LR 2

For questions 48 to 51:

Since, it is given that the number of games won by each person is

distinct, therefore P, Q, R, S, T and U won 5, 4, 3, 2, 1 and 0 games in

the whole tournament not necessarily in that particular order.

As U lost its games only against S and R, U must have won only 3 or

less games. Also, R and S must have won 4 and 5 games not necessarily

in that order.

Also, T did not loose more than 3 games, T must have won 2 or more

games. So, we can conclude that U won 3 games and T won 2

games.

Between P and Q, they have won 0 and 1 game not necessarily in that

order. But, Q won its game against P. So, Q won 1 game and P won 0

game.

As per the information given in the question the possibilities for the

number of games won by each of the mentioned persons is listed in

the table given below.

48. 1 The number of games won by Q is 1.

49. 5 Either R or S won 5 games.

For questions 50 and 51:

Given that R got 11 coins and won exactly four games in the whole

tournament.

So, the number of games won by each of the mentioned six persons

is listed in the table given below.

Since, S won all the five games then he must have necessarily got

10 coins. (Two coins in each game)

50. 2 The minimum number of coins got by U, T, Q and P is 6, 4, 2 and

0 respectively.

So, the total number of coins got by all the mentioned six

persons in the whole tournament cannot be less than 11 + 2 +

6 + 4 + 10 = 33.

51. 3 Given that the total number of coines got by all the mentioned

persons is 47. Now we already known that R, P and S got 11,

0, 10. So, we have 47 = (0, 10, 11) + (T, U, Q)

⇒ (T, U, Q) = 26

which is only possible when T gets 10, Q gets 6 and U gets

10.