Direct proportions and unitary method — lesson. Mathematics CBSE, Class 8.

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Direct proportion detail analysis:

Example:

If the cost of a dress is \(₹\)800, then the price of 1 dress will be \(₹\)800. The price of the dress increases as the number of dresses increases. Proceeding the same way we can find the cost of any number of such dresses.

Consider the above situation, when two quantities, namely the number of dresses and their prices are related to each other. When the number of dresses increases, the price also increases in such a way that their ratio remains constant.

Let us denote the number of the dress as \(X\) and the price of the dress as \(Y\) rupees. Now observe the following table,

Number of dress \(X\)	1	2	4	6	8	10
Price of the dress in \(₹ Y\)	1000	2000	4000	6000	8000	10000

From the table, we can observe that when the values of \(X\) increase the corresponding values of \(₹Y\) also increases in such way that the ratio of

\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 value from the table.

\frac{X}{Y} = \frac{1}{1000} = \frac{2}{2000} = \frac{4}{4000} = \frac{6}{6000} = \frac{8}{8000} = \frac{10}{10000}

and so on.

All the ratios are equivalent, and its simplified form is

\frac{1}{1000}

In general way

\frac{X}{Y}

\(=\)

\frac{1}{1000}

\(= k\) (constant).

When \(X\) and \(Y\) are in direct proportion, we get

\frac{X}{Y}

\(= k\) or

X = kY

Important!

If any two ratios are given above, we should take

X_{1}, X_{2} and Y_{1}, Y_{2}

Their ratio will be

\frac{X_{1}}{Y_{1}} = \frac{X_{2}}{Y_{2}}

[Where, \(Y1\), \(Y2\) are values of \(Y\) corresponding to the values \(X1\), \(X2\) of \(X\)].

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

That is

Number of dress \(X\)	\(X1\)	\(X2\)
Price of the dress in \(₹ Y\)	\(Y1\)	\(Y2\)

Unitary Method:

This is one of the methods 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 what 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 \cdot 10

\(= ₹\) 250.

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3. Direct proportions and unitary method

Theory: