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Let us consider an example to learn how to apply the three methods and find the solution.

A farmer needs $$30 \ kgs$$ of food to feed $$5$$ cows for $$2$$ weeks. Find how many kilograms of food to feed $$10$$ cows for $$5$$ weeks?

Solution:

To determine the type of variation, let us tabulate the data.

 $$Kgs$$ Cows Weeks $$30$$ $$5$$ $$2$$ $$x$$ $$10$$ $$5$$

Compare the unknown value $$x$$ with the known values(Cows and Weeks).

Method 1: Proportion method

Step $$1$$: Compare $$Kgs$$ with Cows. If the number of cows increases then, the quantity of food also increases. So, it is a direct proportion.

Therefore, the proportion is $$30 : x :: 5 : 10$$ ---- ($$1$$)

Step $$2$$: Compare $$Kgs$$ with Weeks. If the number of weeks increases, then the quantity of food also increases. So, it is a direct proportion.

Therefore, the proportion is $$30 : x :: 2 : 5$$ ---- ($$2$$)

Hence, the given problem is in Direct - Direct variation.

Step $$3$$: From equations ($$1$$) and ($$2$$), we have:

$$30:x::\begin{cases}5:10 \\2:5 & \end{cases}$$
We know that The product of the extremes$$=$$ The product of the means
 Extremes Means $$30 : 10$$ $$x : 5$$ $$5$$ $$2$$

$$30 \times 10 \times 5 = x \times 5 \times 2$$

$$\frac{30 \times 10 \times 5}{5 \times 2} = x$$

$$\frac{1500}{10} = x$$

$$150 = x$$

Therefore, the farmer needs $$150 \ kgs$$ of food to feed $$10$$ cows for $$5$$ weeks.

Let us solve the same problem using the multiplicative factor method.

Method 2: Multiplicative factor method

 $$Kgs$$ Cows Weeks $$30$$ $$5$$ $$2$$ $$x$$ $$10$$ $$5$$

Step $$1$$: Compare $$Kgs$$ with Cows. If the number of cows increases then, the quantity of food also increases. So, it is a direct proportion.

Therefore, the multiplying factor is $$\frac{10}{5}$$ ---- ($$1$$)

Step $$2$$: Compare $$Kgs$$ with Weeks. If the number of weeks increases, then the quantity of food also increases. So, it is a direct proportion.

Therefore, the multiplying factor is $$\frac{5}{2}$$ ---- ($$2$$)

Hence, the given problem is in Direct- Direct variation.

Step $$3$$: $$x = 30 \times \frac{10}{5} \times \frac{5}{2}$$

$$x = 150$$

Therefore, the farmer needs $$150 \ kgs$$ of food to feed $$10$$ cows for $$5$$ weeks.

Similarly, we shall solve the problem using the formula method.

Method 3: Formula method

Here, $$P_1 = 5$$, $$D_1 = 2$$, $$W_1 = 30$$

$$P_2 = 10$$, $$D_2 = 5$$, $$W_2 = x$$

Substituting these values in the formula, $$\frac{P_1 \times D_1}{W_1}=\frac{P_2 \times D_2}{W_2}$$, we get:

$$\frac{5 \times 2}{30} = \frac{10 \times 5}{x}$$

$$\frac{10}{30} = \frac{50}{x}$$

$$x = \frac{30 \times 50}{10}$$

$$x = 150$$

Therefore, the farmer needs $$150 \ kgs$$ of food to feed $$10$$ cows for $$5$$ weeks.

Important!
When comparing any of the two quantities and if we get inverse proportion, write the proportion in the reciprocal form(proportion method).

If comparing $$kgs$$ with Weeks, we get inverse variation, and then, we shall write the proportion as: $$30 : x :: 5 : 2$$

Similarly, when comparing any of the two quantities and if we get inverse proportion, there will be no change in the proportion(multiplying factor method).

If comparing $$kgs$$ with Weeks, we get inverse variation, and then the multiplying factor is $$\frac{2}{5}$$.