Question
You arrive at your school 5 minutes late if you walk with a speed of 4 km/h, but you arrive 10 minutes before the scheduled time if you walk with a speed of 5 km/h. The distance of your school from your house (in km) is
Answer: Option B
Answer: (b)If the required distance be = x km, then$x/4 - x/5 = {10 + 5}/60$${5x - 4x}/20 = 1/4$$x/20 = 1/4 ⇒ x= 1/4 × 20$ = 5 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 = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
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Answer: (b)If the required distance be = x km, then$x/4 - x/5 = {10 + 5}/60$${5x - 4x}/20 = 1/4$$x/20 = 1/4 ⇒ x= 1/4 × 20$ = 5 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 = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$
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