Question

A boat goes 21 km upstream and 18 km downstream in 9 hours. In 13 hours, it can go 30
km upstream and 27 km downstream. Determine the speed of the stream and that of the boat in still water. (Reducible)​

Answer 1

Solution :

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

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

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

$\bf{We\:have}\begin{cases}\sf{Upstream\:of\:the\:boat\:=\frac{21}{m-r} }\\ \sf{Downstream\:of\:the\:boat\:=\frac{18}{m+r} }\\ \sf{Time\:(t_1)=9\:hours}\end{cases}$

Using formula :

$\boxed{\bf{Time=\frac{Distance}{Speed} }}}$

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

$\bf{We\:have}\begin{cases}\sf{Upstream\:of\:the\:boat\:=\frac{30}{m-r} }\\ \sf{Downstream\:of\:the\:boat\:=\frac{27}{m+r} }\\ \sf{Time\:(t_2)=13\:hours}\end{cases}$

$\underline{\boldsymbol{According\:to\:the\:question\::}}}$

$\mapsto\sf{\dfrac{21}{m-r} +\dfrac{18}{m+r} =9....................(1)}\\\\\\\mapsto\sf{\dfrac{30}{m-r} +\dfrac{27}{m+r} =13...................(2)}$

$\bigg[\sf{We \:can \:suppose\: \dfrac{1}{m-r} = x \: \:\&\:\: \dfrac{1}{m+r} = y\bigg]}$

$\mapsto\sf{21x + 18y=9........................(3)}\\\\\mapsto\sf{30x + 27y=13.......................(4)}$

$\star\:\underline{\boldsymbol{Using\:Substitution\:Method\::}}}$

From equation (3),we get;

$\longrightarrow\tt{21x+18y=9}\\\\\longrightarrow\tt{21x=9-18y}\\\\\longrightarrow\tt{x=\dfrac{9-18y}{21}...................(5)}$

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

$\longrightarrow\tt{30\bigg(\dfrac{9-18y}{21} \bigg)+27y=13}\\\\\\\longrightarrow\tt{\dfrac{270-540y}{21} +27y=13}\\\\\\\longrightarrow\tt{270-540y+567y=273}\\\\\\\longrightarrow\tt{270+27y=273}\\\\\\\longrightarrow\tt{27y=273-270}\\\\\\\longrightarrow\tt{27y=3}\\\\\\\longrightarrow\tt{y=\cancel{3/27}}\\\\\\\longrightarrow\bf{y=1/9}$

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

$\longrightarrow\tt{x=\dfrac{9-18\bigg(\dfrac{1}{9}\bigg) }{21} }\\\\\\\longrightarrow\tt{x=\dfrac{9-\cancel{\dfrac{18}{9}} }{21}}\\\\\\\longrightarrow\tt{x=\dfrac{9-2}{21} }\\\\\\\longrightarrow\tt{x=\cancel{\dfrac{7}{21} }}\\\\\\\longrightarrow\bf{x=1/3}$

Now;

$\longrightarrow\sf{\dfrac{1}{m-r} =\dfrac{1}{3}}\\\\\longrightarrow\sf{m-r=3}\\\\\longrightarrow\sf{m=3+r...................(6)}$

Putting the value of m other suppose place :

$\longrightarrow\sf{\dfrac{1}{m+r} =\dfrac{1}{9}}\\\\\longrightarrow\sf{m+r=9}\\\\\longrightarrow\sf{3+r+r=9}\\\\\longrightarrow\sf{3+2r=9}\\\\\longrightarrow\sf{2r=9-3}\\\\\longrightarrow\sf{2r=6}\\\\\longrightarrow\sf{r=\cancel{6/2}}\\\\\longrightarrow\bf{r=3\:km/hrs.}$

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

$\longrightarrow\sf{m=3+3}\\\\\longrightarrow\bf{m=6\:km/hrs.}$

Thus;

$\boxed{\sf{The\:speed\:of\:the\:stream\:will\:be\:r=3\:km/hrs}}}\\\boxed{\sf{The\:speed\:of\:the\:boat\:will\:be\:m=6\:km/hrs}}}$