**BOATS AND STREAMS QUANTITATIVE APTITUDE STUDY NOTES**

**FOR BANK AND SSC EXAM**

You know that quantitative

aptitude section is most important in

other competitive exams because if you want good score in bank exam then you

have to score good in maths. In competitive exams the most important thing is

time management, if you know how to manage your time then you can do well in

tricks and formula are comes into action. So continuously we are providing

shortcut tricks on different maths topics.

aptitude section is most important in

**bank**

exams in PO andexams in PO and

**Clerk**and forother competitive exams because if you want good score in bank exam then you

have to score good in maths. In competitive exams the most important thing is

time management, if you know how to manage your time then you can do well in

**Bank Exams.**That’s where maths shortcuttricks and formula are comes into action. So continuously we are providing

shortcut tricks on different maths topics.

Today’s topic is

the most important topics in quantitative aptitude section in Bank and SSC

exam. You should know how to calculate Boats and stream questions and answers

in very short time for bank exam. From this chapter around 1-2 questions are

given in the

this here we are providing shortcut tricks and

also providing the theorems which are elaborate short trick formula’s for

stream boat.

**BOATS AND STREAMS;**this is the one ofthe most important topics in quantitative aptitude section in Bank and SSC

exam. You should know how to calculate Boats and stream questions and answers

in very short time for bank exam. From this chapter around 1-2 questions are

given in the

**SBI and IBPS exams.**Forthis here we are providing shortcut tricks and

**quicker method**to calculate BOATS AND STREAMS in very short time. Wealso providing the theorems which are elaborate short trick formula’s for

stream boat.

**Introduction**

to Boats and Streams: Normally, by speed of the boat or swimmer we mean the speed of

to Boats and Streams

the boat or swimmer in still water. If the boat ( or swimmer ) moves against

the stream then it is called upstream and if it moves with the stream, it is

called downstream.

If the speed of the boat (or the

swimmer) is x and if the speed of the stream is y then, while upstream the

effective speed of the boat = x – y and while downstream the effective speed of

the boat = x + y.

swimmer) is x and if the speed of the stream is y then, while upstream the

effective speed of the boat = x – y and while downstream the effective speed of

the boat = x + y.

**Here**

are the theorems and formula’s with short tricks for boats and streams:

are the theorems and formula’s with short tricks for boats and streams:

**Theorem 1:**

if

x km per hour be the man’s rate in still water, and y km per hour the rate of

the current. Then

x km per hour be the man’s rate in still water, and y km per hour the rate of

the current. Then

**x + y = man’s rate with current**

**x – y = man’s rate against current.**

**Adding and subtracting and then dividing by 2.**

**x = ½ ( man’s rate with current + his rate against current)**

**y = ½ ( man’s rate with current – his rate against current)**

hence, we have the following two

facts:

facts:

(i)

A man’s rate in still water is half the sum of his rates with and

against the current.

A man’s rate in still water is half the sum of his rates with and

against the current.

(ii)

The rate of the current is half the difference between the rates

with and against the current.

The rate of the current is half the difference between the rates

with and against the current.

**Ex. 1:**A man can row upstream at 10

km/hr and downstream at 16 km/hr. Find the man’s rate in still water and the

rate of the current.

**Solution:**Rate in still water = ½ (10 + 6)

= 13 km/hr

Rate of current = ½ (16 – 10) = 3

km/hr

km/hr

**Ex. 2:**A man swims downstream 30 km

and upstream 18 km, taking 3 hrs each time. What is the velocity of current?

**Solution:**Man’s rate downstream = 30/3

km/hr = 10 km/hr

Man’s rate upstream = 18/3 km/hr

= 6 km/hr

= 6 km/hr

Velocity of stream = (10 – 6) / 2

= 2 km/hr

**Ex. 3:**A man can row 6 km/hr in still

water. It takes him twice as long to row up as to row down the river. Find the

rate of the stream.

