UPSC Prelims 2010 Q9: Science & Technology – Quantitative Aptitude

UPSC Prelims 2010, General Studies Paper I · Science & Technology

In a tournament 14 teams play league matches. If each team plays against every other team once only then how many matches are played?

Answer: (b) 91

Explanation

League match counting: 14 teams, play each other once

This is a classic combinations problem: how many unique pairs can be formed from 14 teams?

Step	Calculation
Formula	Number of matches = n(n-1)/2, where n = number of teams
Substitute n = 14	14 × 13 / 2
Compute	182 / 2 = 91

Why this formula works:

Point	Detail
Each team plays	13 other teams once each
Total team-games	14 × 13 = 182 (but each match counted twice, once per team)
Unique matches	Divide by 2: 91

Answer: (b) 91.

(a) Incorrect. 105 would be the result of a different team count (n=15: 15×14/2=105), not 14 teams.
(b) CORRECT. Using n(n-1)/2 with n=14 gives 14×13/2 = 91 unique matches.
(c) Incorrect. 85 does not correspond to the correct n(n-1)/2 formula applied to 14 teams.
(d) Incorrect. 78 corresponds to n=13 (13×12/2=78), not 14 teams.

For n teams each playing every other team once, total matches = n(n-1)/2. With n=14, that's 14×13/2 = 91 matches.

14 teams, single round-robin: 14×13/2 = 91 matches.

Forgetting to divide by 2 after multiplying n by (n-1), double-counting each match.
Using the wrong value of n (e.g., 15 or 13) instead of the given 14 teams.
Confusing a 'league' (each pair plays once) with a double round-robin (each pair plays twice).

Round-robin match count formula: n(n-1)/2. Memorise it — a frequent CSAT quant shortcut.

NCERT Class XI Mathematics — Permutations and Combinations
R. S. Aggarwal — Quantitative Aptitude for Competitive Examinations (Permutations and Combinations)