SSC CGL 2025 Tier-I · 12 September 2025, Shift 3 · General Intelligence and Reasoning · Question 25 of 100

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 11 September 2026.

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SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

In a tournament of 7 players, each player plays every other player once. How many matches are there?

(a)21
(b)42
(c)36
(d)28

Answer

Why

Correct — A. Rule: matches = pairs of players = n(n − 1) ÷ 2.

Each of the 7 players faces the other 6: 7 × 6 = 42
Every match has two players, so it is counted twice: 42 ÷ 2 = 21

That is ⁷C₂ = 21 → option (a)

Why the others are wrong

(b)
42 — 42 is 7 × 6, which counts every match once from each player's side. A against B and B against A are the same match, so halve it to 21.
(c)
36 — 36 is the number of matches among 9 players, 9 × 8 ÷ 2. With 7 players it is 7 × 6 ÷ 2 = 21.
(d)
28 — 28 is 1 + 2 + … + 7, one term too many. Player 1 has 6 new opponents, player 2 has 5, and so on, so the sum is 6 + 5 + … + 1 = 21.

Concept

In a round robin every pair of players meets exactly once, so the number of matches is the number of pairs you can pick from the players: ⁿC₂ = n(n − 1) ÷ 2.

Another way to see it: the first player has 6 opponents, the second has 5 new ones, and so on, giving 6 + 5 + 4 + 3 + 2 + 1 = 21.

Both routes avoid the double count that 7 × 6 makes.

The stem says each player plays every other player once. If each pair met twice, home and away, the count would be 42.

Key facts

Matches in a single round robin of n players = n(n − 1) ÷ 2.
For 7 players: 7 × 6 ÷ 2 = 21.
6 + 5 + 4 + 3 + 2 + 1 = 21 counts the same matches player by player.

Study next

Handshake problems, where every pair of people shakes hands onceLines through n points with no three in a line, also n(n − 1) ÷ 2Selections such as a committee of 3 from 7 (⁷C₃ = 35)

Common traps

Stopping at 7 × 6 = 42 and forgetting that each match has two players.
Summing 1 + 2 + … + 7 = 28 instead of 1 + 2 + … + 6.

Here the count is asked directly: the number of players is given and every pair meets once. The same n(n − 1) ÷ 2 answers handshake and lines-through-points questions.

Related PYQs

No directly related past PYQ was found.