A train 150 m long running at 72 kmph crosses a platform in 25 sec. What is the length of the platform?
Answer: C
\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]
\[ 72 \text{ kmph} = 20 \text{ m/s} \]
\[ \text{total distance} = 20 \times 25 = 500 \text{ m} \]
\[ \text{platform} = 500 - 150 = 350 \text{ m} \]
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A car covers 240 km at 60 km/hr and returns over the same route at 40 km/hr. What is its average speed for the whole journey ?
Answer: A
Correct answer: 48 km/hr
Average speed is total distance divided by total time, not the mean of the two speeds.
\[\text{Time} = \frac{240}{60} + \frac{240}{40} = 4 + 6 = 10 \text{ hours}\]
\[\text{Average speed} = \frac{480}{10} = 48 \text{ km/hr}\]
It falls below 50 because more of the journey’s time is spent at the slower speed.
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Pratik travels 96 km at a speed of 16 km/hr using a bike, 124 km at 31 km/h by car and another 105 km at 7 km/h in horse cart. Then, find his average speed for the entire distance travelled?
Answer: C
Add the distances and the times separately.
\[ \text{distance} = 96 + 124 + 105 = 325 \text{ km} \]
\[ t = \frac{96}{16} + \frac{124}{31} + \frac{105}{7} = 6 + 4 + 15 = 25 \text{ h} \]
\[ \bar{v} = \frac{325}{25} = 13 \text{ km/hr} \]
Averaging the three speeds gives 18 km/hr, well above the truth - the slow 7 km/hr leg accounts for 15 of the 25 hours.
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A boy has to cover a total distance of 300 kms. in 6 hours. He travels at the rate of 60 kmph. for first 90 minutes and next 100 kms. at the rate of 50 kmph. At what average speed must he travel now in order to complete the journey in 6 hours?
Answer: C
Track distance and time together.
\[ \text{first stage} = 60 \times 1.5 = 90 \text{ km in } 1.5 \text{ h} \]
\[ \text{second stage} = \frac{100}{50} = 2 \text{ h for } 100 \text{ km} \]
\[ \text{covered} = 190 \text{ km in } 3.5 \text{ h} \]
\[ \text{remaining} = 110 \text{ km in } 2.5 \text{ h} \quad\Longrightarrow\quad \frac{110}{2.5} = 44 \text{ kmph} \]
The 90 minutes must become 1.5 hours before it can be combined with the rest.
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A train can travel 50% faster than a car. Both start from point A at the same time and reach point B, 75 km away from A, at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:
Answer: C
Let the car do \( c \) kmph; the train then does \( 1.5c \). Both leave together and arrive together, but the train idles 12.5 minutes.
\[ \frac{75}{c} = \frac{75}{1.5c} + \frac{12.5}{60} \]
\[ \frac{75}{c} - \frac{50}{c} = 0.20833 \quad\Longrightarrow\quad \frac{25}{c} = 0.20833 \]
\[ c = 120 \text{ kmph} \]
Checking: the car takes 37.5 minutes and the train 25 minutes of running plus 12.5 minutes stopped - both 37.5 in total.
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Murali travelled from city A to city B at a speed of 40 kmph and from city B to city C at 60 kmph. What is the average speed of Murali from A to C given that the ratio of distances between A to B and B to C is 2 : 3?
Answer: B
Let the distances between city A to B and B to C be 2x km and 3x km respectively.
Total time taken to cover from A to C
= (2x)/40 + (3x)/60 = (6x + 6x)/120 = 12x/120 = x/10 Average speed = (2x + 3x)/(x/10) = 50 kmph.
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In how much time will a train of length 100 m, moving at 36 kmph cross an electric pole?
Answer: D
\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]
\[ 36 \text{ kmph} = 10 \text{ m/s} \]
An electric pole has no length, so the train covers only its own 100 m:
\[ t = \frac{100}{10} = 10 \text{ sec} \]
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Surendra covers a distance by cycle at 18 km/h. He returns to the starting point in a bus at a speed of 45 km/h. Find the average speed for the entire journey.
Answer: B
Average speed is total distance over total time, which for two equal legs is the harmonic mean, not the arithmetic one:
\[ \bar{v} = \frac{2v_{1}v_{2}}{v_{1} + v_{2}} \]
\[ \bar{v} = \frac{2 \times 18 \times 45}{18 + 45} = \frac{1620}{63} = 25.71 \text{ km/hr} \]
The arithmetic mean of 31.5 km/hr overstates it, because the slower cycle leg takes far longer than the bus leg and so weighs more heavily in the total time.
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A car covers a distance of 624 km in 6 hours. Find its speed.
Answer: A
\[ v = \frac{\text{distance}}{\text{time}} = \frac{624}{6} = 104 \text{ kmph} \]
The mph option is a units trap: the distance is given in kilometres, so the speed must be in kilometres per hour.
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Ramesh sees a thief at a distance of 80 m. Ramesh starts chasing the thief who is running at a speed of 5 m/s. Ramesh is chasing him with a speed of 7 m/s. How much distance does the thief covers before Ramesh catches him?
Answer: D
The gap closes at the difference of the two speeds.
\[ \text{relative speed} = 7 - 5 = 2 \text{ m/s} \]
\[ t = \frac{80}{2} = 40 \text{ s} \]
\[ \text{thief's distance} = 5 \times 40 = 200 \text{ m} \]
Ramesh covers 280 m in the same time - exactly 80 m more, which is the head start he had to make up. The question asks for the thief's distance, not Ramesh's.
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