Thursday, 27 October 2011

Progressions-4 (CAT-2000)

Consider a sequence of seven consecutive integers. The average of first five integers is n. the average of all seven integers is:
(1)n      (2) n+1        (3) K*n, where K is a function of n        (4) n+(2/7)
Solution follows here:

Solution:
Given that it is a sequence of 7 consecutive integers. As the average of first five integers is n, sum of first five integers is 5n.
=> t1+t2+t3+t4+t5 = 5n          ------(1)
As they are consecutive integers, the middle number of the first five numbers is nothing but average of the five numbers ie.,  ‘n’.
=> t3 = n => t4 = n+1 => t5 = n+2 => t6 = n+3 => t7 = n+4
=> t6+t7 = n+3+n+4 = 2n+7 ------(2)
 Adding (1) and (2),
t1+t2+t3+t4+t5+t6+t7= 5n+2n+7 = 7n+7
=> Average of all 7 numbers = (t1+t2+t3+t4+t5+t6+t7)/7 = n+1
Answer (2)

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