BOAT AND STREAMS
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 (ab) 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 = (5510) 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 (1210) 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 (54) 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 = 9x^{2} 225
⇒ 9x^{2} 120x – 225 = 0
⇒ 9x^{2} 135x + 15x – 225 =0 ⇒
9x (n15) + 15(x15) = 0
(x15) (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^{2
}+ 5x – 750 = 0
⇒ x^{2 }+ 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,
∴
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
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