Quantitative Aptitude - SPLessons

Time – Distance Problems

Chapter 18

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

Time – Distance Problems

shape Introduction

An object that is moving at a constant rate is said to be in uniform motion and this chapter involves the calculation of the time and distance of the uniform motion object. Uniform motion problems may involve objects going in same direction, opposite directions and round trips.


shape Methods

Distance: The space between two objects or two points is known as distance.


Time: It is an ongoing sequence of events related to past, future and present and it is measured in seconds, minutes, hours, days, weeks, months and years.


Rules to Quickly Solve Problems Related To Time and Distance:

Rule 1: If the speed is given in km per hour and we want to convert it in to meter per second, we have to multiply the given speed by \(\frac{5}{18}\).

    Eg: 90 km/hr = \(90 \times \frac{5}{18}\) = 25 meters per sec


Rule 2: If the speed is given in meter per sec and we want to convert it in to km per hour, we have to multiply the given speed by \(\frac{18}{5}\).

    Eg: 25 m/sec = \(25 \times \frac{18}{5}\) = 90 km/hr.


Rule 3: If the ratio of speeds of two vehicles in the ratio a : b, then the taken ratio of the two vehicles will be b : a.

    Eg: The ratio of speeds of two vehicles is 2 : 3. Then time taken ratio of the two vehicles to cover the same distance will be 3 : 2.


Rule 4: If the ratio of speeds of two vehicles in the ratio a : b, then the distance covered ratio in the same amount of time will also be a : b.

    Eg: The ratio of speeds of two vehicles is 2 : 3. If each vehicle is given one hour time, then the distance covered by the two vehicles will be in the ratio 2 : 3.


Rule 5: If A is twice as fast as B, then the distance covered ratio of A and B in the same amount of time will be 2 : 1.

    Eg: A is twice as fast as B and each given 1 hour time. If A covers 20 miles of distance in
    one hour, then B will cover 10 miles of distance in one hour.


Rule 6: If one increases or decreases the speed of the vehicle in the ratio a : b, then the new speed is
= \(\frac{b \times original \ speed}{a}\).

    Eg: David travels at a speed of 56 miles per hour. If he reduces his speed in the ratio 7 : 6,
    his new speed = \(\frac{6 \times 56}{7}\) = 48 miles per hour.


Rule 7: If a person covers a particular distance at a speed “a” miles per hour and comes backs to his original position at a speed of “b” miles per hour (distance covered is same). Then the average speed for the total distance covered is = \(\frac{2ab}{(a + b)}\) miles hour.

    Eg: John travels 300 miles at the speed 45 miles per hour and travels another 300 miles at the speed of 55 miles per hour. Find the average speed for the whole journey. Average speed for the whole journey is \(\frac{2 \times 45 \times 55}{45 + 55}\) = 55 miles per hour.


Let’s see some examples for applying these rules:


Example 1:
The length of a train is 300 meter and length of the platform is 500 meter. If the speed of the train is 20 m/sec, find the time taken by the train to cross the platform.

Solution:

    Distances needs to be covered to cross the platform = Sum of the lengths of the train and platform

    ∴ Distance traveled to cross the platform = 300 + 500 = 800 m

    Time taken to cross the platform is = \(\frac{Distance}{Speed}\) = \(\frac{800}{20}\) = 40 sec.

    Hence, time taken by the train to cross the platform is 40 seconds.


Example 2:
A train is running at a speed of 20 m/sec. If it crosses a pole in 30 seconds, find the length of the
train in meters.

Solution:

    The distance covered by the train to cross the pole is = Length of the train

    We know, Distance = Speed × Time

    ∴ Length of the train = Speed × Time = 20 × 30 = 600 m. Hence, length of the train is 600 meters.


Example 3:
Two trains running at 60 km/h and 48 km/h cross each other in 15 seconds when they run in opposite direction. When they run in the same direction, a person in the faster train observes that he crossed the slower train in 36 seconds. Find the length of the two trains (in meters).

Solution:

    When two trains are running in opposite direction, relative speed = 60 + 48 = 108 kmph = \(108 \times \frac{5}{18}\)m/sec = 30 m/sec.

    Sum of the lengths of the two trains is sum of the distances covered by the two trains in the above relative speed. Then, sum of the lengths of two trains is = Speed × Time = 30 × 15 = 450 m.

    When two trains are running in the same direction, relative speed = 60 – 48 = 12 km/h = \(12 \times \frac{5}{18}\) = \(\frac{10}{3}\) m/sec.

    When the two trains running in the same direction, a person in the faster train observes that he crossed the slower train in 36 seconds. The distance he covered in 36 seconds in the relative speed is equal to the length of the slower train.

    ∴ Length of the slower train = \(36 \times \frac{10}{3}\) = 120 m

    and, length of the faster train = 450 – 120 = 330 m

    Hence, the length of the two trains are 330 m and 120 m.


