Test Index

CAT Exam Model Paper 3 with solutions for free online practice

Directions (Q. Nos. 39-42) Read the passage given below and solve the questions based on them.
Five universities-A, B, C, L and K participated in the last year inter-university athlete event. Each of the universities were represented by three members. According to the rules, the total number of athletes participating in the race should be equal to 16. The rank obtained by the individual will be the points obtained. And similarly the points obtained by an individual university will be equal to the sum of the ranks of the member athletes of the team. Lower the points, better the team and the best team wins the trophy.
The five universities tied for the trophy, their sum being equal, however no two athletes tied for the same position. In order to determine the order in which the universities will hold the trophy (they agreed to hold it for 73 days each), they multiplied the athletes positions together in each university. The university with the smallest product—K— will hold the trophy first and so on the university with the largest product —A— will hold the cup last. Unfortunately, universities B and C were still tied and had to be separated by a toss of coin.
It was also found that the no two athletes of a university finish at the consecutive positions. Had the race-track maker did not come in between two athletes of university B. university B would have achieved this feat of two athletes being at consecutive positions.

Question : 39

Total: 100

Which of the following is the position of the race-track maker?

Your Answer:

Validate

Solution:

The position of race track maker is 11.
Solution (Q. Nos 39-42)
Sum of ranks of all the 16 players = 136.
Sum of the ranks of all the universities is same, hence, their sum should be divisible by 5. So, their sum should be either 135, or 130 or 125 or 120. In turn, the rank of the race-track maker will be either 1 or 5, or 11 or 16 respectively.
Since race-trace maker comes in between the two athletes of university B, hence rank cannot be 1 or 16.
Case 1-If the rank of race-track maker = 5
Hence, the sum of the ranks obtained of all the universities = 130, and so the sum of ranks of all the athletes of each of the universities = 26.
Since the race track maker comes in between two athletes of university B, hence the ranks obtained by two of the athletes of university B are 5 and 7. Hence the rank of the third athlete of university B = 26 − (5 + 7) = 14
Now, the product of ranks of all the athletes of university B = Product of ranks of all the athletes of university C.
Product of ranks of university B contains 7 and 2 inter alia and no other exponent of 7 is possible in the whole series of 1 to 16 for university C.
Hence, we can conclude that rank of race-track maker cannot be 6.
Case 2-If the rank of race-track maker = 11
Hence, the sum of the ranks obtained of all the universities = 125,
and so the sum of ranks of all the athletes of each of the universities = 25
Since, the race track maker comes in between two athletes of university B, hence, the ranks obtained by two of the athletes of university B, hence, the ranks obtained by two of the athletes of university B are 10 and 12. Hence the rank of the third athlete of university B = 25 − (10 + 12) = 3
Hence the ranks of athletes of university B = 3, 10, 12
Now the product of ranks of all the athletes of university B = Product of ranks of all the athletes of university C.
Hence, the ranks of athletes of university C = 4, 6, 15
We can observe that the product and sum of the ranks of athletes of university B and C are same. From the remaining ranks now, for the smallest product of the ranks of athletes such that their sum is equal to 25, the ranks should be 1, 8, 16 (University K).
Similarly, for the maximum product of the ranks of the athletes for the university A, rank should be 13, 7, 5.
Hence, the ranks for the athletes of the university L = 2, 9, 14.

Go to Question:

Prev Question

Next Question