Time & Distance Mcqs

A. 36
B. 24
C. 30
D. 18
Let t be his usual time to reach his office and v be his usual speed.
v = d/t ……….(d is the distance Afnan travels while going to his office)
vt = d
At v1 = 4v/5 ; t1 = t + 6
4v/5 = d/(t + 6)
4v/5* (t + 6) = d
4v/5* (t + 6) = vt
On solving we get,
t = 24 minutes

A. 17 hr
B. 14 hr
C. 12 hr
D. 19 hr
Relative speed = 5.5 – 5 = .5 kmph (because they walk in the same direction)
distance = 8.5 km
time = distance/speed=8.5/.5=17 hr

A. 30 kmph
B. 27 kmph
C. 25 kmph
D. 20 kmph
Distance = 24*50/60 = 20 km New Speed = 20/(40/60)) = 30 kmph

A. 9
B. 10
C. 12
D. 20
Due to stoppages, the bus travels only 45 kms in an hour (9 kms less). To cover a distance of 9 km at a speed of 54 kmph, time taken
= 9/54 = 1/6 hrs = 10 mins.

A. (7 , 3) km/hr
B. (6 , 4) km/hr
C. (10 , 4) km/hr
D. None of these
If x: speed of boats man in still water
y: speed of the river
Downstream speed (Ds) = x + y
Upstream speed (Us) = x – y
x = (Ds + Us) / 2
y = (Ds – Us) / 2
In the above problem Ds = 10 ; Us = 4
x = (10 + 4) / 2 = 14/2 = 7 km/hr
y = (10 – 4)/2 = 6/2 = 3 km/hr

A. 2.5 km
B. 7.2 km
C. 7 km
D. 3.6 km
Average speed of Taimoor = (2*2*3)/(2 + 3) = 2.4 km/hr
If d is he distance between the school and Taimoor’s home, then total distance walked by Taimoor = 2d, and the total time taken is 3 hours
2.4 = 2d/3
7.2 = 2d
d = 3.6 km

A. 80 kmph
B. 102 kmph
C. 120 kmph
D. 140 kmph

A. 5
B. 6
C. 7
D. 8
Relative speed = Speed of A + Speed of B (∴ they walk in opposite directions)
= 2 + 3 = 5 rounds per hour
=> They cross each other 5 times in 1 hour and 2 times in 1/2 hour
Time duration from 8 am to 9.30 am = 1.5 hour
Hence they cross each other 7 times before 9.30 am

A. 5 kmph
B. 4 kmph
C. 4.5 kmph
D. 3 kmph
If he travels at 6 kmph,
time taken to reach = 12/6 = 2 hours But he takes 3 hours.
Speed = 12/3 = 4 kmph

A. 50 km
B. 45 km
C. 40 km
D. 30 km