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Question: A motorboat, whose speed in \(15{\text{ km/hr}}\) in still water goes \(30\) km downstream and comes...

A motorboat, whose speed in 15 km/hr15{\text{ km/hr}} in still water goes 3030 km downstream and comes back in a total of 44 hours 3030 minutes. The speed of the stream (in km/hr) is:

Explanation

Solution

Let us suppose the unknown term as any variable “x”. With the help of variables and other known terms, find the unknown value using the formula of speed.

Complete step by step solution:
Given that the speed of motor boat is 15 km/hr15{\text{ }}km/hr and the total distance travelled is 30Km30Km
Let us suppose the speed of the stream be x  km/hrx\;km/hr
Therefore, the speed upstream is =(15 - x) km/hr = \,{\text{(15 - x) km/hr}}
And the speed downstream is =(15+x)km/hr = (15 + x)\,{\text{km/hr}}
Also, given that the total time taken is 4 hours 30 minutes4{\text{ hours 30 minutes}} -

Therefore,
The time taken to row down the stream is =3015+x = \dfrac{{30}}{{15 + x}}
And the time taken to row up the stream is =3015x = \dfrac{{30}}{{15 - x}}
Hence, the total time taken is equal to the four and half hour.
3015+x+3015x=412\dfrac{{30}}{{15 + x}} + \dfrac{{30}}{{15 - x}} = 4\dfrac{1}{2}
Convert the integer on the right hand side of the equation in the form of fraction
3015+x+3015x=92\Rightarrow \dfrac{{30}}{{15 + x}} + \dfrac{{30}}{{15 - x}} = \dfrac{9}{2}
Take LCM on the left hand side of the equation
30(15x)+30(15+x)(15x)(15+x)=92\Rightarrow \dfrac{{30(15 - x) + 30(15 + x)}}{{(15 - x)(15 + x)}} = \dfrac{9}{2}
Apply the property of the difference of two squares in the denominator on the left hand side of the equation –
[a2b2=(a+b)(ab)][\because {a^2} - {b^2} = (a + b)(a - b)]
45030x+450+30x(15)2x2=92\Rightarrow \dfrac{{450 - 30x + 450 + 30x}}{{{{(15)}^2} - {x^2}}} = \dfrac{9}{2}
Like terms with the same coefficient with opposite signs cancels each other.
900225x2=92\Rightarrow \dfrac{{900}}{{225 - {x^2}}} = \dfrac{9}{2}
Take cross-multiplication –
900(2)=9(225x2)\Rightarrow 900(2) = 9(225 - {x^2})
Simplify the above equation
1800=9(225x2)\Rightarrow 1800 = 9(225 - {x^2})
According to the property term multiplicative in the numerator goes in the denominator when changes its side.
18009=225x2\Rightarrow \dfrac{{1800}}{9} = 225 - {x^2}
Simplify –
200=225x2\Rightarrow 200 = 225 - {x^2}
Make the unknown variable as the subject- when the terms changes its side its sign also changes, positive becomes negative and vice-versa
x2=225200 x2=25  \Rightarrow {x^2} = 225 - 200 \\\ \Rightarrow {x^2} = 25 \\\
Taking square root on both the sides of the equation
x2=25\Rightarrow \sqrt {{x^2}} = \sqrt {25}
Square and square root cancels each other.
x=±5\Rightarrow x = \pm 5
Since speed can never be negative.

Therefore, the speed of the stream is 5 km/hr5{\text{ km/hr}}

Note: Always suppose the unknown term taking any reference variable and solve using basic mathematical operations to simplify for the required solution. Double check the sign and mathematical operations when terms are moved from the left hand side of the equation to right or vice-versa.