BOAT AND STREAMS
Important Formulae:
1.     If the speed of a boat in still water be x km/hr and that of stream be y km/hr
a)     Speed of the boat  downstream = (x + y) km/hr
b)     Speed of the boat upstream = (x – y) km/hr

2.     If the speed of a boat downstream is u km/hr and speed upstream is v km/hr
a)     Speed of the boat in still water = ½ (u + v) km/hr
b)     Speed of the current = ½ (u – v) km/hr

3.     If a man rows in still water at x km/hr and the rate of current (or stream) is y km/hr.
a)     Man’s rate with the current = (x + y) km/hr
b)     Man’s rate against the current = (x –y ) km/hr
Solved Examples
1.     A man can row up streams at 6 km/hr and down streams at 11 km/hr.  Find man’s rate in still water and the rate of the current.
Ans: Rate in still water = ½ (6 +11) = 8.5 km/hr
Rate of current = ½ (11 – 6) = 2.5 km/hr.
2.     A man can row 9 km/hr in still water.  It takes him twice as long to row up as to row down the river.  Find the rate of stream.
Ans:   Let man’s rate in upstream = x km/hr.
Man’s rate downstream = 2x km/hr.
Man’s rate in still water = ½ (x + 2x) km/hr.
ie, 3x/2 = 6 => x = 4
Rate upstream = 4 km/hr.
& Rate downstream = 8 km/hr.
Rate of current = ½ (8- 4) km/hr
= 2 km/hr.
3.     A boatman can row 3 km against the stream in 45 minutes and return in 30 minutes.  Find the rate of his rowing in still water and also the speed of the stream.
Ans: Let the speed of the boatman in still water be x km/hr and the speed of the stream be y km/hr.
Time taken to row against the stream = 45/60 = 3/4 hr.
Time taken to row with the stream = 30/60 = 1/2 hr.
Speed against current = x  - y = Distance / Time
= 3/ 3/4 = 4 km/hr.
Speed with current = x + y = Distance /Time
= 3 / 1/2 = 6 km/hr.
x + y = 6 & x – y =4
Ã° x = 5 km/hr. & y = 1 km/hr.
General Rules for Solving Boats and Streams Problem

a)     When an object is moving in the direction in which the water in the stream is flowing, then the object is said to be moving downstream.
b)     When an object is moving against (opposite) direction in which the water in the stream is flowing, then the object is said to be moving upstream.
c)     When an object is moving in water where there is no motion in water, the object can move in any direction with a uniform speed, then the object is said to be moving in still water.
d)     Let the speed of the boat in still water = x km/h and speed of the stream be y km/h, then
Speed of the boat with stream = downstream = (x + y) km/h
Speed of the boat against stream = upstream = (x – y) km/h
As, when the boat is moving downstream the speed of the water aids the speed of the boat and when the boat is moving upstream, the speed of the water reduces the speed of the boat.
e)     If the speed downstreams is a km/h and the upstream is b km/h, then

Speed in still water = ½ (a+b) km/h
Rate of Stream = ½ (a-b) km/h

General Rules for Solving Circular Tracks

a)     When two people are running around a Circular Track Starting at the same point and at the same time, then whenever the  two people meet the person moving with a greater speed covers one round more than the person moving with lesser speed.
b)     When two people with speeds of x km/h and y km/h start at the same time and from the same point in the same direction around a circular track of circumference c km, then
The time taken to meet for the first time anywhere on the track =   hour
The time taken to meet for the first time at the starting point = LCM of  hour

c)     When two people with speeds of x km/h and y km/h respectively start at the same time and from the same point but in opposite direction around a circular track of circumference c km, then
The time taken to meet for the first time anywhere on the track =  hour.
The time taken to meet for the first time at the starting point = LCM of  hour

Solved Examples
Type 1 Based onTime, Speed & Distance

Example 1  In what time can an athlete cover a distance of 500 m if he runs at a speed of 20 km/h?
Solution. Speed = 20 km/h =  m/s =  m/s.  Time taken to cover 500 m = s = 90s

Example 2 A man covers 40 km of a journey at 8 km/h and the remaining 30 km of the journey in 2 h. His average speed for the whole journey is.

