TIME AND DISTANCE PROBLEMS

About "Time and Distance Problems"

Time and Distance Problems :

Problems on time and distance play a major role in competitive exams. It is bit difficult to score marks in competitive exams without knowing the stuff on time and distance.

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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

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

Let us look at the next problem on "Time and distance problems".

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 ⁴⁸⁄₆₀hours

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

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

Let us look at the next problem on "Time and distance problems".

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

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

Let us look at the next problem on "Time and distance problems".

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.

Hence, the speed must be 12 miles per hour.

Let us look at the next problem on "Time and distance problems".

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