Question
A man walks from his house at an average speed of 5 km per hour and reaches his office 6 minutes late. If he walks at an average speed of 6 km/h he reaches 2 minutes early. The distance of the office from his house is
Answer: Option B
Answer: (b)Required distance of office from house = x km. (let)Time = $\text"Distance"/ \text"Speed"$According to the question,$x/5 - x/6 = {6 + 2}/60 = 2/15$${6x - 5x}/30 = 2/15$$x/30 = 2/15$$x = 2/15$ × 30 = 4 km. Using Rule 10,If a man travels at the speed of $s_1$, he reaches his destination $t_1$ late while he reaches $t_2$ before when he travels at $s_2$ speed, then the distance between the two places is D = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
Was this answer helpful ?
Answer: (b)Required distance of office from house = x km. (let)Time = $\text"Distance"/ \text"Speed"$According to the question,$x/5 - x/6 = {6 + 2}/60 = 2/15$${6x - 5x}/30 = 2/15$$x/30 = 2/15$$x = 2/15$ × 30 = 4 km. Using Rule 10,If a man travels at the speed of $s_1$, he reaches his destination $t_1$ late while he reaches $t_2$ before when he travels at $s_2$ speed, then the distance between the two places is D = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution