[Maths]
Topic: Quadratic Equations
The speed of a boat in still water is 11 kmph. It can go 12 km upstream and return downstream to the original point in 2 hours 45 minutes. Find the speed of the stream.
Spams/Plagiarism strictly prohibited.
Share
Verified answer
Answer :-
5 km/h.
To Find :-
Find the speed of the stream.
SolutioN :-
Let the speed of the stream be x km/h.
[tex] \boxed{\tt \bullet \: \: \: Distance = Speed \times Time.}[/tex]
[tex] \tt \rightarrow \dfrac{12}{11 - x} + \dfrac{12}{11 + x} = 2 \times \dfrac{3}{4} \\ \\ [/tex]
[tex] \tt \rightarrow \dfrac{12}{11 - x} + \dfrac{12}{11 + x} = \dfrac{11}{4} \\ \\ [/tex]
[tex] \boxed{\tt \bullet \: \: \: {a}^{2} + {b}^{2} = (a + b)(a - b)} \\ \\ [/tex]
[tex] \tt \rightarrow \dfrac{12(11 + x) + 12(11 - x)}{ {11}^{2} - {x}^{2} } = \dfrac{11}{4} \\ \\ [/tex]
[tex] \tt \rightarrow \dfrac{132 + \cancel{12x} + 132\cancel{- 12x}}{ {11}^{2} - {x}^{2} } = \dfrac{11}{4} \\ \\ [/tex]
[tex] \tt \rightarrow \dfrac{ \cancel{264}}{ {11}^{2} - {x}^{2} } = \dfrac{ \cancel{11}}{4} \\ \\ [/tex]
[tex] \tt \rightarrow \dfrac{24}{ {11}^{2} - {x}^{2} } = \dfrac{ 1}{4} \\ \\ [/tex]
[tex] \tt \rightarrow {11}^{2} - {x}^{2} = 24\times 4. \\ \\ [/tex]
[tex] \tt \rightarrow 121- {x}^{2} = 96. \\ \\ [/tex]
[tex] \tt \rightarrow 25- {x}^{2} = 0. \\ \\ [/tex]
[tex] \tt \rightarrow x = \sqrt{25} . \\ \\ [/tex]
[tex] \tt \rightarrow x = \pm5 . \\ \\ [/tex]
Here, x assumes is speed of the stream and speed always be positive.
so,
[tex] \tt \rightarrow x \neq - 5 . \\ \\ [/tex]
Therefore, the speed of the stream is 5 km/h.
Given :-
The speed of a boat in still water is 11 kmph. It can go 12 km upstream and return downstream to the original point in 2 hours 45 minutes
To Find :-
Speed of stream
Solution :-
Let us assume
Speed of stream be s km/h
Speed of boat in downstream be (11 + s)
Speed of boat on upstream be (11 - s)
1 hr = 60 min
2 hr 45 min = 2 3/4 hrs = 11/4 hrs
[tex]\sf \dfrac{11}{4}=\dfrac{12}{11-s}+\dfrac{12}{11+s}[/tex]
[tex]\sf \dfrac{11}{4}=\dfrac{12(11+s+11-s)}{(11-s)(11+s)}[/tex]
[tex]\sf\dfrac{11}{4}=\dfrac{12(11+11)}{(11)^2-s^2}[/tex]
[tex]\sf\dfrac{11}{4}=\dfrac{12\times 22}{121 -s^2}[/tex]
[tex]\sf\dfrac{11}{4}\times\dfrac{1}{22}=\dfrac{12}{121-s^2}[/tex]
[tex]\sf\dfrac{1}{8}=\dfrac{12}{121-s^2}[/tex]
[tex]\sf 121-s^2=12(8)[/tex]
[tex]\sf 121-s^2=96[/tex]
[tex]\sf -s^2=96-121[/tex]
[tex]\sf -s^2=-25[/tex]
[tex]\sf s^2=25[/tex]
[tex]\sf s=\sqrt{25}[/tex]
[tex]\sf s = \pm 5[/tex]
Either
s = 5
or,
s = -5
As speed of stream can't be negative. s = 5 km/h