Time, Speed and Distance is a significant topic in the quantitative aptitude section. Practice the Time, Speed and Distance for bank Clerk prelims exam provided here. Solving the questions from this topic requires applying the formula correctly and solving with good speed. The Time, Speed and Distance for bank Clerk prelims exam will be asked in the application problems. This topic is also expected in the data interpretation section. The Time, Speed and Distance questions for bank exams are provided in the quiz as well as PDF format. Understand the basics and solve these questions with good speed and accuracy. Aspirants should try to understand the interrelationship between the factors speed, distance and time. Time, Speed and Distance for bank Clerk prelims is one of the most popular and important topics in the Quantitative Aptitude section of any competitive exam. Speed is represented in meters (m), kilometers (km), miles, feet. Time is represented in seconds(s), minutes (min), hours (hr). Speed is represented in m/s, km/hr.
Download the Time, Speed and Distance for bank Clerk prelims exam pdf and solve the questions. Now, time speed and distance is something that we come across each day, but we don’t really think about the relationship until we take an actual test. The Time, Speed and Distance for bank Clerk prelims exam pdf is provided with solutions and detailed explanations. Candidates should analyse the scores and learn the new shortcuts and tricks provided. Score good marks in this section and clear your basic concepts with this Time, Speed and Distance for bank Clerk prelims exam pdf provided here. Candidates can attempt the questions by setting a specific time in the form of a quiz or understand the concepts by downloading the PDFs and solving the Time, Speed and Distance questions.
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1) Sanjay runs at a speed of 45kmph from Chennai to Mangalore. Vijay runs at a speed of 35 kmph from Mangalore to Chennai. But Vijay starts after 3 hours when Sanjay start. Vijay and Sanjay decided to meet each other. What is the distance between meeting point and Chennai if the distance between Chennai and Mangalore is 535 km?
A.400 km
B.260 km
C.360 km
D.300 km
E.None of these
2) A car has travelled from place A to place B. There are two toll gates T1 and T2 between A and B. The time taken by car to travel between A and T1 is 2 hrs and between T1 and T2 is 4 hours and between T2 and B is 3 hours. The ratio of speeds in these paths is 12:9:10. Find the distance between the toll gates, if the average speed of the car is 50 km/hr.
A.60km
B.90km
C.240km
D.180km
E.120km
3) Bike A moving with the speed of 45 km/h covers x km distance in 6 hours and bike B covers (x + 50) km distance in 8 hours. If bike A decreased its speed by 5 km/h and bike B increases its speed by 10 km/h, then find the difference between the time taken by bike A and bike B to cover 200km.Â
A.2 hours
B.1 hour
C.2.5 hours
D.1.5 hours
E.0.5 hour
4) A bike travelled from Chennai to Bangalore at the speed of (x + 10) kmph and returned from Bangalore to Chennai at the speed of (x – 10) kmph. Distance between Chennai and Bangalore is 600 km. If the time taken to complete the whole journey is 25 hours and the usual speed is x kmph, then find the value of x.
A.50 kmph
B.60 kmph
C.70 kmph
D.80 kmph
E.None of these
5) If the sum of the speed of bike and bus is 160 kmph and the ratio of the time taken by bike to cover the whole distance and bus to cover 90% of the distance is 10: 7. What is the average speed of the bus and bike?
A.65.75 kmph
B.72.75 kmph
C.78.75 kmph
D.87.75 kmph
E.None of these
6) Banu leaving the school at 5 pm at the speed of 75% of her original speed and after 40 minutes she reached the fruits shop bought some fruits, then travel towards home. If the time taken by Banu to reach the school is one hour at the regular speed, then the distance between fruit shop and home is approximately what percent of the distance between school and home?
A.40%
B.50%
C.60%
D.70%
E.None of these
7) A person leaves the home by 7.55 am and travelled with a speed of 5 km/hr and reaches office 7 minutes late. Had the speed increased by 1 km/hr, he would reach the office 5 minutes earlier. Find at what time he wants to reach the office.
A.9.05 am
B.9 pm
C.9 am
D.8.55 am
E.8.50 am
8) A bike travels a distance of 360 km with certain speed. If the bike travels 25% of the distance at speed of 60% of the usual speed, 80% of the remaining distance at the speed of 80% of the usual speed and the remaining distance at 90% of the usual speed. If the total time taken by the bike is 8 hours, then find the usual speed of the bike?
A.40 kmph
B.50 kmph
C.60 kmph
D.80 kmph
E.None of these
9) If the sum of the time taken by Rahul travels x/2 km at the speed of 20 kmph and 3x/4 km at the speed of y kmph is equal to the time taken by him to travel x km at the speed of 20 kmph, then find the value of y?
