Man cycles from his office to his house at a speed of 5 km/hr and reaches 10 minutes late. if he cycles at a speed of 6 km/hr, he reaches 10 minutes early. what is the distance
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Let the original distance and time be D & t
Case 1 - If man travels at 5kmph, he reaches late 20 min or 1/3 hr
We can write the same as
D/5= t + ⅓-----(1)
Case 2 - If man travels at 7.5kmph, he reaches early 12 min or 1/5 hr
We can write the same as
D/7.5= t - 1/5 -----(2)
Solving both we get D = 8 km
hope this helps
Case 1 - If man travels at 5kmph, he reaches late 20 min or 1/3 hr
We can write the same as
D/5= t + ⅓-----(1)
Case 2 - If man travels at 7.5kmph, he reaches early 12 min or 1/5 hr
We can write the same as
D/7.5= t - 1/5 -----(2)
Solving both we get D = 8 km
hope this helps
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