In this lesson we learn to solve problems on speed, distance, time, relative speed and boats and streams etc

1. Distance = Speed × Time

2. 1 km/hr = $\displaystyle\frac{5}{{18}}$ m/s

3. If the ratios of speed is a : b : c, then the ratios of time taken is : $\displaystyle\frac{1}{a}:\frac{1}{b}:\frac{1}{c}$

To distance, Speed and Time both are directly proportional and To speed, Time is inversely proportional. $S\, \propto \,D$ and $T\, \propto \,D$ i.e. if speed is doubled, distance covered in a given time also gets doubled and ,$S\, \propto \,\displaystyle\frac{1}{T}$ i.e. if speed is doubled, time taken to cover a distance will be half.

Average speed is defined as =.$\displaystyle\frac{{{\text{Total distance travelled}}}}{{{\text{Total time taken}}}}$

If two bodies are moving (in the same direction or in the opposite direction), then the speed of one body with respect to the other is called its relative speed.

Relative speed is a phenomenon that we observe everyday. Suppose you are travelling in college bus and there is a second bus coming in the opposite direction, then it seems that the second Bus is moving much faster than actual. If both the Buses were moving in the same direction at same speeds, they seem to be stationary if seen from one of these Busses, even though they might actually be at a speed of 100 km/hr each. So what you actually observe is your speed relative to the other.

Concepts

1. If two objects are moving in opposite directions towards each other at speeds u and v, then relative speed = Speed of first + Speed of second = u + v.

This is also the speed at which they are moving towards each other or the speed at which they may be moving away from each other.

2. If the two objects move in the same direction with speeds u and v, then

relative speed = difference of their speeds = u – v.

This is also the speed at which the faster object is either drawing closer to the slower object or moving away from the slower object as the case may be.

3. If the two objects start from A and B with speeds u and v respectively, and after crossing each other take a and b hours to reach B and A respectively, then u : v = $\sqrt {\displaystyle\frac{b}{a}} $

Note: In case of Trains moving in the opposite directions or in the same direction, the total distance required to be travelled before they cross each other completely is equal to the sum of the lengths of the two trains. This distance is covered at the relative speeds of the trains.

D = S X T

1. 1 Pole and I Train:

Length of The Train (m) = Speed of the Train (m/s) X Time taken to cross the pole (s)

2. 1 Train and 1 Platform

Length of the Train + Length of the Platform (m) = Speed of the Train (m/s) X Time taken to cross the platform (s)

3. 1 Train with speed speed ${V_1}$ and 1 moving person with speed ${V_2}$

Length of The Train (m) = [Speed of the Train - Speed of the Man] (m/s) X Time taken to cross the man (s)

Length of The Train (m) = [Speed of the Train + Speed of the Man] (m/s) X Time taken to cross the man (s)

4. 2 Trains with speeds ${V_1},{V_2}$

[Length of The Train 1 + Length of the Train 2](m) = [Speed of the Train1 - Speed of the Train 2] (m/s) X Time taken to cross (s)

[Length of The Train 1 + Length of the Train 2](m) = [Speed of the Train1 + Speed of the Train 2] (m/s) X Time taken to cross (s)

Downstream motion of a boat is its motion in the same direction as the flow of the Stream.

Upstream motion is exactly the opposite.

There are two parameters in these problems.

1. Speed of the Stream (S): This is the speed with which the river flows.

2. Speed of the boat in still water (B): If the river is still, this is the speed at which the boat would be moving.

The effective speed of a boat in upstream = B – S

The effective speed of a boat in downstream = B + S

3. The speed of the boat in still water is given as B = $\displaystyle\frac{1}{2}(d\; + \;u)$, and the speed of the Stream S = $\displaystyle\frac{1}{2}(d\; + \;u)$,

where d and u are the downstream and upstream speeds, respectively.

The problems in circular motion deal with races on a circular track to calculate the time of meeting at the starting point and anywhere on the track.

First time meeting of A and B at the starting point = (LCM of x and y)

Note: This formula would remain the same even if they move in the opposite directions.

2. If two people A and B start from the same point with speeds m km/hr and n km/hr respectively, at the same time and move in the same direction along a circular track, then the two would meet for the first time by the formula given below:

Time of the first meeting = .$\displaystyle\frac{{{\rm{Circumference}}\,\,{\rm{of}}\,\,{\rm{the}}\,\,{\rm{track}}}}{{{\rm{Relative}}\,\,{\rm{speed}}}}$

1. A scooterist covers a certain distance at 36 km/hr. How much distance does he cover in 3 min?

Speed = 36 km/hr = 36 × $\displaystyle\frac{5}{{18}}$ × m/s = 10 m/s

Thus, the distance covered in 3 min = (10 x 3 x 60) = 1,800 m.

2. Walking at $\displaystyle\frac{3}{4}$ of his usual speed a man is $1\displaystyle\frac{1}{2}$ hr late. Find his usual travel time.

Let usual time be t hours. We know that S X T = D, But now his speed is 3/4 (S). So time to cover the same distance becomes 4/3 (T).

$\displaystyle\frac{4}{3}$ × t = t + $\displaystyle\frac{3}{2}$

t = 4.5 hr.

Alternative method

We know that when his speed got reduced his time went up by 1 1/2 hr.

st = $\displaystyle\frac{{3s}}{4}$(t +$\displaystyle\frac{3}{2}$ )

$ \Rightarrow $ 4st = 3st + 4.5 s

$ \Rightarrow $ t = 4.5 hr.

