Find the average speed of a bus if it moves from point A to B at a speed of 30 km/h and returns at a speed of 50 km/h
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from a to b time taken is t1
and from b to a time taken is t2
distance d = time × speed
d = t1 × 30 = 30t1
d = t2 × 50 = 50t2
also , 30t1 = 50t2
t1 = 5/3 t2
total distance covered = 2d
total time = t1 + t2
so speed = dist /time
= d + d / t1 + t2
= 30t1 + 50t2 / t1 + t2
= (30 × 5/3 t2 + 50t2 ) / 5/3t2 + t2
= (100 t2) / 8t2/3
= 100 t2 ×3 / 8 t2
37.5 km / hr
and from b to a time taken is t2
distance d = time × speed
d = t1 × 30 = 30t1
d = t2 × 50 = 50t2
also , 30t1 = 50t2
t1 = 5/3 t2
total distance covered = 2d
total time = t1 + t2
so speed = dist /time
= d + d / t1 + t2
= 30t1 + 50t2 / t1 + t2
= (30 × 5/3 t2 + 50t2 ) / 5/3t2 + t2
= (100 t2) / 8t2/3
= 100 t2 ×3 / 8 t2
37.5 km / hr
pravinsir:
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0
Answer:
from a to b time taken is t1
and from b to a time taken is t2
distance d = time × speed
d = t1 × 30 = 30t1
d = t2 × 50 = 50t2
also , 30t1 = 50t2
t1 = 5/3 t2
total distance covered = 2d
total time = t1 + t2
so speed = dist /time
= d + d / t1 + t2
= 30t1 + 50t2 / t1 + t2
= (30 × 5/3 t2 + 50t2 ) / 5/3t2 + t2
= (100 t2) / 8t2/3
= 100 t2 ×3 / 8 t2
37.5 km / hr
Explanation:
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