We have seen how to calculate quantities that are dependent on a single variable. Today we will see how to calculate quantities that are dependent on multiple variables.
If you have understood the concept of dependency well, this is pretty simple. When your dependent variable is dependent on multiple independent variables, then for each one of the independent variable you need to decide whether the dependent variable is directly proportional or inversely proportional to the independent variable.
After, this the steps involve calculating the value of the independent variable for a value of $1$ for all the independent variables. Then calculate the value of the dependent variable for the given values of the independent variables.
Now we get onto an example.
If $5$ farmers take $12$ days to cultivate a farm of size $3000\ m^2$. How many days it would take for $8$ farmers to cultivate a farm of $4000\ m^2$
Let's first identify the variables and dependencies.
Dependent variable - Number of days.
Independent variable 1 - Number of farmers, inversely proportional, more the number of farmers, lesser will be the number of days.
Independent variable 2 - Area of farm, directly proportional, more the area more is the number of days to cultivate.
Now we are ready to write down our solution. As before, we write the independent variables first, the dependent variable last to the right.
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To cultivate $3000\ m^2 $ field $5$ farmers take $12$ days.
To cultivate $1\ m^2$ field $5$ farmers will take $\dfrac{10}{3000}$ days.
To cultivate $1\ m^2$ field $1$ farmer will take $\dfrac{12 \times 5}{3000}$ days.
To cultivate $4000\ m^2$ field $1$ farmer will take $\dfrac{12 \times 5 \times 4000}{3000}$ days
To cultivate $4000\ m^2$ field $8$ farmers will take $\dfrac{12 \times 5 \times 4000}{3000 \times 8}$ days = $30$ days.
Now let us take a look at another type of problem. In this case the independent variable is modified more than once in the problem.
For example:
Tap $A$ can fill up a tank in $3$ hours. Tap $B$ can empty the tank in $6$ hours. With the tank completely empty, tap $A$ is opened at $9:00\ am$. Tap $B$ is opened at $10:00\ am$. At what time will the tank become full?
Let us see what will happen. The tank will fill up partially till $10:00\ am$, after which it will continue to fill up, but at a different rate.
For the first one hour only tap $A$ is open.
In $3$ hours tap $A$ can fill up the whole tank.
In $1$ hour tap $A$ can fill up $\dfrac{1}{3}$ of the tank.
$\texttip{\therefore}{therefore}$ At $10:00\ am$ $\dfrac{1}{3}$ of the tank will be full and $\dfrac{2}{3}$ of the tank will be empty.
Let us see what happens after $10:00\ am$ and for that we need to find out what happens when both the taps are open.
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Tap $A$ can fill the tank in $3$ hours.
Every hour tap $A$ can fill $\dfrac{1}{3}$ of the tank
Tap $B$ can empty the tank in $6$ hours.
Every hour tap $B$ can empty $\dfrac{1}{6}$ of the tank.
With both taps open,
In $1$ hour $\dfrac{1}{3} - \dfrac{1}{6} = \dfrac{2 - 1}{6} = \dfrac{1}{6}$ of the tank will fill up.
Now we can use simple unitary method.
To fill up $\dfrac{1}{6}$ of the tank it takes $1$ hour.
To fill up $1$ (whole) of the tank it takes $1 \times 6 = 6$ hours
To fill up $\dfrac{2}{3}$ of the tank it will take $6 \times \dfrac{2}{3} = 4$ hours.
$\texttip{\therefore}{therefore}$ The tank will become full at $10:00\ am + 4\ hours = 2:00\ pm$
Tip:
Do not calculate the values at the first step, first complete the final result expression, then do your calculations. You will observe that many of the values will cancel out and it will make your calculations much simpler and faster.