Solution.  Total distance = (40 + 30) km = 70 km
Total time taken =  h = 7h
Average speed =  km/h = 10 km/h

Example 3 A man covers half of his journey at 12 km/h and the remaining half at 36 km/h.  His average speed is

Solution.  Average speed =  km/h =  km/h = 18 km /h

Example 4  If a man covers the first 30 min of his journey at 9 km/h and the next 30 min of his journey at 12 km/h his average speed is

Solution.  Average speed =  km/h =  km/h = 10.5 km/h

Type 2 Based on Trains

Example 5  150m long train is running at 36 km/h.  In how much time, will it cross on electric pole?
Solution. Speed of the train =  m/s = 10 m/s
Time taken to cross an electric pole =  =  = 15 s

Example 6  A 125 m long train is running at 45 km/h.  In how much time will it cross a tunnel 375 m long?

Solution.  Speed of the train =  m/s =  m/s
Time taken to cross the tunnel =  =  = s  = 40s
Example 7 A train 150 m long is running at 55 km/h.  In what time will it pass a man running of 10 km/h

Solution.  Speed of the train relative to the man = (55-10) km/h = 45 km/h =  m/s =  m/s.
Time taken by the train to pass the man  =  =  s = s  = 12s

Type 3 Based on Boats & Stream

Example 8 A man can row upstream at 10 km/h and downstream at 12 km/h. Find the man’s rate in still water and the rate of the current.

Solution. Speed upstream      = 10 km/h
Speed downstream  = 12 km/h
Rate of the current = ½ (12-10) km/h = 1km/h
Rate in still water = ½ (12+10) km/h = 11 km/h

Example 9 A man rows 50 km downstream and 40 km upstream taking 10 h each time.  What is the man’s rate in still water and the rate of the current?

Solution.  Speed downstream =  km / h = 5 km/h
Speed upstream  =   km/h  = 4km/h
Rate of current = ½ (5-4) km/h = ½ km/h
Rate in still water = ½ (5+4) km/h =  km/h

Example 10 In a steam running at 5 km/h a motor boat goes 15 km upstream and back again to the starting point in 2h 15 min.  Find the speed of the motor boat in still water.

Solution.  Let the speed of the motor boat in still water be x km/h.
Speed downstream =  (x+5) km/h
Speed upstream = (x -5) km/h
+  = 2   = 2
=   =
120x = 9x2- 225               9x2- 120x – 225 = 0
9x2- 135x + 15x – 225 =0      9x (n-15) + 15(x-15) = 0
(x-15) (9x + 15) = 0       x – 15=0  x = 15 =  15 km/h

Type 4 Based on Circular Track

Example 11 Two men A and B start together from the same point to walk round a circular path 12 km long.  A walks 2 km and B walks 4 km an hour.  When will they next meet at the starting point if they walk in the same direction?

Solution.   Time to complete one revolution by A and B is  h and  h or 6 h and 3 h.
The required time is the LCM of 6 and 3 which is 6 h.
Thus, they will next meet at the starting point after 6 h.

Example 12 Two friends C and D start together from the same point to walk round a circular path 6 km long.  C walks 2 km and D walks 3 km an hour.  If they walk in the opposite direction when will they
(i) first meet at the starting point?                              (ii) meet for the first time?

