Question 03.
In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?

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Solution

Let the parts of red pigment mix with 1800 mL base be x
$\begin{matrix} P a r t s o f r e d p i g m e n t & 1 & x P a r t s o f b a s e & 75 & 1800 \end{matrix}$

Since it is in direct proportion
$∴ \frac{1}{75} = \frac{x}{1800} \Rightarrow 75 \times x = 1 \times 1800 \Rightarrow x = \frac{1 \times 1800}{75} = 24 p a r t s$
Hence, with base 1800 mL, 24 parts red pigment should be mixed.

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In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?

Question 03.
In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?

A mixture of paint is prepared by mixing 1 part of red pigment with 8 parts of base. In the following table, find the parts of base that need to be added.
If 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?

$\begin{matrix} P a r t s o f r e d p i g m e n t & 1 & 4 & 7 & 12 & 20 P a r t s o f b a s e & 8 \end{matrix}$ [3 MARKS]