A.12.5 kmph
B.13.5 kmph
C.14.5 kmph
D.15.5 kmph
E.None of these
10) A car and a bike started together. The ratio of the speed of the car and the bike is 5:8 for first hour. After that, the speed of the car is increased by 10% and the speed of the bike is decreased by 5%. If the distance covered by the car in 5 hours is 324km, then find the distance covered by bike in four hours.
A.369.6km
B.353.6km
C.323.4km
D.342.6km
E.352.6km
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Answers :
1) Answer: C
At the time when Vijay starts, Sanjay travelled 45 * 3 = 135 km
Remaining distance, 535 – 135 = 400 km
400 = (35 + 45) * t
t = 5
Distance between meeting point and Chennai = 45 * (5+3) = 360 km
2) Answer: D
Ratio of speed = 12:9:10
Let the speeds be 12x, 9x and 10x
Distance covered by car = 12x*2 + 9x*4 + 10x*3
= 24x+36x+30x = 90x
Total time = 2+4+3 = 9hrs
Average speed = 50 km/hr
Total distance/Total time = 50
90x/9 = 50
x = 5
Distance between the tolls = 36x = 180km
3) Answer: B
Speed of bike A = 45 kmph
Distance covered by bike A = 45 * 6 = 270 km
Speed of bike B = (270 + 50)/8 = 320/8 = 40 kmph
New speed of bike A = (45 – 5) kmph = 40 kmph
New speed of bike B = (40 + 10) kmph = 50 kmph
Time taken by bike A to cover 200km = 200/40 = 5 hours
Time taken by bike B to cover 200km = 200/50 = 4 hours
Required difference in time = (5 – 4) hours = 1 hour
4) Answer: A
Let Usual speed = x
600/(x +10)+600/ (x -10)=25
24(x – 10) + 24(x + 10) = (x2 – 100)
24x – 240 + 24x + 240 = x2 – 100
x2 – 48x – 100 = 0
x2 – 50x + 2x – 100 = 0
x(x – 50) + 2(x – 50) = 0
(x + 2)(x – 50) = 0
x = -2, 50(negative value neglected)
5) Answer: C
Speed of bus = x
Speed of bike = y
Distance = d
Time taken by bike = t
Time taken by bus to cover 90% of the distance = 7t/10
x = (90/100 * d) / (7t/10)
= 9d/7t
y = d/t
x: y = 9d/7t: d/t
= 9: 7
Speed of bus = 9/16 * 160 = 90 kmph
Speed of bike = 7/16 * 160 = 70 kmph
Average speed = 2 * x * y/(x + y)
= 2 * 90 * 70/160 = 78.75 kmph
6) Answer: B
Let the speed be x km/hr
Total distance between school and Home = x * 1 hr = x km
75% of the original speed = x * 75/100 = 3x/4
Distance covered in 40 minutes with new speed = 3x/4 * 40/60
= 3x/4 * 2/3 = x/2
Distance between fruit shop and home = x – x/2 = x/2
Required percentage = (x/2)/x * 100
= ½ * 100 = 50%
7) Answer: C
Let d and t be the distance and desired time.
Given, (d/5) = t + (7/60)
d= 5t + (7/12)
(d/6) = t – (5/60)
d= 6t – (6/12)
5t + 7/12 = 6t – (6/12)
t = 13/12 hours = 65 minutes
Therefore desired time = 7.55 am + 65 minutes = 9 am
8) Answer: C
Usual speed of the bike = x
25% of the distance = 360 * 25/100 = 90 km
Time taken by travels 25% of the distance = 90/(60 * x/100) = 150/x
80% of the remaining distance = 270 * 80/100 = 216 km
Time taken by travels 216 km = 216/(80/100 * x) = 270/x
Remaining distance = 270 – 216 = 54 km
Time taken by remaining distance = 54/(90/100 * x) = 60/x
150/x + 270/x + 60/x = 8
x = 60 kmph
9) Answer: E
((x/2)/20 + (3x/4)/y) = x/20
x/40 + 3x/4y = x/20
3x/4y = x/40
y = 30
10) Answer: A
Let the initial speed of speed of car = 5x km/hr
And initial speed of bike = 8x km/hr
Increased speed of car = 5x*{(100+10)/100} = 5x*(110/100) = 5.5x
Decreased speed of bike = 8x*{(100-5)/100} = 8x*(95/100) = 7.6x
Distance covered by car in 5 hours = 5x*1 + 5.5x*4 = 5x + 22x = 27x
27x = 324
x = 12
Distance covered by bike in 4 hours = 8x*1 + 7.6x*3 = 30.8x = 30.8*12 = 369.6 km
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