# Let x̄ be the mean of x1 , x2 , ... , xn and ȳ the mean of y1 , y2 , ... , yn . If z̄ is the mean of x1 , x2 , ... , xn , y1 , y2 , ... , yn , then z̄ is equal to

a. x̄ + ȳ

b. (x̄ + ȳ)/2

c. (x̄ + ȳ)/n

d. (x̄ + ȳ)/2n

**Solution:**

It is given that

\(\sum_{i=1}^{n}x_{i}=n\overline{x}\: and\: \sum_{i=1}^{n}y_{i}=n\overline{y}\) …. (1)

As \(\overline{x}=\frac{\sum_{i=1}^{n}x_{i}}{n}\)

z̄ = (x1 + x2 + ….. +xn) + (y1 + y2 + …. + yn)/n + n

So we get

z̄ = \(\frac{\sum_{i=1}^{n}x_{i}+\sum_{j=1}^{n}y_{i}}{2n}\)

z̄ = (nx̄ + nȳ)/2n [From equation (1)]

Taking out n as common

z̄ = (x̄ + ȳ)/2

Therefore, z̄ is equal to (x̄ + ȳ)/2.

**✦ Try This: **The mean of five numbers is 16. If one number is excluded, their mean becomes 12. The excluded number is :

**☛ Also Check:** NCERT Solutions for Class 9 Maths Chapter 14

**NCERT Exemplar Class 9 Maths Exercise 14.1 Problem 15**

## Let x̄ be the mean of x1 , x2 , ... , xn and ȳ the mean of y1 , y2 , ... , yn . If z̄ is the mean of x1 , x2 , ... , xn , y1 , y2 , ... , yn , then z̄ is equal to a. x̄ + ȳ, b. (x̄ + ȳ)/2, c. (x̄ + ȳ)/n, d. (x̄ + ȳ)/2n

**Summary:**

Let x̄ be the mean of x1 , x2 , ... , xn and ȳ the mean of y1 , y2 , ... , yn . If z̄ is the mean of x1 , x2 , ... , xn , y1 , y2 , ... , yn , then z̄ is equal to (x̄ + ȳ)/2

**☛ Related Questions:**

- If x̄ is the mean of x1 , x2 , ... , xn , then for a ≠ 0, the mean of ax1 , ax2 , ..., axn , x1/a, x . . . .
- If x̄1, x̄2, x̄3, ….., x̄n are the means of n groups with n1 , n2 , ... , nn number of observations . . . .
- The mean of 100 observations is 50. If one of the observations which was 50 is replaced by 150, the . . . .

visual curriculum