Time and Distance
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Quiz for Banking Exams
No. Of Questions :15
Time Limit : 15 minutes
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Question 1 of 15
1. Question
1 pointsA man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
Correct
Let time taken to travel the first half = x hr
Then time taken to travel the second half = (10 – x) hrDistance covered in the the first half = 21x [because, distance = time*speed]
Distance covered in the the second half = 24(10 – x)Distance covered in the the first half = Distance covered in the the second half
So,
21x = 24(10 – x)
=> 45x = 240
=> x = 16/3Total Distance = 2*21(16/3) = 224 Km [multiplied by 2 as 21x was distance of half way]
Incorrect
Let time taken to travel the first half = x hr
Then time taken to travel the second half = (10 – x) hrDistance covered in the the first half = 21x [because, distance = time*speed]
Distance covered in the the second half = 24(10 – x)Distance covered in the the first half = Distance covered in the the second half
So,
21x = 24(10 – x)
=> 45x = 240
=> x = 16/3Total Distance = 2*21(16/3) = 224 Km [multiplied by 2 as 21x was distance of half way]

Question 2 of 15
2. Question
1 pointsA Man travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot?
Correct
Let the time in which he travelled on foot = x hour
Time for travelling on bicycle = (9 – x) hrDistance = Speed * Time, and Total distance = 61 km
So,
4x + 9(9x) = 61
=> 5x = 20
=> x = 4So distance traveled on foot = 4(4) = 16 km
Incorrect
Let the time in which he travelled on foot = x hour
Time for travelling on bicycle = (9 – x) hrDistance = Speed * Time, and Total distance = 61 km
So,
4x + 9(9x) = 61
=> 5x = 20
=> x = 4So distance traveled on foot = 4(4) = 16 km

Question 3 of 15
3. Question
1 pointsA car starts at 10 am with a speed of 50 km/hr. Due to the problem in engine it reduces its speed as 10 km/hr for every 2 hours. After 11 am, the time taken to covers 10 km is:
Correct
Initial speed of the car = 50km/hr
Due to engine problem, speed is reduced to 10km for every 2 hours(i.e., 5 km per hour).Speed of the car at 11 am = (50 – 5) = 45km/hr
Time to cover 10 km at 45 km/hr = distance/speed = 10/45 hours. = 2/9 hours
= 2/9 x 60 minutes = 40/3 minutes = 13 minutes + 1/3 minutes
= 13 minutes + 1/3 x 60 seconds
= 13 minutes and 20 seconds.Incorrect
Initial speed of the car = 50km/hr
Due to engine problem, speed is reduced to 10km for every 2 hours(i.e., 5 km per hour).Speed of the car at 11 am = (50 – 5) = 45km/hr
Time to cover 10 km at 45 km/hr = distance/speed = 10/45 hours. = 2/9 hours
= 2/9 x 60 minutes = 40/3 minutes = 13 minutes + 1/3 minutes
= 13 minutes + 1/3 x 60 seconds
= 13 minutes and 20 seconds. 
Question 4 of 15
4. Question
1 pointsA bike rider starts at 60 km/hr and he increases his speed in every 2 hours by 3 km/hr. Then the maximum distance covered by him in 24 hours is:
Correct
Speed of the rider = 60km/hr.
Distance covered in 1st 2 hours = 60 km.He increased his speed in every 2 hours by 3 km/hr.
Distance covered in every 2 hours will be, 60, 63, 66,… upto 12 terms.(for 24 hours).
The above series is an A.P series;
Sum of first n terms = (n/2)(2a+(n1)d)
Here, a = 60, d = 3 and n = 12.
Sum of first 12 terms = (12/2)(2(60)+(11)3) = 6(120 + 33) = 6(153) = 918.
Hence, he covers 918 km in 24 hoursIncorrect
Speed of the rider = 60km/hr.
Distance covered in 1st 2 hours = 60 km.He increased his speed in every 2 hours by 3 km/hr.
Distance covered in every 2 hours will be, 60, 63, 66,… upto 12 terms.(for 24 hours).
The above series is an A.P series;
Sum of first n terms = (n/2)(2a+(n1)d)
Here, a = 60, d = 3 and n = 12.
Sum of first 12 terms = (12/2)(2(60)+(11)3) = 6(120 + 33) = 6(153) = 918.
Hence, he covers 918 km in 24 hours 
Question 5 of 15
5. Question
1 pointsIf a cyclist starts at 7 km/hr and he increases his speed in every 3 hours by 1 km/hr then the time taken by the cyclist to cover 113 km is:
Correct
Initial speed of the cyclist = 7km/hr.
Distance covered in 1st 3 hours = 7 x 3 = 21 km.
After increasing, 1km/hr for every 3 hours period,
Distance covered in 2nd 3 hours period = 8 x 3 = 24 km.
Distance covered in 3rd 3 hours period = 9 x 3 = 27 km.
Distance covered in 4th 3 hours period = 10 x 3 = 30 km.Total distance covered = 21 + 24 + 27 + 30 = 102 km.
Remaining km to cover = 113 – 102 = 11 km.Speed in 5th 3 hours period = 11 km/hr.
Time to cover 13 km at 11km/hr = 11/11 hours = 1 hour.Now, the total time taken by him for 113 km = (3 + 3 + 3 + 3 + 1) = 13 hours.
Incorrect
Initial speed of the cyclist = 7km/hr.
Distance covered in 1st 3 hours = 7 x 3 = 21 km.
After increasing, 1km/hr for every 3 hours period,
Distance covered in 2nd 3 hours period = 8 x 3 = 24 km.
Distance covered in 3rd 3 hours period = 9 x 3 = 27 km.
Distance covered in 4th 3 hours period = 10 x 3 = 30 km.Total distance covered = 21 + 24 + 27 + 30 = 102 km.
Remaining km to cover = 113 – 102 = 11 km.Speed in 5th 3 hours period = 11 km/hr.
Time to cover 13 km at 11km/hr = 11/11 hours = 1 hour.Now, the total time taken by him for 113 km = (3 + 3 + 3 + 3 + 1) = 13 hours.

