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The J&K Express from Delhi to Srinagar was delayed by snowfall for 16 minutes and made upfor the delay on a section of 80 km travelling with a speed 10 km per hour higher than its normal speed. Find the original speed of the J&K Express (according to the schedule)
By travelling at 10 kmph higher than the original speed, the train is able to make up 16 minutes
while traveling 80 km.
This condition is only satisfied at an initial speed of 50 (and a new speed of 60 kmph).
A pedestrian and a cyclist start simultaneously towards each other from Aurangabad and Paithanwhich are 40 km apart and meet 2 hours after the start. Then they resumed their trips and thecyclist arrives at Aurangabad 7 hours 30 minutes earlier than the pedestrian arrives at Paithan.Which of these could be the speed of the pedestrian?
The relative speed is 20 kmph. Also, the pedestrian should take 7:30 hours more than the cyclist.
Using option (a) the speeds of the two people are 4km/hr and 16 km/hr respectively. At this speed,
the respective times would be 10 hrs and 2:30 hours, giving the required answer.
Two cyclists start simultaneously towards each other from Aurangabad and Ellora, which are 28km apart. An hour later they meet and keep pedalling with the same speed without stopping. Thesecond cyclist arrives at Ellora 35 minutes later than the first arrives at Aurangabad. Find thespeed of the cyclist who started from Ellora.
Since the two motorists meet after an hour, their relative speed is 28 kmph. Use options to check
out the values. Since the speed of the faster cyclist is asked for it has to be greater than 14 kmph.
Hence only check options > 14 kmph.
A bus left point X for point Y. Two hours later a car left point X for Y and arrived at Y at the same time as the bus. If the car and the bus left simultaneously from the opposite ends X and Y towards each other, they would meet 1.33 hours after the start. How much time did it take the bus to travelfrom X to Y?
In this question consider the total distance as 100%. Hence the sum of their speeds will be 75%
per hour. Checking option (c)
If the bus took 6 hours, it would cover 16.66% distance per hour and the car would cover 25%
distance per hour. (as it takes 2 hours less than the bus.)
This gives an addition of only 41.66%. Hence, the answer is not correct.
Option (b) is the correct answer.
Two planes move along a circle of circumference 1.2 km with constant speeds. When they move in different directions, they meet every 15 seconds and when they move in the same direction, oneplane over takes the other every 60 seconds. Find the speed of the slower plane.
The sum of speeds would be 0.08 m/s (relative speed in opposite direction). Also if we go by
option (b), the speeds will be 0.03 and 0.05 m/s respectively.
At this speed the overlapping would occur every 60 seconds.
Two Indian tourists in the US cycled towards each other, one from point A and the other from point B. The first tourist left point A 6 hrs later than the second left point B, and it turned out ontheir meeting that he had travelled 12 km less than the second tourist. After their meeting, they kept cycling with the same speed, and the first tourist arrived at B 8 hours later and the second arrivedat A 9 hours later. Find the speed of the faster tourist.
This is a complex trial and error based question and the way you would have to think in this is:
From the figure above, it is clear that A is faster as he takes only t + 2 hours while B has taken t +
9 hours to complete the journey.
Then, we get: (t – 6)/9 = 8/t
Solving for t, we get t = –6 (not possible)
Or t = 12. Putting this value of t in the figure it changes to:
We also get ratio of speeds = 3 : 2 (inverse of ratio of times)
The next part of the puzzle is to think of the 12 km less traveled by the first person till the meeting
point.
If the speed of the faster person is 3s, that of the slower person = 2s.
Further
12 × 2s – 6 × 3s = 12 km
s = 2 kmph.
The speed of the faster tourist is 3 × 2 = 6 kmph
Two ports A and B are 300 km apart. Two ships leave A for B such that the second leaves 8 hours after the first. The ships arrive at B simultaneously. Find the time the slower ship spent on the trip if the speed of one of them is 10 km/h higher than that of the other.
If the slower ship took 20 hours (option d) the faster ship would take 12 hours and their respective
speeds would be 15 and 25 kmph. This satisfies the basic condition in the question.
The Sabarmati Express left Ahmedabad for Mumbai. Having travelled 300 km, which constitutes 66.666 per cent of the distance between Ahmedabad and Mumbai, the train was stopped by a red signal. Half an hour later, the track was cleared and the enginedriver, having increased the speedby 15 km per hour, arrived at Mumbai on time. Find the initial speed of the Sabarmati Express.