3. A man starts from L to M, another from M to L at the same time. After passing each other they complete their journey in 3 1/3 hours and 4 4/5 hours respectively. Find the speed of the second man if the speed of the first is 24 km/hr?

Speed of first man : Speed of second man = $\displaystyle\sqrt {\frac{b}{a}} $ where a and b are the time taken by 1st and 2nd man respectively.

Speed of first man : Speed of second man = $\displaystyle\sqrt {\frac{{\frac{{24}}{5}}}{{\frac{{10}}{3}}}} = \sqrt {\frac{{24}}{5} \times \frac{3}{{10}}} = \sqrt {\frac{{36}}{{25}}} = \frac{6}{5}$

Thus second man’s speed = $\displaystyle\frac{5}{6}$ × 24 = 20 km/hr.

4. Two cyclists cover the same distance in 15 km/hr and 16 km/hr, respectively. Find the distance travelled by each, if one takes 16 min longer than the other does.

Let the required distance be x km.

$\displaystyle\frac{{\rm{x}}}{{{\rm{15}}}}{\rm{ - }}\frac{{\rm{x}}}{{{\rm{16}}}}{\rm{ = }}\frac{{{\rm{16}}}}{{{\rm{60}}}}$ or 16x – 15x = 64 or x = 64

Hence, the required distance = 64 km

Alternative Method:

If the speeds are in the ratio 15 : 16 then the ratio of the times to cover the same distance would be in the ratio 16 : 15 or 16x and 15x respectively. But we know that difference in the times is 16 min or 16x - 15x = 16 min so x = 16 min

From the above derivation, second person takes 15 x 16 min = 4 hours to cover the distance at the speed of 16 km/hr. So distance = 4 X 16 = 64 km

5. A thief is spotted by a policeman at a distance of 400 m. If the speed of the thief be 10 km/hr and that of the policeman be 12 km/hr, at what distance will the policeman catch the thief?

Relative speed of the policeman = 2 km/hr. This means, every hour the distance between police man and thief get reduces by 2000 meters. To cover 400 meters, Police needs 400/2000 Hrs or 1/5th hour. But in 1/5th hour police covers 12/5 km = 2.4 km

6. What is the average speed if a person travels from A to B and back at the speeds of 10 km/hr and 20 km/hr, respectively

a. for equal intervals of time,

b. for equal distance,

a. the concept of weighted average can be used here.

The average speed = $\displaystyle\frac{{\left( {10 + 20} \right)}}{2} = 15{\rm{ }}km/hr$

Note:

For equal intervals of time, the average speed is given as $\displaystyle\frac{{{{\rm{S}}_{\rm{1}}}{\rm{ + }}{{\rm{S}}_{\rm{2}}}{\rm{ + }}{{\rm{S}}_{\rm{3}}}{\rm{ + }}\,...{\rm{ + }}\,{{\rm{S}}_{\rm{n}}}}}{{\rm{n}}}$

, where S1, S2, S3, ... , Sn are the speeds and n is the number of observations.

b. The weighted average concept cannot be applied here, because we do not know the fractions of time spent travelling the two distances. It would be a mistake to calculate the average speed as $\displaystyle\frac{{\left( {{\rm{10 + 20}}} \right)}}{{\rm{2}}}{\rm{ = 15 Km/hr}}$

Total time taken = $\left( {\displaystyle\frac{{\rm{D}}}{{{\rm{10}}}}} \right){\rm{ + }}\left( {\displaystyle\frac{{\rm{D}}}{{{\rm{20}}}}} \right){\rm{ = T}}$

Total distance travelled = 2D.

So, average speed = $\displaystyle\frac{{{\rm{2D}}}}{{\rm{T}}}{\rm{ = }}\displaystyle\frac{{{\rm{2D}}}}{{\left( {\displaystyle\frac{{\rm{D}}}{{{\rm{10}}}}{\rm{ + }}\frac{{\rm{D}}}{{{\rm{20}}}}} \right)}}{\rm{ = }}\displaystyle\frac{{{\rm{2D}}}}{{{\rm{D}}\left( {\displaystyle\frac{{\rm{1}}}{{{\rm{10}}}}{\rm{ + }}\displaystyle\frac{{\rm{1}}}{{{\rm{20}}}}} \right)}}{\displaystyle\rm{ = 13}}\displaystyle\frac{{\rm{1}}}{{\rm{3}}}{\rm{ }}k{\rm{m/hr}}$

Note:

The physical interpretation of average speed of km/hr is that if the person had moved from A to B and back at a constant speed of km/hr, he would have taken the same total time as in that case when he travelled with two different speeds of 10 km/hr and 20 km/hr (may be he was travelling with the car or a bullock cart!)

If you travel equal distances with speeds u and v, then the average speed over the entire journey is given as

7. A hare makes 9 leaps in the same time as a dog makes 5. But the dog’s leap is 2m while hare’s is only 1 m. How many leaps will the dog have to make before catching up with the hare if the hare has a head start of 16 m?

Distance covered by dog in 5 leaps = 5 × 2 = 10 m

Distance covered by hare in 9 leaps = 9 × 1 = 9 m

Distance gained by the dog in 5 leaps = 1 m. Hence, for 1 m gain he has to make 5 leaps.

Number of leaps required by the dog to gain 16 m = 5 × 16 = 80 leaps.

14. A train running at 54 km/hr takes 20 s to cross a platform and 12 s to pass a man walking in the same direction at a speed of 6 km/hr. Find the length of the train and the platform.

Let the length of the train = x m. Let the length of the platform = y m.