Question 6 of 15
6. Question
1 pointsA man covers X km in t hours at S km/hr; another man covers X/2 km in 2t hours at R km/hr. Then the ratio S:R equals
Correct
Distance covered by 1st man at S km/hr = X km
Time taken by him = t hours.
Therefore, Speed = S = X/t km/hr.Distance covered by 2nd man at R km/hr = X/2 km
Time taken by him = 2t hours.
Therefore, Speed = R = (X/2)/2t km/hr = X/4t km/hr.Required ratio = S:R = X/t : X/4t = 1 : 1/4 = 4:1
Incorrect
Distance covered by 1st man at S km/hr = X km
Time taken by him = t hours.
Therefore, Speed = S = X/t km/hr.Distance covered by 2nd man at R km/hr = X/2 km
Time taken by him = 2t hours.
Therefore, Speed = R = (X/2)/2t km/hr = X/4t km/hr.Required ratio = S:R = X/t : X/4t = 1 : 1/4 = 4:1

Question 7 of 15
7. Question
1 pointsExcluding stoppages, the speed of a bus is 54 km/hr and including stoppages, it is 45 km/hr. For how many minutes does the bus stop per hour?
Correct
Due to stoppages, it covers 9km less per hour.
Time is taken to cover 9 km = (9/54×60)min
= 10 min
So, the bus stops for 10 min. per hr.Incorrect
Due to stoppages, it covers 9km less per hour.
Time is taken to cover 9 km = (9/54×60)min
= 10 min
So, the bus stops for 10 min. per hr. 
Question 8 of 15
8. Question
1 pointsIn a kilometer race, A beats B by 100 meters. B beats C by 100 meters. By how much meters does A beat C in the same race ?
Correct
⇒ While A covers 1000 meters, B can cover 900 meters
⇒ While B covers 1000 meters, C can cover 900 meters
⇒ Lets assume that all three of them are running same race. So when B runs 900 meters, C can run 900 × 9/10 =810
⇒ So A can beat C by 190 meters.Incorrect
⇒ While A covers 1000 meters, B can cover 900 meters
⇒ While B covers 1000 meters, C can cover 900 meters
⇒ Lets assume that all three of them are running same race. So when B runs 900 meters, C can run 900 × 9/10 =810
⇒ So A can beat C by 190 meters. 
Question 9 of 15
9. Question
1 pointsIn a 100 m race A runs at a speed of 1.66 m/s. If A gives a start of 4m to B and still beats him by 12 seconds. What is the speed of B ?
Correct
⇒Time taken by A to cover 100 meters = 60 seconds
⇒ Since A gives a start of 4 seconds then time takes by B = 72 seconds
⇒ B takes 72 seconds to cover 96 meters
⇒ Speed of B = 96/72 = 1.33 m/sIncorrect
⇒Time taken by A to cover 100 meters = 60 seconds
⇒ Since A gives a start of 4 seconds then time takes by B = 72 seconds
⇒ B takes 72 seconds to cover 96 meters
⇒ Speed of B = 96/72 = 1.33 m/s 
Question 10 of 15
10. Question