Example 4:
It takes 20 seconds for a train running at 54 km/h to cross a platform.And it takes 12 seconds for the same train in the same speed to cross a man walking at the rate of 6 km/h in the same direction in which the train is running. What is the length of the train and length of platform (in meters)?

Solution:

    Relative speed of the train to man = 54 – 6 = 48 km/h = \(48 \times \frac{5}{8}\) m/sec = \(\frac{40}{3}\) m/sec

    When the train passes the man, it covers the distance which is equal to its own length in the above relative speed. We have 12 seconds for the train to cross the man. So, the length of the train = Relative Speed × Time = \(\frac{40}{3} \times 12\) = 160 m.

    Speed of the train = 54 km/h = \(54 \times \frac{5}{18}\) m/sec = 15 m/sec. When the train crosses the platform, it covers the distance which is equal to the sum of lengths of the train and platform. We have the train taking 20 seconds to cross the platform.

    So, the sum of lengths of train and platform = Speed of the train × Time = 15 × 20 = 300 m

    ∴ Length of train + Length of platform = 300 ⇒ 160 + Length of platform = 300 ⇒ Length of platform = 300 – 160 = 140 m

    Hence, the lengths of the train and platform are 160 m and 140 m, respectively.


Example 1:
A man walks 20 km in 4 hours. Find his speed.

Solution:

    Distance covered = 20 km

    Time taken = 4 hours

    We know, Speed = \(\frac{Distance}{Time}\) = \(\frac{20}{4}\) km/hr

    Therefore, speed = 5 km/hr


Example 2:
A car covers a distance of 450 m in 1 minute whereas a train covers 69 km in 45 minutes. Find the ratio of their speeds.

Solution:

    Speed of car = \(\frac{Distance \ covered}{Time \ taken}\) = \(\frac{450}{60}\) m/sec = \(\frac{15}{2}\)

    = \(\frac{15}{2}\) × \(\frac{18}{5}\) km/hr

    = 27 km/hr

    Distance covered by train = 69 km

    Time taken = 45 min = \(\frac{45}{60}\) hr = \(\frac{3}{4}\) hr

    Therefore, speed of trains = \(\frac{69}{\frac{3}{4}}\) km/hr

    = \(\frac{69}{1}\) × \(\frac{4}{3}\) km/hr

    = 92 km/hr

    Therefore, ratio of their speed i.e., speed of car/speed of train = \(\frac{27}{92}\) = 27 : 92



Example 1:
A car travel 60 km in 30 minutes. In how much time will it cover 100 km?

Solution:

    Using the unitary method;

    Time taken to cover 60 km = 90 minutes

    Time taken to cover 1 km = \(\frac{90}{60}\)90/60 minutes

    Time taken to cover 100 km = \(\frac{90}{60} \times 100\) = 150 minutes

    Formula of Speed = \(\frac{Distance}{Time}\)

    = \(\frac{60}{\frac{3}{2}}\) km/hr [given 1 hour 30 min = 1 \(\frac{30}{60}\) = 1 \(\frac{1}{2}\) hours = \(\frac{3}{2}\) hours]

    = \(\frac{60}{1}\) × \(\frac{2}{3}\) km/hr = 40 km/hr

    Now, using the formula of Time = \(\frac{Distance}{Speed}\) = \(\frac{100}{40}\) km/hr = \(\frac{5}{2}\)hours

    = \(\frac{5}{2}\) × 60 minutes, (Since 1 hour = 60 minutes)

    = 150 minutes


Example 2:
Victor covers 210 km by car at a speed of 70 km/hr. find the time taken to cover this distance.

Solution:

    Using the unitary method;

    70 km is covered in 1 hour.

    1 km is covered in \(\frac{1}{70}\) hours.

    210 km is covered in \(\frac{1}{70}\) × 210 hours = 3 hours

    Given: speed = 70 km/hr, distance covered = 210 km

    Time = \(\frac{Distance}{Speed}\) = \(\frac{210}{70}\) hours = 3 hours



Example 1:
A train moves at a speed of 80 km/hr. How far will it travel in 36 minutes?

Solution:

    Using the unitary method;

    In 60 minutes, distance covered = 80 km

    In 1 minute, distance covered = \(\frac{80}{60}\) km

    Speed = 80 km/hr

    Time = 36 minutes or \(\frac{36}{60}\) hours

    We know, formula of Distance = Speed x Time

    = 80 × \(\frac{36}{60}\)

    = 48 km

    Therefore, the train will travel 48 km in 36 minutes.


Example 2:
A student goes to school at the rate of 7 ½ km/hr and reaches 10 minutes late. If he travels at the speed of 10 km/hr he is 10 minutes early. What is the distance to the school?

