Time and distance problems play a major role in quantitative aptitude test. It is bit difficult to score marks in competitive exams without knowing the shortcuts related to time and distance. We might have already learned this topic in our lower classes.Even though we have been already taught this topic in our lower classes, we need to learn some more short cuts which are being used to solve the problems in the above topic.

The only thing we have to do is, we need to apply the appropriate short cut and solve the problems in a limited time. This limited time will be one minute or less than one minute in most of the competitive exams.

1. Speed or Velocity = Distance / Time |

2. Distance= Speed x Time |

3. Time = Distance / Speed |

4. a km/hr= (ax5/18) m/sec |

5. b m/sec = (bx18/5) km/hr. |

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

7. If a certain distance is covered at p km/hr and an equal distance is covered at q km/hr, then the average speed of the whole journey is = (2pq)/(p+q) km/hr |

Students who are preparing to improve their aptitude skills and those who are preparing for this type of competitive test must prepare this topic in order to have better score. Because, today it is bit difficult to score marks in competitive exams without knowing time and distance problems. Whether a person is going to write placement exam to get placed or a student is going to write a competitive exam in order to get admission in university, they must be knowing to solve time and distance problems. This is the reason for why people must study this topic.

As we mentioned in the above paragraph, a person who wants to get placed in a company and a students who wants to get admission in university for higher studies must write competitive exams like placement test and entrance exam. To meet the above requirement, it is very important to score more marks in the above mentioned competitive exams. To score more marks, they have to prepare this topic. Preparing this topic would definitely improve their marks in the above exams. Preparing this topic is not difficult task. We are just going to remember the stuff that we have already learned in our lower classes

Students will learn, how and when they have to apply shortcuts to solve the problems which are related to time and distance. Apart from the regular shortcuts, students can learn some additional tricks in this topic time and distance shortcuts. Already we are much clear with the four basic operations which we often use in math. They are addition, subtraction, multiplication and division. Even though we are much clear with these four basic operations, we have to be knowing some more stuff to solve the problems which are being asked from the topic time and distance in competitive exams. The stuff which I have mentioned above is nothing but the tricks and shortcuts which need to solve the problems in a very short time.

Short cut is nothing but the easiest way to solve problems related to time and distance. In competitive exams, we will have very limited time to solve each problem. Then only we will be able to attend all the questions. If we do problems in competitive exams in perfect manner with all the steps, it will definitely take much time and we may not able to attend the other questions. So we need some other way in which the problems can be solved in a very short time. The way we need to solve the problem quickly is called as shortcut.

**Here, we are going to have some ****time and distance problems **such that how time and distance shortcuts can be used. You can check your answer online and see step by step solution.

1. A person covers a certain distance at a certain speed. If he increases his speed by 33 1/3%, he takes 15 minutes less to cover the same distance. Find the time taken by him initially to cover the distance at the original speed.

If the original speed is 100%, speed after increment is 133 1/3%.

Ratio of the speeds is 100%: 133 1/3%-->100%:(400/3)%

So, ratio of the speeds is 1:4/3

If the ratio of the speed is 1:4/3, ratio of time taken would be 1:3/4

When the speed is increased by 33 1/3%, 3/4 of the original time is enough to cover the same distance.

That is,when the speed is increased by 33 1/3%, 1/4 of the original time will be decreased.

The question says that when speed is increased by 33 1/3%, time is decreased by 15 minutes.

Therefore, 1/4 of the original time = 15 minutes

Original time = 4X15 = 60 minutes

Hence, time taken by him initially = 60 mins or 1 hour

2. A man traveled from the village to the post office at the rate of 25 k mph and walked back at the rate of 4 kmph. If the entire journey had taken 5 hours 48 minutes, find the required distance of the post office from the village.

Average speed = 2pq/(p+q), here p=25 q=4

= (2X25X4)/(25+4)

Therefore, average speed = 200/29 km/hr

And 5 hour 48 min = 5^{48}⁄_{60}hrs = 29/5 hours

Distance = Speed X Time

Distance = (200/29)X(29/5) = 40 km

Distance covered in 5 hrs 48 min = 40 km

Hence,distance of the post office from the village

= 40/2 = 20km

= (2X25X4)/(25+4)

Therefore, average speed = 200/29 km/hr

And 5 hour 48 min = 5

Distance = Speed X Time

Distance = (200/29)X(29/5) = 40 km

Distance covered in 5 hrs 48 min = 40 km

Hence,distance of the post office from the village

= 40/2 = 20km

3. If a man walks at the rate of 5 km/hr, he misses a train by 7 minutes. However, if he walks at the rate of 6 km/hr, he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.

Let "x" be the distance which has to be found.

Difference between the times in walking at different speed 12 minutes = 12/60 hr = 1/5 hr

When the speed is 5kmph, time = x/5

When the speed is 6kmph, time = x/6

Difference in time taken = 1/5 hr.

(x/5) - (x/6) = 1/5

By simplification, we get x = 6 km

Hence,the distance covered by him to reach the station is 6 km

4. A person has to cover a distance of 6 km in 45 minutes. If he covers one-half of the distance in two-thirds of the total time; to cover the remaining distance in the remaining time, his speed(in kmph) must be

From the given information, Already 3 km distance (one half of the distance 6km) has been covered in 30 minutes or 1/2 hr. (two third of the total time 45 minutes)

Remaining distance = 3 km

Time available = 15 minutes or 1/4 hr

Speed required = Distance / Time

= 3/(1/4) kmph = 3X4 kmph = 12 kmph

Hence, speed required to cover remaining distance is 12 kmph

Remaining distance = 3 km

Time available = 15 minutes or 1/4 hr

Speed required = Distance / Time

= 3/(1/4) kmph = 3X4 kmph = 12 kmph

Hence, speed required to cover remaining distance is 12 kmph

5. A is faster than B . A and B each walk 24 km. The sum of their speeds is
7 km / hr and the sum of their time taken is 14 hrs. Then A's speed and B's speed (in kmph) are

Let "x" be the speed of A . Then speed of B = 7-x

Time taken by A = 24/x

Time taken by B = 24/(7-x)

Time(A)+Time(B) = 14 hours.

24/x + 24/(7-x) = 14

24(7-x)+24x = 14x(7-x)

14x^{2}-98x+168=0--->x^{2}-7x+12=0

(x-4)(x-3)=0--->x = 4 and x = 3

Hence, the speed of A and B are 4kmph and 3 kmph respectively.

Time taken by A = 24/x

Time taken by B = 24/(7-x)

Time(A)+Time(B) = 14 hours.

24/x + 24/(7-x) = 14

24(7-x)+24x = 14x(7-x)

14x

(x-4)(x-3)=0--->x = 4 and x = 3

Hence, the speed of A and B are 4kmph and 3 kmph respectively.

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