PUMPA - THE SMART LEARNING APP

AI system creates personalised training plan based on your mistakes

Download now on Google PlayDirect proportion detail analysis:

Example:

If the cost of a watch is \(₹\)600, then the price of 1 watch will be \(₹\)600. The price of the watch increases as the number of watches increases. Proceeding the same way, we can find the cost of any number of such watches.

Consider the above situation, when two quantities, namely the

**number of**watch**es**and their**prices, are related**to each other. When the number of watches**increases**, the price also**increases**in such a way that their**ratio remains constant.**Let us denote the number of the watch as \(X\) and the

**price**of the watch as \(Y\)**rupees**. Now observe the following table.Number of watch \(X\) | \(1\) | \(2\) | \(4\) | \(6\) | \(8\) | \(10\) |

Price of the watch in \(₹Y\) | 200 | 400 | 800 | 1200 | 1600 | 2000 |

From the table, we can observe that when the values of \(X\) increase, the corresponding values of \(₹Y\) also increase in such a way that the ratio $\frac{X}{Y}$ in each case has the same value, which is a constant (say \(k\)).

Now let us find the ratio for each of the values from the table.

$\frac{X}{Y}=\frac{1}{200}=\frac{2}{400}=\frac{4}{800}=\frac{6}{1200}=\frac{8}{1600}=\frac{10}{2000}$ and so on.

All the ratios are equivalent, and their simplified form is $\frac{1}{200}$.

In a general way, $\frac{X}{Y}$ \(=\) $\frac{1}{200}$ \(= k\) (constant).

When \(X\) and \(Y\) are in direct proportion, we get $\frac{X}{Y}$ \(= k\) or $X=\mathit{kY}$.

Important!

If any two ratios are given above, we should take them ${X}_{1},\phantom{\rule{0.147em}{0ex}}{X}_{2}\phantom{\rule{0.147em}{0ex}}\mathit{and}\phantom{\rule{0.147em}{0ex}}{Y}_{1},{Y}_{2}$.

Their ratio will be $\frac{{X}_{1}}{{Y}_{1}}=\frac{{X}_{2}}{{Y}_{2}}$.

[Where \(Y1\) and \(Y2\) are \(Y\) values that correspond to \(X1\) and \(X2\) values of \(X\)].

From the above table, we should take \(X1\) and \(X2\) from the values of \(X\). Similarly, \(Y1\) and \(Y2\) from \(Y\) values.

Number of watch \(X\) | \(X1\) | \(X2\) |

Price of the watch in \(₹ Y\) | \(Y1\) | \(Y2\) |

**Unitary Method**:

- This is one method to find out the values.
- First, the value of one unit will be found. It will be useful to find the value of the required number of units.

Example:

Consider that \(4\) apples cost \(₹100\). Then how will be the cost of \(10\) apples?

To find this first, we have to determine the cost of one apple (price per unit).

Then we can use this single quantity value to find our required quantity.

Therefore, the cost of \(4\) apples \(= ₹100\).

Then the cost of \(1\) apple \(= ₹\)$\frac{100}{4}$ \(= ₹\)25.

That is the cost of \(10\) apples \(= ₹\)$25\xb710$ \(= ₹\) 250.