Q2. 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.

D Divya Prakash Singh

Suppose the parts of red pigment is x and parts of base, is 'y'.

As the requirement of the number of parts of base for 1 part of red pigment is 8,

and as the parts of red pigment is increases, parts of base also increases in the same ratio. It is a case of direct proportion.

we can assume other parts of the base that will be required for red pigments as y1, y2, y3 and y4 for parts of pigment 4, 7, 12 and 20 respectively.

We make use of the relation of type   $\frac{1}{8}=\frac{4}{y1}$

That gives for the (i) case  $y1\times 1=8\times 4$ or  $y1=32$ .

32 parts of the base will be required for the 4 parts of the red pigment.

(ii) If parts of red pigment used is 7 then parts of base used will be $\frac{1}{8}=\frac{7}{y2}$

that gives $y2\times 1=8\times 7$ or    $y2= 56$.

56 parts of the base will be required for the 7 parts of the red pigment.

(iii) for 12 parts of red pigment : $\frac{1}{8}=\frac{12}{y3}$

we have   $y3\times 1 = 8\times 12$  or   $y3=96$.

so, 96 parts of the base will be required for the 12 parts of the red pigment.

(iv) for 20 parts of red pigment: $\frac{1}{8}=\frac{20}{y4}$

we have  $y4\times 1 = 8\times 20$   or  $y4 = 160$.

Hence for the following parts of red pigment, parts of base are given :

 Parts of red pigment 1 4 7 12 20 Parts of base 8 32 56 96 160
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