Speed of the train relative to the man = 54 - 6 = 48 km/hr X $\displaystyle\frac{5}{{18}}$ = $\displaystyle\frac{{40}}{3}$ m/s

In passing the man, the train covers its own length with relative speed.

Length of the train =$\displaystyle\frac{{40}}{3}$ × 12 = 160 m.

Now Length of the Train + Length of the Platform (m) = Speed of the Train (m/s) X Time taken to cross the platform (s)

Speed of train = 54 km/hr X $\displaystyle\frac{5}{{18}}$ = 15 m/s

160 + y = 15 X 20

Length of the platform = 140 m.

20. Amar and balu as a warm up exercise are jogging on a circular track. Balu is a better athelete and jogs at 18 km/hr, while Amar jogs at 9 km/hr. The circumference of the track is 500 m. They start from the same point and in the same direction. When will they be together again for the first time?

Since Balu is faster than Amar he will take a lead and as they keep running, the gap between them will also keep widening, unlike on a straight track they would meet again even if Balu is faster than Amar.

The same problem could be rephrased as: In what time would Balu take a lead of 500 m over Amar?

Every second, Balu is taking a lead of m = 2.5 m over Amar.

Hence, he takes to take a lead of 500 m over Amar. Hence, they would meet for the first time after 200 s.

For every round that Amar makes, Balu would have made 2 rounds because the ratio of their speeds is 1 : 2. Hence, when Amar has made one full round, Balu would have taken a lead of one round.

Therefore, they would meet after $\displaystyle\frac{{{\rm{500}}}}{{{\rm{2}}{\rm{.5}}}}\,{\rm{s,}}\,\,{\rm{i}}{\rm{.e}}{\rm{.}}\,\,\,\left[ {\displaystyle\frac{{{\rm{1}}\,\,{\rm{round}}}}{{{\rm{Amar's}}\,\,{\rm{speed}}}}{\rm{ = }}\displaystyle\frac{{{\rm{500}}}}{{{\rm{2}}{\rm{.5}}}}} \right]{\rm{ = 200}}\,\,{\rm{s}}$

[Here, ${\rm{9 \times }}\displaystyle\frac{{\rm{5}}}{{{\rm{18}}}}{\rm{ = 2}}{\rm{.5}}\,\,{\rm{m/s}}$ is Amar’s speed.]

21. If the speeds of Dhoni and Sachin were 8 km/hr and 5 km/hr, then after what time will the two meet for the first time at the starting point if they start simultaneously? The length of the circular track is 500 m.

Let us first calculate the time Dhoni and Sachin take to make one full circle.

Time taken by Dhoni = $\displaystyle\frac{{500}}{{\left( {8 \times \displaystyle\frac{5}{{18}}} \right)}} = 225\,s$

Hence, after every 225 s, Dhoni would be at the starting point and after every 360 s, Sachin would be at the starting point. The time, when they will be together again at the starting point simultaneously for the first time, would be the smallest multiple of both 225 and 360, which is the LCM of 225 and 360.

Hence, they would both be together at the starting point for the first time after LCM (225, 360) = 1800 s. Thus, every half an hour, they would meet at the starting point.

From the solution you could realize that it is immaterial whether they move in the same direction or in the opposite.

**Key Concepts**1. Distance = Speed × Time

2. 1 km/hr = $\displaystyle\frac{5}{{18}}$ m/s

3. If the ratios of speed is a : b : c, then the ratios of time taken is : $\displaystyle\frac{1}{a}:\frac{1}{b}:\frac{1}{c}$

**Relation between variables:**To distance, Speed and Time both are directly proportional and To speed, Time is inversely proportional. $S\, \propto \,D$ and $T\, \propto \,D$ i.e. if speed is doubled, distance covered in a given time also gets doubled and ,$S\, \propto \,\displaystyle\frac{1}{T}$ i.e. if speed is doubled, time taken to cover a distance will be half.

Average speed is defined as =.$\displaystyle\frac{{{\text{Total distance travelled}}}}{{{\text{Total time taken}}}}$

**Relative speed**If two bodies are moving (in the same direction or in the opposite direction), then the speed of one body with respect to the other is called its relative speed.

Relative speed is a phenomenon that we observe everyday. Suppose you are travelling in college bus and there is a second bus coming in the opposite direction, then it seems that the second Bus is moving much faster than actual. If both the Buses were moving in the same direction at same speeds, they seem to be stationary if seen from one of these Busses, even though they might actually be at a speed of 100 km/hr each. So what you actually observe is your speed relative to the other.

Concepts

1. If two objects are moving in opposite directions towards each other at speeds u and v, then relative speed = Speed of first + Speed of second = u + v.

This is also the speed at which they are moving towards each other or the speed at which they may be moving away from each other.

2. If the two objects move in the same direction with speeds u and v, then

relative speed = difference of their speeds = u – v.

This is also the speed at which the faster object is either drawing closer to the slower object or moving away from the slower object as the case may be.

3. If the two objects start from A and B with speeds u and v respectively, and after crossing each other take a and b hours to reach B and A respectively, then u : v = $\sqrt {\displaystyle\frac{b}{a}} $

Note: In case of Trains moving in the opposite directions or in the same direction, the total distance required to be travelled before they cross each other completely is equal to the sum of the lengths of the two trains. This distance is covered at the relative speeds of the trains.

