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 Author: Bunuel [ 09 Jul 2012, 04:53 ] Post subject: In City X last April, was the average (arithmetic mean) 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.

 Author: Bunuel [ 09 Jul 2012, 04:53 ] Post subject: Re: In City X last April, was the average (arithmetic mean) SOLUTIONIn 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.

 Author: cyberjadugar [ 09 Jul 2012, 05:57 ] Post subject: Re: In City X last April, was the average (arithmetic mean) 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: 600Is 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,

 Author: sanjoo [ 11 Jul 2012, 11:50 ] Post subject: Re: In City X last April, was the average (arithmetic mean) 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..

 Author: aeglorre [ 17 Dec 2013, 11:51 ] Post subject: Re: In City X last April, was the average (arithmetic mean) Bunuel wrote:The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions ProjectIn 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 TestQuestion: 43Page: 26Difficulty: 650GMAT Club is introducing a new project: The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide for GMAT® Review, 13th Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.We'll be glad if you participate in development of this project:1. Please provide your solutions to the questions;2. Please vote for the best solutions by pressing Kudos button;3. Please vote for the questions themselves by pressing Kudos button;4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.Thank you!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.

 Author: abid1986 [ 08 Apr 2014, 00:50 ] Post subject: Re: In City X last April, was the average (arithmetic mean) Bunuel wrote:SOLUTIONIn 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 fromt1 to t9 and t21 to t30then these wont accomodate the median(t15 and t16)

 Author: Bunuel [ 08 Apr 2014, 02:39 ] Post subject: Re: In City X last April, was the average (arithmetic mean) abid1986 wrote:Bunuel wrote:SOLUTIONIn 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 fromt1 to t9 and t21 to t30then these wont accomodate the median(t15 and t16)No, since $$t_1\leq{t_2}...\leq{t_{15}}\leq{t_{16}}\leq...\leq{t_{29}}\leq{t_{30}}$$.

 Author: stonecold [ 25 Dec 2016, 23:13 ] Post subject: Re: In City X last April, was the average (arithmetic mean) 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 notStatement 1=>Mean 72No 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

 Author: woohoo921 [ 10 Sep 2022, 16:54 ] Post subject: Re: In City X last April, was the average (arithmetic mean) Bunuel wrote:SOLUTIONIn 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!

 Author: RPGXKHX [ 13 Nov 2022, 23:35 ] Post subject: In City X last April, was the average (arithmetic mean) Bunuel wrote:abid1986 wrote:Bunuel wrote:SOLUTIONIn 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 fromt1 to t9 and t21 to t30then 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?

 Author: egmat [ 08 Feb 2023, 06:07 ] Post subject: Re: In City X last April, was the average (arithmetic mean) 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 Quant Expert, e-GMAT

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