The mean of six positive integers is 15 The median is 18 and the only mode of the integers is less t...........

Quantitative Aptitude (CDS) Statistics

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

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Explanation:

Since the mean is 18, the total is 90. Let the numbers be a, b, c, d, e and f in ascending order. a+b+c+d+e+f=90

The median is 18, so there are three numbers less than 18 and three numbers more than 18. Therefore, a,b,c <18 and d,e,f >18. The median is the average of c and d. Therefore, (c+d)/2=18 => c+d=36

Here we need to maximize f, so we need to minimize a, b, c, d and e.

The minimum value a and b can take is 1, the least positive integer.

To minimize e, e should be equal to d+1. It cannot be equal to d as the mode is less than 18.

Hence, the minimum value that d can take such that c+d=36 and d>18 is d=19. Therefore, c=17 and e=20

Hence, a = 1, b = 1, c = 17, d = 19, e = 20. Therefore, f = 90 -(1+1+17+19+20) = 32.