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In City X last April, was the average (arithmetic mean) [#permalink]
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09 Jul 2012, 04:53
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Re: In City X last April, was the average (arithmetic mean) [#permalink]
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09 Jul 2012, 04:53
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.
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Re: In City X last April, was the average (arithmetic mean) [#permalink]
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09 Jul 2012, 05:57
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]
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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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17 Dec 2013, 11:51
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 Test Question: 43 Page: 26 Difficulty: 650 GMAT 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.



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Re: In City X last April, was the average (arithmetic mean) [#permalink]
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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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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 DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: In City X last April, was the average (arithmetic mean) [#permalink]
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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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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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