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

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Question

Q2. 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 ... ... ... ...

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.