# COGNIZANT Speed, Time and Distance Questions with Solutions – MeritTrack

1. A man completes his journey in 8 hours. He covers half the distance at 40 kmph and the rest at 60 kmph. The length of the journey is?

A) 384 km
B) 420 km
C) 450 km
D) Cannot be determined
Explanation: Since, he travelled equal distance at 40 kmph and 60 kmph, the ratio of time taken should be 60 : 40
Ie, 3 : 2
Hence he took 23+2=25th23+2=25thtime of the 8 hours, travelling half the distance at 60 kmph.
Hence he travelled at 60 kmph for 25×8=3.2 hours25×8=3.2 hours
Distance travelled at 60 kmph = 3.2 × 60
= 192 km
Total distance travelled = 192 × 2= 384 km

1. A man performs 2/15 of the total journey by train. 2/20 by bus and the remaining 10 km on foot. His total journey in km is

A)15.6
B) 13.04
C) 16.4
D) 12.8
Explanation: Let total journey be ‘x’ km.
A man performs 2/15 of the total journey by train. 2/20 by bus and the remaining 10 km on foot.
∴(2/15)x+(2/20)x+10=xx
⇒ x = 13.04 km

1. A farmer travelled a distance of 61 Km in 9hrs. He travelled partly on foot at the rate of 4km/hr and partly on bicycle at the rate of 9km/hr. The distance travelled in km on foot is :

A)17
B)16
C)14
D)15
Explanation: Time = distance/speed
Given,
Total distance = 61 Km
Let the distance travelled on foot = d km
∴ Distance travelled on bicycle = (61 – d) km
Given, speed on foot = 4km/hr
∴ Time taken to travel on foot = (d/4) hr
Speed on cycle = 9 km/hr
∴ Time taken to travel on cycle = (61-d)/9
Total time taken = (frac{d}{4} + frac{{61 – d}}{9})
Given, Total time = 9hrs
(therefore frac{d}{4} + frac{{61 – d}}{9} = 9)
⇒ 9d + 244 – 4d = 324
⇒ 5d = 80
⇒ d = 16 Km

1. A,B and C start from the same place to walk around a circular path of length 12 km. A walks at the rate of 4 kmph, B at 8 kmph and C at 3/2 kmph. They will meet together at the starting place at the end of :

A) 10 hours
B) 12 hours
C) 15 hours
D) 24 hours
Explanation: Time taken by A, B, C to cover circular path is 12/4, 12/8, 12/(1.5) respectively
Time interval after which they will all meet at starting point = LCM of 3, 8, 1.5
= LCM (3, 8, 1.5)
= 24 hours

1. Two cars travel from city A to city B at a speed of 42 and 60 km/hr respectively. If one car takes 2 hours lesser time than the other car for the journey, then the distance between City A and City B is:

A) 336 km
B) 280 km
C) 420 km
D) 224 km
Explanation: Let the distance between the city A and city B be ‘x’ km.
Speed of car 1 = 42 km/hr
Speed of car 2 = 60 km/hr
We know that,
Time = Distance/Speed
Time taken by car 1 to travel from city A to city B = x/42
Time taken by car 2 to travel from city A to city B = x/60
(x/42) – (x/60) = 2                (∵ one car takes 2 hours lesser time than the other car)
x = 280 km
∴ The distance between City A and City B = 280 km

1. Pihu and Aayu are running on a circular track of diameter 28 m. Speed of Pihu is 48 m/s and that of Aayu is 40 m/s. They start from the same point at the same time in the same direction. When will they meet again for the first time?

A) 8 seconds
B) 11 seconds
C) 13 seconds
D) 14 seconds
Explanation: As per the given data,
Length of the circular track = π × d = π × 28 m
Given that Pihu and Aayu are running in the same direction at the same point
We know that when two persons A and B running around a circular track of length L mts with speed of a, b m/s in the same direction
They meet each other at any point on the track = L/(a – b) sec
= 28 π/(48 – 40)
= 11 sec

1. A car driver leaves Bangalore at 8.30 A.M. and expects to reach a place 300 km from Bangalore at 12.30 P.M. At 10.30 he finds that he has covered only 40% of the distance. By how much he has to increase the speed of the car in order to keep up his schedule?

