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If z is an odd integer, is 300z > 1500?

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Mascarfi
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If z is an odd integer, is 300z > 1500? [#permalink] New post   Updated on: 08 Nov 2015, 08:35
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If z is an odd integer, is 300z > 1500?

(1) \(0<\sqrt{z}<8\)
(2) \(1<z^2<45\)
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B

Originally posted by Mascarfi on 07 Nov 2015, 16:36.
Last edited by Bunuel on 08 Nov 2015, 08:35, edited 2 times in total.
Edited the question.
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   07 Nov 2015, 17:02
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Mascarfi wrote:
If z is an odd integer, is 300z>1500?

Statement 1, 0<\(\sqrt{z}\)<8
Statement 2, 1<\(z^2\)<45


Z= odd integer, is 300z>1500 ---> z>5 ?

Statement 1, \(0 < \sqrt{z} < 8\) ---> 1<z<64 so if z=3 or 5, answer is "no" but if z=7,9,11... the answer is "yes". Statement is not sufficient.

Statement 2, \(1<z^2<45\) --> only possible values are z=3 or 5, giving you a "no" . Sufficient.

B is the correct answer.
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   01 Jun 2017, 10:16
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rocko911 wrote:
AK125 wrote:
Mascarfi wrote:
If z is an odd integer, is 300z > 1500?

(1) \(0<\sqrt{z}<8\)
(2) \(1<z^2<45\)



Please correct if wrong!!

z > 5?

1) 0 < z < 64 --> not clear weather z is bigger than 5. #insufficient
2) 1 < z < 6.7 --> possible number 5 and 3, which means that z is not > 5! #sufficient

B)




Why don't we consider z= -3? the square of -3 will be 3 and its an odd number also (i.e -3)


The question asks whether z > 5?

From (2) \(1<z^2<45\) (and the fact that we know that z is odd) it follow that z can be -5, -3, 3, and 5. For each of these values we have a NO answer to the question: z is NOT greater than 5. Sufficient.

Hope it's clear.
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   12 Nov 2015, 04:19
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z>5? Yes/No

St1. 0<z<64, can be Y&N. INSUFF

St2. 1<|z|<3*sqrt5 OR -1>z>-3*sqrt5. In both case z=<5. SUFF

B
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   06 Jun 2017, 09:48
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goalMBA1990 wrote:
Hi Bunuel,

I agree if I square both sides I will 0 to 64 values for Z not including 0 and 64.

But what is wrong with my method. Suppose I don't square both side and keep statement intact. I mean we are given that Z= odd integer and square root of Z is between 0 and 8. So I can selected Z=1 and square root Z is 1 which between 0 and Z. 300(1) is not greater than 1500. So statement alone is sufficient to answer.


But you can select many other values for z. Why not 3, 5, 7, ... ? Different z's give different answers.
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   08 Nov 2015, 05:32
300z>1500 is just z>5

Using statement 1, we know that Z is the square of either 1, 3, 5, or 7. Meaning Z can be 1, 9, 25, or 49. Since 1<5 while all the others are >5, INSUFFICIENT.

Using statement 2, we know that Z is the square root of either 9 or 25. Z can then be positive or negative 3 or 5. No matter what the sign, all values of Z are not greater than 5, SUFFICIENT
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   09 Dec 2016, 08:38
Mascarfi wrote:
If z is an odd integer, is 300z > 1500?

(1) \(0<\sqrt{z}<8\)
(2) \(1<z^2<45\)



Please correct if wrong!!

z > 5?

1) 0 < z < 64 --> not clear weather z is bigger than 5. #insufficient
2) 1 < z < 6.7 --> possible number 5 and 3, which means that z is not > 5! #sufficient

B)
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   08 Apr 2017, 08:01
AK125 wrote:
Mascarfi wrote:
If z is an odd integer, is 300z > 1500?

(1) \(0<\sqrt{z}<8\)
(2) \(1<z^2<45\)



Please correct if wrong!!

z > 5?

1) 0 < z < 64 --> not clear weather z is bigger than 5. #insufficient
2) 1 < z < 6.7 --> possible number 5 and 3, which means that z is not > 5! #sufficient

B)




Why don't we consider z= -3? the square of -3 will be 3 and its an odd number also (i.e -3)
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   04 Jun 2017, 08:51
z is an odd integer - but where it is mentioned that 'z' is +ve?
z can be negative, so 'B' is eliminated from the option.
But if 'A' is true then 'z' has to be +ve.

So the right answer will be 'C'.
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   06 Jun 2017, 01:38
Engr2012 wrote:
Mascarfi wrote:
If z is an odd integer, is 300z>1500?

Statement 1, 0<\(\sqrt{z}\)<8
Statement 2, 1<\(z^2\)<45


Z= odd integer, is 300z>1500 ---> z>5 ?

Statement 1, \(0 < \sqrt{z} < 8\) ---> 1<z<64 so if z=3 or 5, answer is "no" but if z=7,9,11... the answer is "yes". Statement is not sufficient.

Statement 2, \(1<z^2<45\) --> only possible values are z=3 or 5, giving you a "no" . Sufficient.

