Question

A boat goes 35 km. upstream and 55 km down-stream in 12 hours. In 10
hours it can go 30 km. upstream and 44 km. down-stream. Determine the
speed of the stream and that of the boat in still water.​

Answer 1

S O L U T I O N :

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

Let the speed of the stream be m km/hrs.

$\underbrace{\bf{1_{st}\:Case\::}}}}$

$\bf{We\:have}\begin{cases}\sf{A\:boat\:speed\:for\:upstream=\frac{35}{r-m} }\\ \sf{A\:boat\:speed\:for\:downstream=\frac{55}{r+m} }\\ \sf{Time\:(t_1)=12\:hours}\end{cases}$

Formula use :

$\tt{Time=\dfrac{Distance}{Speed} }$

So;

$\mapsto\bf{\dfrac{35}{r-m} +\dfrac{55}{r+m} =12...................(1)}$

$\underbrace{\bf{2_{nd}\:Case\::}}}}$

$\bf{We\:have}\begin{cases}\sf{A\:boat\:speed\:for\:upstream=\frac{30}{r-m} }\\ \sf{A\:boat\:speed\:for\:downstream=\frac{44}{r+m} }\\ \sf{Time\:(t_2)=10\:hours}\end{cases}$

So;

$\mapsto\bf{\dfrac{30}{r-m} +\dfrac{44}{r+m} =10...................(2)}$

We are replacing method use :

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

So;

$\mapsto\sf{35x+55y=12.....................(3)}\\\\\mapsto\sf{30x+44y=10.....................(4)}$

$\underbrace{\sf{Using\:Substitution\:method\::}}}}$

From equation (3),we get;

$\longrightarrow\tt{35x+55y=12}\\\\\longrightarrow\tt{35x=12-55y}\\\\\longrightarrow\tt{x=\dfrac{12-55y}{35}.....................(5) }$

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

$\longrightarrow\tt{30\bigg(\dfrac{12-55y}{35} \bigg)+44y=10}\\\\\\\longrightarrow\tt{\dfrac{360-1650y}{35} +44y=10}\\\\\\\longrightarrow\tt{360-1650y+1540y=350}\\\\\\\longrightarrow\tt{360-110y=350}\\\\\\\longrightarrow\tt{-110y=350-360}\\\\\\\longrightarrow\tt{-110y=-10}\\\\\\\longrightarrow\tt{y=\cancel{\dfrac{-10}{-110} }}\\\\\\\longrightarrow\bf{y=1/11}$

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

$\longrightarrow\tt{x=\dfrac{12-55\bigg(\dfrac{1}{11}\bigg) }{35} }\\\\\\\longrightarrow\tt{x=\dfrac{12-\dfrac{55}{11} }{35} }\\\\\\\longrightarrow\tt{x=\dfrac{\dfrac{132-55}{11} }{35} }\\\\\\\longrightarrow\tt{x=\dfrac{\dfrac{77}{11} }{35} }\\\\\\\longrightarrow\tt{x=\cancel{\dfrac{77}{11}} \times \dfrac{1}{35} }\\\\\\\longrightarrow\tt{x=\cancel{7}\times \dfrac{1}{\cancel{35}} }\\\\\\\longrightarrow\bf{x=\dfrac{1}{5}}$

Now;

$\longrightarrow\sf{\dfrac{1}{r-m} =\dfrac{1}{5}} \\\\\longrightarrow\sf{r-m=5}\\\\\longrightarrow\sf{r=5+m....................(6)}$

&

$\longrightarrow\sf{\dfrac{1}{r+m} =\dfrac{1}{11} }\\\\\longrightarrow\sf{r+m=11}\\\\\longrightarrow\sf{5+m+m=11\:\:\:[from(6)]}\\\\\longrightarrow\sf{5+2m=11}\\\\\longrightarrow\sf{2m=11-5}\\\\\longrightarrow\sf{2m=6}\\\\\longrightarrow\sf{m=\cancel{6/2}}\\\\\longrightarrow\bf{m=3\:km/hrs}$

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

$\longrightarrow\sf{r=5+3}\\\\\longrightarrow\bf{r=8\:km/hrs}$

Thus;

$\underbrace{\sf{The\:speed\:of\:the\:Boat\:in\:still\:water\:will\:be\:r=\boxed{\bf{8\:km/hrs}}}}}\\\underbrace{\sf{The\:speed\:of\:the\:Stream\:will\:be\:m=\boxed{\bf{3\:km/hrs}}}}}$