question_answer

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 is needed to be added.

Part of red pigment	1	4	7	12	20
Part of base	8	---	---	---	---

Answer:

As the part of red pigment increases/ part of base also increases in the same ratio. It is a case of direct proportion, we make use of the relation of the type \[\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}=\frac{{{x}_{3}}}{{{y}_{3}}}\] ...

(a) Here, \[{{x}_{1}}=1,{{y}_{1}}=8\] and \[{{x}_{2}}=4\]

Therefore,               \[\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}\]

\[\frac{1}{8}=\frac{4}{{{y}_{2}}}\]

\[\Rightarrow \,\] \[{{y}_{2}}=4\times 8=32\]

(b)   Here,          \[{{x}_{2}}\,4,{{y}_{2}}=32\,\,and\,\,{{x}_{3}}=7\]

\[\frac{{{x}_{2}}}{{{y}_{2}}}=\frac{{{x}_{3}}}{{{y}_{3}}}\]

\[\Rightarrow \]               \[\frac{4}{32}=\frac{7}{43}\]

\[\Rightarrow \]               \[{{y}_{3}}=\frac{7\times 32}{4}=56\]

(c)   Here, \[{{x}_{3}}=7,{{y}_{3}}=56\]and \[{{x}_{4}}=12,{{y}_{4}}=?\]

\[\frac{{{x}_{3}}}{{{y}_{3}}}=\frac{{{x}_{4}}}{{{y}_{4}}}\]

\[\Rightarrow \]   \[\frac{7}{56}=\frac{12}{{{y}_{4}}}\]

\[\Rightarrow \]   \[{{y}_{4}}=\frac{12\times 56}{7}\]

\[\Rightarrow \]   \[{{y}_{4}}=96\]

(d) Here, \[{{x}_{4}}=12,{{y}_{4}}=96\]and \[{{x}_{5}}=20,{{y}_{5}}=?\]

\[\frac{12}{96}=\frac{202}{{{y}_{5}}}\]

\[\Rightarrow \]   \[{{y}_{5}}=\frac{20\times 96}{12}\]

\[\Rightarrow \]   \[{{y}_{5}}=160\]

The table is

Part of pigment	1	4	7	12	20
Part of Base	8	32	56	96	160