Solution:

    Let the distance to school be 1 km, then time taken to cover 1 km at the rate of 7 \(\frac{1}{2}\) km/hr

    = \(\frac{Distance}{speed}\) = \(\frac{1}{\frac{15}{2}}\) = \(\frac{2}{15}\) hr = \(\frac{2}{15}\) × 60 minutes = 8 minutes

    Time taken to cover 1 km at the rate of 10 km/hr

    = \(\frac{Distance}{speed}\) = \(\frac{1}{10}\) hr = \(\frac{1}{10}\) × 60 minutes = 6 minutes

    Therefore, difference in time taken = (8 – 6) minutes = 2 minutes

    But actual difference in time is 20 minutes

    When the difference in time is 2 minutes, distance to school = 1 km

    When the difference in time is 1 minute, distance to school = \(\frac{1}{2}\) km

    When the difference in time is 20 minutes, distance to school = \(\frac{1}{2}\) × 20 km

    Therefore, the distance to school is 10 km.

shape Formulae

1. Speed = \(\frac{Distance}{Time}\), Time = \(\frac{Distance}{Speed}\), Distance = Speed x Time


2. \(x\) km/hr = \(x\) x \(\frac{5}{18}\) m/sec


3. \(x\) m /sec = \(x\) x \(\frac{18}{5}\)


4. If the ratio of the speeds of A and B is a: b, then the ratio of the times taken by them to cover the same distance = \(\frac{1}{a}\) : \(\frac{1}{b}\) (or)
a : b


5. Suppose a man covers a certain distance at \(x\) km/hr and an equal distance at \(y\) km/hr. Then, the average speed during the whole journey is \(\frac{2xy}{x + y}\)


6. An object that moves at a constant rate is said to be in uniform motion. Then,
d = r x t

It gives the relationship between time (t), rate (r) and distance (d).

shape Samples

1. Walking at the rate of 4 km/hr a man covers certain distance in 2 hrs 45 minutes. Running at a speed of 16.5 km/hr, by how much time the man will cover the same distance?

Solution:

    Given that,

    Rate = 4 km/hr

    Time = 2 hrs 45 minutes

    New Speed = 16.5 km/hr

    Now, Distance = speed x time

    Initial distance = 4 x \(\frac{11}{4}\) = 11 km/hr

    Then, Time = \(\frac{initial distance}{Speed}\)

    ⇒ Time = \(\frac{11}{16.5}\)

    ⇒ Time = 40 minutes

    Therefore, by 40 minutes the man will cover the same distance.


2. A person travels 600 meters long distance in 5 minutes. Calculate the speed of the person in km/hr?

Solution:

    Given that,

    Distance travelled by the person = 600 m

    Time = 5 minutes

    Now, Speed = \(\frac{Distance}{Time}\)

    Speed = \(\frac{600}{5 * 60}\)

    Speed = 2 m/sec

    Convert m/sec in kmph, i.e.

    ⇒ Speed = 2 x \(\frac{18}{5}\)

    ⇒ Speed = 7.2 km/hr

    Therefore, Speed of the person = 7.2 km/hr


3. Walking at \(\frac{4}{3}\) of its usual speed, a trian in 10 minutes too late. Find the usual time to cover the journey?

Solution:

    Given,

    Time of train is 10 minutes late

    Let,

    The new speed = \(\frac{4}{3}\) of its usual speed

    As, Speed = \(\frac{Distance}{Time}\) ⇔ Time = \(\frac{Distance}{speed}\)

    ⇒ Time = \(\frac{4}{3}\) of its usual time

    So, (\(\frac{4}{3}\) of usual time) – (usual time)= 10 minutes

    ⇒ \(\frac{1}{3}\) of usual time = 10

    ⇒ usual time = 10 x 3 = 30 minutes

    Therefore, usual time to cover the journey = 30 minutes


4. A man travelled from the village to the post – office at the rate of 20 kmph and walked back at the rate of 5 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post – office from the village?

Solution:

    Let,

    \(x\) = 20 kmph

    \(y\) = 5 kmph

    Given, time took for the whole journey = 5 hours 48 minutes

    Now, Average speed = \(\frac{2xy}{x + y}\)

    ⇒ Average speed = \(\frac{2 * 20 * 5}{20 + 5}\)

    ⇒ Average speed = \(\frac{200}{25}\)

    ⇒ Average speed = 8 kmph

    So, Distance travelled in 5 hours 48 minutes i.e 5 x \(\frac{4}{5}\) hours = 8 x \(\frac{29}{5}\) = 46.4 km

    Therefore, Distance between the post – office from the village = \(\frac{46.4}{2}\) = 23.2 km


5. A goods train leaves a station at a certain time and at a fixed speed. After 6 hours, an express train leaves the same station and moves in the same direction at a uniform speed of 90 kmph. This train catches up the goods train in 4 hours. Find the speed of the goods train?

Solution:

    Given that,

    Distance covered by goods train in (6 + 4) = 10 hours

    Distance covered by express train in 4 hours

    Speed of express train = 90 kmph

    Let,

    The speed of the goods train be \(x\) kmph

    Now,

    Distance = Speed x Time

    Distance covered by goods train in 10 hours = Distance covered by express train in 4 hours

    ⇒ 10 \(x\) = 4 x 90

    ⇒ \(x\) = \(\frac{360}{10}\)

    ⇒ \(x\) = 36

    Therefore, speed of goods train = 36 kmph