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.
Please log in or register to report this question.
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.
Please log in or register to report this question.
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.
Please log in or register to report this question.
Walking 7/6 of his usual rate, a boy reaches his school 4 min early. Find his usual time to reach the school?
Answer: D
Speed and time are inversely proportional over a fixed distance, so walking at \( \tfrac{7}{6} \) of his rate takes \( \tfrac{6}{7} \) of the time.
\[ t - \frac{6}{7}t = \frac{1}{7}t = 4 \text{ min} \]
\[ t = 28 \text{ minutes} \]
Checking: the faster walk takes 24 minutes, exactly 4 less.
Please log in or register to report this question.
Two cars cover the same distance at the speed of 60 and 64 kmps respectively. Find the distance traveled by them if the slower car takes 1 hour more than the faster car.
Answer: C
The slower car takes one hour longer over the same distance.
\[ \frac{d}{60} - \frac{d}{64} = 1 \]
\[ d\left(\frac{64 - 60}{3840}\right) = 1 \quad\Longrightarrow\quad \frac{4d}{3840} = 1 \]
\[ d = 960 \text{ km} \]
The unit matters: the answer is in kilometres, since the speeds were given in kmph.
Please log in or register to report this question.
A man leaves a point P at 6 a.m. and reaches the point Q at 10 a.m. another man leaves the point give at 8 a.m. and reaches the point P at 12 noon. At what time do they meet?
Answer: C
Each man takes four hours, so their speeds are equal - call each \( \tfrac{d}{4} \).
By 8 a.m. the first has already covered half the distance:
\[ 2 \times \frac{d}{4} = \frac{d}{2} \]
The gap is then \( \tfrac{d}{2} \), closing at their combined speed \( \tfrac{d}{2} \) per hour:
\[ t = \frac{d/2}{d/2} = 1 \text{ hour} \]
So they meet at 9 a.m. The distance cancels, so it never needs a value.
Please log in or register to report this question.
A thief goes away with a SANTRO car at a speed of 40 kmph. The theft has been discovered after half an hour and the owner sets off in a bike at 50 kmph when will the owner over take the thief from the start?
Answer: A
The thief's half-hour head start puts him
\[ 40 \times 0.5 = 20 \text{ km} \]
ahead when the owner sets off. The gap then closes at the difference of the speeds:
\[ 50 - 40 = 10 \text{ kmph} \]
\[ t = \frac{20}{10} = 2 \text{ hours} \]
That is two hours from the owner's start - two and a half from the theft itself.
Please log in or register to report this question.
If I walk at 3 kmph, I miss the train by 2 min, if however, I walk at 4 kmph. I reach the station 2 min before the arrival of the train. How far do I walk to reach the station?
Answer: A
The two walks differ by 4 minutes in arrival: 2 late against 2 early.
\[ \frac{d}{3} - \frac{d}{4} = \frac{4}{60} = \frac{1}{15} \text{ h} \]
\[ \frac{d}{12} = \frac{1}{15} \quad\Longrightarrow\quad d = \frac{12}{15} = \frac{4}{5} \text{ km} \]
Late and early add to give the gap of 4 minutes; taking their difference would give zero.
Please log in or register to report this question.
Two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph respectively. In what time will they cross each other completely?
Answer: C
Trains in the same direction close at the difference of their speeds; in opposite directions, at the sum.
\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]
\[ \text{relative speed} = 80 + 70 = 150 \text{ kmph} = \frac{125}{3} \text{ m/s} \]
\[ t = \frac{250 + 250}{125/3} = \frac{500 \times 3}{125} = 12 \text{ sec} \]
Both lengths are covered, since the trains must clear each other completely.
Please log in or register to report this question.
Two trains of equal length, running with the speeds of 60 and 40 kmph, take 50 seconds to cross each other while they are running in the same direction. What time will they take to cross each other if they are running in opposite directions?
Answer: A
Running the same way, they close at \( 60 - 40 = 20 \) kmph \( = \tfrac{50}{9} \) m/s, and cover both lengths:
\[ 2L = \frac{50}{9} \times 50 = \frac{2500}{9} \text{ m} \]
Running opposite ways they close at \( 100 \) kmph \( = \tfrac{250}{9} \) m/s over the same total length:
\[ t = \frac{2500/9}{250/9} = 10 \text{ sec} \]
The combined length cancels, so it never has to be found - the answer is simply 50 seconds scaled by the ratio of the two relative speeds.
Please log in or register to report this question.
