A certain kickball team named “The good guys” won 1/3 of their first 24 games. If the teams lose no more than 1/9 of their remaining games, what is the least number of games they must play to ensure they win more than they loose?
a) 9
b) 10
c) 11
d) 12
e) 13

May someone tell me where is my mistake?

I tried backward solving by starting with choice a=9
(1/9)∗9+16=17(1/9)∗9+16=17
and the winning will be (8/9)∗9+8=16(8/9)∗9+8=16 option a wrong
For the rest, all of the options will give us the first option 1.1 or 1.114, 1.22 … I round it to 2.
Until I reach 13, where the winning games are more than the lost games. So, I chose 13 but the answer is wrong. The answer is 11

If they won \frac{1}{3} of their first 24 games = They won 8 games.

Now, from their remaining games they only lost \frac{1}{9}.

Thus, Using the B D technique:-

Option B )
If they only lost \frac{1}{9} of 10 games it means they lost 1.11 games out of 10 or they won 8.89 out of 10 games.

Now, wining percentage equal \frac{8+8.89}{10+24} or \frac{16.89}{34}= 49.67

49.67 is very close to 50%

Option D )
If they only lost \frac{1}{9} of 12 games it means they lost 1.33 games out of 12 or won 10.67 games out of 12.

Now, wining percentage equal \frac{8+10.67}{12+24} or \frac{18.67}{36}= 51.86

51.86 is above 50 % and would have been our ans but we have Option C 11. Thus, we can definitely can eyeball that C)11 will be the least number of games they must play to ensure they win more than they loose(above 50%).