# Quantitative Aptitude on Times and Distance

Q31. By how much percent must a motorist increases his speed in order to reduce the time by 10%, taken to cover a certain distance ?

(a) 10^{1}/_{9} %

(b) 11^{1}/_{9} %

(c) 13^{4}/_{9} %

(d) 14^{5}/_{9} %

**Answer:** (b) 11^{1}/_{9} %

Let the required increase in speed be P %

Then 90% of time X | 100 + P | X speed = time X speed |

100 |

⇒ | 90 | X time X | 100 + P | X speed = time X speed |

100 | 100 |

⇒ | 90(100 + P) | = 1 |

100 X 100 |

⇒ 900 + 9P = 1000

⇒ 9P = 100

⇒ p = 100/9 = 11^{1}/_{9} %

Q32. A car covers four successive 2 km stretches at 20 km/hr, 30 km/hr, 40 km/hr and 60 km/km respectively. Its average speed over this distance is

(a) 22 km/hr

(b) 25 km/hr

(c) 29 km/hr

(d) 32 km/hr

**Answer:** (d) 32 km/hr

Total distance = 2 X 4 km = 8 km

Total time taken = ( | 2 | + | 2 | + | 2 | + | 2 | ) h |

20 | 30 | 40 | 60 |

= | 12 + 8 + 6 + 4 | h |

120 |

= | 30 | h |

120 |

= | 1 | h |

4 |

Average speed = | 8 | km/h = 32 km/h |

^{1}/_{4} |

Q33. A car travels a distance of 840 km at a uniform speed. If the speed of the car is 10 km/h more, it takes 2 hours less to cover the same distance. The original speed of the car was

(a) 53 km/hr

(b) 57 km/hr

(c) 60 km/hr

(d) 63 km/hr

**Answer:** (c) 60 km/hr

Let, the original speed of the car be x km/hr

A/Q | 840 | = 2 + | 840 |

x | x + 10 |

⇒ | 840 | = | 2x + 20 + 840 |

x | x + 10 |

⇒ 840x + 8400 = 2x^{2} + 860x

⇒ 2x^{2} + 20x - 8400 = 0

⇒ x^{2} + 10x - 4200 = 0

⇒ x^{2} + 70x - 60x - 4200 = 0

⇒ x (x + 70) - 60 (x + 70) = 0

⇒ (x + 70) (x - 60) = 0

∴ x = 60 km/h

Q34. The ratio between the rates of walking of x and y is 3:4, If the time taken by x to cover a certain distance is 36 minutes, the time taken by Y to cover the certain distance is

(a) 25 minutes

(b) 27 minutes

(c) 32 minutes

(d) 35 minutes

**Answer:** (b) 27 minutes

Ratio of the time taken by x and y = | 1 | : | 1 | = 4 : 3 |

3 | 4 |

Let, the time taken by x and y be 4x and 3x minutes

A/Q 4x = 36

⇒ x = 9

∴ time taken by y = 3 X 9 minutes = 27 minutes

Q35. A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500 km away in time, it had to increase the speed by 250 km/hr from the usual speed. Its ususal speed is

(a) 478 km/h

(b) 550 km/h

(c) 600 km/h

(d) 750 km/h

**Answer:** (d) 750 km/h

Let, the usual speed be x km/hr

A/Q | 1500 | - | 1500 | = | 1 |

x | x + 250 | 2 |

⇒ | 1 | - | 1 | = | 1 |

x | x + 250 | 3000 |

⇒ | x + 250 - x | = | 1 |

x (x + 250) | 3000 |

⇒ | 250 | = | 1 |

x (x + 250) | 3000 |

⇒ x^{2} + 250x - 750000 = 0

⇒ x^{2} + 1000x -750x - 750000 = 0

⇒ x (x + 1000) - 750 (x + 1000) = 0

⇒ (x + 1000) (x - 750) = 0

∴ x = 750 km/h

Q36. Two cars starts at the same time from one point and move along two roads at right angles to each other. Their speeds are 18 km/hr and 36 km/hr respectively. After 10 seconds the distance between them will be

(a) 48√5 m

(b) 50√5 m

(c) 53√5 m

(d) 55√5 m

**Answer:** (b) 50√5 m

18 km/h = | 18000 m | = 5 m/sec |

60 X 60 sec |

36 km/h = | 36000 m | = 10 m/sec |

60 X 60 sec |

Distance covered in 10 sec OA = 5 X 10 m = 50 m

Distance covered in 10 sec OB = 10 X 10 m = 100 m

Now AB = √(OA^{2} + OB^{2}) = √(50^{2} + 100^{2}) = √(2500 + 10000) = √12500 = 50√5 m

Q37. A man on tour travels first 150 km at 55 km/hr and the next 150 km at 65 km/hr. The average speed of the whole journey is

(a) 57.48 km/hr

(b) 57.82 km/hr

(c) 58.12 km/hr

(d) 59.58 km/hr

**Answer:** (d) 59.58 km/hr

Average speed = | 2xy | km/hr |

x + y |

= | 2 X 55 X 65 | km/hr |

55 + 65 |

= | 2 X 55 X 65 | km/hr |

120 |

= | 55 X 65 | km/hr |

60 |

= 59.58 km/hr

Q38. Ram has to cover 8 km in 45 minutes. If he covers one half of the distance in 1/3 rd of time, what should be his speed to cover the remaining distance in the remaining time

(a) 6 km/hr

(b) 7 km/hr

(c) 8 km/hr

(d) 9 km/hr

**Answer:** (c) 8 km/hr

Time left = ( | 2 | X 45) minutes = 30 minutes = ½ h |

3 |

Distance left = 8 X ½ = 4 km

∴ Required Speed = (4 ÷ ½ ) km/h = 8 km/hr

Q39. In what time can Naren cover a distance of 500 m, if he runs at a speed of 25 km/hr

(a) 1^{1}/_{5} minutes

(b) 2^{1}/_{3} minutes

(c) 2^{1}/_{5} minutes

(d) 3^{1}/_{5} minutes

**Answer:** (a) 1^{1}/_{5} minutes

Distance = 500 m = | 500 | km = ½ km |

1000 |

Speed = 25 km/hr

∴ Time = | ½ km | = | 1 | hr = | 60 | minutes = 1 | 1 | minutes |

25 km/hr | 50 | 50 | 5 |

Q40. The distance between two stations A and B is 200 km. A train leaves A towards B at 80 km/hr. After half an hour, another train leaves B towards A at 100 km/hr. The distance of the point where the two trains meet, from A is

(a) 111.11 km

(b) 112.22 km

(c) 112.25 km

(d) 115.55 km

**Answer:** (a) 111.11 km

Let the required distance be x km

A/Q | x | - | 200 - x | = ½ |

80 | 100 |

⇒ | 5x - 800 + 4x | = ½ |

400 |

⇒ 9x - 800 = 200

⇒ 9x = 1000

⇒ x = 111.11 km

## Practice Test Exam