**Tips for solving questions on trains:**D = S X T

1. 1 Pole and I Train:

Length of The Train (m) = Speed of the Train (m/s) X Time taken to cross the pole (s)

2. 1 Train and 1 Platform

Length of the Train + Length of the Platform (m) = Speed of the Train (m/s) X Time taken to cross the platform (s)

3. 1 Train with speed speed ${V_1}$ and 1 moving person with speed ${V_2}$

**Case 1:**If both are moving in same directionLength of The Train (m) = [Speed of the Train - Speed of the Man] (m/s) X Time taken to cross the man (s)

**Case 2:**If both are moving in opposite directionLength of The Train (m) = [Speed of the Train + Speed of the Man] (m/s) X Time taken to cross the man (s)

4. 2 Trains with speeds ${V_1},{V_2}$

**Case 1:**If both are moving in same direction[Length of The Train 1 + Length of the Train 2](m) = [Speed of the Train1 - Speed of the Train 2] (m/s) X Time taken to cross (s)

**Case 2:**If both are moving in opposite direction[Length of The Train 1 + Length of the Train 2](m) = [Speed of the Train1 + Speed of the Train 2] (m/s) X Time taken to cross (s)

**Downstream and Upstream Motion:**Downstream motion of a boat is its motion in the same direction as the flow of the Stream.

Upstream motion is exactly the opposite.

There are two parameters in these problems.

1. Speed of the Stream (S): This is the speed with which the river flows.

2. Speed of the boat in still water (B): If the river is still, this is the speed at which the boat would be moving.

The effective speed of a boat in upstream = B – S

The effective speed of a boat in downstream = B + S

3. The speed of the boat in still water is given as B = $\displaystyle\frac{1}{2}(d\; + \;u)$, and the speed of the Stream S = $\displaystyle\frac{1}{2}(d\; + \;u)$,

where d and u are the downstream and upstream speeds, respectively.

**Circular motion**The problems in circular motion deal with races on a circular track to calculate the time of meeting at the starting point and anywhere on the track.

**Concepts****1. If two people A and B start from the same point, at the same time and move in the same direction along a circular track and take x minutes and y minutes respectively to come back to the starting point, then they would meet for the first time at the starting point according to the formula given below:**

First time meeting of A and B at the starting point = (LCM of x and y)

Note: This formula would remain the same even if they move in the opposite directions.

2. If two people A and B start from the same point with speeds m km/hr and n km/hr respectively, at the same time and move in the same direction along a circular track, then the two would meet for the first time by the formula given below:

Time of the first meeting = .$\displaystyle\frac{{{\rm{Circumference}}\,\,{\rm{of}}\,\,{\rm{the}}\,\,{\rm{track}}}}{{{\rm{Relative}}\,\,{\rm{speed}}}}$

**Solved Examples**

1. A scooterist covers a certain distance at 36 km/hr. How much distance does he cover in 3 min?

Speed = 36 km/hr = 36 × $\displaystyle\frac{5}{{18}}$ × m/s = 10 m/s

Thus, the distance covered in 3 min = (10 x 3 x 60) = 1,800 m.

2. Walking at $\displaystyle\frac{3}{4}$ of his usual speed a man is $1\displaystyle\frac{1}{2}$ hr late. Find his usual travel time.

Let usual time be t hours. We know that S X T = D, But now his speed is 3/4 (S). So time to cover the same distance becomes 4/3 (T).

$\displaystyle\frac{4}{3}$ × t = t + $\displaystyle\frac{3}{2}$

t = 4.5 hr.

Alternative method

We know that when his speed got reduced his time went up by 1 1/2 hr.

st = $\displaystyle\frac{{3s}}{4}$(t +$\displaystyle\frac{3}{2}$ )

$ \Rightarrow $ 4st = 3st + 4.5 s

$ \Rightarrow $ t = 4.5 hr.

3. A man starts from L to M, another from M to L at the same time. After passing each other they complete their journey in 3 1/3 hours and 4 4/5 hours respectively. Find the speed of the second man if the speed of the first is 24 km/hr?

Speed of first man : Speed of second man = $\displaystyle\sqrt {\frac{b}{a}} $ where a and b are the time taken by 1st and 2nd man respectively.

Speed of first man : Speed of second man = $\displaystyle\sqrt {\frac{{\frac{{24}}{5}}}{{\frac{{10}}{3}}}} = \sqrt {\frac{{24}}{5} \times \frac{3}{{10}}} = \sqrt {\frac{{36}}{{25}}} = \frac{6}{5}$

Thus second man’s speed = $\displaystyle\frac{5}{6}$ × 24 = 20 km/hr.

4. Two cyclists cover the same distance in 15 km/hr and 16 km/hr, respectively. Find the distance travelled by each, if one takes 16 min longer than the other does.

Let the required distance be x km.

$\displaystyle\frac{{\rm{x}}}{{{\rm{15}}}}{\rm{ - }}\frac{{\rm{x}}}{{{\rm{16}}}}{\rm{ = }}\frac{{{\rm{16}}}}{{{\rm{60}}}}$ or 16x – 15x = 64 or x = 64

Hence, the required distance = 64 km

Alternative Method:

If the speeds are in the ratio 15 : 16 then the ratio of the times to cover the same distance would be in the ratio 16 : 15 or 16x and 15x respectively. But we know that difference in the times is 16 min or 16x - 15x = 16 min so x = 16 min

From the above derivation, second person takes 15 x 16 min = 4 hours to cover the distance at the speed of 16 km/hr. So distance = 4 X 16 = 64 km

5. A thief is spotted by a policeman at a distance of 400 m. If the speed of the thief be 10 km/hr and that of the policeman be 12 km/hr, at what distance will the policeman catch the thief?

