A) 2/3 hrs | B) 3/4 hrs |

C) 1/3 hrs | D) 1/4 hrs |

Explanation:

Walking at 3/4th of usual rate implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time

so 1/3rd of the usual time = 15min

or usual time = 3 x 15 = 45min = 45/60 hrs = 3/4 hrs.

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) 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) 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.**

**1. How to find Speed(s) if distance(d) & time(t) is given:**

$\mathbf{Speed}\mathbf{}\mathbf{\left(}\mathbf{S}\mathbf{\right)}\mathbf{}\mathbf{=}\frac{\mathbf{}\mathbf{Distance}\mathbf{}\mathbf{\left(}\mathbf{D}\mathbf{\right)}}{\mathbf{Time}\mathbf{}\mathbf{\left(}\mathbf{T}\mathbf{\right)}}\mathbf{}$

**Ex:** Find speed if a person travels 4 kms in 2 hrs?

**Speed = D/T = 4/2 = 2 kmph.**

**2. ** Similarly, we can find distance (d) if speed (s) & time (t) is given by

**Distance (D) = Speed (S) x Time (T)**

**Ex :** Find distance if a person with a speed of 2 kmph in 2 hrs?

**Distance D = S X T = 2 x 2 = 4 kms. **

**3. **Similarly, we can find time (t) if speed (s) & distance (d) is given by

**Time (T) = **$\frac{\mathbf{Distance}\mathbf{}\mathbf{\left(}\mathbf{D}\mathbf{\right)}}{\mathbf{Speed}\mathbf{}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$

**Ex :** Find in what time a person travels 4 kms with a speed of 2 kmph?

**Time T = D/S = 4/2 = 2 hrs.**

**4. **How to convert** km/hr into m/sec :**

**$\mathbf{P}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{hr}\mathbf{}\mathbf{=}\mathbf{}\left(\mathbf{P}\mathbf{}\mathbf{x}\mathbf{}\frac{\mathbf{5}}{\mathbf{18}}\right)\mathbf{}\mathbf{m}\mathbf{/}\mathbf{sec}$**

**Ex : **Convert 36 kmph into m/sec?

**36 kmph = 36 x 5/18 = 10 m/sec **

**5. **How to convert** m/sec into km/hr :**

**$\mathbf{P}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{sec}\mathbf{}\mathbf{=}\mathbf{}\left(\mathbf{P}\mathbf{}\mathbf{x}\mathbf{}\frac{\mathbf{18}}{\mathbf{5}}\right)\mathbf{}\mathbf{kmph}$.**

**Ex : **Convert 10 m/sec into km/hr?

**10 m/sec = 10 x 18/5 = 36 kmph.**

**6. **If the ratio of the speeds of A and B is **a:b**, then the ratio of the times taken by them to cover the same distance is **b : a.**

**7.** Suppose a man covers a certain distance at** x km/ hr** and an equal distance at **y km/hr .** Then, the average speed during the whole journey is $\frac{\mathbf{2}\mathbf{xy}}{\mathbf{x}\mathbf{+}\mathbf{y}}$ **km/hr.**

A) 105 mts | B) 115 mts |

C) 120 mts | D) 140 mts |

Explanation:

Speeds of two trains = 30 kmph and 58 kmph

=> Relative speed = 58 - 30 = 28 kmph = 28 x 5/18 m/s = 70/9 m/s

Given a man takes time to cross length of faster train = 18 sec

Now, required Length of faster train = speed x time = **70/9 x 18 = 140 mts.**