A has twice as much money as B. They play together and at the end of the first game B wins one-third of A's money from A, what fraction of the sum that B now has, must A win black in the second game, so that they may have exactly equal money?

Let A has ₹ 200 at the starting of game, then B has ₹ 100 at the starting of game.
After the game, money left with A = 200 - 200/3 = ₹ 400/3
∴ Total money with B = 100 + 200/3 = ₹ 500/3
Let the fraction of money lost by B = N
Then, 500/3 - N x 500/3 = 400/3 + N x 500/3
⇒ 500/3 - 400/3 = 500N/3 + 500N/3
⇒ 100/3 = 1000N/3
∴ N = (100/3) x (3/1000) = 1/10
so, the fraction of money lost is 1/10.