Ages of father and son add up to 50. Six years back father's age...

Let the age of father be F and that of son be S . ($$\begin{aligned} \text{ and } F+S=50 \text { (Given) }\\ \text{ and } S=50-F \quad ...(i) \end{aligned}$$) Six years back father's age was 6 more than thrice his son's age. According to problem: $(F-6)=3(S-6)+6$ ...(ii) Substituting the value of equation (i) in (ii), we get: \$($$\begin{aligned}& F-6=3(50-F-6)+6 \\ & \Rightarrow F-6=3(44-F)+6 \\ & \Rightarrow F-6=132-3 F+6 \\ & \Rightarrow F+3 F=132+6+6 \\ & \Rightarrow 4 F=144 \\ & \Rightarrow F=144 / 4 \\ & \Rightarrow F=36 \end{aligned}$$) So, father's age 6 years hence$=(36+6)=42$ Hence, ' 42 ' is the correct answer.