XAT 2011 Question 91

Question 91

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?

Solution

Marks = 71, 76, 80, 82 and 91

For average to be an integer each time, the sum of numbers entered after 2nd entry should be divisible by 2, after third entry should be divisible by 3 and so on.

The first two numbers have to be both odd or both even, so that their sum is even and can be divisible by 2.

Also, the sum of first four numbers should also be even.

=> Numbers entered are either $$OOEEE$$ or $$EEOOE$$   (where O -> odd and E -> even)

Case 1 : First two numbers are 71 and 91 in any order.

Average = $$\frac{71 + 91}{2} = 81$$

Now, sum of 71 and 91 is multiple of 3, so the third number has to be a multiple of 3, which is not possible.

Case 2 : First two numbers can be = $$(76,80) , (76,82) , (80,82)$$

Now, following above criteria, only 2nd option is possible 

So, third number has to be 91, average = $$\frac{76 + 82 + 91}{3} = 83$$

$$\therefore$$ The fourth and fifth marks that Prof. Bee entered = 71 and 80



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