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,

• A. 1/10
• B. 1/20
• C. 1/5
• D. 1/15

Answer: (A)

Let A has Rs. 200 at the starting of game, then B has Rs. 100 at the starting of game.After the game, money left with A = 200 - 200/3 = Rs. 400/3 ∴ Total money with B = 100 + 200/3 = Rs. 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.