What is the speed of a train if it overtakes two persons who are walking in the same direction at the rate of a m/s and (a + 1) m/s and passes them completely in b seconds and (b + 1) seconds respectively?
a) (a + b) m/s
b) (a + b + 1) m/s
c) (2a + 1) m/s
d) $$\frac{{2{\text{a}} + 1}}{2}$$ m/s

Answer: b
Explanation:
$$\eqalign{ & {\text{Let the length of the train be }}x{\text{ metres}} \cr & {\text{and its speed be }}y{\text{ m/s}} \cr & {\text{ }}\frac{x}{{y – a}}{\text{ = b}}\,\,{\text{and}}\, \cr & \,\frac{x}{{y – \left( {a + 1} \right)}} = \left( {b + 1} \right) \cr & \Leftrightarrow {\text{ }}x{\text{ = }}b\left( {y – a} \right){\text{ and}} \cr & \,\,\,\,\,\,\,\,\,\,{\text{ }}x = \left( {b + 1} \right)\left( {y – a – 1} \right) \cr & \Leftrightarrow b\left( {y – a} \right) = \left( {b + 1} \right)\left( {y – a – 1} \right) \cr & \Leftrightarrow by – ba = by – ba – b + y – a – 1 \cr & y = \left( {a + b + 1} \right) \cr} $$.