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Direct and Inverse Proportions | Exercise 13.1

Question 3

Q2) A mixture of paint is prepared by mixing 1 pan of red pigments with 8 parts of the base. In the following table, find the parts of the base that need to be added.

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Solution:

The given mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of the base. For more parts of red pigments, the parts of the base will also be more.

Therefore, the parts of red pigments and the parts of the base are in direct proportion. The given information in the form of a table is as follows.

According to direct proportion,

\frac{^x1}{4}=\frac{8}{1}\Rightarrow^x1=\ 4\ \times8=32

\frac{^x2}{7}=\frac{8}{1}\Rightarrow^x2=\ 7\times8=56

\frac{^x3}{12}=\frac{8}{1}\Rightarrow^x2=8\times12\ =\ 96

\frac{^x4}{20}=\frac{8}{1}\Rightarrow^x4=8\times20=160

The table can be drawn as follows.

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Our top 5% students will be awarded a special scholarship to Lido.

subject-cta

Question 3

Q2) A mixture of paint is prepared by mixing 1 pan of red pigments with 8 parts of the base. In the following table, find the parts of the base that need to be added.

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Solution:

The given mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of the base. For more parts of red pigments, the parts of the base will also be more.

Therefore, the parts of red pigments and the parts of the base are in direct proportion. The given information in the form of a table is as follows.

According to direct proportion,

\frac{^x1}{4}=\frac{8}{1}\Rightarrow^x1=\ 4\ \times8=32

\frac{^x2}{7}=\frac{8}{1}\Rightarrow^x2=\ 7\times8=56

\frac{^x3}{12}=\frac{8}{1}\Rightarrow^x2=8\times12\ =\ 96

\frac{^x4}{20}=\frac{8}{1}\Rightarrow^x4=8\times20=160

The table can be drawn as follows.

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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