• # question_answer 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 is needed to be added. Part of red pigment 1 4 7 12 20 Part of base    8 --- --- --- ---

As the part of red pigment increases/ part of base also increases in the same ratio. It is a case of direct proportion, we make use of the relation of the type $\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}=\frac{{{x}_{3}}}{{{y}_{3}}}$ ...
 (a) Here, ${{x}_{1}}=1,{{y}_{1}}=8$ and ${{x}_{2}}=4$ Therefore,               $\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}$ $\frac{1}{8}=\frac{4}{{{y}_{2}}}$ $\Rightarrow \,$ ${{y}_{2}}=4\times 8=32$ (b)   Here,          ${{x}_{2}}\,4,{{y}_{2}}=32\,\,and\,\,{{x}_{3}}=7$ $\frac{{{x}_{2}}}{{{y}_{2}}}=\frac{{{x}_{3}}}{{{y}_{3}}}$ $\Rightarrow$               $\frac{4}{32}=\frac{7}{43}$ $\Rightarrow$               ${{y}_{3}}=\frac{7\times 32}{4}=56$ (c)   Here, ${{x}_{3}}=7,{{y}_{3}}=56$and ${{x}_{4}}=12,{{y}_{4}}=?$ $\frac{{{x}_{3}}}{{{y}_{3}}}=\frac{{{x}_{4}}}{{{y}_{4}}}$ $\Rightarrow$   $\frac{7}{56}=\frac{12}{{{y}_{4}}}$ $\Rightarrow$   ${{y}_{4}}=\frac{12\times 56}{7}$ $\Rightarrow$   ${{y}_{4}}=96$ (d) Here, ${{x}_{4}}=12,{{y}_{4}}=96$and ${{x}_{5}}=20,{{y}_{5}}=?$ $\frac{12}{96}=\frac{202}{{{y}_{5}}}$ $\Rightarrow$   ${{y}_{5}}=\frac{20\times 96}{12}$ $\Rightarrow$   ${{y}_{5}}=160$