**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 **

**= 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

**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 - 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

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

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