For which of the above lists is the average of the numbers less than t

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

niks18 · Answer

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 A

for 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

chetan2u · Answer

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 = MEAN 
so III is out

ONLY A and B are without III 
which 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 correctly

But lets see II 
II.  \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 median 
mean = 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 A

but 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

MathRevolution · Answer

=>
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

bumpbot · Answer

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For which of the above lists is the average of the numbers less than t

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For which of the above lists is the average of the numbers less than t

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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