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