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M28-44

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M28-44 [#permalink]

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New post 16 Sep 2014, 01:31
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Was the average (arithmetic mean) temperature in degrees Celsius in city A in March less than the average (arithmetic mean) temperature in degrees Celsius in city B in March?


(1) The median temperature in degrees Celsius in City A in March was less than the median temperature in degrees Celsius in city B.

(2) The ratio of the average temperatures in degrees Celsius in A and B in March was 3 to 4, respectively.

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Re M28-44 [#permalink]

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New post 16 Sep 2014, 01:31
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Official Solution:


Was the average (arithmetic mean) temperature in degrees Celsius in city A in March less than the average (arithmetic mean) temperature in degrees Celsius in city B in March?

(1) The median temperature in degrees Celsius in City A in March was less than the median temperature in degrees Celsius in city B. Clearly insufficient.

(2) The ratio of the average temperatures in degrees Celsius in A and B in March was 3 to 4, respectively. Temperatures can be negative, thus this statement is also not sufficient. Consider \(T(A)=3\) and \(T(B)=4\) AND \(T(A)=-3\) and \(T(B)=-4\).

(1)+(2) We have no additional useful info. Not sufficient.


Answer: E
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Re: M28-44 [#permalink]

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New post 03 Aug 2015, 22:50
cant believe I got this wrong. Very basic of ratio and yet 700+ level question because everyone doing same mistake in looking option B

Thx for this question :)
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Re: M28-44 [#permalink]

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New post 13 Oct 2015, 11:27
I think this is a poor-quality question and I don't agree with the explanation. unlikely to meet this kind of q in real gmat
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Re: M28-44 [#permalink]

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New post 16 Oct 2015, 07:19
Trickiest question ever! Is disguised as an easy one. Very important to improve focus while doing the test.
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Re: M28-44 [#permalink]

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New post 15 Nov 2015, 06:18
This is more of a logic based then quant based question.

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Re: M28-44 [#permalink]

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New post 20 Feb 2016, 22:29
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I think this question could be improved if it was mentioned that the temperatures were in DECEMBER instead of MARCH since where I live, we never see negative temperatures in March! Or if it was mentioned that the cities were in Alaska or something. I feel like throwing March in the question stem implies positive numbers and gives an unfair advantage to test takers in Antarctica.
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Re: M28-44 [#permalink]

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New post 23 May 2016, 00:54
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Bunuel wrote:
Official Solution:


(1) The median temperature in City A in March was less than the median temperature in city B. Clearly insufficient.

(2) The ratio of the average temperatures in A and B in March was 3 to 4, respectively. Temperatures can be negative, thus this statement is also not sufficient. Consider \(T(A)=3\) and \(T(B)=4\) AND \(T(A)=-3\) and \(T(B)=-4\).

(1)+(2) We have no additional useful info. Not sufficient.


Answer: E


I'm just salty I got tricked on this question too, but technically temperatures can/cannot be negative depending on which scale you use? (i.e. never specified it wasn't temperature in kelvins)
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Re: M28-44 [#permalink]

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New post 31 May 2016, 07:14
davidwu wrote:
Bunuel wrote:
Official Solution:


(1) The median temperature in City A in March was less than the median temperature in city B. Clearly insufficient.

(2) The ratio of the average temperatures in A and B in March was 3 to 4, respectively. Temperatures can be negative, thus this statement is also not sufficient. Consider \(T(A)=3\) and \(T(B)=4\) AND \(T(A)=-3\) and \(T(B)=-4\).

(1)+(2) We have no additional useful info. Not sufficient.


Answer: E


I'm just salty I got tricked on this question too, but technically temperatures can/cannot be negative depending on which scale you use? (i.e. never specified it wasn't temperature in kelvins)


I got tricked as well ! Technically, you're right also davidwu - however most of normal temperature now in daily life use either Celsius of Fahrenheit scale - Kevin is usually used in physics, lab or special environment - so I think the question still valid!
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Re: M28-44 [#permalink]

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New post 25 Jun 2016, 11:20
Tricky one indeed ! Thanks for this question !
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Re: M28-44 [#permalink]

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New post 06 Oct 2016, 14:23
How do I make a decision between C and E? It took me some time to see all possible cases.

Bunuel Can you throw some light!
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Re: M28-44 [#permalink]

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New post 03 Jan 2017, 20:47
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I think if you mentioned the unit of measurement, it would be better. Because if the temperatures are in Kelvin, there would be no negative temperatures.
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New post 06 Mar 2017, 01:19
hi brunel

can u clarify how S1 & S2 together are insufficient with an example
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M28-44 [#permalink]

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New post Updated on: 05 Apr 2017, 14:21
sidagar wrote:
hi brunel

can u clarify how S1 & S2 together are insufficient with an example


See screenshot. Edited the table so the logic can be seen more clearly. Basically, overall values can be relatively stable but extreme values (outliers) can throw the averages off while keeping the median the same.
>> !!!

You do not have the required permissions to view the files attached to this post.


Originally posted by mbadude2017 on 01 Apr 2017, 14:04.
Last edited by mbadude2017 on 05 Apr 2017, 14:21, edited 2 times in total.
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Re: M28-44 [#permalink]

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New post 05 Apr 2017, 08:06
Bunuel wrote:
Official Solution:


Was the average (arithmetic mean) temperature in degrees Celsius in city A in March less than the average (arithmetic mean) temperature in degrees Celsius in city B in March?

(1) The median temperature in degrees Celsius in City A in March was less than the median temperature in degrees Celsius in city B. Clearly insufficient.

(2) The ratio of the average temperatures in degrees Celsius in A and B in March was 3 to 4, respectively. Temperatures can be negative, thus this statement is also not sufficient. Consider \(T(A)=3\) and \(T(B)=4\) AND \(T(A)=-3\) and \(T(B)=-4\).

(1)+(2) We have no additional useful info. Not sufficient.


Answer: E


I tried substituting the values for A and somehow was convinced that A is sufficient. Can you explain on what grounds this is insufficient?
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Re: M28-44 [#permalink]

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New post 05 Apr 2017, 14:21
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ezzo wrote:
Bunuel wrote:
Official Solution:


Was the average (arithmetic mean) temperature in degrees Celsius in city A in March less than the average (arithmetic mean) temperature in degrees Celsius in city B in March?

(1) The median temperature in degrees Celsius in City A in March was less than the median temperature in degrees Celsius in city B. Clearly insufficient.

(2) The ratio of the average temperatures in degrees Celsius in A and B in March was 3 to 4, respectively. Temperatures can be negative, thus this statement is also not sufficient. Consider \(T(A)=3\) and \(T(B)=4\) AND \(T(A)=-3\) and \(T(B)=-4\).

(1)+(2) We have no additional useful info. Not sufficient.


Answer: E


I tried substituting the values for A and somehow was convinced that A is sufficient. Can you explain on what grounds this is insufficient?


See example above.
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Re M28-44 [#permalink]

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New post 29 Jul 2017, 10:14
I think this is a poor-quality question and I agree with explanation. Although i agree with the explanation, i dont think this question should be included in the quant section. This question does not assume anything related to quant. As we understand that we should not put in any extra subject matter expertise, this question requires it. Someone who does not know if temperature in celsius can be +ve or -ve (theoretically or practically) can not solve this question with "100% confidence" ( This is the key)
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Re M28-44 [#permalink]

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New post 16 Jan 2018, 21:46
I think this is a poor-quality question and I agree with explanation.
Re M28-44   [#permalink] 16 Jan 2018, 21:46

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