A) 45 km/hr
b) 40 km/hr
C) 35 km/hr
D) 30 km/hr
Explanation: Speed = distance/time
Let the original speed be ‘v’ km/hr
Given, car driver leaves Bangalore at 8.30 A.M. and expects to reach a place 300 km from Bangalore at 12.30 P.M.
Time of journey = 4 hours
Given, at 10:30 he found that he has covered 40% of the distance.
Distance covered = 40% of 300 = 120 km
Time to cover this 120 km = 2 hours
∴ Speed = 120/2 = 60 km/hr
Distance to be covered = (300 – 120) = 180 km
Time left = 4 – 2 = 2 hours
∴ Speed at which he has to travel = 180/2 = 90 km/hr
∴ Increase in speed = 90 – 60 = 30 km/hr

1. In a 100 m race, A beats B by 15 mand C by 18 m. In a race of 170 m, B will beat C by

A) 6 m
B) 10 m
C) 8 m
D) 9 m
Explanation: Let, x, y and z are the distances covered by A,B and C respectively
Given, In a 100 m race, A beats B by 15 mand C by 18 m
∴ Ratio of distance covered by A and B,
x : y = 100: 85⇒ x/y = 100/85
Ratio of distance covered by A and C,
x: z = 100: 82⇒ x/z = 100/82
∴ y : z = 85: 82
∴ When B covers 85 m, distance covered by C = 82 m
∴ When B covers 170 m, distance covered by C = (82/85) × 170 = 164 m
∴ B will beat C by = 170 m – 164 m = 6 m

1. A boat goes 15 km an hour in still water, and takes thrice the time to cover the same distance upstream. The speed of the current (in km / hr) is –

A) 10 km/hr
B) 12 km/hr
C) 13 km/hr
D) 14 km/hrs
Explanation: Let the speed of stream be ‘a’ and speed of boat be ‘b’.
In still water, speed of boat = b
In upstream speed of boat relative to stream = b – a
Given, boat goes 15 km an hour in still water, and takes thrice the time to cover the same distance upstream.
Speed = distance/time
b = 15/1
⇒ b = 15 km/hr
b – a = 15/3
⇒ b – a = 5
⇒ a = b – 5 = 10 km/hr

10.A man can row at 7 km/hour in still water. He finds that it takes twice the time to row upstream than the time to row downstream. The speed of the stream is

A) 2.6 km/hour
B) 7 km/hour
C) 2.3 km/hour
D) 4 km/hour
Explanation: The speed of the stream is y km/hour (say)
Hence upstream speed will be (7 – y) km/hour
And downstream speed will be (7 + y)km/hour
it takes twice the time to row upstream than the time to row downstream. Hence, speed of downstream is twice than the speed of upstream.
According to the problem, 7 + y = 2(7 – y)
⇒7 + y = 14 – 2y
⇒3y = 7
⇒y = 2.33

Speed of the stream = 2.33 km/hour

11.A boat goes 10kms an hour in still water, but takes twice as much time in going the same distance against the current. The speed of the current (in km/hr) is –

A) 2 km/hr
B) 4 km/hr
C) 3 km/hr
D) 5 km/hr
Explanation: Speed of Boat in Current = Speed of Boat in still water – Speed of Current
A boat goes 10 kms an hour in still water
∴ Speed of Boat in still water = 10km/hr
Let it travels for 1 hr and covered 10 km
It takes twice as much time in going the same distance against the current
Thus, it will take 2 hr to cover the same distance of 10 km
∴ Speed of Boat against Current = 10km / 2hrs = 5km/hr
Speed of Current = 10km/hr – 5km/hr = 5km/hr

12.A boat goes downstream in one-third the time it takes to go upstream. Then the ratio between the speed of boat in still water and speed of the stream is

A) 5 : 1
B) 3 : 2
C) 1 : 2
D) 2 : 1
Explanation: Let the speed of boat be ‘a’ and speed of stream be ‘b’.
Relative speed of boat going upstream = a – b
Relative speed of boat going downstream = a + b
Given, boat goes downstream in one-third the time it takes to go upstream.
Time = distance/speed
Distance is same in both cases.
(therefore frac{d}{{a + b}} = frac{1}{3} times frac{d}{{a – b}})
⇒ 3a – 3b = a + b
⇒ a = 2b
⇒ a : b = 2 : 1

13.The speed of a boat along the stream is 12 km/h and against the stream is 8 km/h. the time taken by the boat to sail 24 km in still water is

A) 2 h
B) 3 h
C) 2.4 h
D) 1.2 h
Explanation: Let the speed of the boat be C and the velocity of the stream be V.
Hence,
C + V = 12 and
C – V = 8
Adding both the equations to eliminate V, we have
2C = 20
⇒ C = 10 kmph
Hence, the time taken by the boat to cover 24 km in still water = 24/10 = 2.4 h

14.A boat running at a speed of 34 km/h downstream covers a distance of 4.8 km in 8 minutes. The same boat while running upstream at same speed covers the same distance in 9 minutes. What is the speed of the current?