B is the correct answer.


Hi,
This is what I did. But I don't know where I missed. Please correct me.

Statement 1: Given z= odd integer and 0<sqrt(Z)<8. So only possible value of Z is 1 and 300*1=300 is not greater than 1500. Answer is definitely NO. So sufficient.

Statement 2: Z= 3,5 and answer from this statement is also definite NO. So sufficient.

That is why I got D.

Thanks.
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   06 Jun 2017, 09:17
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goalMBA1990 wrote:
Engr2012 wrote:
Mascarfi wrote:
If z is an odd integer, is 300z>1500?

Statement 1, 0<\(\sqrt{z}\)<8
Statement 2, 1<\(z^2\)<45


Z= odd integer, is 300z>1500 ---> z>5 ?

Statement 1, \(0 < \sqrt{z} < 8\) ---> 1<z<64 so if z=3 or 5, answer is "no" but if z=7,9,11... the answer is "yes". Statement is not sufficient.

Statement 2, \(1<z^2<45\) --> only possible values are z=3 or 5, giving you a "no" . Sufficient.

B is the correct answer.


Hi,
This is what I did. But I don't know where I missed. Please correct me.

Statement 1: Given z= odd integer and 0<sqrt(Z)<8. So only possible value of Z is 1 and 300*1=300 is not greater than 1500. Answer is definitely NO. So sufficient.

Statement 2: Z= 3,5 and answer from this statement is also definite NO. So sufficient.

That is why I got D.

Thanks.


(1) \(0 < \sqrt{z} < 8\) --> square: \(0 < z < 64\). Any odd z will work here: 1, 3, 5, 7, 9, ..., 63. z's less than or equal to 5, give NO answer while greater than 5, give an YES answer. Not sufficient.
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   06 Jun 2017, 09:45
Hi Bunuel,

I agree if I square both sides I will 0 to 64 values for Z not including 0 and 64.

But what is wrong with my method. Suppose I don't square both side and keep statement intact. I mean we are given that Z= odd integer and square root of Z is between 0 and 8. So I can selected Z=1 and square root Z is 1 which between 0 and Z. 300(1) is not greater than 1500. So statement alone is sufficient to answer.
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If z is an odd integer, is 300z > 1500? [#permalink] New post   Updated on: 06 Jun 2017, 10:07
aveek86 wrote:
z is an odd integer - but where it is mentioned that 'z' is +ve?
z can be negative, so 'B' is eliminated from the option.
But if 'A' is true then 'z' has to be +ve.

So the right answer will be 'C'.


Hi aveek86

as per the question we need to find 300z>1500 or z>5 (dividing both sides by 300)

As per statement 1 0<z^(1/2)<8, So squaring both sides will give 0<z<64. So z can be 1 and 3 or z can be 5, 7....So insufficient

Statement 2: 1<z^2<45. Now as z is odd so what all values will satisfy this inequality? "z" could be -3, -5, 3 or 5. Each of these numbers definitely shows that "z" is not greater than 5. So we have a definite NO hence Sufficient to answer the question.

So Option B

Originally posted by niks18 on 06 Jun 2017, 09:58.
Last edited by niks18 on 06 Jun 2017, 10:07, edited 1 time in total.
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   06 Jun 2017, 10:02
Bunuel wrote:
goalMBA1990 wrote:
Hi Bunuel,

I agree if I square both sides I will 0 to 64 values for Z not including 0 and 64.

But what is wrong with my method. Suppose I don't square both side and keep statement intact. I mean we are given that Z= odd integer and square root of Z is between 0 and 8. So I can selected Z=1 and square root Z is 1 which between 0 and Z. 300(1) is not greater than 1500. So statement alone is sufficient to answer.


But you can select many other values for z. Why not 3, 5, 7, ... ? Different z's give different answers.


Got your point. I was restricting my self for some values only and not understanding the question.

Thanks for clarification.
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   07 Jun 2017, 00:05
Engr2012 wrote:
Mascarfi wrote:
If z is an odd integer, is 300z>1500?

Statement 1, 0<\(\sqrt{z}\)<8
Statement 2, 1<\(z^2\)<45


Z= odd integer, is 300z>1500 ---> z>5 ?

Statement 1, \(0 < \sqrt{z} < 8\) ---> 1<z<64 so if z=3 or 5, answer is "no" but if z=7,9,11... the answer is "yes". Statement is not sufficient.

Statement 2, \(1<z^2<45\) --> only possible values are z=3 or 5, giving you a "no" . Sufficient.

B is the correct answer.


300*5 =1500, and 1500 is not greater than 1500.

Had the question asked, is 300z greater than equal to 1500, the answer would have been B. In this case, there is ambiguity and hence both the statements are insufficient.
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   01 Jan 2020, 09:43
A good question that looks innocent but is tricky enough to have me choose E!

That is not a really good start of a new year. :-(
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Re: If z is an odd integer, is 300z > 1500? [#permalink] New post   01 Jan 2024, 08:52
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Re: If z is an odd integer, is 300z > 1500? [#permalink]
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