$$

How do cube roots behave on the right of 0? If x lies between 0 and 1, the cube root will be more than x. If x is greater than 1, it will be less than x. We don't know whether x lies in between 0 and 1 or is greater than 1. Not sufficient.

Using both,
x will lie between x^3 and $$\sqrt[3]{x}$$ in either case. Since y is greater than both, y must be greater than x. Sufficient.

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Karishma
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Re: If x and y are positive numbers, is x > y?  [#permalink]

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30 Jun 2018, 22:58
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If x and y are positive numbers, is x > y? &nbs [#permalink] 30 Jun 2018, 22:58
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