Relative speed of the policeman = 2 km/hr. This means, every hour the distance between police man and thief get reduces by 2000 meters. To cover 400 meters, Police needs 400/2000 Hrs or 1/5th hour. But in 1/5th hour police covers 12/5 km = 2.4 km

6. What is the average speed if a person travels from A to B and back at the speeds of 10 km/hr and 20 km/hr, respectively

a. for equal intervals of time,

b. for equal distance,

a. the concept of weighted average can be used here.

The average speed = $\displaystyle\frac{{\left( {10 + 20} \right)}}{2} = 15{\rm{ }}km/hr$

Note:

For equal intervals of time, the average speed is given as $\displaystyle\frac{{{{\rm{S}}_{\rm{1}}}{\rm{ + }}{{\rm{S}}_{\rm{2}}}{\rm{ + }}{{\rm{S}}_{\rm{3}}}{\rm{ + }}\,...{\rm{ + }}\,{{\rm{S}}_{\rm{n}}}}}{{\rm{n}}}$

, where S1, S2, S3, ... , Sn are the speeds and n is the number of observations.

b. The weighted average concept cannot be applied here, because we do not know the fractions of time spent travelling the two distances. It would be a mistake to calculate the average speed as $\displaystyle\frac{{\left( {{\rm{10 + 20}}} \right)}}{{\rm{2}}}{\rm{ = 15 Km/hr}}$

Total time taken = $\left( {\displaystyle\frac{{\rm{D}}}{{{\rm{10}}}}} \right){\rm{ + }}\left( {\displaystyle\frac{{\rm{D}}}{{{\rm{20}}}}} \right){\rm{ = T}}$

Total distance travelled = 2D.

So, average speed = $\displaystyle\frac{{{\rm{2D}}}}{{\rm{T}}}{\rm{ = }}\displaystyle\frac{{{\rm{2D}}}}{{\left( {\displaystyle\frac{{\rm{D}}}{{{\rm{10}}}}{\rm{ + }}\frac{{\rm{D}}}{{{\rm{20}}}}} \right)}}{\rm{ = }}\displaystyle\frac{{{\rm{2D}}}}{{{\rm{D}}\left( {\displaystyle\frac{{\rm{1}}}{{{\rm{10}}}}{\rm{ + }}\displaystyle\frac{{\rm{1}}}{{{\rm{20}}}}} \right)}}{\displaystyle\rm{ = 13}}\displaystyle\frac{{\rm{1}}}{{\rm{3}}}{\rm{ }}k{\rm{m/hr}}$

Note:

The physical interpretation of average speed of km/hr is that if the person had moved from A to B and back at a constant speed of km/hr, he would have taken the same total time as in that case when he travelled with two different speeds of 10 km/hr and 20 km/hr (may be he was travelling with the car or a bullock cart!)

If you travel equal distances with speeds u and v, then the average speed over the entire journey is given as

7. A hare makes 9 leaps in the same time as a dog makes 5. But the dog’s leap is 2m while hare’s is only 1 m. How many leaps will the dog have to make before catching up with the hare if the hare has a head start of 16 m?

Distance covered by dog in 5 leaps = 5 × 2 = 10 m

Distance covered by hare in 9 leaps = 9 × 1 = 9 m

Distance gained by the dog in 5 leaps = 1 m. Hence, for 1 m gain he has to make 5 leaps.

Number of leaps required by the dog to gain 16 m = 5 × 16 = 80 leaps.

8. Two men A and B walk from P to Q, a distance of 21 km at 3 km/h and 4 km/h respectively. B reaches at point Q and returns immediately and he meets A at point R. Find the distance from P to R.

Both of them together have walked twice the distance from P to Q i.e. 42 km.

Ratio of speeds of A and B = 3 : 4.

Therefore, Distance travelled by A = PR = $\displaystyle\frac{3}{7}$ x 42 = 18 km.

9. Buses take 12 hr to cover the distance of 120 km between A and B. A bus starts from point A at 8.00 a.m. and another bus starts from point B at 10.00 a.m. on the same day.

When do the two buses meet?

The distance between A and B is 120 km.

Speed of the buses = $\displaystyle\frac{{{\rm{120}}}}{{{\rm{12}}}}$ = 10 km/hr

By 10.00 a.m., the bus from A would have covered 20 km.

Hence, the distance between the buses at 10.00 a.m. = 120 – 20 = 100 km

Relative speed of the buses = 20 km/hr. Time taken to meet = $\displaystyle\frac{{100}}{{20}}$ = 5 hr after B starts, i.e.the buses will meet at 3 p.m.

9. Buses take 12 hr to cover the distance of 120 km between A and B. A bus starts from point A at 8.00 a.m. and another bus starts from point B at 10.00 a.m. on the same day.

When do the two buses meet?

The distance between A and B is 120 km.

Speed of the buses = $\displaystyle\frac{{{\rm{120}}}}{{{\rm{12}}}}$ = 10 km/hr

By 10.00 a.m., the bus from A would have covered 20 km.

Hence, the distance between the buses at 10.00 a.m. = 120 – 20 = 100 km

Relative speed of the buses = 20 km/hr. Time taken to meet = $\displaystyle\frac{{100}}{{20}}$ = 5 hr after B starts, i.e.the buses will meet at 3 p.m.

