Join 60,000+ competitive exam aspirants
or with email
Your Cart
Subtotal₹0
Total₹0
Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
Join 60,000+ competitive exam aspirants
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 = 2×60=120 km or 3×40=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=v1+v2v1×v2×T
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 = 2×60=120 km or 3×40=120 km.
Students often calculate the average speed 240+60=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 t1=40d and return journey is t2=60d.
t1+t2=40d+60d=5
Solve for d
Find a common denominator for 40 and 60, which is 120. Convert fractions: 1203d+2d=5.
1205d=5
Calculate the final distance
Multiply both sides by 120 and divide by 5 to isolate d.
d=55×120=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.