Question

sahil rides his bicycle at the speed of 15 km/hr and reaches school late by 2 minutes .on increasing the speed by 5km/hr,he reaches the school 8 minutes early.find the distance between the school and his house

Answer 1

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.

Alternate method:
D = S1* S2 * (Change in time)/Change in Speed
S1= 15 km/hr... S2 = 20 Km/hr... Change in time = (T+2)-(T-8) = 10 mins = 10/60 hrs... Speed change = 5 Km/hr
D = 15*20*10/60*5 = 10 Km.
Hope it helps

Answer 2

Answer:

Step-by-step explanation: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.

Alternate method:

D = S1* S2 * (Change in time)/Change in Speed

S1= 15 km/hr... S2 = 20 Km/hr... Change in time = (T+2)-(T-8) = 10 mins = 10/60 hrs... Speed change = 5 Km/hr

D = 15*20*10/60*5 = 10 Km.

Hope it helps