Ex 13.1, 3 - Chapter 13 Class 8 Direct and Inverse Proportions

Transcript

Ex 13.1, 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.