# Simple way for solving generic work and time problems

I was looking for a general way of formulating solutions for work and time problems.

For example,

30 soldiers can dig 10 trenches of size 8*3*3 ft in half a day working 8 hours per day. How many hours will 20 soldiers take to dig 18 trenches of size 6*2*2 ft working 10 hours per day?

Now i know that Work = Efficiency * Time, but i get confused sometimes in selecting which factor in the given problem will contribute directly to the work i.e. increase it and which factors will result in the work being done faster.

I've seen the method in which one uses a table to write all the parameters given in the problem e.g. making columns titled work, number of soldiers, volume of trench, number of days required, number of hours per day, efficiency, wages etc and uses the direct and inverse proportionality to write the equation for solving a given unknown. However i face the same problem i.e. finding out which factors are directly related and which are in an inverse relation.

Is there a simple way of solving these general problems?

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$30$ soldiers can dig $10$ trenches of size $8\cdot 3\cdot 3$ cube fit in $4$ hours.

$1$ soldier can dig $10$ trenches of size $8\cdot 3\cdot 3$ cube fit in $4\cdot 30$ hours.

$1$ soldier can dig $1$ trench of size $8\cdot 3\cdot 3$ cube fit in $\frac{4\cdot 30}{10}$ hours.

$1$ soldier can dig $1$ trench of size $1\cdot 1\cdot 1$ cube fit in $\frac{4\cdot 30}{10\cdot 8\cdot 3\cdot 3}$ hours.

$20$ soldiers can dig $18$ trenches of size $6\cdot 2\cdot 2$ cube fit in $$\frac{4\cdot 30\cdot 18\cdot 6\cdot 2\cdot 2}{10\cdot 8\cdot 3\cdot 3\cdot 20}=\frac{36}{10}=3.6$$ hours which is clearly $<10$ hours.

The number of trenches and the size of trenches are directly proportional to the time, but the number of soldiers is inversely roportional to the time, the more the number of soldiers, the lesser is the time.

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@BrianM.Scott, thanks for your observation. – lab bhattacharjee Oct 10 '12 at 5:00

Identify the basic assumptions. In this problem it’s clear that soldiers are considered interchangeable: they all work at the same rate. (Recall that in some problems we have different workers working at different rates and have to keep track of the work rate of each worker. And it’s the work rates that are important, because they’re additive: if $A$ and $B$ have work rates of $r$ and $s$ amount of work per unit of time, then $A$ and $B$ working together have a combined work rate of $r+s$.) It’s also clear the amount of work is being measured in cubic feet dug, and that cubic feet are to be considered interchangeable: they’re all equally hard to dig. Finally, the basic unit of time here is the hour, since we’re dealing with working days of two different lengths in hours.

Now convert the initial data to basic units. We have $30$ soldiers doing the work. They dig $10$ trenches of $8\cdot3\cdot3$ cubic feet each, for a total of $720$ cubic feet of earth dug, and they do it in half of an $8$-hour day, or $4$ hours.

At this point we can calculate their combined work rate: $720$ cubic feet in $4$ hours is $\frac{720}4=180$ cubic feet per hour. But that’s the combined work rate of $30$ interchangeable soldiers, so each soldier is digging only $\frac1{30}$-th of that, or $\frac{180}{30}=6$ cubic feet per hour.

Now let’s take a look at the question (which we should have had in the backs of our minds all along as a guide to what’s relevant and what’s likely to be useful). We have $20$ soldiers available. We want to dig $18$ trenches, each $6\cdot2\cdot2$ cubic feet in size, so we want to move $18\cdot6\cdot2\cdot2=432$ cubic feet of earth. Finally, we want to know how long this will take. We know that one soldier digs $6$ cubic feet per hour, so $20$ soldiers dig $20\cdot6=120$ cubic feet per hour. It will therefore take them $\frac{432}{120}=3.6$ hours, considerably less than one $10$-hour working day.

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Thanks for the explanation. – Karan Oct 10 '12 at 14:59