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Average of fractions

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by Anindya Madhudor » Thu Nov 22, 2012 8:17 pm
Is there any simple way of solving the following?

Which of the following series has average greater than median?
i) 3/4, 4/5, 5/6, 6/7, 7/8
ii) 8/7, 7/6, 6/5, 5/4, 4/3
iii)5/4, 6/4, 7/4, 8/4, 9/4

a) None
b) I only
c) II only
d) I and II
e) I, II, and III
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by nisagl750 » Thu Nov 29, 2012 11:03 pm
Anindya Madhudor wrote:Is there any simple way of solving the following?

Which of the following series has average greater than median?
i) 3/4, 4/5, 5/6, 6/7, 7/8
ii) 8/7, 7/6, 6/5, 5/4, 4/3
iii)5/4, 6/4, 7/4, 8/4, 9/4

a) None
b) I only
c) II only
d) I and II
e) I, II, and III
Take LCM of the denominators

Series I = LCM=840
630/840 , 672/840 , 700/840 , 720/840 , 735/840

Series II = LCM=420
480/420 , 490/420 , 504/420 , 525/420 , 560/420

Series III = LCM=4
5/4 , 6/4 , 7/4 , 8/4 , 9/4

You just need to take avg of Numerators and check if it is greater than 700, 504, 7 (Middle term i.e. Median) in each series.

The answer is C
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by GMATGuruNY » Sat Dec 01, 2012 5:57 am
Anindya Madhudor wrote:Is there any simple way of solving the following?

Which of the following series has average greater than median?
i) 3/4, 4/5, 5/6, 6/7, 7/8
ii) 8/7, 7/6, 6/5, 5/4, 4/3
iii)5/4, 6/4, 7/4, 8/4, 9/4

a) None
b) I only
c) II only
d) I and II
e) I, II, and III
Since the fractions increase in a PATTERN, examine how the values at the EXTREMES relate to the median.

I: 3/4....5/6...7/8
18/24....20/24....21/24
The smallest value is further from the median than is the greatest value.
Thus, the average will be weighted toward the SMALLEST value, with the result that average < median.
Eliminate B, D and E.

II: 8/7...6/5...4/3
120/105...126/105...140/105
The greatest value is further from the median than is the smallest value.
Thus, the average will be weighted toward the GREATEST value, with the result that average > median.
Eliminate A.

The correct answer is C.
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