Time & Distance Mcqs

A. 1 hr 42 min
B. 1 hr
C. 2 hr
D. 1 hr 12 min
New speed = 6/7 of usual speed
Speed and time are inversely proportional.
Hence new time = 7/6 of usual time
Hence, 7/6 of usual time – usual time = 12 minutes
=> 1/6 of usual time = 12 minutes
=> usual time = 12 x 6 = 72 minutes
= 1 hour 12 minutes

A. 12
B. 11
C. 10
D. 9
speed of the bus excluding stoppages = 54 kmph
speed of the bus including stoppages = 45 kmph
Loss in speed when including stoppages = 54 – 45 = 9kmph
=> In 1 hour, bus covers 9 km less due to stoppages
Hence, time that the bus stop per hour = time taken to cover 9 km
=distance/speed=9/54 hour=1/6 hour = 60/6 min=10 min

A. 39 seconds
B. 40 seconds
C. 39.38 seconds
D. 39.5 seconds
The first pole is crossed at the zeroth second. The timer starts after the car crosses the first pole. Let the distance between each pole be x m. To reach the 16th pole, the distance travelled by the car is 15x. The speed of the car is 15x/30 = x/2 m/s To reach the 21st pole, it has to travel a distance of 20x. Time taken
= 20x÷(x/2)
= 40 seconds

A. 3 hours 20 minutes
B. 3 hours 15 minutes
C. 3 hours 30 minutes
D. 3 hours
Let his usual speed be x kmph, the distance to his office be y km and his usual travel time be t hrs. y = xt = (5/6)x * (t+(40/60)) Solving the equation,
t = 3.33 hrs = 3 hrs 20 minutes

A. 12 km
B. 14 km
C. 16 km
D. 18 km

A. 8.2
B. 4.2
C. 6.1
D. 7.2

A. 20 minutes
B. 30 minutes
C. 40 minutes
D. 50 minutes
Let his
usual speed be x kmph
usual travel time be t hours
distance to office be d km d=xt d=(0.8x)*[t+(10/60)] xt=0.8xt+(2x/15) Solving,
t=40 minutes

A. 38
B. 40
C. 44
D. 48
Let the speed of the faster train be x miles per hour and
the distance travelled by it when it meets the slower train be y miles. Time taken by the faster train to cover y miles
= Time taken by the slower train to cover (287-y) miles
= 3.5 hours (y/x) = (287-y)/(x-6) = 3.5 Solving, x = 44 miles/hr

A. 2:5
B. 5:3
C. 3:5
D. 1:2
X ———— Y ———— Z
If ‘d’ is the distance between X and Y, then ‘d’ is the distance between Y and Z.
Now the total time for the batsman to row from X to Z is 4 hours. Therefore, time to row from X to Y is 2 hours.
Also the time for the boats man to row from X to Y and back is 10 hours. Hence, time required to row from Y to X is 8 hours.
If, a: speed of boats man in still water
b: speed of the river
d/(a + b) = 2; d/(a – b) = 8
2*(a + b) = 8*(a – b)
a + b = 4a – 4b
3a = 5b
a:b = 5:3

A. 22 km/hr
B. 33 km/hr
C. 66 km/hr
D. 44 km/hr
The relative speed of A and B is 6 km/hr = 1.67 m/s
As the car passes A after 10s, the distance between A and B after 10s (i.e. at 11th second) is the distance covered by car in 1 second.
Therefore, at t = 11, d = 1.67 * 11
d = 18.33 m
v = d/t = 18.33/1 = 18.33m/s
v = 66 km/hr