A) 100 mts | B) 96 mts |

C) 98 mts | D) 120 mts |

Explanation:

Let the distance between home and school is 'x'.

Let actual time to reach be 't'.

Thus, x/4 = t + 4 ---- (1)

and x/6 = t - 30 -----(2)

Solving equation (1) and (2)

=> x = 98 mts.

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.**

A) 200 mts | B) 180 mts |

C) 160 mts | D) 145 mts |

Explanation:

Length of train A = 48 x 9 x 5/18 = 120 mts

Length of train B = 48 x 24 x 5/18 - 120

=> 320 - 120 = 200 mts.

A) 132 km | B) 264 km |

C) 134 km | D) 236 km |

Explanation:

Let the distance travelled by Tilak in first case or second case = d kms

Now, from the given data,

d/20 = d/22 + 36 min

=> d/20 = d/22 + 3/5 hrs

=> d = 132 km.

Hence, the total distance travelled by him = d + d = 132 + 132 = 264 kms.

A) 15 kms | B) 9 kms |

C) 6 kms | D) 18 kms |

Explanation:

The first train covers 180 kms in 5 hrs

=> Speed = 180/5 = 36 kmph

Now the second train covers the same distance in 1 hour less than the first train => 4 hrs

=> Speed of the second train = 180/4 = 45 kmph

Now, required difference in distance in 1 hour = 45 - 36 = 9 kms.