Two children A and B are playing a game. A can draw a picture in,30 minutes and B can erase it in 40 minutes. If A starts drawing, and if the drawing sheet is passed on to these two alternately for every one minute, then the time (in minutes) required to complete a picture for the first time is

A can do full picture in 30 min.

B can erase it in 40 min.

A's rate =$$\frac{1\ }{30}$$ per min.

B's rate=$$\frac{1\ }{40}$$ per min.

So, 1st min A will do=$$\frac{1\ }{30}$$.

2nd min work done after B erase=$$\frac{1\ }{30}-\frac{1\ }{40}=\frac{40-30\ }{120}=\frac{10\ }{120}.$$

3rd min,work done=$$\frac{10\ }{120}+\frac{1\ }{30}\frac{10+4}{120}=\frac{14\ }{120}.$$

4th min=$$\frac{14\ }{120}-\frac{1}{40}=\frac{14-3}{120}=\frac{11}{120}.$$

5th min=$$\frac{11}{120}+\frac{1}{30}=\frac{11+4}{120}=\frac{15}{120}.$$

6thmin=$$\frac{15}{120}-\frac{1}{40}=\frac{15-3}{120}=\frac{12}{120}.$$

So, every even number minutes it gain by 1 unit of work.

So, 2,4,6,......,232nd min it will complete ==> 10,11,12,.......,116th part of work.

But in 233rd min it will complete$$\frac{116\ }{120}+\frac{1\ }{30}=\frac{\ 116+4}{120}=\frac{120\ }{120}=1.$$

So, full job will done in 233 min.

D is correct choice.