# Quantitative Aptitude on Times and Distance

#### Home⇒Quantitative Aptitude⇒ Quantitative Questions on Time 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