Question

Question 02.
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.
$\begin{array}{cccccc}Partsofredpigment& 1& 4& 7& 12& 20\\ Partsofbase& 8& \dots & \dots & \dots & \dots \end{array}$

Answer 1

Let the ratio of parts of red pigment and parts of base be
$\frac{a}{b}$.
$\begin{array}{cccc}Here& a_1=1,b_1=8& \Rightarrow & \frac{a_1}{b_1}=\frac{1}{8}=k\left(say\right)\\ When& a_2=4,b_2=?& & \\ & k=\frac{a_2}{b_2}& \Rightarrow & b_2=\frac{a_2}{k}=\frac{4}{\frac{1}{8}}=4\times 8=32\\ When& a_3=7,b_3=?& & \\ & k=\frac{a_3}{b_3}& \Rightarrow & b_3=\frac{a_3}{k}=\frac{7}{\frac{1}{8}}=7\times 8=56\\ When& a_4=12,b_4=?& & \\ & k=\frac{a_4}{b_4}& \Rightarrow & b_4=\frac{a_4}{k}=\frac{12}{\frac{1}{8}}=12\times 8=96\\ When& a_5=20,b_5=?& & \\ & k=\frac{a_5}{b_5}& \Rightarrow & b_5=\frac{a_5}{k}=\frac{20}{\frac{1}{8}}=20\times 8=160\end{array}$

$\begin{array}{cccccc}Partsofredpigment& 1& 4& 7& 12& 20\\ Partsofbase& 8& 32& 56& 96& 160\end{array}$