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In City X last April, was the average (arithmetic mean)

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Bunuel
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In City X last April, was the average (arithmetic mean) [#permalink] New post   09 Jul 2012, 04:53
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In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160°.
(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature.
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Re: In City X last April, was the average (arithmetic mean) [#permalink] New post   09 Jul 2012, 04:53
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SOLUTION

In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

Let's consider the temperatures in April in ascending order: \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\). The median temperature would be the average of 15th and 16th greatest temperatures. The question asks: is this value less than the average temperature, (which would be the sum of temperatures divided by 30).

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160° --> gives the average temperature, but not info about the median. Not sufficient.

(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature --> each temperature (term) from the 18 lowest ones (from \(t_1\) to \(t_{18}\)) given to be less than average, hence the median (average of 15th and 16th terms) would also be less than the average. Sufficient.

Answer: B.
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Re: In City X last April, was the average (arithmetic mean) [#permalink] New post   09 Jul 2012, 05:57
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Bunuel wrote:
In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160°.
(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature.

Hi,

Difficulty level: 600

Is AM > Median?
Median = Middle value of a sequence.
Arthimetic meam (AM) = Central tendency of a sequence.

Using (1),
Sum of 30 daily high temperature = 2160. We can find the mean, but no median. Insufficient.

Using (2),
60% of the daily high temperatures were less than the average daily high temperature.
xxxxxxxxxxxxxxxxxx, the median will be in the green region and on the left side of mean.
Thus, Median < AM. Sufficient.

Answer (B)

Regards,
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Re: In City X last April, was the average (arithmetic mean) [#permalink] New post   11 Jul 2012, 11:50
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I think its B.. bunuel plz post the ans as soon as possible because i have exam in few days ..so i have to make sure m doing rite or wrong..
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Re: In City X last April, was the average (arithmetic mean) [#permalink] New post   17 Dec 2013, 11:51
Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160°.
(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature.

Diagnostic Test
Question: 43
Page: 26
Difficulty: 650


GMAT Club is introducing a new project: The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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I solved this in a rather unconventional way, and I do not know if it works generally:

1) is clearly insufficient. But 2) basically is telling us that "there are outliers to the right that jack up the average", and a key rule of thumb in statistics is that if there are observations in the sample that jack up the arithmetic mean, then the median is a better representation of the "real" population value. The median is in that case always higher/lowr than the average (in our case, lower because 60% of the observations are lower than the average value) and thus through inference and common sense I concluded that the arithmetic mean is higher than the median. B is therefore sufficient.
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Re: In City X last April, was the average (arithmetic mean) [#permalink] New post   08 Apr 2014, 00:50
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Bunuel wrote:
SOLUTION

In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

Let's consider the temperatures in April in ascending order: \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\). The median temperature would be the average of 15th and 16th greatest temperatures. The question asks: is this value less than the average temperature, (which would be the sum of temperatures divided by 30).

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160° --> gives the average temperature, but not info about the median. Not sufficient.

(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature --> each temperature (term) from the 18 lowest ones (from \(t_1\) to \(t_{18}\)) given to be less than average, hence the median (average of 15th and 16th terms) would also be less than the average. Sufficient.

Answer: B.


Hi Bunuel,

Cant 60% of daily temperature less than average daily temperature range from
t1 to t9 and t21 to t30
then these wont accomodate the median(t15 and t16)
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Re: In City X last April, was the average (arithmetic mean) [#permalink] New post   08 Apr 2014, 02:39
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abid1986 wrote:
Bunuel wrote:
SOLUTION

In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

Let's consider the temperatures in April in ascending order: \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\). The median temperature would be the average of 15th and 16th greatest temperatures. The question asks: is this value less than the average temperature, (which would be the sum of temperatures divided by 30).

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160° --> gives the average temperature, but not info about the median. Not sufficient.

(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature --> each temperature (term) from the 18 lowest ones (from \(t_1\) to \(t_{18}\)) given to be less than average, hence the median (average of 15th and 16th terms) would also be less than the average. Sufficient.

Answer: B.


