Practice MCQ Questions and Answer on Problems on Trains

Solution: $$\eqalign{ & {\text{Relative speed}} \cr & {\text{ = 60 + 50}} \cr & {\text{ = 110 km/h}} \cr & {\text{Time taken}} \cr & {\text{ = }}\frac{{220}}{{110}} \cr & {\text{ = 2 hr}} \cr} $$

Solution: $$\eqalign{ & {\text{Total}}\,{\text{distance}}\,{\text{covered}} \cr & = \left( {\frac{7}{2} + \frac{1}{4}} \right)\,{\text{miles}} \cr & = \frac{{15}}{4}\,{\text{miles}} \cr & \therefore {\text{Time}}\,{\text{taken}} \cr & = \left( {\frac{{15}}{{4 \times 75}}} \right)\,{\text{hrs}} \cr & = \frac{1}{{20}}\,{\text{hrs}} \cr & = \left( {\frac{1}{{20}} \times 60} \right)\,\min \cr & = 3\,\min \cr} $$

Solution: $$\eqalign{ & {\text{Speed}} \cr & {\text{ = }}\left( {\frac{{132}}{6}} \right){\text{m/sec}} \cr & {\text{ = }}\left( {22 \times \frac{{18}}{5}} \right){\text{km/sec}} \cr & {\text{ = 79}}{\text{.2 km/hr}} \cr} $$

Solution: $$\eqalign{ & {\text{Distance travelled in 14 sec}} \cr & {\text{ = 50 + }}l \cr & {\text{Distance travelled in 10 sec}} \cr & {\text{ = }}l \cr & {\text{So speed of train}} \cr & {\text{ = }}\frac{{50}}{{14 - 10}}{\text{m/sec}} \cr & {\text{ = }}\frac{{50}}{4} \times \frac{{18}}{5}{\text{km/hr}} \cr & {\text{ = 45 km/hr}} \cr} $$

Solution: $$\eqalign{ & {\text{In these type of questions use the given}} \cr & {\text{below formula to save your valuable time}} \cr & \frac{{{{\text{S}}_1}}}{{{{\text{S}}_2}}}{\text{ = }}\sqrt {\frac{{{{\text{T}}_2}}}{{{{\text{T}}_1}}}} {\text{ }} \cr & {\text{Where }}{{\text{S}}_1}{\text{,}}{{\text{S}}_2}{\text{ and }}{{\text{T}}_1}{\text{, }}{{\text{T}}_2}{\text{ are the respective}} \cr & {\text{speeds and times of the objects}} \cr & \Rightarrow \frac{{45}}{{{{\text{S}}_2}}} = \sqrt {3\frac{1}{3} \div 4\frac{4}{5}} \cr & {\text{ = }}{{\text{S}}_2}{\text{ = 45}} \times \frac{6}{5}{\text{ = 54 km/hr}} \cr & \therefore {\text{Required speed = 54 km/hr}} \cr} $$

Solution: $$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speeds}}\,{\text{of}}\,{\text{the}}\,{\text{two}}\,{\text{trains}}\,{\text{be}}\,x\,{\text{m/sec}} \cr & {\text{and}}\,y\,{\text{m/sec}}\,{\text{respectively}}. \cr & {\text{Then,}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}} = 27x\,{\text{metres}}, \cr & {\text{and}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}} = 17y\,{\text{metres}}. \cr & \therefore \frac{{27x + 17y}}{{x + y}} = 23 \cr & \Rightarrow 27x + 17y = 23x + 23y \cr & \Rightarrow 4x = 6y \cr & \Rightarrow \frac{x}{y} = \frac{3}{2} \cr} $$

Solution: $$\eqalign{ & {\text{Length of train}} \cr & {\text{ = 280 m }} \cr & {\text{Length of platform}} \cr & {\text{ = (3}} \times {\text{280) m = 840m}} \cr & \therefore {\text{Speed of train}} \cr & {\text{ = }}\left( {\frac{{280 + 840}}{{50}}} \right)m/\sec \cr & = \frac{{1120}}{{50}}m/\sec \cr & = \left( {\frac{{1120}}{{50}} \times \frac{{18}}{5}} \right)km/hr \cr & = 80.64\,km/hr \cr} $$

Solution: Speed of the train relative to first man $$\eqalign{ & = \frac{{75}}{{7.5}}{\text{m/sec}} = 10\,{\text{m/sec}} \cr & = \left( {10 \times \frac{{18}}{5}} \right){\text{km/hr}} = 36\,{\text{km/hr}} \cr} $$ Let the speed of the train be x km/hr. Then, relative speed = (x - 6) km/hr ∴ x - 6 = 36 ⇒ x = 42 km/hr Speed of the train relative to second man $$\eqalign{ & {\text{ = }}\frac{{75}}{{6\frac{3}{4}}}\,{\text{m/sec}} \cr & = \left( {75 \times \frac{4}{{27}}} \right){\text{m/sec}} \cr & = \frac{{100}}{9}{\text{m/sec}} \cr & = \left( {\frac{{100}}{9} \times \frac{{18}}{5}} \right){\text{km}} \cr & = 40\,{\text{km/hr}} \cr} $$ Let the speed of the second man be y kmph. Then, relative speed = (42 - y) kmph ∴ 42 - y = 40 ⇒ y = 2 km/hr

Solution: $$\eqalign{ & {\text{At the time of meeting ,}} \cr & {\text{let the distance travelled by the}} \cr & {\text{first train be }}x{\text{ km}}{\text{.}} \cr & {\text{Then distance travelled by the }} \cr & {\text{second train is (}}x{\text{ + 60) km}} \cr & \therefore \frac{x}{{16}} = \frac{{x + 60}}{{21}} \cr & \Rightarrow 21x = 16x + 960 \cr & \Rightarrow 5x = 960 \Rightarrow x = 192 \cr & {\text{Hence,}} \cr & {\text{distance between two stations}} \cr & {\text{ = (192 + 192 + 60) km}} \cr & {\text{ = 444 km}}{\text{.}} \cr} $$

Solution: $$\eqalign{ & {\text{Relative}}\,{\text{speed}} \cr & = \left( {45 + 30} \right)\,{\text{km/hr}} \cr & = \left(75 \times \frac{5}{18} \right)\, \text{m/sec} \cr & = {\frac{{125}}{6}} \,{\text{m/sec}} \cr & {\text{We}}\,{\text{have}}\,{\text{to}}\,{\text{find}}\,{\text{the}}\,{\text{time}}\,{\text{taken}}\,{\text{by}}\,{\text{the}} \cr & {\text{slower}}\,{\text{train}}\,{\text{to}}\,{\text{pass}}\,{\text{the}}\,{\text{DRIVER}}\,{\text{of}}\, \cr & {\text{The}}\,{\text{faster}}\,{\text{train}}\,{\text{and}}\,{\text{not}}\,{\text{the}}\,{\text{complete}}\,{\text{train}}{\text{.}} \cr & \cr & {\text{So,}}\,{\text{distance}}\,{\text{covered = Length}}\,{\text{of}}\,{\text{the}}\,{\text{slower}}\,{\text{train}}. \cr & {\text{Therefore,}}\,{\text{Distance}}\,{\text{covered = 500}}\,{\text{m}}. \cr & \therefore {\text{Required}}\,{\text{time}} \cr & = {500 \times \frac{6}{{125}}} \cr & = 24\,\sec \cr} $$