• # question_answer 3)                 In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?

Let the number of parts of red pigment be $x$and the amount of base be $y\,mL$. As the number of parts of red pigment increases, amount of base also increases in the same ratio. It is a case of direct proportion. We make use of the relation of the type. $\frac{{{x}_{1}}}{{{y}_{1}}}\,=\frac{{{x}_{2}}}{{{y}_{2}}}$ Here      ${{x}_{1}}=1$ ${{y}_{1}}=75$ and ${{y}_{2}}=1800$ Therefore, $\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}$ gives $\frac{1}{75}=\frac{{{x}_{2}}}{1800}$ $\Rightarrow$               $75\,{{x}_{2}}=1800$ $\Rightarrow$               ${{x}_{2}}=\frac{1800}{75}$ $\Rightarrow$               ${{x}_{2}}=24$ Hence, 24 parts of the red pigment should be mixed.

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