UPSC Prelims 2010 · General Studies
Question 9: Basic Numeracy
In a tournament 14 teams play league matches. If each team plays against every other team once only then how many matches are played ?
AnswerThe answer is (b).
Each pair of teams meets exactly once, so the number of matches equals the number of ways to choose 2 teams from 14: 14 x 13 / 2 = 91. Check: each of the 14 teams plays 13 others, which counts every match twice, so 14 x 13 = 182 halved gives 91. Remember: in a league where every team meets every other once, matches = n(n - 1) / 2.
How you could have got this
No shortcut here: in a “how many” question, ruling out options doesn’t help. Every statement has to be judged.
This follows an unofficial answer: UPSC has not published its key for this year, and if that answer is wrong, so is this path.
Which options go is worked out from UPSC’s key and the options, not by a model.
Why each option is right or wrong
- (a) 105 is 15C2, the figure for 15 teams.
- (b) Right: 14C2 = 91.
- (c) 85 does not match any pairing count here.
- (d) 78 is 13C2, the figure for 13 teams.
Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2010, General Studies, Series A. The paper on upsc.gov.in.