Question
A train 75 m long overtook a person who was walking at the rate of 6 km/hr in the same direction and passed him in $$7\frac{1}{2}$$ seconds. Subsequently, it overtook a second person and passed him in $$6\frac{3}{4}$$ seconds. At what rate was the second person travelling?
Answer: Option B
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
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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
Was this answer helpful ?
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