CAT 2019 Shift 2 Paper

Initial amount of salt in vessel $A=10\ \text{gms}$ per $100\ \text{ml}$, therefore in $500\ \text{ml}$ the amount of salt $=50\ \text{gms}$ Initial amount of salt in vessel $B=22\ \text{gms}$ per $100\ \text{ml}$, therefore in $500\ \text{ml}$ the amount of salt $=110\ \text{gms}$ Initial amount of salt in vessel $C=32\ \text{gms}$ per $100\ \text{ml}$, therefore in $500\ \text{ml}$ the amount of salt $=160\ \text{gms}$ When $100\ \text{ml}$ is trasfered from $A$ to $B$, the amount of salt now in $B=10+110=120\ \text{gms}$ in $600\ \text{ml}.$ The new concentration of salt in $B=\frac{120}{600}=20\ \text{gms}$ per $100\ \text{ml}$. Also, the amount of salt lef in $A=50-10=40\ \text{gms}$ in $400\ \text{ml}$. Now, when $100\ \text{ml}$ is trasfered from $B$ to $C$, the amount of salt now in $C=20+160=180\ \text{gms}$ in $600\ \text{ml}$. The new concentration of salt in $C=\frac{180}{600}=30\ \text{gms}$ per $100\ \text{ml}$. Finally, when $100\ \text{ml}$ is trasfered from $C$ to $A$, the amount of salt now in $A=30+40=70\ \text{gms}$ in $500\ \text{ml}.$ Therefore, the strength of salt in $A=\;\frac{70}{500} \times 100=14\%$