வெல்லுங்கள் CSAT 2025 - 10: Q and A on Time & Distance

Questions & Answer (Multiple Choice) - TIME & DISTANCE

கொள்குறி வினா- விடைகள் - காலமும் தூரமும்

1. A school boy walks from his house to school at the rate of 4kmph. He reaches the school 20 minutes earlier than the scheduled time. If he walks at the rate of 3kmph, he reaches the school 20 minutes late. What is the distance of the school from his house?
a. 5km b. 6km c.7km d. 8km

2. A rectangular ground of dimensions 150m × 75m. Arun runs around the four sides of the ground eight times. How much total distance in kilometers he covered during his running?
a. 2km b. 1.8km
c.1.6km d. 3.6km

3. A man walks at 3kmph for 6 hours and at 4kmph for 4 hours. What is his average speed?
a. 3.5kmph b. 0.7kmph
c. 3.4kmph d. 7kmph

4. A man can cover a distance in 1hr 24min by covering 2/3 of the distance at 4 km/hr and the rest at 5km/hr. What is the total distance?
a. 8.4km b. 5.6km c.7.2km d. 9km

5. In covering a distance of 30 km, A takes 2 hours more than B. If A doubles his speed, then he would take 1 hour less than B to cover the same distance. What is A's speed?
a. 7.5 kmph b. 5 kmph
c. 4 kmph d. 6 kmph

6. Walking at the rate of 3 kmph a man cover certain distance in 105 minutes. Running at a speed of 10.5 kmph the man will cover the same distance in how many minutes?
a. 50mts b. 40mts
c. 42mts d. 30mts

7. Two persons starting from the same place walk at a rate of 4kmph and 4.5kmph respectively. What time will they take to be 6.5km apart, if they walk in the same direction?
a. 12hrs b. 13hrs
c. 9hrs d. 10hrs

8. Excluding stoppages, the average speed of a bus is 54 kmph and including stoppages, the average speed is 45 kmph. For how many minutes does the bus stop per hour?
a. 50mts b. 12mts
c. 10mts d. 20mts

9. A motorist covers a distance of 39km in 45min by moving at a speed of s kmph for the first 15min, then moving at double the speed for the next 20 min, and then again moving at his original speed for the rest of the journey. Find s.
a. 36 km b. 45 km
c. 32 km d. 42 km

10. A & B are two towns. A person covers the distance from A to B on cycle at 14kmph and returns to A by a bus running at a uniform speed of 66kmph. What is his average speed for the whole journey?
a. 40 kmph b. 32.1 kmph
c. 20 kmph d. 23.1 kmph

11. A car covers 4 successive 12km stretches at speed of 10kmph, 20kmph, 30kmph & 40kmph respectively. What is its average speed?
a. 25 kmph b. 19.2 kmph
c. 20 kmph d. 24.1 kmph

12. A person crosses a 900 m long bridge in 6 minutes. What is his speed in km per hour?
a. 6 kmph b. 8 kmph
c. 9 kmph d. 10 kmph

13. A man takes 3hr 45min. in walking to certain place and riding back. He would have gained 1hr 30min by riding both ways. What is the time he would take to walk both ways?
a. 5hrs 15min. b. 1hr 30min.
c. 3hrs 30 min. d. 4hrs 30min.

14. The ratio between the speeds of the A& B is 2 : 3 and therefore A takes 10 minutes more than the time taken by B to reach the destination. If A had walked at double the speed, in how many minutes he would have covered the same distance?
a. 30mts b. 10mts
c. 20mts d. 15mts

15.. A bullock cart has to cover a distance of 60km in 6hrs. If it covers half of the journey in 2/5th time. What should be its speed to cover the remaining distance in the time left.?
a. 10 kmph b. 8.33 kmph
c. 9 kmph d. 8 kmph

Answers

1. d 2. d 3. c 4. a 5. b

6. d 7. b 8. c 9. a 10. d

11. b 12. c 13. a 14. d 15. b

Explanation to answers

1. Let the distance of the school from house be 'd ' k.m. and the usual time be 't' hours.
Given
d/4 = t - 20/60 = t - 1/3
d/3 = t +20/60 = t + 1/3
So d/3 - d/4 = 1/3 + 1/3 = 2/3
Therefore d/12 = 2/3
On simplification, d = 8
Hence the required distance is 8 k.m.

2. Perimeter of the ground
= 2(150+75) = 450 metres
Total distance covered = 8×450 = 3600m
The distance covered in kilometres
= 3.6 km.

3. Walked distance at 3kmph for 6 hours
= 3 × 6 = 18km.
Walked distance at 4kmph for 4 hours
= 4 × 4 = 16km.
Average speed
= (Total distance)/(Total time taken)
= (18+16)/(6+4) = 3.4kmph.

4. 1 hr 24 minutes = 84/60
= 7/5 = 1.4hr.
Two third distance = 4 × 1.4 = 5.6 km
Total distance (5.6)/(2/3) = 8.4km.

Extra:
One third distance = 5.6 ÷2 = 2.8 km.
Time taken to cover 2.8km @ 5km/hr
= (2.8)/5 = 0.56hr.

5. Let the time taken by B to cover 30km be 'h' hours. So the time taken by A to cover 30km is 'h+2' hours.
The speed of A = (30)/(h+2)
If he doubled the speed the time taken by A = (h - 1) hours.
-> 2×{(30)/(h+2)} = (30)/(h - 1)
-> 2 (h - 1) = h + 2
-> h = 4
A's original speed = 30/(4+2) = 5kmph

6. Distanced travelled by walk
= 3×(105/60) km
Time in minutes to travel by running the same distance
= 3×(105/60) ÷ (10.5/60) = 30 minutes

7. Distance between two persons after one hour = 4.5 - 4 = 0.5 km.
Time taken for them to be 6.5 km. apart
= (6.5)/(0.5) = 13 hrs.

8. Time in minutes to cover 45km at the speed of 54kmph
= (60)/(54) × 45 = {(6×10)/(6×9)}×(5×9)
= 50 minutes.
Hence the number of minutes the bus stpos per hour = 60 - 50 = 10 mts.

9. The total distance covered at a speed of s kmph in 25 minutes and 2s kmph in 20 minutes
(25/60) × s + (20/60) × (2s) = 39
-> (65/60)s = 39 -> s = 36

10. The harmonic speed formula
(2xy)/(x + y).
Here x = 14 and y = 66
So the average speed is
(2×14×66)/(14 + 66) = 23.1kmph

11. Average speed =
Total distance/Totaltime taken
Total distance = (4×12) = 48
Total time taken
= {(12/10)+(12/20)+(12/30)+(12/40)}
= 12×{(12+6+4+3)/120} = 5/2 = 2.5
Average speed = 48/(2.5) = 19.2

12. 1 hour = 60 minutes
1 km = 1000metres
So
(900/6)×(60/1000) = 9 kmph.

13. Let 2p and 2q be the time taken by walk and riding on both ways respectively.
Here p + q = 3hr 45min
p - q = 1hr 30min
So 2p = 5hrs 15 min.

14. Let the speeds of A and B be 2s and 3s respectively.
So the ratio of time taken by them to reach a destination is 3s : 2s
Given 3s - 2s = 10min. -> s = 10min.

The time taken by A to reach the destination by original speed is 30 min.
If he doubled the speed the time taken is halved.
Hence the required time is 15min.

15.The remaining half of the distance to be covered by 3/5 th of 6hrs is 30km.
The required speed is
(30)/{(3/5)×6} = (30×5)/(3×6)
= 25/3 = 8.33 kmph.

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