10. Train A took 35 minute to cover a distance of 50 km. If the speed of train B is 25% faster than train A, it will cover the same distance in:

Let, speed of train A = 100 km/h

Then, speed of train B = 125 km/h

Ratio of speeds of trains A and B = 100 : 125 = 4 : 5

Therefore, Ratio of time taken by them to cover equal distance = 5 : 4

Given, time taken by train A = 35 minute

Time take by train B = $\displaystyle\frac{4}{5}$ x 35 minute = 28 minute.

Alternative Method:

Speed of train B is faster by 25% = $\displaystyle\frac{1}{4}$

Therefore, Decrease in time taken by train B = $\displaystyle\frac{1}{5}$ x 35 minute = 7 minute.

Therefore, Time taken by train B = 35 - 7 = 28 minute.

11. Two persons start walking in opposite directions at 5 km/h and 4 km/h respectively. After how many hours will they be 45 km apart?

Relative speed of two persons = 5 + 4 = 9 km/h

Therefore, Time taken = $\displaystyle\frac{{45}}{9}$ = 5 hours

12. A man is walking at a speed of 10 km/h. After every km, he takes rest for 4 minutes. How much time will he take to cover a distance of 10 km?

He covers 10 km in 1 hour (i.e. in 60 minutes)

Therefore He will take 6 minutes in covering 1 km.

He rests for 4 minutes after every km.

Time taken = (6 + 4) minutes = 10 minutes for every km.

Therefore, Time taken (for first 9 km) = 9 x 10 = 90 minutes.

Time taken to cover 10th km = 6 minutes

Therefore, Total time taken = 90 + 6 = 96 minute.

Hint: Rest time after 10th km is not added as he has reached his destination.

13. A theft is reported to a policeman. The thief starts running and the policeman chases him. When the policeman starts chasing, the thief was at a distance of 250 metre. The thief and the policeman run at the speed of 8 km/h and 9 km/h respectively. Find the time the policeman will take to catch the thief.

Policeman gains = (9 - 8) km.h = 1 km/h

Therefore, He will gain 250 metre in 15 minutes.

He will catch the thief in 15 minutes.

**Problems on Trains:**14. A train running at 54 km/hr takes 20 s to cross a platform and 12 s to pass a man walking in the same direction at a speed of 6 km/hr. Find the length of the train and the platform.

Let the length of the train = x m. Let the length of the platform = y m.

Speed of the train relative to the man = 54 - 6 = 48 km/hr X $\displaystyle\frac{5}{{18}}$ = $\displaystyle\frac{{40}}{3}$ m/s

In passing the man, the train covers its own length with relative speed.

Length of the train =$\displaystyle\frac{{40}}{3}$ × 12 = 160 m.

Now Length of the Train + Length of the Platform (m) = Speed of the Train (m/s) X Time taken to cross the platform (s)

Speed of train = 54 km/hr X $\displaystyle\frac{5}{{18}}$ = 15 m/s

160 + y = 15 X 20

Length of the platform = 140 m.

15. A man standing on a railway platform notices that a train going in one direction takes 10 seconds to pass him and other train of the same length takes 15 seconds to pass him. Find the time taken by the two trains to cross each other when they are running in the opposite directions.

Time taken = $\displaystyle\frac{{{\rm{2ab}}}}{{{\rm{a + b}}}}{\rm{ = }}\displaystyle\frac{{{\rm{2 \times 10 \times 15}}}}{{{\rm{25}}}}$ = 12 seconds

16. A man standing on a railway platform notices that a train going in one direction takes 9 seconds to pass him and other train of the same length takes 6 seconds to pass him. Find the time taken by the two trains to cross each other when they are running in the same direction.

Time taken = $\displaystyle\frac{{{\rm{2ab}}}}{{{\rm{a - b}}}}{\rm{ = }}\displaystyle\frac{{{\rm{2 \times 9 \times 6}}}}{{\rm{3}}}$ = 36 seconds

17. Two trains start at the same time from two stations and proceed towards each other at the speed of 15 km and 20 km per hour respectively. When the trains meet, it is found that one train has travelled 50 km more than the other. Find the distance between the two stations.

In one hour, faster train travels (20 - 15) = 5 km more than the shower.

In 10 hours, faster train travels 50 km more than the slower.

In one hour, distance covered by the two trains = (15 + 20) = 35 km

Therefore, Total distance covered in 10 hours = 10 x 35 = 350 km.

18. The distance between two stations A and B is 300 km. One train leaves station A towards station B at the average speed of 40 km/h. At the same time, another train left station B towards A at the average speed of 80 km/h. The distance from station A where the two trains meet is:

Ratio of distance covered = Ratio of speeds = 40 : 80 = 1 : 2

Therefore, Distance covered are 100 km and 200 km respectively.

Therefore, Distance from station a = 100 km.

19. A train leaves station A towards B at an average speed of 60 km/h at 9 a.m., after 2 hours another train leaves station A at an average speed of 100 km/h. When the two trains will meet?

Distance covered by the first train in 2 hours = 2 x 60 = 120 km.

Second train gains = 100 - 60 = 40 km/h

Therefore, Second train will gain 120 km in $\displaystyle\frac{{120}}{{40}}$ = 3 hours.

Therefore, They will meet at 9 a.m. + 2 hours + 3 hours = 2 p.m.

**Problems on circular Tracks:**

20. Amar and balu as a warm up exercise are jogging on a circular track. Balu is a better athelete and jogs at 18 km/hr, while Amar jogs at 9 km/hr. The circumference of the track is 500 m. They start from the same point and in the same direction. When will they be together again for the first time?

Since Balu is faster than Amar he will take a lead and as they keep running, the gap between them will also keep widening, unlike on a straight track they would meet again even if Balu is faster than Amar.