Hi Bunuel,

Cant 60% of daily temperature less than average daily temperature range from
t1 to t9 and t21 to t30
then these wont accomodate the median(t15 and t16)


No, since \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\).
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Re: In City X last April, was the average (arithmetic mean) [#permalink] New post   25 Dec 2016, 23:13
Great Quality Official Question.
Here is what i did in this question=>
We need to see if the median is less than the mean or not

Statement 1=>Mean 72
No clue of median -> Not sufficient

Statement 2->
60 percent of the values are less than the mean => Median must be less than the mean
Sufficient

Hence B
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Re: In City X last April, was the average (arithmetic mean) [#permalink] New post   10 Sep 2022, 16:54
Bunuel wrote:
SOLUTION

In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

Let's consider the temperatures in April in ascending order: \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\). The median temperature would be the average of 15th and 16th greatest temperatures. The question asks: is this value less than the average temperature, (which would be the sum of temperatures divided by 30).

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160° --> gives the average temperature, but not info about the median. Not sufficient.

(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature --> each temperature (term) from the 18 lowest ones (from \(t_1\) to \(t_{18}\)) given to be less than average, hence the median (average of 15th and 16th terms) would also be less than the average. Sufficient.

Answer: B.


Experts - KarishmaB ScottTargetTestPrep
"60% of the daily high temperatures were less than the average daily high temperature." Does that essentially mean there were a few really large numbers that were far above the average for 60% of the temperatures to be less than the average --> e.g., the few high numbers were high enough to tip the scale...

thank you!
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In City X last April, was the average (arithmetic mean) [#permalink] New post   13 Nov 2022, 23:35
Bunuel wrote:
abid1986 wrote:
Bunuel wrote:
SOLUTION

In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?

Let's consider the temperatures in April in ascending order: \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\). The median temperature would be the average of 15th and 16th greatest temperatures. The question asks: is this value less than the average temperature, (which would be the sum of temperatures divided by 30).

(1) In City X last April, the sum of the 30 daily high temperatures was 2,160° --> gives the average temperature, but not info about the median. Not sufficient.

(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature --> each temperature (term) from the 18 lowest ones (from \(t_1\) to \(t_{18}\)) given to be less than average, hence the median (average of 15th and 16th terms) would also be less than the average. Sufficient.

Answer: B.


Hi Bunuel,

Cant 60% of daily temperature less than average daily temperature range from
t1 to t9 and t21 to t30
then these wont accomodate the median(t15 and t16)


No, since \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\).


Bunuel Why should we consider that the temperatures are in ascending order?
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Re: In City X last April, was the average (arithmetic mean) [#permalink] New post   08 Feb 2023, 06:07
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RPGXKHX wrote:
Bunuel Why should we consider that the temperatures are in ascending order?


Hello RPGXKHX,

We are not assuming that the temperatures are in ascending order. Rather, we are arranging the temperatures in ascending order.
Let me explain what I mean:

  • April has 30 different days. So, we get daily high temperatures for these 30 days.
  • Now, let us call the lowest daily high temperature t\(_1\), the next subsequently higher daily high temperature as t\(_2\), and so on such that the highest daily high temperature is marked as t\(_3\)\(_0\).
    • t\(_1\) is not the high temperature on April 1st. It CAN be so, but it does NOT have to be.
    • So, the temperatures are marked as t\(_1\), t\(_2\)…t\(_3\)\(_0\) on the basis of their values (from lowest to highest) and not on the day of the month they correspond to.


Now, you may ask why we do all this?

Well, this is how we find the median of any set. Here’s the process:
  • Arrange the set in ascending or descending order.
  • Find the middle term (in case there are odd number of terms) or the average of 2 middles terms (in case there are even number of terms). This is the median of the set.

Similarly, to find the median daily high temperature for the set of 30 daily high temperatures, we must:
  • Arrange the daily high temperatures in ascending or descending order. Bunuel arranged them in ascending order from t\(_1\), t\(_2\)…t\(_3\)\(_0\).
  • Since 30 is an even number, there will not be one middle term. So, we take the average of the two middle terms, 15\(^t\)\(^h\) and 16\(^t\)\(^h\) terms. Hence, Median = (t\(_1\)\(_5\) + t\(_1\)\(_6\))/2.


Hope this helps!

Best Regards,
Ashish
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Re: In City X last April, was the average (arithmetic mean) [#permalink] New post   23 Feb 2024, 12:36
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