A mixture of paint is prepared by mixing $1$ part of red pigments with $8$ parts of the base. In the following table, find the parts of the base that need to be added.
$\begin{matrix} Parts of red pigment & 1 & 4 & 7 & 12 & 20 Parts of base & 8 & . . . & . . . & . . . & . . . \end{matrix}$

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Solution

The given mixture of paint is prepared by mixing $1$ part of red pigments with $8$ parts of base. For more parts of red pigments, the parts of the base will also be more. Therefore, the parts of red pigments and the parts of the base are in direct proportion. The given information in the form of a table is as follows.
$\begin{matrix} Parts of red pigment & 1 & 4 & 7 & 12 & 20 Parts of base & 8 & x_{1} & x_{2} & x_{3} & x_{4} \end{matrix}$
According to direct proportion, we have
$\frac{x_{1}}{4} = \frac{8}{1} \Rightarrow x_{1} = 4 \times 8 = 32$
$\frac{x_{2}}{7} = \frac{8}{1} \Rightarrow x_{2} = 7 \times 8 = 56$
$\frac{x_{3}}{12} = \frac{8}{1} \Rightarrow x_{3} = 8 \times 12 = 96$
$\frac{x_{4}}{20} = \frac{8}{1} \Rightarrow x_{4} = 8 \times 20 = 160$
The table can be drawn as follows.
$\begin{matrix} Parts of red pigment & 1 & 4 & 7 & 12 & 20 Parts of base & 8 & 32 & 56 & 96 & 160 \end{matrix}$

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