Question

1. If a student cycles down to school at the rate of 12 km h, he is early by 10 minutes. If he cycles a
the speed of 9 km/h, he reaches the school late by 5 minutes. Find the distance of the school for
his home.​

Answer 1

Step-by-step explanation:

Let the distance between school and home be x km/hr

Let the travel time be t hours Speed from home to school is 12 km/hr

Recall that Distance = Speed x time d = 12[t - 10 min] = 12[t - (10/60)] = 12[6t - 1]/6 = 12t – 2 ---

(1) When speed is 9 km/hr he reaches 5 minutes late d = 9[t - 5 min] = 9[t - (5/60)] = 9[12t - 1]/12 = (36t – 3)/4 --- (2) From (1) and (2), we get 12t – 2 = (36t – 3)/4 48t – 8 = 36t – 3 12t = 5 t = (5/12) Put value of t in (1) we get d = 12t – 2 = 12(5/12) – 2 5 – 2 = 3 Hence the distance between the home and school is 3 km.

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