Time & Distance - Solved Problem
Time & Distance - Solved Problem - 1
Question 1
A cyclist moves at the speed of 20km/hr. Suppose he travelled at a speed of 25km/hr, he would have travelled 10km more then What will be the actual distance travelled by the cyclist?
a)80km b)79km c)60km d)50km
Answer : a) 80km
Solution:
Let the actual distance travelled be x km.
From the given question, we have
distance speed
x 20
x+20 25
then, x/20 = x+20 / 25
25x = 20x + 400
5x = 400
x = 80
Hence the answer is 80 km.
Question 2
A bike at a speed of 60km/hr, it covers certain distance in 3 hours and the train can cover the same distance in 1 and half hours then the speed of train is
a)80km/hr b)120km/hr c)160km/hr d)50km/hr
Answer: b)120km/hr
Solution:
The diatance covered by the bike with speed 60km/hr in 3 hours = (60 x 3) = 180 km
Now, Speed = Distance/Time
since the train covers the same 180 km in 3/2 hours (we can write 1 and half hours as 3/2 hours)
Then the speed of the train = 180/ (3/2) = 120 km/hr.
Hence the train travelled at the speed of 120km/hr to cross 180km in 1 and half hour.
Question 3
A car starts at A and reaches a destination in 8 hours. If it travels first and second half at the speed of 80 km/hr and 100 km/hr respectively then the distance between A and destination is:
a)818km b)979km c)711km d)850km
Answer: c)711km
Solution:
Let the distance between A and destination be X.
The total time taken by the car to cover X = 8 hours
Since X/2 by 80km/hr and remaining X/2 by 100km/hr
Then by Time = distance/speed, we have
(X/2)/80 + (X/2)/100 = 8
X/80 + X/100 = 16
5X + 4X = 6400
9X = 6400
X = 6400/9 = 711
Hence 711km is the required answer.
Question 4
Albert takes 3 hours more than Shahul to cover 50 km. Suppose the time taken by Albert is 30 minutes less than shahul he must triple his speed. Then the speed of Albert will be:
a)8km/hr b)9.5km/hr c)9km/hr d)5.5km/hr
Answer: b)9.5km/hr
Solution:
Let Albert's speed be X km/hr.
And let the time taken by Shahul be Y.
Since Albert takes 3+Y hours to cross 50km at X km/hr.
i.e., 50 / X = 3+Y ---eqn1
And he takes Y - 1/2 hours to cross 50km at 3X km/hr.
i.e., 50/3X = Y- 1/2 ---eqn2
Subtract eqn2 from eqn1, we have
50/X - 50/3X = 3+Y - Y + 1/2 = 7/2
21X = 200
X = 9.5
Hence Albert's speed is 9.5km/hr.
Question 5
Mr.X left for village 1 from Village 2 at 1.00 pm. He travelled at the speed of 40km/hr for 1 and half hours. After that the speed was reduced to 30 km/hr. If the distance between two villages is 80 kms, at what time did Mr.X reach village 1 ?
a) 2 pm b) 3.10 pm c) 2.40 pm d) 1.55 pm
Answer : b) 3.10 pm
Solution :
Mr.X travelled at 40 km/hr for 1 1/2 hours (3/2 hours).
Distance covered in 3/2 hours = 40 x 3/2 = 60 km.
Therefore, remaining distance = 80 - 60 = 20 km
After 3/2 hours, the speed was reduced to 30 km/hr.
Time taken to cover the remaining 20 km = 20/30 = 2/3 hour = 40 minutes.
Total time taken = 3/2 + 2/3 hours = 13/6 hours = 2 1/6 = 2 hours 10 minutes.
So, Mr.X reached village 2 at 1.00 pm + 2 hours 10 minutes = 3.10pm
Hence the required answer is 3.10 pm.
Question 6
A bike rider increases his speed by 1 km after every 30 minutes. If the distance travelled in the 1st 30 minutes was 30 km then what was the total distance travelled in 6 hours ?
a) 426 km b) 213 km c) 639 km d) none of these
Answer : a) 426 km
Solution :
Given that distance travelled in 1st 30 minutes = 30 km
Speed of the bike increases by 1 km after every 30 minutes.
Distance travelled in 2nd 30 minutes = 31 km
Distance travelled in 3rd 30 minutes = 32km
6 hours contains 12 thirty minutes.
Total Distance Travelled = [30 + 31 + 32 + ... (12 terms)]
This is an Arithmetic Progression(AP) with
first term a = 35, number of terms n = 12 and common difference d = 1.
Sum of the first n terms of an Arithmetic Progression(AP),Sn = (n/2)[2a + (n-1)d] where n = number of terms
Here, [30 + 31 + 32 +... (12 terms)]
S12 = (12/2)[2 x 30 + (12-1)1] = 6[60 + 11] = 6 x 71 = 426
Hence the total distance travelled = 426 km
Question 7
A man runs at a speed of 7 km per hour and he increases his speed in every hour by 1 km per hour. In how many hours will he covers 19.5 km ?
a) 1 1/4 hours b) 2 1/2 hours c) 3 1/3 hours d) 3 hours
Answer : b) 2 1/2 hours
Solution :
The man starts with 7 kmph.
Distance covered in first 1 hour = 7 km
He increases his speed in every hour by 1 km.
Speed in 2nd hour = 8 km/hr
Distance covered in 2nd hour = 8 km
Remaining distance = 19.5 - (7 + 8) = 4.5 km
Speed in the third hour = 9 km/hr
Time taken to cover 4.5 km at 9 km/hr = 4.5/9 = 1/2 hour.
Therefore, total time = 1 + 1 + 1/2 hours = 2 1/2 hours.
Hence the answer is 2 1/2 hours.
Question 8
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 9
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 10
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 11
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.