## Speed Time And Distance

Question 1 |

3 : 4 | |

4 : 3 | |

2 : 3 | |

3 : 2 |

Question 2 |

^{4}/

_{5}

^{th}of its original speed. As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached 36 minutes late. The original speed of the car (in km/h) is:

25 | |

35 | |

20 | |

30 |

^{4}/

_{5}S where S is his usual speed means

^{1}/

_{5}decrease in speed which will lead to

^{1}/

_{4}increase in time. Now the main difference comes in those 12km (30-18) and the change in difference of time = (45-36) min = 9 min

Thus,

^{1}/_{4} * T = 9 min where T is the time required to cover the distance of (30 - 18) = 12 km

T = 36 min = ^{36}/_{60} hours = 0.6 hours.

Speed of the car = ^{12}/_{0.6} = 20 kmph.

Question 3 |

15 km | |

16 km | |

14 km | |

17 km |

Question 4 |

12 kmph | |

11 kmph | |

14 kmph | |

8 kmph |

Question 5 |

624 km | |

600 km | |

584 km | |

684 km |

A.....X km.....M.....X km.....B

Total time taken in the journey = 17.5 hours

Time taken to cover X km at 30 km/h = ^{X}/_{30}.

Time taken to cover X km at 40 km/h = ^{X}/_{40}

Now, \begin{align} & \left( {\frac{X}{{30}}} \right) + \left( {\frac{X}{{40}}} \right) = 17.5 \cr & \left[ {\frac{{\left( {40X + 30X} \right)}}{{1200}}} \right] = 17.5 \cr & 70X = 17.5 \times 1200 \cr & X = 300\,km \cr & {\text{Total}}\,{\text{distance}},\,2x \cr & = 600\,km \cr\end{align}

Question 6 |

8 seconds | |

7 seconds | |

5 seconds | |

6 seconds |

^{st}train = 105 m;

Length of the 2^{nd} train = 90 m.

Relative speed of the trains,

= 45+72 = 117kmph

= ^{117*5}/_{18} = 32.5 m/sec;

Time taken to cross each other,

= (Length of 1^{st} train + length of 2^{nd} train) /relative speed of the trains.

Time taken = ^{195}/_{32.5} = 6 secs.

Question 7 |

1000 m | |

848 m | |

1675 m | |

800 m |

= 20 * 20

= 400 m.

Distance traveled by dog when he goes towards the child,

= 400 * ^{60}/_{40}

= 600 m and time required is 10 minutes.

In 10 min remaining distance between man and child,

= 400 - (20 * 10)

= 200 m.

Time taken by dog to meet the man,

= ^{200}/_{100}

= 2 min

**(Relative speed of dog with child is 40 m/min and same with man is 100 m/min.)**

Remaining distance in 2 min,

= 200 - (2 * 20),

= 160 m.

Now, the time taken by dog to meet the child again,

= ^{160}/_{40}

= 4 min.

In 4 min he covers = 4 * 60 = 240 m.

Now, remaining distance in 4 min = 160 - (4 * 20) = 80 m.

Time required by dog to meet the man once again = ^{80}/_{100} = 0.8 min.

Now remaining distance = 80 - (0.8 * 20) = 64 m.

Question 8 |

30 h | |

24 h | |

25 h | |

15 h |

Let

Usual Speed = X kmph.

Now,

Speed of bus after increasing the speed = (X +14)kmph. -------------- (1)

**A .....1680km .....B** In first case,

Time taken to covered the distance 1680 km = 1680/X ----------- (2)

In Second Case,

Time Taken to covered the distance 1680 km = 1680/(X + 14).

Time difference = 6 Hours.

So,

\begin{align} & \Rightarrow \left[ {\frac{{1680}}{x} - \frac{{1680}}{{\left( {x + 14} \right)}}} \right] = 6 \cr & \Rightarrow 1680\left[ {\frac{{14}}{{\left( {{x^2} + 14x} \right)}}} \right] = 6 \cr & \Rightarrow 280 \times 14 = {x^2} + 14x \cr & \Rightarrow {x^2} + 70x - 56x - 3920 = 0 \cr & \Rightarrow x\left( {x + 70} \right) - 56\left( {x + 70} \right) = 0 \cr & \Rightarrow x = - 70,56. \cr & {\text{hence,}}\,{\text{speed}}\,{\text{of}}\,{\text{minibus}}\,{\text{is}}\,56\,km/h \cr & {\text{put}}\,x = 56\,{\text{in}}\,{\text{equation}}\,(2) \cr & T = \frac{{1680}}{{56}} \cr & \,\,\,\,\,\, = 30\,{\text{hours}} \cr\end{align}

Question 9 |

262.4 km | |

283.33 km | |

260 km | |

275 km |

^{45}/

_{60}hour =

^{3}/

_{4}hour.

Let the distance be x km from Delhi where the two trains will be together.

Time taken to cover x km with speed 136 kmph be t hour

and time taken to cover x km with speed 100 kmph (As the train take 45 mins. more) be

(t+^{3}/_{4})

= (^{(4t+3)}/_{4});

Now,

100 * ^{(4t+3)}/_{4} = 136t;

Or, 25(4t+3) = 136t;

Or, 100t+75 = 136t;

Or, 36t = 75;

Or, t = ^{75}/_{36} = 2.08 hours;

Then, distance x km = 136 * 2.084 = 283.33 km.

Question 10 |

60 m | |

45 m | |

95 m | |

50 m |

Hence, when A runs 500 m, B runs 450 m.

Again, when B runs 400 m, C runs 360 m.

And, when B runs 450 m, C runs = 360 * ^{450}/_{400} = 405 m.

Required distance = 500 - 405 = 95 meter.

That means when A runs 500 meter then B can run 450m then C runs 405 m.