Solution.  (i)  Time taken to meet for the first time at the starting point
= LCM of  hour = LCM of  h = LCM of 3 and 2h = 6h
(ii)  Time taken to meet for the first time anywhere
on the track =  h =  h=  min = 72 min

Type 5 Data Sufficiency

Directions (Examples 13 to 15) Each of the questions below consists of a question and two statements. A and B given below it.  You have to decide whether the data provided in the statements are sufficient to answer the question.  Read the questions and both statements and give answer
a)     if statement A is alone sufficient to answer the questions
b)     if statement B is alone sufficient to answer the question
c)     if both the statements are independently sufficient to answer the questions
d)     if both statement are not sufficient to answer the question
e)     if both the statements together are sufficient to answer the question but neither statement
is sufficient alone

Example 13 What is the speed of a bus?
A.    The bus covers a distance of 120 km in 8h
B.    The bus covers a distance of 37.5 km in 2h 30 min

Solution.     c) From statement A, we get speed = km/h = 15km/h
From statement B, we get speed =  =  =  = 15km/h
Both statements A and B alone are sufficient to answer the question.

Example 14  What is the speed of the boat in still water?
A.     The boat takes 8 h to go 40 km upstream
B.     The boat takes, 6 h to go km downstream

Solution.    e)  From statement A, Speed upstream  =  = 5km/h
From statement B, speed downstream =  = 6km /h
From A and B speed of the boat in still water =  (5 + 6) km/h = 5.5 km/h
So, both statements A and B together are required to answer the question.

Example 15  What is the length of the train?
A.    The Speed of the train is 60 km/h
B.    B. It crosses a telegraph post in 6 s

Solution.      e)  From statements A and B, we get
Speed of the train =
Length of the train =  m = 100 m
Hence, required both statements

Example 16 A car covers a distance of 300 km at a constant speed. If the speed of the car is 5 miles per hour more it takes 2 h less to cover the same distance.  The original speed of the car was.

Solution.  Let the constant speed be x km/h.  Then,  –   = 2
-   =  =  =
=   x+ 5x – 750 = 0
x+ 30x –  25x - 750 = 0                      x(x+ 30) –  25(x +30) = 0
(x+ 30)  (x -25) = 0                 x  = 25      [x = -30 is not possible]
Original speed of the car = 25 km/h

Example 17 If a man walks at 5 km/h he misses the train by 7 min.  If the walks at 6 km/h reaches 5 min before the departure of the train what is the distance of the station from the mans house?

Solution.    Let the required distance be x km.  Then,  –  =  [ Difference between two times is 12 min]
= x =   = 6
Required distance = 6 km

Example 18  Walking with 4/5 of his usual speed a man covers a certain distance in 30 min more than the time he takes to cover the distance at his usual  speed.  The time taken by him to cover the distance with usual speed is

Solution.     Let the distance be x km and usual speed be y km/h.  Then,
–   =   =   x    =   = 2
Take taken to cover the distance with usual speed = 2h

Example 19 The speeds of three cars are in the ratio 1 : 2 : 5 The ratio of times taken by these cars to travel the same distance is.

Solution.   Speed is inversely proportional to time
Ratio of times taken = 1 :   :    = 10 : 5 : 2

Example 20  A train leaves from A to B 16.30 h travelling at a speed of 60 km/h and another train leaves on the same day from B to A at the same time with a speed of 100 km/h.  if the distance between A to B is 480 km. how far from A do the two trains meet?

Solution.     Let the two trains meet after 1 hour
Then,                        60 + 100 =   =   =  3h.
Thus, the distance from A = (60 x 3) km = 180 km

Example 21 Two trains start from city A and city B towards each other at 8 am with speeds of 75 kmph and 60 kmph respectively.  If they cross each other at 4 pm what is the distance between the two cities?

Solution.  Let the distance between the two cities be x km
Time taken to meet each other = 8 h
Then,                          (75+60)   = x = (75 + 60) x 8 = 1080 km

Example 22 A and B start simultaneously from a certain point in East and West directions on bicycles.  The speed of A is 30 km/h and that of B is 40 km/h.  What is the distance between A and B after 24 min?

Solution.    Required distance = Sum of the distance covered by A and B
=  km = (12 + 16) km = 28 km