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M40-33

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Bunuel
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M40-33 [#permalink] New post   31 Dec 2023, 06:48
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If \(-9 < m < -4\) and \(7 < n < 11\), which of the following can be a value of \(\sqrt[4]{-mn}\)?

I. 2

II. 3

III. 4


A. I only
B. II only
C. III only
D. I and II only
E. II and III only
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B
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Re M40-33 [#permalink] New post   31 Dec 2023, 06:48
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Official Solution:


If \(-9 < m < -4\) and \(7 < n < 11\), which of the following can be a value of \(\sqrt[4]{-mn}\)?

I. 2

II. 3

III. 4


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


\(mn\) must be more than \(-4*7 = -28\) and less than \(-9*11 = -99\):

\(-99 < mn < -28\)

Multiplying by -1 gives:

\(28 < -mn < 99\)

Taking the fourth root gives:

\(\sqrt[4]{28} < \sqrt[4]{-mn} < \sqrt[4]{99}\)

Now, considering that \(2^4 < 28 < 3^4 < 99 < 4^4\), it follows that \(\sqrt[4]{-mn}\) can only be 3.


Answer: B
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Re: M40-33 [#permalink] New post   31 Jan 2024, 02:47
Bunuel

This is what i donot understand.

You say that mn must be more than −4∗7=−28, meaning mn >-28 or -28<mn

and that mn must be less than −9∗11=−99, meaning mn<-99 or -99>mn.

Then you go on and say: −99<mn<−28. How is this possible?

How can we go from mn>-28 to mn<-28 and mn <-99 to -99<mn?

What is it that i donot see?

Can you please elaborate more on the above?


Thanks in advance!
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Re: M40-33 [#permalink] New post   31 Jan 2024, 03:52
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Rebaz wrote:
Bunuel

This is what i donot understand.

You say that mn must be more than −4∗7=−28, meaning mn >-28 or -28<mn

and that mn must be less than −9∗11=−99, meaning mn<-99 or -99>mn.

Then you go on and say: −99<mn<−28. How is this possible?

How can we go from mn>-28 to mn<-28 and mn <-99 to -99<mn?

What is it that i donot see?

Can you please elaborate more on the above?


Thanks in advance!


Multiply \(-99 < mn < -28\) by -1 and flip the signs:

\(99 > -mn > 28\), which is the same as \(28 < -mn < 99\).
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M40-33 [#permalink] New post   05 Feb 2024, 06:10
Bunuel,

Now i understand where i did go wrong, Thanks a lot!
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