Question

Average of ‘m’ numbers is ‘2x’ and average of ‘n’ numbers is ‘3y’. Then, what will be the average of (m + n) numbers?

Answer 1

Step-by-step explanation:

Given :-

Average of ‘m’ numbers is ‘2x’

Average of ‘n’ numbers is ‘3y’.

To find :-

The average of (m+n) numbers

Solution :-

Given that

Average of ‘m’ numbers is ‘2x’

We know that

Average = Sum of observations / Number of observations

Average of m numbers

=> 2x = Sum of m numbers / m

=> Sum of m numbers = 2x × m

=> Sum of m numbers = 2xm

and

Average of ‘n’ numbers is ‘3y'

We know that

Average = Sum of observations / Number of observations

Average of m numbers

=> 3y = Sum of n numbers / n

=> Sum of n numbers = 3y×n

=> Sum of n numbers = 3yn

Now,

Sum of m and n numbers = 2xm + 3yn

Number of all numbers = m+n

We know that

Average = Sum of observations / Number of observations

Average of (m+n) numbers

= (2xm+3yn)/(m+n)

Answer :-

The average of (m+n) numbers is

(2xm+3yn)/(m+n)

Used formulae:-

→ Average = Sum of observations / Number of observations

Answer 2

Answer:

2xm + 3yn/m + n

Step-by-step explanation:

Given :-

Average of ‘m’ numbers is ‘2x’ and average of ‘n’ numbers is ‘3y’.

To Find :-

Then, what will be the average of (m + n) numbers?

Solution :-

Average = Sum of terms/Total no. of terms

For m numbers

2x = Sum/m

2x × m = Sum

2xm = Sum

For n numbers

3y = Sum/n

3y × n = Sum

3yn = Sum

Now,

Average of m + n

Average = (2xm + 3yn)/m + n

Average = 2xm + 3yn/m + n