The same problem could be rephrased as: In what time would Balu take a lead of 500 m over Amar?

Every second, Balu is taking a lead of m = 2.5 m over Amar.

Hence, he takes to take a lead of 500 m over Amar. Hence, they would meet for the first time after 200 s.

**Alternative method**

For every round that Amar makes, Balu would have made 2 rounds because the ratio of their speeds is 1 : 2. Hence, when Amar has made one full round, Balu would have taken a lead of one round.

Therefore, they would meet after $\displaystyle\frac{{{\rm{500}}}}{{{\rm{2}}{\rm{.5}}}}\,{\rm{s,}}\,\,{\rm{i}}{\rm{.e}}{\rm{.}}\,\,\,\left[ {\displaystyle\frac{{{\rm{1}}\,\,{\rm{round}}}}{{{\rm{Amar's}}\,\,{\rm{speed}}}}{\rm{ = }}\displaystyle\frac{{{\rm{500}}}}{{{\rm{2}}{\rm{.5}}}}} \right]{\rm{ = 200}}\,\,{\rm{s}}$

[Here, ${\rm{9 \times }}\displaystyle\frac{{\rm{5}}}{{{\rm{18}}}}{\rm{ = 2}}{\rm{.5}}\,\,{\rm{m/s}}$ is Amar’s speed.]

21. If the speeds of Dhoni and Sachin were 8 km/hr and 5 km/hr, then after what time will the two meet for the first time at the starting point if they start simultaneously? The length of the circular track is 500 m.

Let us first calculate the time Dhoni and Sachin take to make one full circle.

Time taken by Dhoni = $\displaystyle\frac{{500}}{{\left( {8 \times \displaystyle\frac{5}{{18}}} \right)}} = 225\,s$

Hence, after every 225 s, Dhoni would be at the starting point and after every 360 s, Sachin would be at the starting point. The time, when they will be together again at the starting point simultaneously for the first time, would be the smallest multiple of both 225 and 360, which is the LCM of 225 and 360.

Hence, they would both be together at the starting point for the first time after LCM (225, 360) = 1800 s. Thus, every half an hour, they would meet at the starting point.

From the solution you could realize that it is immaterial whether they move in the same direction or in the opposite.

22. A and B walk around a circular path of 900 metre in
circumference, starting together from the same point in the same direction. If
their speeds are 150 metre per minute and 200 metre per minute respectively,
after how many minutes will they be again at the starting point?

They will be together at the starting point when B (the
faster one) has gained 900 metre (i.e. one round) over A.

In 1 minute B gains (200 - 150) metre = 50 metre over A.

Therefore, B will gain 900 metres in $\displaystyle\frac{{900}}{{50}}$
= 18 minute

23. A man rows 27 km downstream and 18 km upstream taking 3 hr each time. What is the velocity of the current?

Rate downstream = $\displaystyle\frac{{{\rm{27}}}}{{\rm{3}}}$ = 9 km/hr

Rate upstream = $\displaystyle\frac{{{\rm{18}}}}{{\rm{3}}}$ = 6 km/hr

Velocity of current = $\displaystyle\frac{{\rm{1}}}{{\rm{2}}}$ (9 – 6) = 1.5 km/hr.

24. A man can row upstream at 7 km/hr and downstream at 10 km/hr. Find his rate in still water and the rate of the current.

Rate in still water =$\displaystyle\frac{1}{2}(d\; + \;u)$ = 0.5 (10 + 7) = 8.5 km/hr

Rate of current = $\displaystyle\frac{1}{2}(d\; - \;u)$ 0.5 (10 – 7) = 1.5 km/hr

25. A man rows a boat from point A against current for 10 minutes and then come back with the current for next 10 minutes and reaches to a point B. If distance between A and B is 1 km, find the speed of the current.

Going with current and against current for the same time means speed of the boat in still water

is neutralized.

Speed of current only moves the boat.

Boat is moved 1 km by the current in (10 + 10) minutes = 20 minute = $\displaystyle\frac{1}{3}$ hour.

Therefore, Speed of current = 1 x 3 = 3 km/h.

26. A man can row a boat at 6 km/h in still water and speed of the current is 2 km/h. If he

takes 45 minutes to row the boat to a place and back. Find the distance between the two places.

Speed of the boat downstream = 6 + 2 = 8 km/h

Sped of the boat upstream = 6 - 2 = 4 km/h

Therefore, Average speed = $\displaystyle\frac{{2 \times 8 \times 4}}{{8 + 4}} = \displaystyle\frac{{2 \times 8 \times 4}}{{12}}$ km/h

Time taken = 45 minute = $\displaystyle\frac{{45}}{{60}}$ hour = $\displaystyle\frac{3}{4}$ hour

27. A man rows boat to a place covering 72 km distance and back in 15 hours. He finds that he

can row 3 km with the stream in the same time as 2 km against the stream. Find the speed of the stream.

Speed downstream : Speed upstream = 3 : 2

Therefore, Time taken to row downstream : Time taken to row upstream = 2 : 3

But, total time taken = 15 hours.

Therefore, Time taken to row downstream = $\displaystyle\frac{2}{5}$ x 15 hours = 6 hours

And, time taken to row upstream = 15 - 6 = 9 hours

Therefore, Speed downstream = $\displaystyle\frac{{72}}{6}$ = 12 km/h

And, speed upstream = $\displaystyle\frac{{72}}{9}$ = 8 km/h

Therefore, Speed of the stream = $\displaystyle\frac{{12 - 8}}{2}$ = 2 km/h

Alternative Method:

Let, downstream and upstream speeds are 3 km/h and 2 km/h respectively.

