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A traveller moves from City A to City B at 40 km/h and returns at 60 km/h. If the total time taken for the whole journey is 5 hours, find the distance between A and B.
100 km
120 km
140 km
150 km
120 km
Use the ratio of speeds (40:60=2:3). The ratio of time taken will be inverse (3:2). Since total time 3+2=5 units equals 5 hours, 1 unit = 1 hour. Distance = 2├Ч60=120 km or 3├Ч40=120 km.
Speed from A to B (v1) = 40 km/h, Speed from B to A (v2) = 60 km/h, Total time (T) = 5 hours.
d=v1тАЛ+v2тАЛv1тАЛ├Чv2тАЛтАЛ├ЧT
Use the ratio of speeds (40:60=2:3). The ratio of time taken will be inverse (3:2). Since total time 3+2=5 units equals 5 hours, 1 unit = 1 hour. Distance = 2├Ч60=120 km or 3├Ч40=120 km.
Students often calculate the average speed 240+60тАЛ=50 km/h and multiply by 5 to get 250 km, forgetting that the total distance covers the path twice (round trip).
Define the relationship between distance and speed
Let d be the distance between A and B. Time taken for forward journey is t1тАЛ=40dтАЛ and return journey is t2тАЛ=60dтАЛ.
t1тАЛ+t2тАЛ=40dтАЛ+60dтАЛ=5
Solve for d
Find a common denominator for 40 and 60, which is 120. Convert fractions: 1203d+2dтАЛ=5.
1205dтАЛ=5
Calculate the final distance
Multiply both sides by 120 and divide by 5 to isolate d.
d=55├Ч120тАЛ=120
B is correct because the distance between the two cities is 120 km.
This concept of harmonic mean for speeds is frequently tested in 'Average Speed' problems where total distance is constant.
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