TIME AND DISTANCE PROBLEMS

Problem 1 : 

A person covers a certain distance at a certain speed. If he increases his speed by 33⅓ %, 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.

Solution :

Let the original speed be 100%.

Given : The speed is increased by 33⅓ %.

Then, the speed after increment is 

133⅓ %

Ratio of the speeds is

100 %  :  133⅓ %

100  :  133⅓

100  :  400/3

Divide both sides by 100. 

1  :  3/4

So, ratio of the speeds is

1 : 3/4

If the ratio of the speeds is 1 : 4/3, then the ratio of time taken to cover the same distance would be

1 : 3/4

When the speed is increased by 33⅓ %, 3/4 of the original time is enough to cover the same distance. 

That is, when the speed is increased by 33⅓ %, 1/4 of the original time will be decreased. 

The question says that when speed is increased by 33⅓ %, time is decreased by 15 minutes. 

So, we have

1/4 of the original time  =  15 minutes

Multiply both sides by 4.  

4 ⋅ (1/4 of the original time)  =  (15 minutes) ⋅ 4

Original time  =  60 minutes

Original time  =  1 hour

So, the time taken by the person initially is 1 hour. 

Problem 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 distance of the post office from the village.

Solution :

Here, the distance covered in both the ways is same. 

So, the formula to find the average speed is

=  2pq / (p + q)

Plug p  =  25 and q  =  4.  

=  (2 ⋅ 25 ⋅ 4) / (25 + 4) 

=  200 / 29

The average speed is 200/29 km/hr.  

Given :The entire journey had taken 5 hours 48 minutes

5 hour 48 min  =  5 ⁴⁸⁄₆₀ hours

5 hours 48 min  =  5 ⅘ hours

5 hours 48 min  =  29 / 5 hours

The formula to find the distance is

=  Speed ⋅ Time

Then, the distance covered in (29/5) hours at the average speed (200/29) kmph is

=  (200 / 29) ⋅ (29 / 5)

=  40 km 

So, the distance covered in the whole journey is 40 km.

(Whole journey : Village to post office + Post office to village)

Then the distance between the post office and village is 

=  40 / 2

=  20

So, the distance of the post office from the village is 20 km. 

Problem 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.

Solution :

Let "x" be the distance to be covered by the person to reach the station.  

The formula to find the time is

=  Distance / Speed

When the speed is 5 kmph, time is 

=  x/5  hrs

When the speed is 6 kmph, time is

=  x/6  hrs

Let "t" be the actual time required to cover the distance x.

And also,

7 minutes  =  7/60  hrs

5 minutes  =  5/60  =  1/12  hrs

Given : If the man walks at the rate of 5 km/hr, he misses the train by 7 minutes.

That is, he takes 7 minutes more than actual time. 

So, we have

t  =  x/5 - 7/60 ------(1)

Given : If he walks at the rate of 6 km/hr, he reaches the station 5 minutes before.

That is, he takes 5 minutes less than actual time. 

So, we have

t  =  x/6 + 1/12 ------(2)

From (1) and (2), we get

x/5  -  7/60  =  x/6  +  1/12

Solving for x :

12x/60  -  7/60  =  2x/12  +  1/12

(12x - 7) / 60  =  (2x + 1) / 12

L.C.M of (60, 12) is 60. 

Multiply both sides by 60.

12x - 7  =  5(2x + 1)

12x - 7  =  10x + 5

Simplify. 

2x  =  12

Divide both sides by 2. 

x  =  6

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

Problem 4 : 

A person has to cover a distance of 6 miles in 45 minutes. If he covers one-half of the distance in two-thirds of the total time. What must his speed be to cover the remaining distance in the remaining time ?

Solution :

Given : Total distance is 6 miles and total time is 45 minutes. And also, he covers one-half of the distance in two-thirds of the total time.

One-half of the total distance 6 miles is 

=  ½ ⋅ 6

=  3 km

Two-thirds of the total time 45 minutes is 

=  ⅔ ⋅ 45

=  30 minutes

From the above calculations, we have 

Remaining distance  =  6 - 3  =  3 miles

Remaining time  =  45 - 30  =  15 minutes

The formula to find the speed is 

=  Distance / Time

Plug, Distance  =  3 and Time  =  15.

=  3 / 15

=  1 / 5 miles per minute

=  (1/5) ⋅ 60 miles per hour

=  12 miles per hour. 

So, the speed must be 12 miles per hour.

Problem 5 :

A is faster than B . A and B each walk 24 miles. The sum of their speeds is 7 miles per hour and the sum of their time taken is 14 hrs. Find A's speed and B's speed (in mph).

Solution :

Let "x" be the speed of A .

Then speed of B is

=  7 - x 

The formula to find time is

=  Distance / Speed

Then, time taken by A is 

=  24/x  hrs

Time taken by B is 

=  24/(7 - x)  hrs 

Given : Sum of time taken is 14 hours. 

So, we have 

24/x  +  24/( 7 - x)  =  14

L.C.M of x and (7 - x) is x(7 - x). 

Multiply both sides by x(7 - x). 

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

168 - 24x + 24x  =  98x - 14x²

14x² - 98x + 168  =  0

Divide both sides by 14.

x² -  7x + 12  =  0

(x - 4)(x - 3)  =  0

x  =  4 or x  =  3

So, A's speed can be 4 mph or 3 mph. 

If A's speed is 4 mph, then B's speed

7 - x  =  7 - 4

7 - x  =  3 mph

If A's speed is 3 mph, then B's speed

7 - x  =  7 - 3

7 - x  =  4 mph

Given : A is faster than B

So, the speed of A is 4 miles per hour and B is 3 miles per hour. 

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