A mixture of paint is prepared by mixing $1$ part of red pigment with $8$ parts of base. In the following table, find the parts of base that need to be added for the respective parts of red pigment.

\begin{tabular}{|c|c|c|c|c|c|} \hline Parts of red pigment & 1 & 4 & 7 & 12 & 20 \\ \hline Parts of base & 8 & & & & \end{tabular}

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Solution

Let the ratio of the parts of red pigment and the parts of base be $\frac{a}{b}$ .
So, according to direct proportion,
$\frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} = \frac{a_{3}}{b_{3}} = \frac{a_{4}}{b_{4}} = \frac{a_{5}}{b_{5}}$
Let $\frac{a_{1}}{b_{1}} = \frac{1}{8}$ , $a_{2}$ = 4 and $b_{2}$ be its respective parts of base and so on.
$∴$ $\frac{a_{2}}{b_{2}} = \frac{a_{3}}{b_{3}} = \frac{a_{4}}{b_{4}} = \frac{a_{5}}{b_{5}} = \frac{1}{8}$
$\frac{a_{2}}{b_{2}} = \frac{4}{b_{2}} = \frac{1}{8}$
$\Rightarrow b_{2} = 32$

$\frac{a_{3}}{b_{3}} = \frac{7}{b_{3}} = \frac{1}{8}$
$\Rightarrow b_{3} = 56$

$\frac{a_{4}}{b_{4}} = \frac{12}{b_{4}} = \frac{1}{8}$
$\Rightarrow b_{4} = 96$

$\frac{a_{5}}{b_{5}} = \frac{20}{b_{5}} = \frac{1}{8}$
$\Rightarrow b_{5} = 160$

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