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For which of the above lists is the average of the numbers less than t
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14 Nov 2017, 01:16
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[GMAT math practice question] I. \(\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6}\) II. \(\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}\) III. \(\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6}\) For which of the above lists is the average of the numbers less than the median of numbers? A. I only B. II only C. III only D. II and III E. I, II and III
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For which of the above lists is the average of the numbers less than t
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14 Nov 2017, 03:48
MathRevolution wrote: [GMAT math practice question]
I. \(\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6}\) II. \(\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}\) III. \(\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6}\) For which of the above lists is the average of the numbers less than the median of numbers?
A. I only B. II only C. III only D. II and III E. I, II and III LCM of 2, 3, 4, 5, 6 = 60 So multiply numerator and denominator of each fraction by 60 I. becomes => 30, 40, 45, 48, 50 Average of above numbers = 42.6 and Median=45. clearly this satisfies the required condition II. becomes => 30, 20, 15, 12, 10 Average of above numbers = 17.4 and Median = 15. Does not satisfy our condition. We can Stop here and eliminate all option. so our answer is Afor the sake of testing  III becomes => 10, 20, 30, 40, 50 Average of above numbers = median of above set = 30. Hence does not satisfy given condition. Option A



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For which of the above lists is the average of the numbers less than t
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14 Nov 2017, 07:34
MathRevolution wrote: [GMAT math practice question]
I. \(\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6}\) II. \(\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}\) III. \(\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6}\) For which of the above lists is the average of the numbers less than the median of numbers?
A. I only B. II only C. III only D. II and III E. I, II and III Hi... If you don't want to get into lengthy calculations as it would surely eat into your time, some observations...here is a method...III. \(\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6}\) this is clearly an AP with difference \(\frac{1}{6}\), so MEDIAN = MEANso III is out ONLY A and B are without IIIwhich means either I is correct or II, but only one here you have been able to get down to TWO choices out of 5, so 50% chance to answer correctlyBut lets see IIII. \(\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}\) median is 1/4 and highest value is 1/2, which is1/4 away from the median 1/4, so lowest should be 0 to reach medianmean = median at 0,1/4,1/2 but it is 1/6,1/4,1/2 BUT the lowest value is >0, so HIGHER values are MORE farther from the MEDIAN, so MEAN will be above median so mean>median.... so NO ans is I or the choice Abut lets see why I is correct..I. \(\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6}\) these can be written as I. \(1\frac{1}{2},1\frac{1}{3},1\frac{1}{4},1\frac{1}{5},1\frac{1}{6}\) MEAN will be 1(average of 1/2,1/3,1/4,1/5,1/6)this is OPPOSITE of B, here we are subtracting higher values from 1 in 1/2 and 2/3 as compared to 1/5 and 1/6 so m MEAN<median
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Re: For which of the above lists is the average of the numbers less than t
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16 Nov 2017, 01:41
=> I. The differences between the consecutive terms are \(\frac{1}{6} (= \frac{2}{3} – \frac{1}{2} )\), \(\frac{1}{12} (= \frac{3}{4} – \frac{2}{3} )\), \(\frac{1}{20} (= \frac{4}{5} – \frac{3}{4} )\), and \(\frac{1}{30} (= \frac{5}{6} – \frac{4}{5})\). As these are decreasing, the average is smaller than the median. II. The differences between consecutive terms are  \(\frac{1}{6}(= \frac{1}{3} – \frac{1}{2})\),\(\frac{1}{12}(= \frac{1}{4} – \frac{1}{3} )\),  \(\frac{1}{20}(= \frac{1}{5} – \frac{1}{4})\), and \(\frac{1}{30}(= \frac{1}{6} – \frac{1}{5})\). As these are increasing, the average is larger than the median. III. As the data are symmetric, the average and the median are equal. Therefore, the answer is A Answer: A
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