A) 10.30 | B) 10 |

C) 8.45 | D) 9.30 |

Explanation:

Assume both trains meet after x hours after 7 am

Distance covered by train starting from P in x hours = 20x km

Distance covered by train starting from Q in (x-1) hours = 25(x-1)

Total distance = 110

=> 20x + 25(x-1) = 110

=> 45x = 135

=> x= 3

Means, they meet after 3 hours after 7 am, ie, they meet at 10 am

A) 42 kmph | B) 43 kmph |

C) 40 kmph | D) 45 kmph |

Explanation:

We know that, **Speed = Distance/time**

Let the speed of the train be = x kmph** NORMAL SPEED SPEED 5 KMPH MORE**

Distance = 360 Km Distance =360

speed = X Kmph Speed = (x+5)

Time =360/X Time =360/(x+15)

For the same journey if the speed increased 10 kmph more it will take 1 hour less

Time with original speed - time with increased speed = 1

=> (360/x) - (360/x+10) = 1

=> (360(x+5)-360x)/x(x+5) =1

=> 360X + 1800 -360X = X(X+5)

=> X^2 + 5x - 1800 = 0

X = 40 or -45

X cannot be negetive value

**X = 40 kmph.**

A) 50 kms | B) 48 kms |

C) 46 kms | D) 44 kms |

Explanation:

Let the distance be 'd' kms.

According to the given data,

$\frac{\mathbf{d}}{\mathbf{30}}\mathbf{}\mathbf{-}\mathbf{}\frac{\mathbf{d}}{\mathbf{50}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{40}}{\mathbf{60}}\mathbf{}\mathbf{hrs}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{}\mathbf{}\mathbf{6}\mathbf{d}\mathbf{}\mathbf{=}\mathbf{}\mathbf{300}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{}\mathbf{}\mathbf{d}\mathbf{}\mathbf{=}\mathbf{}\mathbf{50}\mathbf{}\mathbf{kms}\mathbf{.}\phantom{\rule{0ex}{0ex}}$

A) 42 kmph | B) 72 kmph |

C) 36 kmph | D) 44 kmph |

Explanation:

Given speed of the first train = 60 km/hr = 60 x 5/18 = 50/3 m/s

Let the speed of the second train = x m/s

Then, **the difference in the speed** is given by

$\frac{\mathbf{50}}{\mathbf{3}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{120}}{\mathbf{18}}$

=> x = 10 m/s

=> 10 x 18/5 = **36 km/hr**

A) 27 7/9 mts | B) 25 8/7 mts |

C) 21 1/4 mts | D) 22 mts |

Explanation:

Relative speed = (40 - 20) km/hr = | 20 x | 5 | m/sec = | 50 | m/sec. | ||||

18 | 9 |

Length of faster train = | 50 | x 5 | m = | 250 | m = 27 | 7 | m. | ||

9 | 9 | 9 |

A) 202 mts | B) 188 mts |

C) 165 mts | D) 156 mts |

Explanation:

Let the length of the train = L mts

Relative speed of train and man = 74 - 8 = 66 kmph = 66 x 5/18 m/s

=> 66 x 5/18 = L/9

=> L = 165 mts.

A) 250 mts | B) 400 mts |

C) 320 mts | D) 390 mts |

Explanation:

Let the length of the 1st train = L mts

Speed of 1st train = 48 kmph

Now the length of the 2nd train = L/2 mts

Speed of 2nd train = 42 kmph

Let the length of the bridge = D mts

Distance = L + L/2 = 3L/2

Relative speed = 48 + 42 = 90 kmph = 90 x 5/18 = 25 m/s(opposite)

Time = 12 sec

=> 3L/2x25 = 12

=> L = 200 mts

Now it covers the bridge in 45 sec

=> distance = D + 200

Time = 45 sec

Speed = 48 x5/18 = 40/3 m/s

=> D + 200/(40/3) = 45

=> D = 600 - 200 = 400 mts

Hence, the length of the bridge = **400** mts.

A) 36 kmph | B) 30 kmph |

C) 34 kmph | D) 40 kmph |

Explanation:

Let the speed of the faster train be 'S' kmph

Then speed of the slower train will be '(S-5)' kmph

Time taken by faster train = 350/S hrs

Time taken by slower train = 350/(S-5) hrs

$\frac{350}{s-5}-\frac{350}{s}=2hrs20min=2\frac{1}{3}=\frac{7}{3}$

=> S = 30 km/hr.

A) 180 mts | B) 190 mts |

C) 200 mts | D) 210 mts |

Explanation:

Let the length of the train be 'L' mts

let the speed of the train be 'S' m/s

Given it crosses a pole in 10 sec=> L/S = 10 ......(1)

Given it takes 20 sec (double of pole) to cross a platform of length 200 mts

=> (L + 200)/S = 20

=> L/S + 200/S = 20

But from (1) L/S = 10

=> 200/S = 20 - 10

=> S = 20 m/s

Then, from (1)

=> L = 10 x 20 = 200 mts.

Hence, the length of the train = 200 mts.