A 180 meter long train crosses a man standing on the platform in 6 sec. What is the speed of the train?
Answer: B
A man standing is a point, so the train covers only its own length.
\[ v = \frac{180}{6} = 30 \text{ m/s} \]
\[ 30 \times \frac{18}{5} = 108 \text{ kmph} \]
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A car travelling with 5/7th of its actual speed covers 42km in 1hr. 40mins. 48 secs. Find the actual speed of the car.
Answer: B
Convert the time to hours first.
\[ 1 \text{ h } 40 \text{ min } 48 \text{ s} = 1 + \frac{40}{60} + \frac{48}{3600} = 1.68 \text{ h} \]
\[ \text{reduced speed} = \frac{42}{1.68} = 25 \text{ km/hr} \]
That 25 is \( \tfrac{5}{7} \) of the actual speed:
\[ \frac{5}{7}v = 25 \quad\Longrightarrow\quad v = 35 \text{ km/hr} \]
Multiplying 25 by \( \tfrac{5}{7} \) instead of dividing gives \( 17\tfrac{6}{7} \) - the first distractor.
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Walking with 4/5 of my usual speed, I miss the bus by 5 minutes. What is my usual time?
Answer: D
Walking at \( \tfrac{4}{5} \) of the usual speed takes \( \tfrac{5}{4} \) of the usual time, since speed and time are inversely proportional.
\[ \frac{5}{4}t - t = \frac{1}{4}t = 5 \text{ min} \]
\[ t = 20 \text{ minutes} \]
Checking: the slower walk takes 25 minutes, exactly 5 more.
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