# Ex.13.1 Q2 Direct and Inverse Proportions Solution - NCERT Maths Class 8

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## Question

A mixture of paint is prepared by mixing $$1$$ part of red pigments with $$8$$ parts of base. In the following table find the parts of base that needed to be added.

 Parts of red pigment 1 4 7 12 20 Parts of base 8 ... ... ... ...

Video Solution
Direct And Inverse Proportions
Ex 13.1 | Question 2

## Text Solution

What is Known?

Parts of red pigments used.

What is Unknown?

Parts of base used.

Reasoning:

Two numbers $$x$$ and $$y$$ are said to vary in direct proportion if

$\frac{x}{y} = k,x = yk$

Where, $$k$$ is a constant.

Steps:

Let the parts of red pigments used be $$x$$ and parts of base used be $$y.$$

\begin{align}\therefore \frac{{{x_1}}}{{{y_1}}} = \frac{{{x_2}}}{{{y_2}}}\end{align}

Here

\begin{align} \quad{x_1} &= 1, \qquad {x_2} = 4 \\ {y_1} &= 8,\qquad {y_2} = ?\\ \frac{{{x_1}}}{{{y_1}}} &= \frac{{{x_2}}}{{{y_2}}} \\ \frac{1}{8} &= \frac{4}{{y2}}\\ \,{y_2} &= 8 \times 4 \\ &= 32{\text{ parts }}\end{align}

$$32$$ parts of base is needed for $$4$$ parts of red pigment.

Here

\begin{align} {{x}_{1}}&=1,\quad \ {{x}_{2}}=7 \\ {{y}_{1}}&=8,\quad{{y}_{2}}=? \\ \frac{{{x}_{1}}}{{{y}_{1}}}&=\frac{{{x}_{2}}}{{{y}_{2}}} \\ \,\,\frac{1}{8}&=\frac{7}{{{y}^{2}}} \\ {{y}_{2}}&=8\times 7 \\ \,\,\,\,\,\,&=56\,\ \text{parts} \end{align}

$$56$$ parts of base is needed for $$7$$ parts of red pigment.

Here

\begin{align} \,\,\,\,\,\,\,\,{{x}_{1}}&=1,\quad{{x}_{2}}=12 \\ {{y}_{1}}&=8,\quad{{y}_{2}}=? \\ \frac{{{x}_{1}}}{{{y}_{1}}}&=\frac{{{x}_{2}}}{{{y}_{2}}} \\ \frac{1}{8}&=\frac{12}{{{y}_{2}}} \\ {{y}_{2}}&=8\times 12\quad \\ &=96 \end{align}

$$96$$ parts of base is needed for $$12$$ parts of red pigment.

Here

\begin{align} {{x}_{1}}&=1,\quad{{x}_{2}}=20 \\ {{y}_{1}}&=8,\quad{{y}_{2}}=? \\ \frac{{{x}_{1}}}{{{y}_{1}}}&=\frac{{{x}_{2}}}{{{y}_{2}}} \\ \frac{1}{8}&=\frac{20}{{{y}_{2}}} \\ {{y}_{2}}&=8\times 20 \\ &=160 \end{align}

$$160$$ parts of base are needed for $$20$$ parts of red pigment.

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