Let’s denote Ajit’s average for the first 9 innings as

$A$A. The average is calculated by dividing the total runs by the number of innings.

The total runs from the first 9 innings would be

$9A$9A (average multiplied by the number of innings).

Now, in the 10th inning, he scores 100 runs, so the new total runs become

$9A+100$9A+100.

The new average is then calculated by dividing the new total runs by the total number of innings (10 innings now).

So, the new average (

$B$B) is given by:

$B=\frac{9A+100}{10}$B=109A+100

It is mentioned that the new average is 8 runs more than the previous average (

$A$A).

$B=A+8$B=A+8

Now, we can set up an equation and solve for

$A$A:

$A+8=\frac{9A+100}{10}$A+8=109A+100

To solve for

$A$A, you can follow these steps:

Multiply both sides of the equation by 10 to eliminate the fraction:

$10(A+8)=9A+100$10(A+8)=9A+100

Distribute the 10 on the left side:

$10A+80=9A+100$10A+80=9A+100

Subtract $9A$

$9A$ from both sides:

$A+80=100$A+80=100

Subtract 80 from both sides:

$A=20$A=20

So, Ajit’s average for the first 9 innings (

$A$A) is 20 runs. Now, his new average after the 10th inning is

$A+8=20+8=28$A+8=20+8=28 runs.