Practice Questions
Concept Explanation
Average speed = Total distance / Total time
It is NOT the simple average of speeds unless equal distances are covered at each speed.
Key Formulas
- Total time = d₁/s₁ + d₂/s₂ + ...
- Equal distances at speeds a and b: average speed = 2ab/(a + b)
- Round trip same distance at a and b: average speed = 2ab/(a + b)
Common Trap
60 km at 40 km/h and 60 km at 60 km/h → average speed = 120/2.5 = 48 km/h, not 50 km/h.
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Q1 EasyA motorist covers a distance at 40 km/h and returns at 60 km/h. The average speed for the whole journey is:ExplanationSame distance each way → average speed = 2ab/(a + b) = 2 × 40 × 60 / (40 + 60) = 4800/100 = 48 km/h. (Arithmetic mean 50 is wrong for equal distances.)
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Q2 EasyGoing to a place at 40 km/h and returning at 60 km/h, the average speed for the whole journey is:ExplanationEqual one-way distances → harmonic mean: 2 × 40 × 60 / (40 + 60) = 4800/100 = 48 km/h.
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Q3 EasyA cyclist covers 30 km at 15 km/h and another 30 km at 30 km/h. What is the average speed for the whole journey?ExplanationEqual distances: average = 2 × 15 × 30 / (15 + 30) = 900/45 = 20 km/h. Time check: 2 + 1 = 3 h for 60 km → 20 km/h.
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Q4 EasyEqual distances are covered at 20 km/h and 30 km/h. The average speed is:ExplanationFor equal distances, average = 2ab/(a + b) = 2 × 20 × 30 / (20 + 30) = 1200/50 = 24 km/h (not the arithmetic mean 25).
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Q5 EasyIf equal distances are covered at speeds a and b, which expression gives the average speed?ExplanationFor equal distances, average speed is the harmonic mean of the two speeds: 2ab/(a + b). The arithmetic mean (a + b)/2 applies only when times (not distances) are equal.
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Q6 MediumA car owner travels 160 km at 80 km/h and the next 160 km at 60 km/h. The average speed for the whole journey is:ExplanationEqual distances → average speed = 2ab/(a + b). Here a = 80, b = 60. Average = 2 × 80 × 60 / (80 + 60) = 9600/140 = 68.571… ≈ 68.57 km/h. Check: time = 160/80 + 160/60 = 2 + 8/3 = 14/3 h; speed = 320 ÷ (14/3) = 960/14 ≈ 68.57 km/h.
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Q7 MediumA man on tour travels first 160 km at 64 km/h and the next 160 km at 80 km/h. The average speed for the first 320 km of the tour is:ExplanationEqual distances → average = 2ab/(a + b) = 2 × 64 × 80 / (64 + 80) = 10240/144 = 71.111… ≈ 71.11 km/h. Check: time = 160/64 + 160/80 = 2.5 + 2 = 4.5 h; speed = 320/4.5 ≈ 71.11 km/h.
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Q8 MediumA car travels 2 hours at 60 km/h and 3 hours at 40 km/h. The average speed for the entire trip is:ExplanationAverage speed = total distance / total time. Distance = 60 × 2 + 40 × 3 = 120 + 120 = 240 km. Time = 5 h. Average = 240/5 = 48 km/h.
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Q9 MediumA journey of 90 km is covered at 30 km/h for the first 45 km and at 45 km/h for the remaining 45 km. Average speed is:ExplanationEqual distances: average = 2 × 30 × 45 / (30 + 45) = 2700/75 = 36 km/h. Time = 45/30 + 45/45 = 1.5 + 1 = 2.5 h; 90/2.5 = 36 km/h.
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Q10 MediumA train goes from P to Q at 50 km/h and returns from Q to P at 75 km/h. The average speed for the round trip is:ExplanationRound trip on the same route → equal distances: average = 2 × 50 × 75 / (50 + 75) = 7500/125 = 60 km/h.
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Q11 MediumA person walks 12 km at 4 km/h and then 12 km at 6 km/h. The average speed for the whole walk is:ExplanationEqual distances: average = 2 × 4 × 6 / (4 + 6) = 48/10 = 4.8 km/h. Time = 3 + 2 = 5 h for 24 km → 4.8 km/h.
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Q12 HardA man travels 600 km partly by train and partly by car. If he travels 400 km by train and the rest by car, it takes 6 hours 30 minutes. If he travels 200 km by train and the rest by car, it takes 7 hours. Find the speed of the train and the car.ExplanationLet train speed = t km/h and car speed = c km/h. Then 400/t + 200/c = 6.5 and 200/t + 400/c = 7. Divide by 200: 2/t + 1/c = 0.0325 and 1/t + 2/c = 0.035. Solving: 1/t = 0.01 ⇒ t = 100; 1/c = 0.0125 ⇒ c = 80. So train = 100 km/h and car = 80 km/h.
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Q13 HardThe distance between two stations A and B is 450 km. A train starts from A towards B at 20 km/h. Another train starts from B 2 hours later towards A at 30 km/h. The average speed of the trains until they meet is:ExplanationIn 2 hours the first train covers 40 km; remaining distance = 410 km. Relative speed = 50 km/h, so time after the second starts = 410/50 = 8.2 h. Distances: first = 20 × 10.2 = 204 km; second = 30 × 8.2 = 246 km. Average speed of both until meeting = total distance covered / total time taken = 450 / (10.2 + 8.2) = 450/18.4 ≈ 24.46 km/h.
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Q14 HardDuring a screening test, 100 km is covered at 25 km/h and another 100 km at 50 km/h. Average speed for 200 km is:ExplanationEqual distances: average = 2 × 25 × 50 / (25 + 50) = 2500/75 = 33.333… ≈ 33.33 km/h. Time = 4 + 2 = 6 h; 200/6 ≈ 33.33 km/h.
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DS1 Easy
What is the average speed of the journey?
Statement ITotal distance is 120 km and total time is 2.5 hours.
Statement IITwo equal legs of 60 km each are covered at 40 km/h and 60 km/h.
ExplanationStatement I: average = 120/2.5 = 48 km/h. Statement II: average = 2 × 40 × 60 / 100 = 48 km/h. Either alone is sufficient. -
DS2 Easy
During a screening round, is the average speed less than 50 km/h?
Statement IEqual distances are covered at 40 km/h and 60 km/h.
Statement IIThe total distance of the journey is 120 km.
ExplanationStatement I alone: average = 48 km/h, which is less than 50. Statement II alone gives no speeds.
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DS3 Medium
What is the average speed for equal distances at two constant speeds?
Statement IThe first part is covered at 30 km/h.
Statement IIThe second part is covered at 60 km/h.
ExplanationBoth speeds are required for the harmonic mean. Together: average = 2 × 30 × 60 / 90 = 40 km/h. -
DS4 Hard
In a quantitative interview, what is the total time taken for the journey?
Statement I100 km is covered at 25 km/h and another 100 km at 50 km/h.
Statement IIThe average speed for the whole journey is 33.33 km/h approximately.
ExplanationStatement I: time = 100/25 + 100/50 = 4 + 2 = 6 hours. Statement II lacks distance, so time cannot be found from II alone.