**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.

Before look at the problems, if you would like to know the shortcuts related to speed, distance and time,

**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 **

**= 2****p****q / (p + q)**

**Plug p = 25 and q = 4. **

**= (2 **⋅ **25 **⋅ **4) / (25 + 4) **

**= 200 / 29**

**The average speed is 200/29 km/hr. **

**Given :****T**he 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

**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

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 - 14x^{2}

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

Divide both sides by 14.

x^{2 }- 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

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

After having gone through the stuff given above, we hope that the students would have understood "Time and distance problems"

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