A can contain a mixture of two liquids A and B in the ratio \[7:5\]. When 9 liters of mixture are drawn off and the can is filled with B the ratio of A and B becomes \[7:9\]. How many liters of liquid A was contained by the can initially?
A.10
B.20
C.21
D.25

Hint: We will be using the concepts of allegation and mixtures to solve the problem. We will also be using the traditional approach of solving word problems to make equations and solve them.

Complete step-by-step answer:
We have been given that a can contains a mixture of two liquids A and B in the ratio \[7:5\]. So, we will let the quantity of liquid \[\text{A = 7}x\] and \[\text{B = 5}x\].
Now, we have been given that 9 liters of amount of mixture are drawn off so the amount in which liquid A and B drawn off will be \[7:5\] and therefore the amount of A left and B left liquids are
Liquid A left \[\text{= 7}x-\dfrac{7\times 9}{12}\]
Now since B is filled after remaining 9L
Liquid B left \[\text{= 5}x-\dfrac{5\times 9}{12}+9\]
Now, we have been given that the ratio of liquid A and liquid B after the 9 liters of mixture are drawn off. So, we have the ratios as
\[\dfrac{\text{7}x-\dfrac{7\times 9}{12}}{5x-\dfrac{5\times 9}{12}+9}\ =\ \dfrac{7}{9}\]
\[\dfrac{\text{7}x-\dfrac{21}{4}}{5x-\dfrac{15}{4}+9}\ =\ \dfrac{7}{9}\]
We will cross-multiply and evaluate the expression to get x.
\[63x-\dfrac{21}{4}\times 9\ =\ 35x-\dfrac{15}{4}\times 7+63\]
\[28x\ =\ \dfrac{21}{4}\times 9-\dfrac{15}{4}\times 7+63\]
\[28x\ =\ \dfrac{189}{4}-\dfrac{105}{4}+63\]
\[28x\ =\ \dfrac{189-105}{4}+63\]
\[28x\ =\ \dfrac{84}{4}+63\]
\[28x\ =\ 21+63\]
\[28x\ =\ 84\]
\[x\ =\ \dfrac{84}{28}\]
\[x\ =\ 3\]
Now, since we have x, we can substitute this to find initial value of A. Therefore, initial value of \[\text{A}\ =\ 7\times 3\ =\ 21\ l\]
Hence option C is the correct answer.

Note: To solve this type of question it is important to note that the B has been filled after remaining 9L of mixture. So, it is understood that 9L of B has been added to fill up the container.