# WORKSHEET ON 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.

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.

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.

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 ?

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).

Let the original speed be 100%.

Given : The speed is increased by 33⅓ %.

Then, the speed after increment is 133 %.

Ratio of the speeds is

100 % : 133⅓ %

100 : 133

100 : ⁴⁰⁰⁄₃

Divide both the terms by 100.

1 : ⁴⁄₃

So, ratio of the speeds is 1 : ⁴⁄₃.

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

1 : ¾

When the speed is increased by 33%, ¾ of the original time is enough to cover the same distance.

That is, when the speed is increased by 33⅓ %, ¼ of the original time will be decreased.

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

So, we have

¼ of the original time = 15 minutes

Multiply both sides by 4.

⋅ (¼ 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.

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

So, the formula to find the average speed is

Substitute p = 25 and q = 4.

The average speed is ²⁰⁰⁄₂₉ km/hr.

Given :The entire journey had taken 5 hours 48 minutes

5 hour 48 min = 5⁴⁸⁄₆₀ hours

5 hours 48 min = 5 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.

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

The formula to find the time is

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 = ⁵⁄₆₀¹⁄₁₂ 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

tˣ⁄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

tˣ⁄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 kms.

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

⋅ 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

Substitute distance = 3 and time = 15.

= ³⁄₁₅

= miles per minute

⋅ 60 miles per hour

= 12 miles per hour.

So, the speed must be 12 miles per hour.

Let x be the speed of A .

Then speed of B is

= 7 - x

The formula to find time is

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

14x- 98x + 168 = 0

Divide both sides by 14.

x-  7x + 12 = 0

x-  3x - 4x + 12 = 0

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

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

x = 3 or x = 4

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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