Dear Reader,
Below are four problems dealing with time, distance and speed.
Question 1
A 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
a) 4:1 b) 2:1 c) 1:4 d) 1:2
Answer : a) 4:1
Solution :
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 2
If 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:
a) 27/2 hours b) 20/3 hours c) 12 hours d) 13 hours
Answer : d) 13 hours
Solution :
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 3
A 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:
a) 1000km b) 918km c) 899 km d) none of these
Answer : b) 918 km.
Solution :
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+(n-1)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 4
A 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:
a) 12 minutes and 10 seconds
b) 15 minutes and 09 seconds
c) 13 minutes and 20 seconds
d) none of these
Answer : c) 13 minutes and 20 seconds
Solution :
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.
Below are four problems dealing with time, distance and speed.
Question 1
A 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
a) 4:1 b) 2:1 c) 1:4 d) 1:2
Answer : a) 4:1
Solution :
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 2
If 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:
a) 27/2 hours b) 20/3 hours c) 12 hours d) 13 hours
Answer : d) 13 hours
Solution :
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 3
A 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:
a) 1000km b) 918km c) 899 km d) none of these
Answer : b) 918 km.
Solution :
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+(n-1)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 4
A 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:
a) 12 minutes and 10 seconds
b) 15 minutes and 09 seconds
c) 13 minutes and 20 seconds
d) none of these
Answer : c) 13 minutes and 20 seconds
Solution :
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.
