Question

[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.

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Answer 1

Answer :-

5 km/h.

To Find :-

Find the speed of the stream.

SolutioN :-

Let the speed of the stream be x km/h.

• so, speed of boat in downstream is ( 11 + x )
• and also speed of boat in upstream is ( 11 - x )
• Distance cover by boat is 12 km.
• The speed of boat in still water is 11 km/h

$\boxed{\tt \bullet \: \: \: Distance = Speed \times Time.}$

$\tt \rightarrow \dfrac{12}{11 - x} + \dfrac{12}{11 + x} = 2 \times \dfrac{3}{4}$

$\tt \rightarrow \dfrac{12}{11 - x} + \dfrac{12}{11 + x} = \dfrac{11}{4}$

$\boxed{\tt \bullet \: \: \: {a}^{2} + {b}^{2} = (a + b)(a - b)}$

$\tt \rightarrow \dfrac{12(11 + x) + 12(11 - x)}{ {11}^{2} - {x}^{2} } = \dfrac{11}{4}$

$\tt \rightarrow \dfrac{132 + \cancel{12x} + 132\cancel{- 12x}}{ {11}^{2} - {x}^{2} } = \dfrac{11}{4}$

$\tt \rightarrow \dfrac{ \cancel{264}}{ {11}^{2} - {x}^{2} } = \dfrac{ \cancel{11}}{4}$

$\tt \rightarrow \dfrac{24}{ {11}^{2} - {x}^{2} } = \dfrac{ 1}{4}$

$\tt \rightarrow {11}^{2} - {x}^{2} = 24\times 4.$

$\tt \rightarrow 121- {x}^{2} = 96.$

$\tt \rightarrow 25- {x}^{2} = 0.$

$\tt \rightarrow x = \sqrt{25} .$

$\tt \rightarrow x = \pm5 .$

Here, x assumes is speed of the stream and speed always be positive.

so,

$\tt \rightarrow x \neq - 5 .$

Therefore, the speed of the stream is 5 km/h.

Answer 2

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

$\sf \dfrac{11}{4}=\dfrac{12}{11-s}+\dfrac{12}{11+s}$

$\sf \dfrac{11}{4}=\dfrac{12(11+s+11-s)}{(11-s)(11+s)}$

$\sf\dfrac{11}{4}=\dfrac{12(11+11)}{(11)^2-s^2}$

$\sf\dfrac{11}{4}=\dfrac{12\times 22}{121 -s^2}$

$\sf\dfrac{11}{4}\times\dfrac{1}{22}=\dfrac{12}{121-s^2}$

$\sf\dfrac{1}{8}=\dfrac{12}{121-s^2}$

$\sf 121-s^2=12(8)$

$\sf 121-s^2=96$

$\sf -s^2=96-121$

$\sf -s^2=-25$

$\sf s^2=25$

$\sf s=\sqrt{25}$

$\sf s = \pm 5$

Either

s = 5

or,

s = -5

As speed of stream can't be negative. s = 5 km/h