Question

The average of 9 consecutive natural numbers is 18.
The highest of these numbers will be
(a) 24 (b) 18
(c) 20
(d) 22
pls let me know the steps for solving​

Answer 1

Step-by-step explanation:

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Answer 2

The highest of these numbers will be 22

Given :

The average of 9 consecutive natural numbers is 18

To find :

The highest of these numbers will be

(a) 24

(b) 18

(c) 20

(d) 22

Solution :

Step 1 of 2 :

Form the equation to calculate the highest of these numbers

Let 9 consecutive natural numbers are n , n + 1 , n + 2 , n + 3 , n + 4 , n + 5 , n + 6 , n + 7 , n + 8

By the given condition

$\displaystyle \sf \frac{n + ( n + 1 ) + (n + 2 ) + ( n + 3 ) + (n + 4 ) + ( n + 5 ) + ( n + 6 ) + ( n + 7 ) + ( n + 8)}{9} = 18$

Step 2 of 2 :

Calculate highest of these numbers

$\displaystyle \sf \frac{n + ( n + 1 ) + (n + 2 ) + ( n + 3 ) + (n + 4 ) + ( n + 5 ) + ( n + 6 ) + ( n + 7 ) + ( n + 8)}{9} = 18$

$\displaystyle \sf{ \implies } \frac{9n + 36}{9} = 18$

$\displaystyle \sf{ \implies } \frac{9(n + 4)}{9} = 18$

$\displaystyle \sf{ \implies }n + 4 = 18$

$\displaystyle \sf{ \implies }n = 18 - 4$

$\displaystyle \sf{ \implies }n = 14$

So 9 consecutive natural numbers are 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 22

Thus highest of these numbers will be 22

Hence the correct option is (d) 22

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