• # question_answer 2)                 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 need to be added.                 Parts of red pigment 1 4 7 12 20 Parts of base 8 ? ? ? ?

Let the number of parts of red pigment be $x$ and the number of parts of the base be $y$. As the number of parts of red pigment increases, number of parts of the base also increases in the same ratio. It is a case of direct proportion. We make use of the relation of the type $\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}$ (i) Here, ${{x}_{1}}=1$ ${{y}_{1}}=8$ and ${{x}_{2}}=4$ Therefore,                 $\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}$ gives $\frac{1}{8}\,=\frac{4}{{{y}_{2}}}$ $\Rightarrow$               ${{y}_{2}}=8\times 4$ $\Rightarrow$               ${{y}_{2}}=32$ Hence, 32 parts of the base are needed to be added. (ii) Here                 ${{x}_{1}}=1$                 ${{y}_{1}}=8$ and ${{x}_{3}}=7$ Therefore, $\frac{{{x}_{1}}}{{{y}_{1}}}\,=\frac{{{x}_{3}}}{{{y}_{3}}}$ gives $\frac{1}{8}\,=\frac{7}{{{y}_{3}}}$ $\Rightarrow$               ${{y}_{3}}=8\times 7$ $\Rightarrow$               ${{y}_{3}}=56$ Hence, 56 parts of the base are needed to be added. (iii) Here                 ${{x}_{1}}=1$ ${{y}_{1}}=8$ and        ${{x}_{4}}=12$ Therefore, $\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{4}}}{{{y}_{4}}}$ gives $\frac{1}{8}=\frac{12}{{{y}_{4}}}$ $\Rightarrow$               ${{y}_{4}}=12\times 8$ $\Rightarrow$               ${{y}_{4}}=96$ Hence, 96 parts of the base are needed to be added. (iv) Here ${{x}_{1}}=1$ ${{y}_{1}}=8$ and        ${{x}_{5}}=20$ Therefore, $\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{5}}}{{{y}_{5}}}$ gives $\frac{1}{8}=\frac{20}{{{y}_{5}}}$ $\Rightarrow$               ${{y}_{5}}=8\times 20$ $\Rightarrow$               ${{y}_{5}}=160$ Hence, 160 parts of the base are needed to be added.