TheCATExam
Sign in

CAT 2000 — QA Question 51

Forming a committeeEasy
Passage / Data

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.

ABCDEFGH is a regular octagon. A and E are opposite vertices of the octagon. A frog starts jumping from vertex to vertex, beginning from A. From any vertex of the octagon except E, it may jump to either of the two adjacent vertices. When it reaches E, the frog stops and stays there. Let an be the number of distinct paths of exactly n jumps ending in E. Then what is the value of a2n – 1?

Answer & solution

  • Zero

  • B

    Four

  • C

    2n – 1

  • D

    Cannot be determined

Solution

The frog has to jump at least four times to reach E.

i.e. a4 = 2

(i.e. (i) A-B, B-C,C-D and D-E; (ii) A-H, H-G, G-F and F-E).

If the frog keeps jumping on left (right) hand side vertices it will not take more than four jumps.

If it jumps on right vertex and then keeps on jumping left vertices then it will take 6 jumps to reach E.

∴ We will not find any path such that the frog takes 5 jumps and reaches E.

∴ a5 = 0

Similarly, we find that only for even values of n the frog can reach E.

2n − 1 is an odd number

∴ No route is possible with odd number of jumps.

Hence, option (a).

CAT 2000 QA Q51: ABCDEFGH is a regular octagon. A and E are opposite vertices of the octagon. A frog starts jumping from vertex — Solution | TheCATExam