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If a man runs at 6 kmph from his house, he misses the train at the station by 8 minutes. If he runs at 10 kmph, he reaches the station 7 mi…

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If a man runs at 6 kmph from his house, he misses the train at the station by 8 minutes. If he runs at 10 kmph, he reaches the station 7 minutes earlier than the departure of the train. What is the distance of station from his house? (in km)
  • A$$4\frac{3}{4}$$
  • B$$3\frac{1}{2}$$
  • C$$4\frac{1}{4}$$
  • D$$3\frac{3}{4}$$
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CCommunity
Solution
Let the distance of the station from the house of the person = $x$ km
Difference of time = 8 + 7 = 15 minutes = $\frac{1}{4}$ hr
Since, Time = $\frac{Distance}{Speed}$
$\therefore \frac{x}{6} - \frac{x}{10} = \frac{1}{4}$
=> $\frac{10x - 6x}{60} = \frac{1}{4}$
=> $\frac{2x}{30} = \frac{1}{4}$
=> $x = \frac{15}{4} = 3\frac{3}{4}$ km
0
CCommunity
Solution
Let the distance of the station from the house of the person = $x$ km
Difference of time = 8 + 7 = 15 minutes = $\frac{1}{4}$ hr
Since, Time = $\frac{Distance}{Speed}$
$\therefore \frac{x}{6} - \frac{x}{10} = \frac{1}{4}$
=> $\frac{10x - 6x}{60} = \frac{1}{4}$
=> $\frac{2x}{30} = \frac{1}{4}$
=> $x = \frac{15}{4} = 3\frac{3}{4}$ km