JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Permutations and Combinations

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There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84 , then the value of

m

is :

(A)12✓ Correct
(B)11
(C)9
(D)7

Answer: (A)

Solution

Each pair of the

m

men plays two games, giving

2 (2 m) = m (m - 1)

men-against-men games, while each man plays two games with each of the two women, giving

m \times 2 \times 2 = 4 m

mixed games. The condition

m (m - 1) = 4 m + 84

becomes

m^{2} - 5 m - 84 = 0

, i.e.

(m - 12) (m + 7) = 0

, so

m = 12

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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