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Apr 5, 2017

BOATS AND STREAMS QUANTITATIVE APTITUDE STUDY NOTES

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BOATS AND STREAMS QUANTITATIVE APTITUDE STUDY NOTES

FOR BANK AND SSC EXAM
You know that quantitative aptitude section is most important in bank exams in PO and Clerk and for 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 Bank Exams. That’s where maths shortcut tricks and formula are comes into action. So continuously we are providing shortcut tricks on different maths topics.
BOATS AND STREAMS QUANTITATIVE APTITUDE STUDY NOTES
Today’s topic is BOATS AND STREAMS; this is the one of 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 SBI and IBPS exams. For this here we are providing shortcut tricks and quicker method to calculate BOATS AND STREAMS in very short time. We also 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 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.
Here 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 + 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:
(i)                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.
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
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
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;
Solution: method 1:
Let man’s rate upstream = x km/hr
Then, man’s rate downstream = 2x km/hr
Man’s rate in still water = ½ (x + 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

Method 2:
We have,
Up rate + down rate = 2 * rate in 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
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
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(x2 – y2 ) / 2x
Proof: man’s speed up stream = (x – y) km/hr
Man’s speed downstream = (x + y ) km/hr
Let the required distance be ‘A’ km then
BOATS AND STREAMS QUANTITATIVE APTITUDE STUDY NOTES

Or, 2Ax / x2 – y2 = z
The required distance = z(x2 – y2) / 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
Let the required distance be x 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*(62 – (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
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.
And downstream speed = (m + z) km/hr/
Now, we are given that up and 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.

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