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

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

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New post 14 Jun 2016, 02:00
1
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A
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C
D
E

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  95% (hard)

Question Stats:

40% (02:22) correct 60% (01:40) wrong based on 102 sessions

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

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New post 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

E is the answer.
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If x and y are positive numbers, is x > y?  [#permalink]

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New post 14 Jun 2016, 20:49
adiagr 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

E is the answer.


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

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New post 26 Jun 2016, 21:33
rockstar23 wrote:
adiagr 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

E is the answer.


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

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New post 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.

Answer (C)
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Re: If x and y are positive numbers, is x > y?  [#permalink]

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New post 30 Jun 2018, 22:58
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