Question

A group of 30 people can complete a job by working for 10 hours a day in 15 days. The group starts the work. But at the end of every day, starting from the 1st day, 1 person leaves the group and the remaining people work for 20 min less on the next day. On which day will the work be completed?​

Answer 1

Answer:

Actually, a correct answer is "Not possible" for this question, and I can prove it by solving it.

Explanation:

So let's suppose the efficiency of 1 person = p and total work = w units.

The formula for Work and Efficiency is:

Efficiency of 1 person = (Work Done)/Time Taken

So, it is given that,

30*p = w/(10*15)

p = w/4500

Now, variation of the same formula used above,

Efficiency*(Number of persons * Time) = Total Work

(w/4500) * (30*10 + 29*(10-1/3) + 28*(10-2/3) + ... ) = w

Multiplying both sides by 3,

3 * ( 30*10 + 29*(10-1/3) + 28*(10-2/3) + ... ) = 3 * 4500

30^2 + 29^2 + .... = 13500

Now, finding number of terms at LHS will give us number of days taken.

Now, lets take the highest possible value i.e. 30.

30^2 + 29^2 + .... + 1^2 = 13500

Now, we have a standard formula to solve LHS (you can google it if you don't know) and solving it gives:

9455 = 13500, which is still lesser than RHS.

So, it is not possible because number of persons will become negative for more than 30 days.