Ex 13.1, 2 - A mixture of paint is prepared by mixing 1 part of red

Ex 13.1, 2 - Chapter 13 Class 8 Direct and Inverse Proportions - Part 2
Ex 13.1, 2 - Chapter 13 Class 8 Direct and Inverse Proportions - Part 3 Ex 13.1, 2 - Chapter 13 Class 8 Direct and Inverse Proportions - Part 4 Ex 13.1, 2 - Chapter 13 Class 8 Direct and Inverse Proportions - Part 5 Ex 13.1, 2 - Chapter 13 Class 8 Direct and Inverse Proportions - Part 6

  1. Chapter 13 Class 8 Direct and Inverse Proportions
  2. Concept wise

Transcript

Ex 13.1, 2 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 need to be added. Given, 1 part of red pigment is mixed requires 8 parts of base. Finding parts of base If 4 parts of red pigments are added Let the parts of base to be added = y Since, as parts of red pigment increases, the parts of base to be added also increases. ∴ Parts of red pigment and parts of base are in direct proportion. 1/8 = 4/𝑦 𝑦 × 1 = 4 × 8 𝑦 = 32 ∴ 32 parts of base are added with 4 parts of red pigment. If 7 parts of red pigments are added Let the parts of base to be added = y Since, Parts of red pigment and parts of base are in direct proportion. 1/8 = 7/𝑦 𝑦 × 1 = 7 × 8 𝑦 = 56 ∴ 56 parts of base are added with 7 parts of red pigment. If 12 parts of red pigments are added Let the parts of base to be added = y Since, Parts of red pigment and parts of base are in direct proportion. 1/8 = 12/𝑦 𝑦 × 1 = 12 × 8 𝑦 = 96 ∴ 96 parts of base are added with 12 parts of red pigment. If 20 parts of red pigments are added Let the parts of base to be added = y Since, Parts of red pigment and parts of base are in direct proportion. 1/8 = 20/𝑦 𝑦 × 1 = 20 × 8 𝑦 = 160 ∴ 160 parts of base are added with 20 parts of red pigment. Thus, our table looks like

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.