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

Go back to  'Ex.13.1'

Question

Q3. 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?

 

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.

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

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