Question
8 men started a job and worked on it for 12 days. From the 13th day onwards, 1 new worker joined them everyday till the job was finished in a total of 20 days. If the same job has to be done starting with n workers and one worker quits at the end of every day, then what is the minimum value of n?
And if you want a more challenging question …..with the same data as above …. if n is the least possible value, then how many days will it take for the work to get done?
Step 1: Express the total work in man-days. 8 men for 12 days and then 9, 10, 11, 12 …. men on each of the following day, till the 20th day.
Step 2: In second case, we start with n men and each day one man reduces. Thus, men working on successive days is n, (n – 1), (n – 2), …… . Now the tricky part is to realise what is the condition for n to be minimum. Try figuring that out using following random numbers …
If work to be done was 15 man-days. And say we started with ….
…. 8 men. Then work would get done as 8 + 7 = 15 i.e. in 2 days. Can we start with fewer men?
…. 6 men. Then work would get done as 6 + 5 + 4 = 15 i.e. in 3 days. Can we start with even lesser men than 6?
…. 5 men? Then work would get done as 5 + 4 + 3 + 2 + 1 = 15 i.e. in 5 days. Can we start with even lesser men than 6?
…. 4 men? Work done will be 4 + 3 + 2 + 1 i.e. only 10 man-days will be done and work cannot be completed.
So, minimum men to start, in case work is 15 man-days, is 5 men.
Draw your learning from the above and apply it to the original question.
On the 13th day, there will be 9 men; on 14th day, 10 men; … and so on ⇒ on 20th day, there will be 16 men.
Work, in man-days = (8 × 12) + 9 + 10 + 11 + … + 16.
The second part has avergae = (9 + 16)/2 and number of terms is 8 ⇒ sum = 25 × 4 = 100
Work = 196 man-days.
For n to be minimum, we need the following conditions …
Σn ≥ 196, so that by the end when only 1 man remains working, the work gets completed. And …
Σ(n–1) < 196, such that, if we start with 1 fewer man, then the work will not get completed and before this, all men would have quit.
A bit of intelligent guessing … finding perfect square closer to 2 × 196 i.e. closer to 392note, we realise 202 = 400.
Thus, Σ20 = 210 and Σ19 = 190
Answer to the minimum value of n is 20.
This last part of comparing Σn and Σ(n–1) with work to be done, is a little tricky to understand. It is best understood with a number. Now that we have 20 men and 19 men as relevant numbers, the following is very easy.
Had we start with 19 men, then maximum work that can be done, with 1 man quitting at end of each day, will be
19 + 18 + 17 + ….. + 1 i.e. Σ19 i.e. 190 man-days.
The work in question, can thus not be completed if we start with just 19 men.
With 20 men, the maximum work that can be done is 210 units i.e. more than the work in question. While more than required work can be done, yet we cannot start with any fewer men, as illustrated with what happens when we start with 19 men … all men quit with just 190 units done.
And this is where the very difficult part of the question, ‘on which day will the work get done?’, comes into picture.
If we started with 20 men and worked for entire 20 days, then total work done = 210 man-days. But we need to do only 196 man-days. I.e. 14 fewer man-days.
Working backwards, work done on 20th, 19th, 18th, … day will be 1, 2, 3, …. man-days
If we start adding 1 + 2 + 3 …. such that we reach near-abouts 14, then we will have Σ4 = 10 and Σ5 = 15
Work done in last 4 days = 10 man-days ⇒ work done at the end of 20 – 4 = 16th day is 200 man-days.
Work done in last 5 days = 15 man-days ⇒ work done at the end of 20 – 5 = 15th day is 195 man-days.
Thus, our work will get done after the 15th day i.e. on the 16th day. (On 16th day 5 men will be working, but we need to do only 1 man-day of work, so only 1/5th of the 16th day will be needed, not the entire day)
note. Σn is n(n – 1)/2. And this has to be near 196 i.e. n(n–1) has to be near 2×196; further n and (n–1) are consecutive values, so we are searching for a perfect square near 392.
