Question

Ex. 3)
A boat travels 16 km upstream
and 24 km downstream in 6
hours.
The same boat travels 36 km
upstream and 48 km downstream
in 13 hours.
Find the speed of water current and speed of boat in still water.
ln
n Irm/hr and the speed​

Answer 1

Solution :

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

A boat travels 16 km upstream and 24 km downstream in 6 hours.The same boat travels 36 km upstream and 48 km downstream in 13 hours.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The speed of water current and speed of boat in still water.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

Let the speed of the boat in still water be r km/hrs.

Let the speed of the water current be m km/hrs.

$\bf{\orange{\large{\underline{\underline{\bf{1_{st}\:Case\::}}}}}}$

$\bf{We\:have}\begin{cases}\sf{A\:boat\:speed\:for\:Upstream=\dfrac{16}{r-m} }\\ \sf{A\:boat\:speed\:for\:Downstream=\dfrac{24}{r+m} }\\ \sf{Time(t_{1})=6\:hours}\end{cases}}$

Formula use :

$\sf{\pink{Time=\dfrac{Distance}{Speed} }}$

So;

$\mapsto\bf{\dfrac{16}{r-m} +\dfrac{24}{r+m} =6....................(1)}$

$\bf{\orange{\large{\underline{\underline{\bf{2_{nd}\:Case\::}}}}}}$

$\bf{We\:have}\begin{cases}\sf{A\:boat\:speed\:for\:Upstream=\dfrac{36}{r-m} }\\ \sf{A\:boat\:speed\:for\:Downstream=\dfrac{48}{r+m} }\\ \sf{Time(t_{1})=13\:hours}\end{cases}}$

So;

$\mapsto\bf{\dfrac{36}{r-m} +\dfrac{48}{r+m} =13....................(2)}$

We are replacing method use :

$\bf{\dfrac{1}{r-m} =x\:\:\:\:\:\:\:\& \:\:\:\:\:\dfrac{1}{r+m} =y}$

Then;

$\mapsto\sf{16x+24y=6..............(3)}\\\\\mapsto\sf{36x+48y=13............(4)}$

Using Substitution Method :

From equation (3) we get;

$\mapsto\sf{16x+24y=6}\\\\\mapsto\sf{16x=6-24y}\\\\\mapsto\bf{x=\dfrac{6-24y}{16} ................(5)}$

Putting the value of x in equation (4),we get;

$\mapsto\sf{36\big(\dfrac{6-24y}{16} \big)+48y=13}\\\\\\\mapsto\sf{\dfrac{216-864y}{16} +48y=13}\\\\\\\mapsto\sf{216-864y+768y=208}\\\\\\\mapsto\sf{216-96y=208}\\\\\\\mapsto\sf{-96y=208-216}\\\\\\\mapsto\sf{-96y=-8}\\\\\\\mapsto\sf{y=\cancel{\dfrac{-8}{-96}} }\\\\\\\mapsto\sf{\pink{y=\dfrac{1}{12} }}$

Putting the value of y in equation (5),we get;

$\mapsto\sf{x=\dfrac{6-24(\frac{1}{12}) }{16} }\\\\\\\mapsto\sf{x=\dfrac{6-\frac{24}{12} }{16} }\\\\\\\mapsto\sf{x=\dfrac{\frac{72-24}{12} }{16} }\\\\\\\mapsto\sf{x=\dfrac{\frac{48}{12} }{16} }\\\\\\\mapsto\sf{x=\dfrac{\cancel{48}}{12} \times \dfrac{1}{\cancel{16}} }\\\\\\\mapsto\sf{x=\cancel{\dfrac{3}{12}} }\\\\\\\mapsto\sf{\pink{x=\dfrac{1}{4} }}$

Now;

$\mapsto\sf{\dfrac{1}{r-m} =\dfrac{1}{4} }\\\\\mapsto\sf{r-m=4}\\\\\mapsto\bf{r=4+m.......................(6)}$

&

$\mapsto\sf{\dfrac{1}{r+m} =y}\\\\\mapsto\sf{\dfrac{1}{r+m} =\dfrac{1}{12} }\\\\\mapsto\sf{r+m=12}\\\\\mapsto\sf{4+m+m=12\:\:\:\:\:\:[from(1)]}\\\\\mapsto\sf{4+2m=12}\\\\\mapsto\sf{2m=12-4}\\\\\mapsto\sf{2m=8}\\\\\mapsto\sf{m=\cancel{\dfrac{8}{2} }}\\\\\mapsto\sf{\pink{m=4\:km/hrs}}$

Putting the value of m in equation (6),we get;

$\mapsto\sf{r=4+4}\\\\\mapsto\sf{\pink{r=8\:km/hrs}}$

Thus;

$\underbrace{\sf{The \:speed \:of\: the \:boat\: in\: still\: water=r=8km/hrs}}}}}\\\underbrace{\sf{The \:speed \:of\: the\: water\:current=m=4km/hrs}}}}}$

Answer 2

Answer:

ygh

Step-by-step explanation:

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