Question

The average of 9 consecutive numbers is n. If the next two numbers are also included, then the new average will 9

Answer 1

Answer

n + 1

Explanation

Let the 9 consecutive integers be

(x - 4), (x - 3), (x - 2)............................(x + 4)

Their sum = 9x

Average = Sum of observations/number of observations

⇒ n = 9x/9

⇒ x = n

Now, the next two numbers would be (x + 5) and (x + 6)

So, the new sum becomes 9x + x + 5 + x + 6

⇒ 11x + 11

And, number of observations = 11 (as there are a total of 11 numbers now)

So, average = (11x + 11)/11

⇒ 11(x + 1)/11

⇒ (x + 1)

Now, x = n

⇒ n + 1

Hence, the new average is (n + 1)

Answer 2

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❚ QuEstiOn ❚

# The average of 9 consecutive numbers is n. If the next two numbers are also included, then the new average will be ?

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❚ ANsWeR ❚

Let , the 9 consecutive numbers are ,

✏ (N-4) , (N-3) , (N-2) , (N-1) , N , (N+1) , (N+2),(N+3) , (N+4)

✺ Now , The Sum of the numbers :-

➩ S = (N-4)+(N-3)+(N-2)+(N-1)+N+(N+1)+(N+2)+(N+3)+(N+4)

➩S = 9N

✺ Now , The Average of the numbers is :-

✏ $\sf{\ \ {S_{avg}=\dfrac{9N}{9}}}$

✏ $\sf{\ \ {S_{avg}=\dfrac{\cancel{9}N}{\cancel9}}}$

✏ $\sf{\ \ {S_{avg}=N}}$

According to the question The average of 9 consecutive numbers is n

✏ $\sf{\ \ {S_{avg}=N=n}}$

✏ $\sf{\ \ {N=n}}$

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If the next two numbers are also included then the numbers would be (N+5) & (N+6) .

✺ Now , The Sum of the 11 consecutive numbers is :-

➩ S' = (N-4)+(N-3)+(N-2)+(N-1)+N+(N+1)+(N+2)+(N+3)+(N+4)+(N+5)+(N+6)

➩ S' = 11N+11

➩ S' = 11(N+1)

✺ Now , The Average of the 11 consecutive numbers is :-

✏ $\sf{\ \ {S'_{avg}=\dfrac{11(N+1)}{11}}}$

✏ $\sf{\ \ {S'_{avg}=\dfrac{\cancel{11}(N+1)}{\cancel11}}}$

✏ $\sf{\ \ {S'_{avg}=N+1}}$

[ as , N = n ]

✏ $\large{\boxed{\sf{\ \ {S'_{avg}=(n+1)}}}}$

✺ Therefore :-

The new average of 11 consecutive numbers will be = (n+1)

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