Question

office of the Sanjay cycle 6km delayed by a spead of 10 km per hour. if it increase its speed by two km per hour then it can reach 6 minute before. it's knows the distance of the office from his house​

Answer 1

Answer:

Let the distance be 'D' Km

Let the usual time be 'T' hrs

At 15 km/hr, he takes 2 minutes more than the usual time.

D/15 = T + 2/60 hrs ....(i)

Speed increased by 5 km/hr, reaches 8 minutes early

D/20 = T - 8/60 hrs.....(ii)

(i) - (ii)

D/15 - D/20 = T + 2/60 - (T - 8/60)

D[1/15 - 1/20] = 2+8/60

D[4-3/60] = 10/60

D/60 = 10/60

D = 10 Km

Hence distance between school and his house = 10 Km.

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