UPSKILL MATH PLUS

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Learn moreLet 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 \(\frac{2}{5}\).