Then, distance covered in 15 hours = $\displaystyle\frac{{2 \times 3 \times 2}}{5} \times 15$ = 36 km

But, actual distance covered = 2 x 72 km = 144 km (i.e. 4 times 36 km)

Therefore, Speeds are 4 x 3 = and 4 x 2, i.e. 12 and 18 km/h

Therefore, Speed of the stream = $\displaystyle\frac{{12 - 8}}{2}$ = 2 km/h

**Problems on Boats and Streams:**23. A man rows 27 km downstream and 18 km upstream taking 3 hr each time. What is the velocity of the current?

Rate downstream = $\displaystyle\frac{{{\rm{27}}}}{{\rm{3}}}$ = 9 km/hr

Rate upstream = $\displaystyle\frac{{{\rm{18}}}}{{\rm{3}}}$ = 6 km/hr

Velocity of current = $\displaystyle\frac{{\rm{1}}}{{\rm{2}}}$ (9 – 6) = 1.5 km/hr.

24. A man can row upstream at 7 km/hr and downstream at 10 km/hr. Find his rate in still water and the rate of the current.

Rate in still water =$\displaystyle\frac{1}{2}(d\; + \;u)$ = 0.5 (10 + 7) = 8.5 km/hr

Rate of current = $\displaystyle\frac{1}{2}(d\; - \;u)$ 0.5 (10 – 7) = 1.5 km/hr

25. A man rows a boat from point A against current for 10 minutes and then come back with the current for next 10 minutes and reaches to a point B. If distance between A and B is 1 km, find the speed of the current.

Going with current and against current for the same time means speed of the boat in still water

is neutralized.

Speed of current only moves the boat.

Boat is moved 1 km by the current in (10 + 10) minutes = 20 minute = $\displaystyle\frac{1}{3}$ hour.

Therefore, Speed of current = 1 x 3 = 3 km/h.

26. A man can row a boat at 6 km/h in still water and speed of the current is 2 km/h. If he

takes 45 minutes to row the boat to a place and back. Find the distance between the two places.

Speed of the boat downstream = 6 + 2 = 8 km/h

Sped of the boat upstream = 6 - 2 = 4 km/h

Therefore, Average speed = $\displaystyle\frac{{2 \times 8 \times 4}}{{8 + 4}} = \displaystyle\frac{{2 \times 8 \times 4}}{{12}}$ km/h

Time taken = 45 minute = $\displaystyle\frac{{45}}{{60}}$ hour = $\displaystyle\frac{3}{4}$ hour

27. A man rows boat to a place covering 72 km distance and back in 15 hours. He finds that he

can row 3 km with the stream in the same time as 2 km against the stream. Find the speed of the stream.

Speed downstream : Speed upstream = 3 : 2

Therefore, Time taken to row downstream : Time taken to row upstream = 2 : 3

But, total time taken = 15 hours.

Therefore, Time taken to row downstream = $\displaystyle\frac{2}{5}$ x 15 hours = 6 hours

And, time taken to row upstream = 15 - 6 = 9 hours

Therefore, Speed downstream = $\displaystyle\frac{{72}}{6}$ = 12 km/h

And, speed upstream = $\displaystyle\frac{{72}}{9}$ = 8 km/h

Therefore, Speed of the stream = $\displaystyle\frac{{12 - 8}}{2}$ = 2 km/h

Alternative Method:

Let, downstream and upstream speeds are 3 km/h and 2 km/h respectively.

Then, distance covered in 15 hours = $\displaystyle\frac{{2 \times 3 \times 2}}{5} \times 15$ = 36 km

But, actual distance covered = 2 x 72 km = 144 km (i.e. 4 times 36 km)

Therefore, Speeds are 4 x 3 = and 4 x 2, i.e. 12 and 18 km/h

Therefore, Speed of the stream = $\displaystyle\frac{{12 - 8}}{2}$ = 2 km/h

**Level - 2**

25. A train left station A for station B at a certain speed.
After travelling for 100 km, the train meets with an accident and could travel
at $\displaystyle\frac{4}{5}$th of the original speed and reaches 45 minutes
late at station B. Had the accident taken place 50 km further on, it would have
reached 30 minutes late at station B. What is the distance between station A
and B?

Let, initial speed of the train = 5 km/h

Then, speed after the accident = $\displaystyle\frac{4}{5}$
x 5 = 4 km/h

Time taken to cover 50 km @ 5 km/h = 10 hours

Time taken to cover 50 km @ 4 km/h = 12$\displaystyle\frac{1}{2}$
hours

Difference between times taken = 12$\displaystyle\frac{1}{2}$
- 10 = 2$\displaystyle\frac{1}{2}$ hours = 150 minutes

But, actual difference = (45 - 30) minutes = 15 minutes

= $\displaystyle\frac{1}{{10}}$ of 150 minutes

Therefore, Speed is 10 times of assumed speed.

Therefore, Speed before accident = 10 x 5 = 50 km/h

And, Speed after accident = 10 x 4 = 40 km/h

Distance after accident is covered @ 40 km/h instead of 50
km/h

And, time difference = 45 minutes = $\displaystyle\frac{3}{4}$
hour

Therefore, Distance between place of accident and B = $\displaystyle\frac{{50
\times 40}}{{50 - 40}} \times \displaystyle\frac{3}{4}$ = 150 km

Therefore, Distance between A and B = 100 + 150 = 250 km.

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