A) 2.4 km/h
B) 3 km/h
C) 2 km/h
D) 3.2 km/h
Explanation: Let the speed of the water current be ‘x’ km/hr
Downstream:
While going downstream,
Total speed = speed of the boat + speed of the water current
⇒ Total speed = 34 + x
Distance covered = 4.8 km
Time taken = 8 min = 0.133 hrs
We know that
Speed = Distance/Time
⇒ 34 + x = 4.8/0.133
⇒ 34 + x = 36
⇒ x = 2
Thus speed of the Current is 2 km/hr
(The same can be confirmed using the upstream condition where the total speed will be the difference between the speed of the boast and that of the current)

15.A boat is rowed downstream at 15.5 km/hr and upstream 8.5 km/hr. The speed of the stream is

A) 3.5 km/hr
B) 5.75 km/hr
C) 6.5 km/hr
D) 7 km/hr
Explanation: Let the speed of boat in still water be x and that of stream be y, so we have the following equations,
x + y = 15.5 km/hr               …(1)and,
x – y = 8.5 km/hr …(2)
Now we subtract (2) from (1) we get,
⇒ 2y = 7 km/hr
⇒ y = 3.5 km/hr

1. A steamer running downstream covers a distance of 30 km in 2 hours. While coming back the steamer takes 6 hours to cover the same distance. If the speed of the current is half of that of the steamer, then find the speed of the steamer in kmph.

A) 10 kmph
B) 12 kmph
C) 18 kmph
D) 20 kmph
Explanation: Speed of the steamer downstream Sd = 30/2 = 15 kmph
Speed of the steamer upstream Su = 30/6 = 5 kmph
Let speed of the steamer be ‘x’, then speed of the current would be (x/2).
According to the question,
x + (x/2) = 15 ……………..(1)
x – (x/2) = 5 ………………..(2)
From equations (1) and (2) we get,
2x = 20 ⇒ x = 10 kmph
Hence, speed of the steamer is 10 kmph.

17.In a fixed time, a boy swims double the distance along the current that he swims against the current. If the speed of the current is 3 kmph, the speed of the boy in still water is:

A) 6 kmph
B) 9 kmph
C) 10 kmph
D) 12 kmph
Explanation: We know that Speed = Distance/Time
Let speed of boy in still water be g kmph.
Then speed of boy in upstream = g – 3 kmph
Speed of boy in downstream = g + 3 kmph
We have the boy covering double the distance in downstream than in upstream in the same time.
Hence, the speed of boy in downstream should be double his speed in upstream
Hence,
We have
⇒ (g + 3) = 2 × (g – 3)
⇒ g + 3 = 2g –
⇒ g = 9 kmph

18.A bus covers a distance in 6 minutes. If it runs at 30 kmph on an average speed, then the speed at which the bus should run to increase the time of journey to 30 minutes will be –

A) 5 kmph
B) 8 kmph
C) 6 kmph
D) 7 kmph
Explanation: We know that, Distance traveled = Average speed × Time required
Given, Average velocity = 30 kmph
= (30/60) km/min ———– (1 Hour = 60 Minutes)
= 0.5 km/min
Time required = 6 min
∴ Distance traveled = (0.5 × 6) km = 3 km
Now, Time required = 30 min
∴ Average speed = Distance traveled/Time required
= 3 km/30 min
= 0.1 km/min
= 0.1 × 60 kmph ———– (1 Hour = 60 Minutes)
= 6 kmph

19.In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:

A) 1 hour
B) 2 hours
C) 3 hours
D) 4 hours
Explanation: Let, the original duration of flight be t hrs.
∵ Average Speed = Total Distance / Time
Total Distance of the flight is 600 km.
So, original average speed = 600/t
Due to bad weather speed of trip is reduced by 200 km/hr and time of flight is increased by 30 minutes i.e. 0.5 hr.
∴ Reduced average speed = (600/t) – 200
And New duration of flight = (t + 0.5) hrs
So, the new average speed = 600/(t + 0.5)
Equating,
(begin{array}{l} Rightarrow frac{{600}}{t}-200 = frac{{600}}{{t + 0.5}}\ Rightarrow frac{3}{t}-1 = frac{3}{{t + 0.5}}\ Rightarrow frac{{3-t}}{t} = frac{3}{{t + 0.5}} end{array})
⇒ t2 + 0.5t – 1.5 = 0
⇒ 2t2 + t – 3 = 0
⇒ 2t2 – 2t + 3t – 3 = 0
⇒ (t – 1)(2t + 3) = 0
⇒ (t – 1) = 0
∴ t = 1 hr

20.How does a train 220 meters long, running at the rate of 54 km an hour, take to cross a bridge 120 meters in length?

A) 28.33 seconds
B) 22.67 seconds
C) 20.67 seconds
D) 19 seconds