This

questions we can solve by three methods, we will discuss one by one;

questions we can solve by three methods, we will discuss one by one;

**Solution:**

**method 1:**

Let man’s rate upstream = x km/hr

Then, man’s rate downstream = 2x

km/hr

km/hr

Man’s rate in still water = ½ (x

+ 2x) km/hr

+ 2x) km/hr

3x/2 = 6

Or x = 4 km/hr

Thus, man’s rate upstream = 4 km/hr

Man’s rate downstream = 8 km/hr

Rate of stream = ½ (8 – 4) = 2

km/hr

km/hr

**Method 2:**

We have,

Up rate + down rate = 2 * rate in

still water

still water

= 2 * 6 = 12 km/hr

Also, up rate: down rate = 1:2

So, dividing 12 in in the ration

of 1: 2, we get

of 1: 2, we get

Up rate = 4 km/hr

Down rate = 8 km/hr

Rate of stream = 8 – 4 / 2

= 2 km/hr

**Method 3**

(shortest Method):let the rate of stream = x km/hr

(shortest Method):

Then,

6 + x = 2 (6-x)

Or, 3x = 6

X = 6/3 = 2 km/h

**Theorem 2: for boats and streams:**

A

man can row x km/hr in still water. If in a stream which is following at y

km/hr, it takes him z hrs to row to a place and back, the distance between the

two places is z(x

man can row x km/hr in still water. If in a stream which is following at y

km/hr, it takes him z hrs to row to a place and back, the distance between the

two places is z(x

^{2 }– y^{2 }) / 2x**Proof:**man’s speed up stream = (x – y)

km/hr

Man’s speed downstream = (x + y )

km/hr

km/hr

Let the required distance be ‘A’

km then

km then

**Or, 2Ax / x**

^{2}– y^{2}= z**The required distance = z(x**

^{2 }– y^{2}) / 2x**Ex. 4:**A man can row 6 km/hr in still

water. When the river is running at 1.2 km/hr, it takes him 1 hour to row to a

place and back. How far is the place?

**Solution:**Man’s rate downstream = (6 +

1.2) km/hr = 7.2 km/hr

Man’s rate upstream = (6 – 1.2)

km/hr = 4.8 km/hr

km/hr = 4.8 km/hr

Let the required distance be x

km. Then

km. Then

x/7.2 + x/4.8 = 1

or 4.8x + 7.2x = 7.2 * 4.8

x = 7.2 * 4.8 / 12

= 2.88 km.

**By direct formula:**required distance = 1*(6

^{2 }–

(1.2)

^{2}) / 2*6

36 – 1.44 / 12

3 – 0.12

=2.88 km.

**Theorem 3: for boats and streams:**

A

man rows a certain distance downstream in x hours and returns the same distance

in y hrs. If the stream follows at the rate of z km/hr then the speed of the

man in still water is given by

man rows a certain distance downstream in x hours and returns the same distance

in y hrs. If the stream follows at the rate of z km/hr then the speed of the

man in still water is given by

**Z(x+y) / y-x km/hr.**

**Proof:**let the speed of the man in

still water be ‘m’ km/hr.

Then, his upstream speed = (m – z)

km/hr.

km/hr.

And downstream speed = (m + z)

km/hr/

km/hr/

Now, we are given that up and

down journey are equal, therefore x(m + z) = y(m – z)

down journey are equal, therefore x(m + z) = y(m – z)

Or, m(y-x) = z(x+y)

M = z(x + y) / y – x km/hr

**Ex. 5:**Ramesh can row a certain

distance downstream in 6 hours and return the same distance in 9 hours. If the

stream flows at the rate of 3 km per hour, find the speed of Ramesh in still

water.

**Solution:**

By the above formula: Ramesh’s

speed in still water = 3(9+6) / 9-6 km/hr.

speed in still water = 3(9+6) / 9-6 km/hr.

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