Average Practice Test 1

Question : 22 of 40

Marks: +1, -0

The average of 5 consecutive integers starting from‘m’ is n. What is the average of 6 consecutive integers starting from (m + 2)?

Solution:

Average of 5 consecutive integers starting from m = n

{m+(m+1)+(m+2)+(m+3)+(m+2)}/5=n

{5m+10}/5=n

m+2=n

m=n-2

Average of 6 consecutive integers starting from (m + 2)

={(m+2)+(m+3)+(m+4)+(m+5)+(m+6)+(m+7)}/6

={6m+27}/6

where

m = (n-2 )

={6(n-2)+27}/6

={6(n-2)+27}/9

={6n-12+27}/6

={6n+15}/6={2n+5}/2

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