A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides, whereas the remaining n + 1 coins are fair. A coins is picked up at random from the bag and tossed. If the probability that the toss results in a head is 31/42. Then n is equal to:

10

P(toss result is H) = $(\frac{n}{2n+1}).1+(\frac{n+1}{2n+1}).\frac{1}{2}=\frac{31}{42}⇒n=10$