A cricketer has a certain average for 9 innings. In the tenth innings,

We can approach this either algebraically or by PITA (Plugging In The Answers). Since nobody else has already shown the algebra, I'll post both methods.

Algebra:
Let R be the average over the first 9 innings.
\((9R+100)/10 = R+8\)
\(9R+100 = 10R+80\)
R=20
We want the new average, which is 8 more than R.
Answer choice C.

PITA:
Let's start with B. The new average is 24 runs. The average after nine innings was 8 less than that, so 16. He scored 9*16 = 144 in the first nine innings. He scored 10*24 = 240 as of the end of the tenth. That's a difference of 96 runs. Is that the 100 runs we need? Nope. B is wrong.
If you're not confident whether you need something smaller or larger, let's try D. The new average is 32 runs. The average after nine innings was 8 less than that, so 24. He scored 9*24 = 216 in the first nine innings. He scored 10*32 = 320 as of the end of the tenth. That's a difference of 104 runs. Is that the 100 runs we need? Nope. D is wrong.
B gave us 96 and D gave us 104. Did we get farther away? Did we go right past what we wanted? Or did we get closer but not all the way to what we wanted?
We went right past what we wanted, so need an answer between B and D.
Answer choice C.


ThatDudeKnowPITA ... and I guess some algebra...