Two solution of 90% and 97% purity and mixed resulting in 21 liters of mixture of 94% purity. How much is the quantity of the first solution in the resulting mixture?

Method 1 to solve the equation.
Let us assume the number of liters of the 90% purity solution = A
and the number of liters of the 97% purity solution = B.
According to question,
Since there are 21 liters of the solution,
A + B = 21 ...................... (1)
Since after mixing the two solutions the new mixture has 94% purity,
Concentrate of A + Concentrate of B = Concentrate of (A + B)
A x 90% + B x 94% = (A+ B) x 97%
⇒ A x 90/100 + B x 97/100 = (A + B) x 94/100
⇒ 90A + 97B = (A + B) x 94
⇒ 90A + 97B = 94A + 94B
⇒ 94A + 94B - 90A- 97B = 0
⇒ 4A - 3B = 0 ........................(2)
Multiply the 3 with Equation (1) and add with Equation (2),
3A + 3B + 4A - 3B = 63 + 0
⇒ 7A = 63
⇒ A = 63/7 = 9
Put the value of A in Equation (1) , we will get
9 + B = 21
B = 21 - 9
B = 12
The first solution would be A = 9 liters.
Method 2 to solve the equation.
Hit and trail method.
94% is closer to 97% but barely meaning the mixtures will not be equal parts but will be slightly more of the higher purity. Quickly eliminate A and B. Out of the others 9 is the easy choice. If the other choices were closer to half this wouldn't work.