Ex. 3) A boat travels 16 km upstreamand 24 km downstream in 6hours.The

1.	Ex. 3) A boat travels 16 km upstreamand 24 km downstream in 6hours.The same boat travels 36 kmupstream and 48 km downstreamin 13 hours.Find the speed of water current and speed of boat in still water.lnn Irm/hr and the speed
Answer» 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; $\mapsto\sf{16x+24y=6..............(3)}\\\\\mapsto\sf{36x+48y=13............(<klux>4</klux>)}$ 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; $\underbrace{\sf{The \:speed \:of\: the \:boat\: in\: still\: water=r=8km/hrs}}}}}\\\underbrace{\sf{The \:speed \:of\: the\: water\:current=m=4km/hrs}}}}}$

Ex. 3) A boat travels 16 km upstreamand 24 km downstream in 6hours.The same boat travels 36 kmupstream and 48 km downstreamin 13 hours.Find the speed of water current and speed of boat in still water.lnn Irm/hr and the speed

Answer»

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;

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

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;

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