What distance will be covered by a bus moving at 72 kmph in 30 seconds?
Answer: B
\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]
\[ 72 \text{ kmph} = 20 \text{ m/s} \]
\[ d = 20 \times 30 = 600 \text{ m} \]
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Two trains of length 100 m and 200 m are 100 m apart. They start moving towards each other on parallel tracks, at speeds 54 kmph and 72 kmph. After how much time will the trains meet?
Answer: C
Trains in the same direction close at the difference of their speeds; in opposite directions, at the sum.
\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]
\[ \text{relative speed} = 54 + 72 = 126 \text{ kmph} = 35 \text{ m/s} \]
To meet, only the 100 m gap between them has to close:
\[ t = \frac{100}{35} = \frac{20}{7} \text{ sec} \]
Their lengths play no part until they begin to pass each other.
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The distance between Delhi and Mathura is 110 km. A starts from Delhi at 7 a.m. with a speed of 20 kmph towards Mathura, and B starts from Mathura at 8 a.m. with a speed of 25 kmph towards Delhi. When will they meet?
Answer: B
A has a one-hour head start:
\[ 20 \times 1 = 20 \text{ km covered by 8 a.m.} \]
\[ \text{gap remaining} = 110 - 20 = 90 \text{ km} \]
They approach at \( 20 + 25 = 45 \) kmph:
\[ t = \frac{90}{45} = 2 \text{ hours} \]
So they meet at 10 a.m.
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Two trains are moving at 50 kmph and 70 kmph in opposite directions. Their lengths are 150 m and 100 m respectively. The time they will take to pass each other completely is?
Answer: D
Trains in the same direction close at the difference of their speeds; in opposite directions, at the sum.
\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]
\[ \text{relative speed} = 50 + 70 = 120 \text{ kmph} = \frac{100}{3} \text{ m/s} \]
\[ t = \frac{150 + 100}{100/3} = \frac{250 \times 3}{100} = 7.5 = 7\tfrac{1}{2} \text{ sec} \]
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Rohit drives from his home at a speed of 30 km/hr and reaches his bank 20 minutes late. Then the next day he increases his speed by 15 km/hr but still he is late by 8 minutes. How far is his bank from his home?
Answer: A
Both journeys cover the same distance and both are late, so the difference in lateness is the difference in travel time.
\[ \frac{D}{30} - \frac{D}{45} = \frac{20 - 8}{60} = 0.2 \text{ h} \]
\[ D\left(\frac{3 - 2}{90}\right) = 0.2 \quad\Longrightarrow\quad \frac{D}{90} = 0.2 \]
\[ D = 18 \text{ km} \]
The scheduled arrival time never has to be found - only the 12-minute gap between the two lateness figures matters.
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Kiran travels from A to B by car and returns from B to A by cycle in 7 hours. If he travels both ways by car he saves 3 hours. What is the time taken to cover both ways by cycle?
Answer: A
Let the time taken to cover from A to B in car and cycle be x hours and y hours respectively.
x + y = 7 --- (1) ; 2x = 4 --- (2)
solving both the equations, we get y = 5
So, time taken to cover both ways by cycle = 2y hours = 10 hours.
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Express 25 mps in kmph?
Answer: C
\[ 1 \text{ m/s} = \frac{18}{5} \text{ km/hr} \]
\[ 25 \times \frac{18}{5} = 90 \text{ kmph} \]
Going from m/s to kmph the number always grows, since a metre per second is a little over 3.6 kmph - so any option below 25 can be discarded at once.
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A train 240 m in length crosses a telegraph post in 16 seconds. The speed of the train is?
Answer: C
Crossing a telegraph post means covering the train's own length.
\[ v = \frac{240}{16} = 15 \text{ m/s} \]
\[ 15 \times \frac{18}{5} = 54 \text{ kmph} \]
The conversion factor \( \tfrac{18}{5} \) turns m/s into kmph; using \( \tfrac{5}{18} \) by mistake would give 4.17 kmph, far too slow for a train.
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Two trains of length 120 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively. In what time will they be clear of each other from the moment they meet?
Answer: D
Trains in the same direction close at the difference of their speeds; in opposite directions, at the sum.
\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]
\[ \text{relative speed} = 42 + 30 = 72 \text{ kmph} = 20 \text{ m/s} \]
\[ t = \frac{120 + 280}{20} = \frac{400}{20} = 20 \text{ sec} \]
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A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:
Answer: C
Let \( a \) be the hours on foot and \( b \) the hours on the bicycle.
\[ a + b = 9, \qquad 4a + 9b = 61 \]
Substituting \( b = 9 - a \):
\[ 4a + 81 - 9a = 61 \quad\Longrightarrow\quad -5a = -20 \quad\Longrightarrow\quad a = 4 \text{ h} \]
\[ \text{distance on foot} = 4 \times 4 = 16 \text{ km} \]
Setting up the unknowns as times rather than distances is what makes the second equation linear; the question then asks for a distance, so one more multiplication is needed at the end.
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