Question
The average of 5 consecutive integers starting with ‘m’ is n. What is the average of 6 consecutive integers starting with (m + 2) ?
Answer: Option D
Answer: (d)
m + m + 1 + m + 2 + m + 3 + m + 4 = 5n
⇒ 5m + 10 = 5n
⇒ m + 2 = n ....(i)
Required average = ${m+2+m+3+m+4+m+5+m+6+m+7}/6$
= ${6m+27}/6$
= ${2m+9}/2$ = ${2(n-2)+9}/2$= ${2n+5}/2$ By (i) [m = n – 2]
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Answer: (d)
m + m + 1 + m + 2 + m + 3 + m + 4 = 5n
⇒ 5m + 10 = 5n
⇒ m + 2 = n ....(i)
Required average = ${m+2+m+3+m+4+m+5+m+6+m+7}/6$
= ${6m+27}/6$
= ${2m+9}/2$ = ${2(n-2)+9}/2$= ${2n+5}/2$ By (i) [m = n – 2]
Was this answer helpful ?
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