A) 48 kmph | B) 36 kmph |

C) 24 kmph | D) 20 kmph |

Explanation:

Let the initial speed of Akhil be '4p' kmph

Then speed after decrease in speed = '3p' kmph

We know that,

**Change in speed == Change in time**

According to the given data,

$\frac{\mathbf{24}}{\mathbf{3}\mathbf{p}}\mathbf{}\mathbf{-}\mathbf{}\frac{\mathbf{24}}{\mathbf{4}\mathbf{p}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{35}\mathbf{}\mathbf{-}\mathbf{}\mathbf{25}}{\mathbf{60}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{}\mathbf{}\mathbf{p}\mathbf{}\mathbf{=}\mathbf{}\mathbf{12}$

Hence, the initial speed of Akhil = **4p = 4 x 12 = 48 kmph.**

A) 50 kmph | B) 60 kmph |

C) 70 kmph | D) 80 kmph |

Explanation:

Speed of tractor = 360/12 = 30 kmph

Speed of jeep = 250 x 30/100 = 75 kmph

But, given ratio of speed of car, jeep and tractor is 3 : 5 : 2

Speed of car = 3 x 75/5 = 45 kmph

Required, average speed of car and jeep = **75 + 45/2** = **60 kmph**.

A) 10 kms | B) 8 kms |

C) 6 kms | D) 4 kms |

Explanation:

Let the actual distance be t kms

Now, According to the data,

$\mathbf{5}\left(\mathbf{t}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{8}}{\mathbf{60}}\right)\mathbf{}\mathbf{=}\mathbf{}\mathbf{6}\mathbf{t}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=5\left(\frac{60\mathrm{t}+8}{60}\right)=6\mathrm{t}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=60\mathrm{t}+8=72\mathrm{t}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=8=12\mathrm{t}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\mathrm{t}=8/12=2/3\mathrm{hrs}$

Now, the total distance =** 6t = 6 x 2/3 = 2 x 2 = 4 kms.**

A) 75 kmph | B) 80 kmph |

C) 86 kmph | D) 90 kmph |

Explanation:

Let the distance between A and B = **D**

Given the speed of maruthi = **60 kmph**

Maruthi reaches from A to B in 6 hrs 20 min =** 6 + 20/60 = 6 1/3 hrs**

Now, the distance between A & B = D = speed x time = **60 x 19/3 = 380 kms**

As the distance is same for toyato also,

given time for toyato = 4 hrs 45 min = **4 + 45/60 = 4 3/4 = 19/4 hrs**

speed of toyato = distance/time = 380/19/4 = **380 x 4/19 = 80 kmph.**

A) 8 hrs | B) 7.5 hrs |

C) 7 hrs | D) 6.5 hrs |

Explanation:

Given speed of the bike after servicing = 55 kmph

Time taken for travelling some distance at 55 kmph = 5 hrs

Then,

Distance = Speed x Time = 55 x 5 = 275 kms.

Now,

Speed of the bike before servicing = 35 kmph

Distance = 275 kms

Now, **Time = Distance/Speed = 275/35 = 7.85 hrs =~ 8 hrs.**

A) 36 kmph | B) 32 kmph |

C) 29 kmph | D) 24 kmph |

Explanation:

Let the speed of the train be 'S' kmph

From the given data,

Distance = Length of train = D = 200 mts = 200 x 18/5 kms

Time = 10 sec

given speed of the motorcycler = 12 kmph

Relative speed as they are moving in the same direction = (S - 12) kmph

$\mathbf{(}\mathbf{S}\mathbf{-\; 12)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{200}\mathbf{}\mathbf{x}\mathbf{}{\displaystyle \frac{\mathbf{18}}{\mathbf{5}}}}{\mathbf{15}}\phantom{\rule{0ex}{0ex}}\mathbf{S}\mathbf{}\mathbf{-}\mathbf{}\mathbf{12}\mathbf{}\mathbf{=}\mathbf{}\mathbf{48}\phantom{\rule{0ex}{0ex}}\mathbf{S}\mathbf{}\mathbf{=}\mathbf{}\mathbf{60}\mathbf{}\mathbf{kmph}\phantom{\rule{0ex}{0ex}}$

Hence, the speed of the train = **S = 60 kmph**

Now,

Let the speed of the jeep be** 'x'** kmph

$\mathbf{60}\mathbf{}\mathbf{-}\mathbf{}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{200}\mathbf{}\mathbf{x}\mathbf{}{\displaystyle \frac{\mathbf{18}}{\mathbf{5}}}}{\mathbf{20}}\phantom{\rule{0ex}{0ex}}\mathbf{60}\mathbf{}\mathbf{-}\mathbf{}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{36}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{60}\mathbf{}\mathbf{-}\mathbf{}\mathbf{36}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{24}\mathbf{}\mathbf{kmph}$

Therefore, the speed of the jeep =** x = 24 kmph.**

A) 5 kmph | B) 6 kmph |

C) 7.5 kmph | D) 8 kmph |

Explanation:

Let **p** be the speed of man in kmph

According to the given data in the question,

**Distance travelled by bus in 10 min with 20 kmph == Distance travelled by man in 8 min with (20 + p) kmph in opposite direction**

=> 20 x 10/60 = 8/60 (20 + p)

=> 200 = 160 + 8p

=> p = 40/8 = 5 kmph.

A) 70 kms | B) 60 kms |

C) 45 kms | D) 30 kms |

Explanation:

Let the time taken by train be **'t'** hrs.

Then,

$\mathbf{40}\mathbf{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{35}\left(\mathbf{t}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{15}}{\mathbf{60}}\right)$

40t = 35t + 35/4

**t = 7/4 hrs**

Therefore, Required length of the total journey d = s x t

= 40 x 7/4

**= 70 kms.**

A) 330/29 m/s | B) 330 x 30 m/s |

C) 330/14 m/s | D) 330/900 m/s |

Explanation:

Second gun shot take 30 sec to reach rahul imples distance between two.

given speed of sound = 330 m/s

Now, distance = 330 m/s x 30 sec

Hence, speed of the bus **= d/t = 330x30/(14x60 + 30) = 330/29 m/s.**