1 pointsA car running at 65 km/h takes one hour to cover a distance. If the speed is reduced by 15 km/hour then in how much time it will cover the distance ?
Correct
⇒Reduced speed = 6515 = 50 km/h
⇒ Now car will take 65/50 × 60 mins = 78 minsIncorrect
⇒Reduced speed = 6515 = 50 km/h
⇒ Now car will take 65/50 × 60 mins = 78 mins 
Question 11 of 15
11. Question
1 pointsRam walks at a speed of 12 km/h. Today the day was very hot so walked at ⅚ of his average speed. He arrived his school 10 minutes late. Find the usual time he takes to cover the distance between his school and home?
Correct
⇒ If Ram is walking at ⅚ of his usual speed that means he is taking 6/5 of using time.
⇒ 6/5 of usual time – usual time = 10 mins
⇒ 1/5 of usual time = 10 mins
⇒ Usual time = 50 minsIncorrect
⇒ If Ram is walking at ⅚ of his usual speed that means he is taking 6/5 of using time.
⇒ 6/5 of usual time – usual time = 10 mins
⇒ 1/5 of usual time = 10 mins
⇒ Usual time = 50 mins 
Question 12 of 15
12. Question
1 pointsSpeed of a train is 20 meters per second. It can cross a pole in 10 seconds. What is the length of train ?
Correct
⇒ Lenght of train = 20 × 10 = 200 meters
Incorrect
⇒ Lenght of train = 20 × 10 = 200 meters

Question 13 of 15
13. Question
1 pointsJohn travelled from his town to city. John went to city by bicycle at the speed of 25 km/h and came back at the speed of 4 km/h. If John took 5 hours and 48 min to complete his journey, what is the distance between town and city ?
Correct
⇒ Average speed of John = 2xy/x+y = 2 × 25 × 4 / 25 + 4= 200/29 km/h
⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km
⇒ Distance between city and town = 40/2 = 20 kmIncorrect
⇒ Average speed of John = 2xy/x+y = 2 × 25 × 4 / 25 + 4= 200/29 km/h
⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km
⇒ Distance between city and town = 40/2 = 20 km 
Question 14 of 15
14. Question
1 pointsHow many minutes Raman will take to cover a distance of 400 meters if he runs at a speed of 20 km/hr ?
Correct
⇒ Raman’s speed = 20 km/hr = 20 × 5/18 = 50/9 m/sec
⇒ 400 × 9/50 = 1⅕ minsIncorrect
⇒ Raman’s speed = 20 km/hr = 20 × 5/18 = 50/9 m/sec
⇒ 400 × 9/50 = 1⅕ mins 
Question 15 of 15
15. Question
1 pointsA beats B by 20 m in a 100 m race and B beats C by 20 m in a 100 m race. How much start should A give to C in a 100 m race so that both of them reach the winning post at the same time ?
Correct
Ratio of Speed of A:B = 10:8
Ratio of Speed of B:C = 10:8
A : B : C = 100:80:64
So, A should give 36 min heat start to CIncorrect
Ratio of Speed of A:B = 10:8
Ratio of Speed of B:C = 10:8
A : B : C = 100:80:64
So, A should give 36 min heat start to C