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.
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