question_answer 1)

                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.                

Parts of red pigment	1	4	7	12	20
Parts of base	8	?	?	?	?

 

Answer:

                Let the number of parts of red pigment be \[x\] and the number of parts of the base be \[y\]. As the number of parts of red pigment increases, number of parts of the base also increases in the same ratio. It is a case of direct proportion. We make use of the relation of the type \[\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}\] (i) Here, \[{{x}_{1}}=1\] \[{{y}_{1}}=8\] and \[{{x}_{2}}=4\] Therefore,                 \[\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}\] gives \[\frac{1}{8}\,=\frac{4}{{{y}_{2}}}\] \[\Rightarrow \]               \[{{y}_{2}}=8\times 4\] \[\Rightarrow \]               \[{{y}_{2}}=32\] Hence, 32 parts of the base are needed to be added. (ii) Here                 \[{{x}_{1}}=1\]                 \[{{y}_{1}}=8\] and \[{{x}_{3}}=7\] Therefore, \[\frac{{{x}_{1}}}{{{y}_{1}}}\,=\frac{{{x}_{3}}}{{{y}_{3}}}\] gives \[\frac{1}{8}\,=\frac{7}{{{y}_{3}}}\] \[\Rightarrow \]               \[{{y}_{3}}=8\times 7\] \[\Rightarrow \]               \[{{y}_{3}}=56\] Hence, 56 parts of the base are needed to be added. (iii) Here                 \[{{x}_{1}}=1\] \[{{y}_{1}}=8\] and        \[{{x}_{4}}=12\] Therefore, \[\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{4}}}{{{y}_{4}}}\] gives \[\frac{1}{8}=\frac{12}{{{y}_{4}}}\] \[\Rightarrow \]               \[{{y}_{4}}=12\times 8\] \[\Rightarrow \]               \[{{y}_{4}}=96\] Hence, 96 parts of the base are needed to be added. (iv) Here \[{{x}_{1}}=1\] \[{{y}_{1}}=8\] and        \[{{x}_{5}}=20\] Therefore, \[\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{5}}}{{{y}_{5}}}\] gives \[\frac{1}{8}=\frac{20}{{{y}_{5}}}\] \[\Rightarrow \]               \[{{y}_{5}}=8\times 20\] \[\Rightarrow \]               \[{{y}_{5}}=160\] Hence, 160 parts of the base are needed to be added.