Quantitative Aptitude on Times and Distance
Q11. An athlete runs 400 meters race in 25 seconds. His speed is
(a) 16 m/sec
(b) 37 m/sec
(c) 42 m/sec
(d) 45 m/sec
Answer: (a) 16 m/sec
Speed = Distance/Time
= 400m/25sec = 16 m/sec
Q12. A man riding his bicycle covers 150 meters in 25 seconds. What is his speed in km per hour ?
(a) 21.6 km/h
(b) 23.3 km/h
(c) 25.7 km/h
(d) 26.9 km/h
Answer: (a) 21.6 km/h
| Speed = | 150 meters | = | 150/100 km | = | 150 X 60 X 60 | km/h = 21.6 km.h |
| 25 seconds | 25/(60 X 60) hr | 25 X 1000 |
Q13. A student walks from his house at 2½ km/h and reaches his school late by 6 minutes. Next day, he increases his speed by 1 km/hr and reaches 6 minutes before school time. How far is the school from his house ?
(a) 3/5 km
(b) 5/3 km
(c) 7/3 km
(d) 7/4 km
Answer: (d) 7/4 km
Let, the required distance be x km
| In 1st day, time taken = | x | h = | x | h = | 2x | h |
| 2½ | 5/2 | 5 |
| In 2nd day, time taken = | x | h = | x | h = | 2x | h |
| (2½ + 1) | (5/2) + 1 | 7 |
| A/Q ⇒ | 2x | - | 2x | = | 12 |
| 5 | 7 | 60 |
| ⇒ | 2x | - | 2x | = | 1 |
| 5 | 7 | 5 |
| ⇒ | 14x - 10x | = | 1 |
| 35 | 5 |
| ⇒ | 4x | = | 1 |
| 35 | 5 |
⇒ 20x = 35
⇒ x = 35/20 = 7/4 km
Q14. A and B start simultaneously from a certain point in North and South directions on motor cycles. The speed of A is 72 km/hr and that of B is 60 km/hr. What is the distance between A and B after 10 minutes ?
(a) 22 km
(b) 25 km
(c) 27 km
(d) 31 km
Answer: (a) 22 km
| Distance covered by A = 72 X | 10 | km = 12 km |
| 60 |
| Distance covered by B = 60 X | 10 | km = 10 km |
| 60 |
∴ Required distance = 12 km + 10 km = 22 km
Q15. A constable is 160 m behind a thief. The constable runs 20 m and the thief 12 m in a minute. In what time will the constable catch the thief ?
(a) 20 minutes
(b) 23 minutes
(c) 26 minutes
(d) 29 minutes
Answer: (a) 20 minutes
In 1 minute distance covered = 20 m - 12 m = 8 m
8m covered in = 1 minute
∴ 160 m covered in = 1/8 X 160 minute = 20 minute
Q16. A walks at a uniform rate of 5 km an hour and 5 hours after his start, B cycles after him at the uniform rate of 10 km an hour. How far from the starting point will B catch A ?
(a) 39 Km
(b) 42 Km
(c) 47 Km
(d) 50 Km
Answer: (d) 50 Km
Let, B catches A after x hours
Now Distance travelled by A in (x + 5) hrs = Distance travelled by B in x hours
⇒ 5(x + 5) = 10x
⇒ 5x + 25 = 10x
⇒ 5x = 25
⇒ x = 5 hr
Distance traveleld by B in 5 Hrs = 10 X 5 km = 50 km
Q17. By walking at ½ of his usual speed, a man reaches his office 10 minutes later than ususal. His usual time is
(a) 8 minute
(b) 10 minute
(c) 13 minute
(d) 15 minute
Answer: (b) 10 minute
Let, the ususal time be x minute
At a speed of ½ of the ususal speed, time taken = 2 of ususal time
∴ 2 of ususal time - ususal time = 10 minutes
∴ 2x - x = 10
⇒ x = 10 minutes
Q18. A takes 3 hours more than B to walk d km. If A doubles his speed then he can make it in 2 hours less than B. How much time does B required for walking d km ?
(a) 5 hr
(b) 6 hr
(c) 7 hr
(d) 9 hr
Answer: (c) 7 hr
Let the required time be x hour
∴ Time taken by A to walk d km = (x + 3) hr
With double of the speed, time taken by A = (x+3)/2 hr
| A/Q x = | x + 3 | + 2 |
| 2 |
| ⇒ x - | x + 3 | = 2 |
| 2 |
| ⇒ | 2x - x - 3 | = 2 |
| 2 |
| ⇒ | x - 3 | = 2 |
| 2 |
⇒ x - 3 = 4
⇒ x = 7 hr
Q19. If a student walks from his house to school at 4 km/hr, he is late 20 minutes. However, if he walks at 5 km/hr, he is late by 5 minutes only. The distance of his school from his house is
(a) 5 km
(b) 6 km
(c) 8 km
(d) 9 km
Answer: (a) 5 km
Let the required distance be x km
At the speed 4 km/hr , time taken = x/4 hr
At the speed 5 km/hr, time taken = x/5 hr
| A/Q | x | - | x | = | 20 - 5 |
| 4 | 5 | 60 |
| ⇒ | 5x - 4x | = | 15 |
| 20 | 60 |
| ⇒ | x | = | 15 |
| 20 | 60 |
⇒ x = (15/60) X 20 = 5 km
Q20. Bipul coveres a distance by walking for 5 hours. While returning his speed decreases by 1 km/hr and he takes 10 hours to cover the same distance. What was his speed in return journey ?
(a) 0.5 km/hr
(b) 1 km/hr
(c) 1.5 km/hr
(d) 2 km/hr
Answer: (b) 1 km/hr
Let, the required speed be x km/hr
Now 5(x + 1) = 10x
⇒ 5x + 5 = 10x
⇒ 5x = 5
⇒ x = 1 km/hr
Practice Test Exam