Page 285 - Services Selection Board (SSB) Interviews

Games and Tournament 281
Lets sum up the information, 32. (b)
The teams who played semifinal round is B, F, G and H 33. (b) Number of matches in stage 1 is 2(7 × 8/2) = 56,
Given that F lost both the matches to G i.e F and G is in at semifinal stage we have 3 matches (2 semifinal
rd
the same group, then B and H is in the same group. and 1 match for 3 place) and 1 final match, hence
Since number of matches won by A and G is same hence total number of matches is 56 + 3 + 1 = 60
they must belong to two different groups, and C won only 34. (a) Seed 9 will play with seed, 1, 3, 5, 7, 11, 13, and
I match against D hence they must be in the same group. 15 without an upset seed 9 can win with seed 11,
So we can divide the group now- 13, and 15, for minimum number of upset let seed 1
Group 1: F, G, C, D won all the matches and seed 9 won against seed 3
and 5, in that case number of wins of seed 3 and 9 is
Group 2: B, H, A, E 5 but with tie breaker rule seed 9 will advance to the
In one group number of matches is 2(3 × 4/2) = 12 since next stage.
each team won a different number of matches and total 35. (d) Total number of matches is 60 and out of those
number of matches is 12 so we have following possibilities more than 45 matches are upset. But seed 1 need
for number of matches won by 4 teams- only 9 matches to win the tournament hence seed 1
Case (i) 6, 4, 2, 0 may win the tournament.
Case (ii) 6, 3, 2, 1 36. (b) Total number of matches in the 1st stage is 4 × 7

Case (iii) 5, 4, 2, 1 = 28, lets consider group 1 here if seed 1 won all the
Case (iv) 5, 4, 3, 0 matches then remaining 21 matches or points can be
Since number of matches won by A and G is the same but equally distributed to 7 player (3 points each) and
then also A eliminated so only possibility is case (ii) and the lowest possible player would advance to next
(iv) and they won 3 matches. stage with tie breaker rule. In this stage seed 13 can
get 3 points after 2 upsets caused by him. So from
Lets sum up the conclusion till now this group seed 1 and 13 would advance to the next
Group 1: F, G, C, D and their points F = 6, G = 3, C = stage. Similarly from 2 group seed 2 and 14 would
nd
1 and D = 2 advance to the next stage.
Group 2: B, H, A, E and their points B and H 5 and 4 in Now as per the rule seed 1 will play with seed 14 and
any order and A = 3, E = 0 seed 2 will play with seed 13,
Now lets draw the table for Group 1 If seed 13 and 4 meet in the tournament then seed
F G C D 13 will win with 3 upset.
F X W W W W W W 6 37. (a) Here only problem is stage 1, so a player has to
G LL X WW WL 3 qualify for semifinal without any upset. In previous
question we have seen that with 3 points/win a team
C LL LL X WL 1 can advanced to next stage.
D LL LW LW X 2
Consider group 1: Here players are seeded 1, 3,
Now lets draw the table for Group 2 5, 15. The player who can win 3 matches without an

B/H H/B A E upset is seed 9.
B/H X WW LW WW 5 Consider group 2: same as previous case seed
H/B LL X WW WW 4 10 can advanced to next stage without causing an
A WL LL X WW 3 upset.
E LL LL LL X 0 So seed 10 is the answer.
Consider 1 such case semifinal round Group 1: seed
28. (a) option (a) is correct 13 and 15
nd
29. (d) 2 Highest is either B or H so cant determine.
Group 2: seed 2 and 10
30. (c) E lost the maximum number of matches. Final match: seed 10 and 15
31. (d) A lost both the matches to B or H not B and H
hence option C is incorrect.