Test: Speed, Time & Distance - 5 - SSC CGL MCQ

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).

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.

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.

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.

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.

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

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.

When the signal happened distance left was 150 km.
150/(s) – 150/(s + 15) = 1/2 hours Æ s = 60.

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).

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

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.

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.

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

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.

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.