# Quantitative Aptitude Questions On Time and Distance

Quantitative Aptitude Questions On Time and Distance

Time and distance

concept:

Speed:

We have the relation between speed, time and distance as follows:

Speed = distance / time.

So the distance covered in unit time is called speed.

This forms the basis for Time and Distance. It can be re-written as Distance = Speed X Time or

Time = Distance / Speed.

Units of Speed:

The units of speed are kmph (km per hour) or m / s.

1 kmph = 5 / 18 m / s

1 m / s = 18 / 5 kmph

Average Speed:

When the travel comprises of various speeds then the concept of average speed is to be applied.

Average Speed = Total distance covered / Total time of travel

Note: In the total time above, the time of rest is not considered.

Example 1:

If a car travels along four sides of a square at 100 kmph, 200 kmph, 300 kmph and 400 kmph find its average speed.

Soln:

Average Speed = Total distance / Total time.

Let each side of square be x km. Then the total distance = 4x km.

The total time is sum of individual times taken to cover each side.

To cover x km at 100 kmph, time  = x / 100.

For the second side time = x / 200.

Using this we can write average speed = 4x / (x/100 + x/200 + x/300 + x/400) = 192 kmph.

Example 2:

A man if travels at 5/6 th of his actual speed takes 10 min more to travel a distance. Find his usual time.

Soln:

Let s be the actual speed and t be the actual time of the man.

Now the speed is (5/6)s and time is (t+10) min. But the distance remains the same.

So distance 1 = distance 2 => s X t = (5/6)s X (t+10) => t = 50 min.

Example 3:

If a person walks at 30 kmph he is 10 min late to his office. If he travels at 40 kmph then he reaches to his office 5 min early. Find the distance to his office.

Soln:

Let the distance to his office be d. The difference between the two timings is given as 15 min = 1 / 4 hr.

Now if d km are covered at 30 kmph then time = d/30. Similarly second time = d/40.

So, d/30 – d/40 = 1 / 4 => d = 30 km.

Note:

When two objects move with speeds s1 and s2

1. In opposite directions their combined speed = s1 + s2
2. In same direction their combined speed = s1 ~ s2.

Example 4:

Two people start moving from the same point at the same time at 30 kmph and 40 kmph in opposite directions. Find the distance between them after 3 hrs.

Soln:

Speed = 30 + 40 = 70 kmph (since in opposite directions)

Time = 3 hrs

So distance = speed X time = 70 X 3 = 210 km.

Example 5:

A starts from X to Y at 6 am at 40 kmph and at the same time B starts from Y to X at 50 kmph. When will they meet if X and Y are 360 km apart?

Soln:

Distance = 360 km

Speed = 40 + 50 = 90 kmph.

Time = distance / speed = 360 / 90 = 4hrs from 6 am => 10 am.

Example 6:

A starts from X to Y at 6 am at a speed of 50 kmph. After two hours B starts from Y to X at 60 kmph. When will they meet if X and Y are 430 km apart?

Soln:

By the time B started A traveled for 2 hrs => 2 X 50 = 100 km.

So at 8 am, distance = 430 – 100 = 330 km

Speed = 50 + 60 = 110 kmph.

Time = distance / speed = 330 / 110 = 3 hrs from 8 am => 11 am.

Note:

When a train crosses a negligible length object (man / pole / tree) the distance that it has to travel is its own length.

When a train has to cross a lengthy object (train / bridge / platform) the distance it has to travel is the sum of its length and the length of the object.

Example 7:

If a train traveling at 40 kmph crosses another train of length 100m traveling at 14 kmph in opposite direction in 30 s find the length of the train.

Soln:

Let length of train be d.

Distance to be covered = d + 100.

Speed = 40 + 14 = 54 kmph = 54 X 5 / 18 = 15 m / s

Time = 30 s.

Distance = speed X time => d+100 = 15 X 30 => d = 350 m.