If a man can cover 12 metres in one second, how many kilometres can he cover in 3 hours 45 minutes?
Answer: B
Convert the time to seconds and multiply.
\[ 3 \text{ h } 45 \text{ min} = 3.75 \times 3600 = 13{,}500 \text{ s} \]
\[ d = 12 \times 13{,}500 = 162{,}000 \text{ m} = 162 \text{ km} \]
Equivalently, 12 m/s is \( 12 \times \tfrac{18}{5} = 43.2 \) kmph, and \( 43.2 \times 3.75 = 162 \).
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A train 100 m long crosses a platform 125 m long in 15 sec; find the speed of the train?
Answer: C
Crossing the platform covers the train plus the platform.
\[ v = \frac{100 + 125}{15} = 15 \text{ m/s} \]
\[ 15 \times \frac{18}{5} = 54 \text{ kmph} \]
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A man cycles at a rate of 4 kmph. He cycles for 45 minutes and then takes a 15 minute break. Determine the total time taken to cover 12 kms by him.
Answer: D
Speed of man = 4kmph
Time for which he cycled = 45 minutes.
Distance covered = speed*time
=> 4*45/60
=> 3km
Total time to cover 3 km = 45 minutes + 15 minutes = 60 minutes = 1 hour
Thus 3 km is covered in 1 hour
Required distance to be covered = 12 km
Beginning from origin point, he covers 9 km in 3 hours taking 15 minutes rest after every 45 minutes. To complete 12 kms he goes another 3 km in 45 minutes.
Thus, total time he takes is 3 hours 45 minutes.
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In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:
Answer: A
Let the usual speed be \( v \). The slowed trip takes half an hour longer.
\[ \frac{600}{v - 200} - \frac{600}{v} = 0.5 \]
\[ \frac{600v - 600(v - 200)}{v(v-200)} = \frac{120000}{v^{2} - 200v} = 0.5 \]
\[ v^{2} - 200v - 240000 = 0 \]
\[ v = \frac{200 + \sqrt{40000 + 960000}}{2} = \frac{200 + 1000}{2} = 600 \text{ km/hr} \]
\[ \text{duration} = \frac{600}{600} = 1 \text{ hour} \]
The slowed flight would take 1½ hours at 400 km/hr - exactly the 30 minutes more the question describes.
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The speed of a car is 90 km in the first hour and 60 km in the second hour. What is the average speed of the car?
Answer: B
Here the two times are equal - one hour each - not the distances, so the plain arithmetic mean is correct.
\[ \bar{v} = \frac{90 + 60}{1 + 1} = 75 \text{ kmph} \]
Had the two distances been equal instead, the harmonic mean \( \tfrac{2 \times 90 \times 60}{150} = 72 \) kmph would apply - which is exactly the first distractor.
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If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:
Answer: A
Both journeys take the same time; only the speed and the distance differ.
\[ \frac{x}{10} = \frac{x + 20}{14} \]
\[ 14x = 10x + 200 \quad\Longrightarrow\quad 4x = 200 \]
\[ x = 50 \text{ km} \]
Checking: 50 km at 10 km/hr and 70 km at 14 km/hr both take 5 hours. The 70 on the list is the longer distance, not the actual one.
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Amar takes as much time in running 18 meters as a car takes in covering 48 meters. What will be the distance covered by Amar during the time the car covers 1.6 km?
Answer: D
In equal times, distances are in proportion to speeds.
\[ \frac{\text{Amar}}{\text{car}} = \frac{18}{48} = \frac{3}{8} \]
\[ 1.6 \text{ km} = 1600 \text{ m} \]
\[ \text{Amar} = 1600 \times \frac{3}{8} = 600 \text{ m} \]
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A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
Answer: B
Each half of the journey is \( \tfrac{D}{2} \) km.
\[ \frac{D/2}{21} + \frac{D/2}{24} = 10 \]
\[ \frac{D}{42} + \frac{D}{48} = \frac{8D + 7D}{336} = \frac{15D}{336} \]
\[ \frac{15D}{336} = 10 \quad\Longrightarrow\quad D = 224 \text{ km} \]
Equivalently, the average speed is the harmonic mean \( \tfrac{2 \times 21 \times 24}{45} = 22.4 \) kmph, and \( 22.4 \times 10 = 224 \).
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A cab on its way to Bangalore overtakes an auto also going in the same direction at 11 am. The cab reaches Bangalore at 1.30 pm and starts the return journey after taking a break of 1 hour. On its way back it meets the auto at 3 pm. At what time will the auto reach Bangalore?
Answer: D
Let \( D \) be the distance from the overtaking point to Bangalore.
The cab covers it between 11 am and 1:30 pm, so \( v_{\text{cab}} = \tfrac{D}{2.5} \). After a one-hour break it leaves at 2:30 pm and meets the auto at 3 pm, half an hour later:
\[ \text{cab's return distance} = \frac{D}{2.5} \times 0.5 = \frac{D}{5} \]
So they meet \( D - \tfrac{D}{5} = \tfrac{4D}{5} \) from the overtaking point. The auto took from 11 am to 3 pm - four hours - to get there:
\[ v_{\text{auto}} = \frac{4D/5}{4} = \frac{D}{5} \text{ per hour} \]
The auto still has \( \tfrac{D}{5} \) to go, which takes exactly one more hour:
\[ \text{arrival} = 4 \text{ p.m.} \]
The distance \( D \) cancels throughout, so it never needs a value.
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Sourav walks 20 meters towards North. He then turns left and walks 40 meters. He again turns left and walks 20 meters. Further, he moves 20 meters after turning to
right. How far is he from his original position?
Answer: D
Track the two axes separately. Starting north, a left turn faces west, another left faces south, then a right turn faces west again.
\[ \text{north--south} = 20 - 20 = 0 \]
\[ \text{east--west} = 40 + 20 = 60 \text{ m west} \]
The vertical movements cancel exactly, so the displacement is a straight 60 metres - no Pythagoras needed.
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