Q. If a man and a half digs a hole and a half in an hour and a half, how long does it take for one man to dig one hole?
A. Scenario # 1:
As stated before, there is no such thing as half a hole (as half a hole is still a hole) and there is no such thing as half a man (as half a man cannot dig). Making these two assumptions, the original problem simplified to:
If a man digs a hole and a half in an hour and a half, how long does it take to dig one hole?
Assuming constant digging rate and through linear interpolation we can state that the answer is 1 hour
Scenario # 2:
Lets assume:
There is such a thing as half a hole; If a hole is of dept X, then half a hole is of dept [X/2].
There is such a thing as half a man. However, this half-man, does not work at the same rate as a full-man (as explained later).
Both half-man and full-man work at a constant rate.
Making these assumptions the original question still stands.
Solution:
Lets take into consideration the various cases as illustrated in figure 1 below:
[Figure 1: Case breakdown] - In accordance to assumption 2
* Operational efficiency = Rate of work where a full-man is used as control. A full-man's operational efficiency is 1.
Case 1 calculations:
As per the figure 1(case 1), 2 men can dig 1.5 hole in 1.5 hour
Hence, 1 man in 1.5 hour digs 0.75 hole (=1.5/2; linear interpolation and in accordance to assumption 3)
Hence, 1 man in 1 hour digs 0.5 hold (=0.75/1.5 linear interpolation and in accordance to assumption 3)
Therefore, 1 man in 2 hours digs 1 hole (linear interpolation and in accordance to assumption 1 and assumption 3)
Case 2 calculations:
As per the figure 1(case 2), 1 men can dig 1.5 hole in 1.5 hour
Hence, 1 man in 1 hour digs 1 hole (same as answer # 1 and in accordance to assumption 3)
Case 3&4 calculations:
As per the figure 1(case 3 and case 4), 1.5 men can dig 1.5 hole in 1.5 hour
Hence, 1 man in 1.5 hour digs 1 hole (=1.5/1.5; linear interpolation)
Case 5 calculations:
Case specific assumption - this half man is half the size as a full man. Hence, he works at half the rate (or half the operational efficiency).
As per the figure 1(case 5), 1.5 men can dig 1.5 hole in 1.5 hour
Hence, 1 man in 1.5 hour digs 1 hole (=1.5/1.5; linear interpolation)
Overall:
Given that all cases are mutually exclusive we need to take an expected average.
Expected average = 0.2(2 hour) + 0.2(1 hour) + 0.4(1.5 hour) + 0.2(1.5 hour) = 1.50 hour
Final Answer:
Given that scenario # 1 and scenario # 2 are mutually exclusive events but both scenarios are equally likely,
Total expected time = Probability of scenario 1*(time of scenario 1) + Probability of scenario 2*(time of scenario 2)
= 0.5(1hour) + 0.5(1.5hour)
= 1.25 hour
Hence, I expect 1 man to dig 1 hole in 1.25 hour