There are 3 beakers each of which contains saline solution.
Beaker A initially contains 3 liters of 10\% salt solution.
Beaker B initially contains 2 liters of 20\% salt solution.
Beaker C initially contains 4 liters of 0\% salt solution.

Two liters are transferred from A to B and the result is thoroughly mixed. Then one liter is transferred from B to C and the result mixed. Finally two liters are transferred from C back to A. What is the percentage concetration of salt in A after all this?

Percentage concentration of salt in A=____ \%

Assume all the concentrations are expressed in volumetric terms, i.e 10% = 10 liter salt / 100 liter solution, 20% = 20 liter salt / 100 liter soluton, 0% = 0 liters salt.

1) First transformation: 2 liters are from A to B

Solution A:

Concentration: 10 % salt ⇒ 10 liter salt / 100 liter solution

Volume of solution: 3 liters

Volume of salt: 3 liters × 10 liter / 100 liters = 0.300 liter salt

Volume of water: 3 liters - 0.300 liters = 2.700 liters solvent

Solution B:

Concentraion: 20%

Volume of solution: 2 liter

Volume of salt: 2 liter × 0.20 = 0.4 liter salt

Volume of water: 2 liter - 0.4 liter = 1.6 liter water

Resultant mixture in beaker B: 2 liters of solution A plus 2 liters of solution B

Salt: 2 liter × 0.10 + 2 liter × 0.20 = 0.20 + 0.4 = 0.6 liter salt

Water: 2 liter × 0.90 + 2 liter × 0.80 = 3.40 liter water

Solution: 2 liter + 2 liter = 4 liter solution

Concentration: 0.6 liter salt / 4 liter solution = 0.15 = 15%

2) Second transformation: 1 liter transferred from B to C

Salt: 1 liter × 0.15 + 0 = 0.15 liter salt

↑ ↑

(from B) ( in C)

Solution: 1 liter + 4 liter = 5 liter solution

Concentration: 0.15 liter salt / 5 liter solution = 0.03 = 3.0%

3) Third transformation: 2 liters are from C to A.

Salt: 0.03 liter salt × 2 liter solution + 1 liter × 0.10 = 0.16 liter salt

↑ ↑

(from C) (in A)

Solution: 2 liter + 1 liter = 3 liter solution

% of salt in A = (0.16 liter salt / 3 liter solution) × 100 =5.3 %