CAT 2000 — QA Question 42
Answer the following question based on the information given below.
Sixteen teams have been invited to participate in the ABC Gold Cup cricket tournament. The tournament is conducted in two stages. In the first stage, the teams are divided into two groups. Each group consists of eight teams, with each team playing every other team in its group exactly once. At the end of the first stage, the top four teams from each group advance to the second stage while the rest are eliminated. The second stage comprises of several rounds. A round involves one match for each team. The winner of a match in a round advances to the next round, while the loser is eliminated. The team that remains undefeated in the second stage is declared the winner and claims the Gold Cup.
The tournament rules are such that each match results in a winner and a loser with no possibility of a tie. In the first stage, a team earns one point for each win and no points for a loss. At the end of the first stage teams in each group are ranked on the basis of total points to determine the qualifiers advancing to the next stage. Ties are resolved by a series of complex tie-breaking rules so that exactly four teams from each group advance to the next stage.
Consider a circle with unit radius. There are 7 adjacent sectors, S1, S2, S3,...., S7 in the circle such that their total area is (1/8)th of the area of the circle. Further, the area of the jth sector is twice that of the (j –1)th sector, for j = 2, ..., 7. What the angle, in radians, subtended by the arc of S1 at the centre of the circle?
Answer & solution
π/508
- B
π/2040
- C
π/1016
- D
π/1524
Let the area of sector S1 be x units.
The area of the sectors S2, S3, S4, S5, S6, S7 will be 2x, 4x, 8x, 16x, 32x and 64x
∴ The total area of 7 sectors = 127x units = (1/8) × total area of circle = (1/8) π
∴ 127x = π/8 units
A circle subtends an angle of 2π at the centre.
Hence, (1/8)th of the circle will subtend an angle of π/4 at the centre.
i.e. area of the seven sectors i.e. 127x will subtend an angle of π/4 at the centre.
Sector S1, whose area is x will subtend an angle of π/(127 × 4) at the centre.
∴The required angle = π/508 radians
Hence, option (a).