Xavier Aptitude Test 2013 Paper

Question : 62 of 121

Marks: +1, -0

The mean of six positive integers is 15. The median is 18, and the only mode of the integers is less than 18. The maximum possible value of the largest of the six integers is

Validate

Solution:

(d) Let us assume that the integers are in increasing order are a, b, c, d, e and f.
Hence, (a + b + c + d + e + f) = 6 × 15 = 90
Now, median =

{c + d}/2

= 18 or c + d = 36
Now, to maximise of we need to minimize all the others.
Minimum possible values of a and b is 1.
Now, (c + d) = 36, but c and d can’t be equal to 18 as that would make 18 a mode of integers. Hence, least possible value of d is 19.
Now, e must be close as possible to d. But again, it cannot be equal to d which will make e = d = a mode > 18
Hence, e = 20.
Hence, f = 90 – (a + b + c + d + e)
= 90 – (1 + 1 + 36 + 20) = 32

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