### Month of January

### 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

*, what is the possible average weight of the class now?Also given that a number*

**m + n = 50***x*such that following operations are performed on it

**If the sum of digits of**

*(576*261*x) + 4787 = y**y*until we get a single digit as

*9*then

**else**

__40>new average____.__

*new average>40*Answer: 40.56

### Comments on Question 1

### Solution

**Answer: b) 40.56 kgsIf 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 > 40Consequently, 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