Question 1

Consider a class of 40 students whose average weight is 40 kgs.

m new students join this class whose average weight is n kgs. If it is known that m + n = 50,
what is the possible average weight of the class now?

Also given that a number x such that following operations are performed on it(576*261*x) + 4787 = y

If the sum of digits of y until we get a single digit as 9 then 40>new average else new average>40.

Solution

Answer: b) 40.56 kgs

If the number is a multiple of 9 then sum of its digits = 9.Given 4787 is not a multiple of 9. So ‘new average’>40.
If the overall average weight has to increase after the new people are added, the averageweight of the new entrants has to be higher than 40.
So, n > 40

Consequently, m has to be < 10 (as n + m = 50).
We know that the total additional weight addedby m students would be (n - 40) each, above the already existing average of 40.
m*(n - 40) isthe total extra additional weight added, which is shared amongst 40 + m students.
So, m * (n-40)/(m+40) has to be maximum for the overall average to be maximum.

At this point, use the trial and error approach to arrive at the answer.
The maximum average occurs when m = 5,and n = 45 .
And the average is 40 + (45 - 40) * 5/45 .
= 40 + 5/9
= 40.56 kgs