Ex 11.1, 2 - Chapter 11 Class 8 Direct and Inverse Proportions

Last updated at April 16, 2024 by Teachoo

Transcript

Ex 11.1, 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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