Question

The average of 5 consecutive integers starting with m
as the first integer is n. What is the average of 9
consecutive integers that start with m+2?
A) m + 4
B)n +6
C) n + 4
D) m + 5

Answer 1

Answer:

(C). n + 4

Step-by-step explanation:

[ m + ( m + 1 ) + ( m + 2 ) + ( m + 3 ) + ( m + 4 ) ] ÷ 5 = n

( 5m + 10 ) ÷ 5 = n

m + 2 = n ⇒ m = n - 2

[ ( m + 2 ) + ( m + 3 ) + ( m + 4 ) + ( m + 5 ) + ( m + 6 ) + ( m + 7 ) + ( m + 8 ) + ( m + 9 ) + ( m + 10 ) ] ÷ 9 =

( 9m + 54 ) ÷ 9 = m + 6

If m = n - 2 , then m + 6 = ( n - 2 ) + 6 = n + 4

Answer 2

Answer:

C) n + 4

Step-by-step explanation:

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