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 \[\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}=\frac{{{x}_{3}}}{{{y}_{3}}}\] ...
The table is (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 \[{{x}_{5}}=20,{{y}_{5}}=?\] \[\frac{12}{96}=\frac{202}{{{y}_{5}}}\] \[\Rightarrow \] \[{{y}_{5}}=\frac{20\times 96}{12}\] \[\Rightarrow \] \[{{y}_{5}}=160\]
Part of pigment 1 4 7 12 20 Part of Base 8 32 56 96 160
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