When the signal happened distance left was 150 km.
150/(s) – 150/(s + 15) = 1/2 hours Æ s = 60.
Two horses started simultaneously towards each other and meet each other 3 h 20 min later. How much time will it take the slower horse to cover the whole distance if the first arrived at the place of departure of the second 5 hours later than the second arrived at the point of departure of thefirst?
Since the two horses meet after 200 minutes, they cover 0.5% of the distance per minute
(combined) or 30% per hour. This condition is satisfied only if you the slower rider takes 10
hours (thereby covering 10% per hour) and the faster rider takes 5 hours (thereby covering 20%
per hour).
A motorcyclist rode the first half of his way at a constant speed. Then he was delayed for 5minutes and, therefore, to make up for the lost time he increased his speed by 10 km/h. Find the initial speed of the motorcyclist if the total path covered by him is equal to 50 km.
v₁ = speed in first half
D = total distance traveled = 50 km
d₁ = distance of the first half = 25 km
t₁ = time taken in first half
t₁ = d₁ /v₁
t₁ = 25/v₁ (1)
for the second half :
v₂ = speed in second half = v₁ + 10
d₂ = distance of the second half = 25 km
t₂ = time taken in second half
t₂ = d₂ /v₂
t₂ = 25/(v₁ + 10) (2)
given that :
t₁  t₂ = 5/60 (5 min in h = 5/60 h)
(25/v₁ )  (25/(v₁ + 10)) = 5/60
v₁ = 50 km/h
Ravi, who lives in the countryside, caught a train for home earlier than usual yesterday. His wife normally drives to the station to meet him. But yesterday he set out on foot from the station to meet his wife on the way. He reached home 12 minutes earlier than he would have done had he waitedat the station for his wife. The car travels at a uniform speed, which is 5 times Ravi’s speed onfoot. Ravi reached home at exactly 6 O’clock. At what time would he have reached home if hiswife, forewarned of his plan, had met him at the station?
The wife drives for 12 minutes less than her driving on normal days.
Thus, she would have saved 6 minutes each way. Hence, Ravi would have walked for 30 minutes
(since his speed is 1/5th of the car’s speed).
In effect, Ravi spends 24 minutes extra on the walking (rather than if he had traveled the same
distance by car).
Thus, if Ravi had got the car at the station only, he would have saved 24 minutes more and reached
at 5 : 36.
The metro service has a train going from Mumbai to Pune and Pune to Mumbai every hour, thefirst one at 6 a.m. The trip from one city to other takes 4*1/2hours, and all trains travel at the samespeed. How many trains will you pass while going from Mumbai to Pune if you start at 12 noon?
If you start at 12 noon, you would reach at 4:30 PM. You would be able to meet the train which
left Mumbai at 8 AM, 9 AM, 10 AM, 11 AM, 12 Noon, 1 PM, 2 PM, 3 PM and 4 PM – a total of 9
trains.
Two rifles are fired from the same place at a difference of 11 minutes 45 seconds. But a man who is coming towards the place in a train hears the second sound after 11 minutes. Find the speed oftrain.
If we assume the speed of the sound as 330 m/s, we can see that the distance traveled by the sound
in 45 seconds is the distance traveled by the train in 11 minutes.
330 × 45 = 660 × s Æ s = 22.5 m/s = 81 kmph
A dog sees a cat. It estimates that the cat is 25 leaps away. The cat sees the dog and starts running with the dog in hot pursuit. If in every minute, the dog makes 5 leaps and the cat makes 6 leaps andone leap of the dog is equal to 2 leaps of the cat. Find the time in which the cat is caught by thedog (assume an open field with no trees)
Initial distance = 25 dog leaps.
Per minute Æ dog makes 5 dog leaps
Per minute Æ Cat makes 6 cat leaps = 3 dog leaps.
Relative speed = 2 dog leaps/minutes.
An initial distance of 25 dog leaps would get covered in 12.5 minutes.
A wall clock gains 2 minutes in 12 hours, while a table clock loses 2 minutes in 36 hours; bothare set right at noon on Tuesday. The correct time when they both show the same time next would be
In 36 hours, there would be a gap of 8 minutes. The two watches would show the same time when
the gap would be exactly 12 hours or 720 minutes.
The no. of 36 hour time frames required to create this gap = 720/8 = 90.
Total time = 90 × 36 = 3240 hours. Since this is divisible by 24, the watches would show 12
noon.
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