Ex 11.1, 3 - Chapter 11 Class 8 Direct and Inverse Proportions

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Ex 11.1, 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?Given, 1 part of red pigment requires 75 ml of base Let x parts of red pigment require 1800 mL of base. Thus, our table looks like Now, as parts of red pigment increases, the quantity of base also increases. ∴ Parts of red pigment and quantity of base are in direct proportion 𝟏/𝟕𝟓 = 𝒙/𝟏𝟖𝟎𝟎 1/75 × 1800 = 𝑥 24 = 𝑥 𝑥 = 24 ∴ 24 parts of red pigment should be mixed with 1800 mL of base.