# Ex.13.1 Q3 Direct and Inverse Proportions Solution - NCERT Maths Class 8

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## Question

In question ($$2$$), above if $$1$$ part of red pigment requires $$75\,\rm{ml}$$ of base, how much red pigment should we mix with $$1800\,\rm{ml}$$ of base?

Video Solution
Direct And Inverse Proportions
Ex 13.1 | Question 3

## Text Solution

What is Known?

$$1$$ part of red pigment requires $$75\,\rm{ml}$$ of base.

What is Unknown?

$$1800$$ mL of base needed how much red pigment?

Reasoning:

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

\begin{align}\frac{x}{y} = k, \qquad x = y\,k\end{align}

Where $$k$$ is a constant.

Steps:

Let the number of parts of red pigment be $$x$$

As the number of parts of red pigment increases, amount of base also increases in the same ratio. So it is a case of direct proportion.

Here,

\begin{align} \frac{{{x}_{1}}}{{{y}_{1}}}&=\frac{{{x}_{2}}}{{{y}_{2}}} \\ {{x}_{1}}&=1,\qquad {{x}_{2}}=? \\ {{y}_{1}}&=8,\qquad{{y}_{2}}=1800 \\ \frac{1}{75}&=\frac{{{x}_{2}}}{1800} \\ {{x}_{2}}&=\frac{1\ \times \ 1800}{75} \\ {{x}_{2}}&=24 \end{align}

$$24$$ parts of red pigment should be mixed with $$1800\,\rm{ml}$$ of base.

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