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

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In City X last April, was the average (arithmetic mean)  [#permalink]

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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.

Diagnostic Test
Question: 43
Page: 26
Difficulty: 650

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Re: In City X last April, was the average (arithmetic mean)  [#permalink]

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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]

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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)

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Re: In City X last April, was the average (arithmetic mean)  [#permalink]

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New post 11 Jul 2012, 11:50
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]

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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


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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]

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New post 08 Apr 2014, 00:50
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]

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New post 08 Apr 2014, 02:39
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: In City X last April, was the average (arithmetic mean)  [#permalink]

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New post 24 Apr 2014, 04:10
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 \(t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}\).

are u assuming this because we are dealing with median temperature?
because for median, set is ordered from the lowest to highest.
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Re: In City X last April, was the average (arithmetic mean)  [#permalink]

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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]

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