XAT 2011 — QA & DI Question 33
Answer the following question based on the information given below.
From a group of 545 contenders, a party has to select a leader. Even after holding a series of meetings, the politicians and the general body failed to reach a consensus. It was then proposed that all 545 contenders be given a number from 1 to 545. Then they will be asked to stand on a podium in a circular arrangement, and counting would start from the contender numbered 1. The counting would be done in a clockwise fashion. The rule is that every alternate contender would be asked to step down as the counting continued, with the circle getting smaller and smaller, till only one person remains standing. Therefore the first person to be eliminated would be the contender numbered 2.
Prof. Bee noticed something peculiar while entering the quiz marks of his five students into a spreadsheet. The spreadsheet was programmed to calculate the average after each score was entered. Prof. Bee entered the marks in a random order and noticed that after each mark was entered, the average was always an integer. In ascending order, the marks of the students were 71, 76, 80, 82 and 91. What were the fourth and fifth marks that Prof. Bee entered?
Answer & solution
- A
71 and 82
- B
71 and 76
71 and 80
- D
76 and 80
- E
91 and 80
It is convenient to solve this question by evaluating options.
Option 1:
If 71 and 82 are fourth & fifth numbers then, sum of first three numbers is,
76 + 80 + 91 = 247
But, 247 is not divisible by 3.
Hence, they cannot be fourth & fifth numbers.
∴ Option 1 is not possible.
Option 2:
Fourth and fifth numbers are 71 and 76.
Sum of first three numbers will be,
80 + 91 + 82 = 253
It is also not divisible by 3.
Hence, option (b) is also not possible.
Option 3:
Fourth and fifth numbers are 71 and 80.
Hence, sum of first three numbers will be
76 + 82 + 91 = 249, which is divisible by 3.
Hence, they can be fourth and fifth numbers.
Hence, option (c).