Question 115

# The student fest in an Engineering College is to be held in one month’s time and no sponsorship has yet been arranged by the students. Finally the General Secretary (GS) of the student body took the initiative and decided to go alone for sponsorship collection. In fact, he is the only student doing the fund raising job on the first day. However, seeing his enthusiasm, other students also joined him as follows: on the second day, 2 more students join him; on the third day, 3 more students join the group of the previous day; and so on. In this manner, the sponsorship collection is completed in exactly 20 days. If an MBA student is twice as efficient as an Engineering student, the number of days which 11 MBA students would take to do the same activity , is:

Solution

Let 'N' be the total effective student work day required to finish the sponsorship works.

N = 1+(1+2)+(1+2+3)+...+(1+2+3+ ... +20)

N = $$\sum_{n=1}^{n=20} \dfrac{(n)*(n+1)}{2}$$

N = $$\dfrac{1}{2}*[\dfrac{20*21*41}{6}+\dfrac{20*21}{2}]$$

N = 1540

We can say that 1540 engineering students can finish the sponsorship task in one day. It is given that each MBA student is twice as efficient as an engineering student. Hence, we can say that 770 MBA students will finish the sponsorship task in a day.

Therefore, time taken by 11 MBA students would take to do the same activity = $$\dfrac{770}{11}$$ = 70 days