8th Class Mathematics Direct and Inverse Proportions

  • 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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