Question

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 the base that need to be added

Parts of red pigment	1	4	7	12	20
Parts of base	8	…	…	…	…

If 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800mL of base?

Answer 1

Hint: To solve, this question we can use the concept of direct proportion as both the values are dependent and increasing, so the ratio of the both quantity must be a constant, second part of this question, we can use the unitary method as for any specific value corresponds to some other value, so that can be change to a unit value corresponds to what value, effectively we all can multiply with the unit value with the number of times according to the question.

Complete step-by-step answer:
Given, a mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base.
Let ${y_1}$, be the amount of base added with 4 part of red pigment.
As, the part of red pigment is increasing, therefore, the base quantity has to also increase.
So, it can be observed that they both are in direct proportion.
As they are in direct proportion, so,
$
  \dfrac{1}{8} = \dfrac{4}{{{y_1}}} \\
  {y_1} = 4 \times 8 \\
  {y_1} = 32 \\
 $
So, 32 mL of base was added when the 4 parts of red pigment was used.
Let ${y_2}$, be the amount of base added with 7 parts of red pigment.
Since, both base and pigment are in direct proportion.
So,
$
  \dfrac{1}{8} = \dfrac{7}{{{y_2}}} \\
  {y_2} = 7 \times 8 \\
  {y_2} = 56 \\
 $
So, 56 mL of base was added when the 7 part of red pigment was used.
Let ${y_3}$, be the amount of base added with 12 part of red pigment.
Since, both base and pigment are in direct proportion.
So,
$
  \dfrac{1}{8} = \dfrac{{12}}{{{y_3}}} \\
  {y_3} = 12 \times 8 \\
  {y_3} = 96 \\
 $
So, 96 mL of base was added when the 12 part of red pigment was used.
Let ${y_4}$, be the amount of base added with 20 parts of red pigment.
Since, both base and pigment are in direct proportion.
So,
$
  \dfrac{1}{8} = \dfrac{{20}}{{{y_4}}} \\
  {y_4} = 20 \times 8 \\
  {y_4} = 160 \\
 $
So, 160 mL of base was added when the 20 part of red pigment was used.
Therefore, the table will be as follows,

Parts of red pigment	1	4	7	12	20
Parts of base	8	32	56	96	160

If 1 part of a red pigment requires 75 mL of base,
Use unitary method,
75 mL of base uses 1 part of a red pigment.
1 mL of base uses $\dfrac{1}{{75}}$ part of a red pigment.
So,
1800mL of base uses $1800 \times \dfrac{1}{{75}}$ part of a red pigment, that is equal to 24 parts.

Note: The first part of the question can be solved with the unitary method, that is for the unit value we can calculate its part and then multiply the unit value according to the data given. The 2nd part of question can also be solved with the help of the concept of direct proportion, It can be observed that when the value of something is increasing or decreasing, simultaneously the associated value also increases or decreases in such a way that the ratio these two quantities does not change; it remains constant. We say that these two quantities are in direct proportion.