Given that, the strength of a salt solution is $p \%$ if $100\;\text{ml}$ of the solution contains $p$ grams of salt.
Each of three vessels $A,B,C$ contains $500\;\text{ml}$ of salt solution of strengths $ 10 \%, 22\%$ and $33\%$ respectively. So the initial amount of salt in each vessel.
- $ A = 500 \times \frac{10}{100} = 50\;\text{ml} $
- $ B = 500 \times \frac{22}{100} = 110\;\text{ml}$
- $ C = 500 \times \frac{32}{100} = 160\;\text{ml}$
Now, the $100\;\text{ml}$ of the solution is transferred from vessel $A$ to vessels $B.$
- Then, salt in vessel $B = 10 \%\;\text{of}\;100 + 110 = \frac{10}{100} \times 100 + 110 = 120$
- Now, the total salt solution in vessel $B = 500 + 100 = 600\;\text{ml}$
- So, strength of salt solution in vessel $B = \frac{120}{600} \times 100\% = 20\%$
- And, the total salt solution in vessel $A = 500 - 100 = 400\;\text{ml}$
Now, $100\;\text{ml}$ of the solution is transferred from vessel $B$ to vessels $C.$
- Then, salt in vessel $C = 20 \%\;\text{of}\;100 + 160 = \frac{20}{100} \times 100 + 160 = 180$
- Now, the total salt solution in vessel $C = 500 + 100 = 600\;\text{ml}$
- So, strength of salt solution in vessel $C = \frac{180}{600} \times 100\% = 30\%$
- And, the total salt solution in vessel $B = 500 - 100 = 400\;\text{ml}$
Now, $100\;\text{ml}$ of the solution is transferred from vessel $C$ to vessels $A.$
- Then, salt in vessel $A = 30 \%\;\text{of}\;100 + 10\% \;\text{of}\; 400 = 30 + 40 = 70$
- Now, the total salt solution in vessel $A = 400 + 100 = 500\;\text{ml}$
- So, strength of salt solution in vessel $A = \frac{70}{500} \times 100\% = 14\%.$
Correct Answer $:\text B$