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Given that x^3*y > 0 and that x^2*y^3 < 0 which of the following state

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Given that x^3*y > 0 and that x^2*y^3 < 0 which of the following state  [#permalink]

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New post 24 Jun 2017, 00:07
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A
B
C
D
E

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  45% (medium)

Question Stats:

56% (01:23) correct 44% (01:32) wrong based on 186 sessions

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Re: Given that x^3*y > 0 and that x^2*y^3 < 0 which of the following state  [#permalink]

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New post 24 Jun 2017, 01:22
1
Bunuel wrote:
Given that \(x^3*y > 0\) and that \(x^2*y^3 < 0\) which of the following statements must be true?

I. \(x < 0\)
II. \(x < y < 0\)
III. \(y^3 < x^2\)

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III


From the stem we can deduce that y<0 and x<0
(x^3)*y >0 means either both are -ve or both are +ve
(x^2)*(y^3)<0 means y is -ve

Hence y<0 and x<0
I holds true
II does not always (e.g. y = -4 and x = -2)
III will always hold as (y = -ve number)^3 < Positive number (x = -ve number)^2

Hence D
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Re: Given that x^3*y > 0 and that x^2*y^3 < 0 which of the following state  [#permalink]

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New post 24 Jun 2017, 02:02
Bunuel wrote:
Given that \(x^3*y > 0\) and that \(x^2*y^3 < 0\) which of the following statements must be true?

I. \(x < 0\)
II. \(x < y < 0\)
III. \(y^3 < x^2\)

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III


(i) \(x^3*y > 0\)
Either both \(x\) and \(y\) must be negative or both must be positive for the above result to be true.


(ii) \(x^2*y^3 < 0\)
Since square of any number must be positive, \(y\) must be negative for the above result to be true.

From (i) and (ii), it can be concluded that both \(x\) and \(y\) are negative.

I) This is true from the above.
II) This need not necessarily be true. For e.g., let \(x = y = -2\), then \((-2)^{2} * (-2)^{3} = 4 * -8 = -32\), but here \(x\) is not less than \(y\).
III) This must be true as both \(x\) and \(y\) are negative numbers.

So, only I and III are true. Ans - D.
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Given that x^3*y > 0 and that x^2*y^3 < 0 which of the following state  [#permalink]

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New post 24 Jun 2017, 02:27
Given conditions are \(x^3\)∗y>0 and \(x^2\)∗\(y^3\)<0
Assume x=1 and y=1, both conditions cannot be satisfied
Assume x=1 and y=-1, first conditions cannot be satisfied
Assume x=-1 and y=-1, both conditions can be satisfied

Answer D
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Re: Given that x^3*y > 0 and that x^2*y^3 < 0 which of the following state  [#permalink]

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New post 24 Jun 2017, 02:41
Bunuel wrote:
Given that \(x^3*y > 0\) and that \(x^2*y^3 < 0\) which of the following statements must be true?

I. \(x < 0\)
II. \(x < y < 0\)
III. \(y^3 < x^2\)

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III


Solution:

The second question stem says that y is negative. So for the first equation to be true, x has to be negative too.
But the value of x can be greater or lesser than y. We can't answer this with the available information.

Therefore, I and III are true.
Option D.
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Re: Given that x^3*y > 0 and that x^2*y^3 < 0 which of the following state  [#permalink]

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New post 23 May 2019, 20:04
Bunuel wrote:
Given that \(x^3*y > 0\) and that \(x^2*y^3 < 0\) which of the following statements must be true?

I. \(x < 0\)
II. \(x < y < 0\)
III. \(y^3 < x^2\)

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III


From two given statements we can deduce the following-
xy>0
y<0
So, x<0

I Yes, we already deduced.
II Can't say, we don't know which one is greater. We just know that both are -ve numbers.
III y^3 < x^2; definitely. y^3 gives -ve and x^2 gives +ve number.

D. I & III
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Re: Given that x^3*y > 0 and that x^2*y^3 < 0 which of the following state   [#permalink] 23 May 2019, 20:04
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