Important for :

1

C

Relative speed = | Total distance | = | 60 + 90 | = 5 m/s |

Total time | 30 |

∴ 5 m/s = 5 × |
18 | = 18 kmph |

5 |

Now relative speed = speed of auto rickshaw – speed of lorry

or, 18 = 38 – speed of lorry

Hence, option C is correct.

2

D

When B started, A is ahead by 40 × 5 = 200 m

But C will catch A before the B as he is faster than the B.

Since A and C run in the same direction, relative speed of C = 60 − 40 = 20 m/min

But C will catch A before the B as he is faster than the B.

Since A and C run in the same direction, relative speed of C = 60 − 40 = 20 m/min

∴ Time taken by C to actually catch A = |
200 | = 10 mins |

20 |

Distance actually covered by C in this duration = 10 × 60 = 600 m

Hence, option D is correct.

3

E

Distance Travelled by Salman in 3 minutes= | 3 | × 120 = 6 km |

60 |

Distance travelled by Govinda in 3 minutes

= | 3 | × 80 = 4 km |

60 |

Hence Salman will have to travel a distance of 10 km before catching up with Govinda.

Relative speed of Salman to that of Govinda after taking U turn = 120 – 80 = 40 kmph

Total time taken by Salman to meet Govinda = (3 + 15)minutes = 18 minutes

Hence, option E is correct.

4

D

Let the fixed charges = Rs. x (for the first 5 km)

and the additional charges = Rs. y /km

As per the question,

x + 5y = 350 ....(i)

x + 20y = 800 ...(ii)

On solving eqn. (i) and (ii), we get

x = 200, y = 30

= x + 25y = 200 + 30 × 25 = Rs. 950

Hence, option D is correct.

Hence, option D is correct.

5

B

If R is the reduction in maximum speed and N is the number of passengers, then R is proportional to N

⇒ R = NK where K is constant of proportionality,

Given data: R = 20, when N = 5, hence K = 4

Maximum number of passenger that bus can’t move = (80 – N × K) = 80 – N × 4 ⇒ N = 20

So number of passenger for bust just move = 20 – 1 = 19

Hence correct option (B) is correct.