# Ajit has a certain average for 9 innings. In the tenth innings, he scores 100 runs thereby increasing his average 19 people went to a hotel for combine dinner party 13 by 8 runs. His new average is:

A. 20

B. 21

C. 28

D. 32

### Solution(By Apex Team)

Let Ajit's average be x for 9 innings. So, Ajit scored 9x run in 9 innings. In the 10th inning, he scored 100 runs then became average (x+8). And he scored (x + 8) × 10 runs in 10 innings. Now, $\begin{array}{l} \Rightarrow 9 x+100=10 \times(x+8) \\ \text { or, } 9 x+100=10 x+80 \\ \text { or, } x=100-80 \\ \text { or, } x=20 \\ \text { New average }=(x+8) \\ =28 \text { runs } \end{array}$

## Related Questions On Average

A. 20
B. 21
C. 28
D. 32

A. 18
B. 20
C. 24
D. 30

A. 10 years
B. 10.5 years
C. 11 years
D. 12 years

### If the arithmetic mean of 0, 5, 4, 3 is a, that of -1, 0, 1, 5, 4, 3 is b and that of 5, 4, 3 is c, then the relation between a, b, and c is.

A. a = b = c
B. a : b : c = 3 : 2 : 4
C. 4a = 5b = c
D. a + b + c = 12