Question

A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.​

Answer 1

${\pmb{\underline{\sf{ Required \ Solution... }}}}$

Let the Speed of the boat be x and Speed of the stream be y.

$\circ \ {\pmb{\underline{\boxed{\sf{ Time = \dfrac{Distance}{Speed} }}}}}$

Firstly, A boat goes 30 km upstream and 44 km downstream in 10 hours.

• For Upstream = (x - y)
• For Downstream = (x + y)

$\colon\implies{\sf{ \dfrac{30}{(x-y)} + \dfrac{44}{(x+y)} = 10 }}$

Suppose 1/x-y as a and 1/x+y as b

$\colon\implies{\sf{ 30a + 44b = 10 \ \ \ \ \ \cdots(1) }}$

Secondly, Boat can go 40 km upstream and 55 km downstream in 13 hours.

• For Upstream = (x - y)
• For Downstream = (x + y)

$\colon\implies{\sf{ \dfrac{40}{(x-y)} + \dfrac{55}{(x+y)} = 13 }}$

Suppose 1/x-y as a and 1/x+y as b

$\colon\implies{\sf{ 40a + 55b = 13 \ \ \ \ \ \cdots(2) }}$

~ Now, we've to Equalise both Equations by Multiplying Eq.(1) with 4 and Eq.(2) with 3 as that:-

${\sf{ \cancel{120a} +176b = 40 }} \\ {\sf{ - \cancel{120a} -165b = -39 }}$

After Calculating, we've:

$\colon\implies{\sf{11b = 1}} \\ \\ \colon\implies{\sf{ b = \dfrac{1}{11} }}$

So, we can put value of b in any Equation to get the value of a as:-

$\colon\implies{\sf{ 30a+44b = 10 }} \\ \\ \colon\implies{\sf{ 30a + \cancel{44} \times \dfrac{1}{ \cancel{11} } = 10 }} \\ \\ \colon\implies{\sf{ 30a + 4 = 10 }} \\ \\ \colon\implies{\sf{ 30a = 6 }} \\ \\ \colon\implies{\sf{ a = \cancel{ \dfrac{6}{30} } }} \\ \\ \colon\implies{\sf{ a = \dfrac{1}{5} }}$

Finally, We two values that are:-

$\begin{cases} {\sf{ a = \dfrac{1}{5} \ \ \ \ \ \cdots(3)}} \\ \\ {\sf{ b = \dfrac{1}{11} \ \ \ \ \ \cdots(4)}} \end{cases}$

So, Putting values of a and b in Equations (3) and (4) as:-

$\colon\implies{\sf{ a = \dfrac{1}{5} }} \\ \\ \colon\implies{\sf{ \dfrac{1}{x-y} = \dfrac{1}{5} }} \\ \\ \colon\implies{\sf{ x-y = 5 \ \ \ \ \ \cdots(5)}} \\ \\ \colon\implies{\sf{ b = \dfrac{1}{11} }} \\ \\ \colon\implies{\sf{ \dfrac{1}{x+y} = \dfrac{1}{11} }} \\ \\ \colon\implies{\sf{ x+y = 11 \ \ \ \ \ \cdots(6) }}$

~Now, We can compare Equation (5) and (6) to get desired value as:-

${\sf{ x- \cancel{y} = 5}} \\ {\sf{ x+ \cancel{y} = 11}}$

After Calculating, we've

$\colon\mapsto{\sf{ \cancel{2} x = \cancel{16} }} \\ \colon\mapsto{\sf{ x = 8 }}$

So, We can Put value of x in Equation (5) or Equation (6) to get the value of y as:-

$\colon\mapsto{\sf{ x-y = 5 }} \\ \colon\mapsto{\sf{ 8 - y = 5 }} \\ \colon\mapsto{\sf{ 8-5 = y }} \\ \colon\mapsto{\sf{ y = 3 }}$

Hence,

$\\ {\pmb{\underline{\sf\red{ The \ Speed \ of \ the \ Boat \ and \ Stream}}}} \\ {\pmb{\underline{\sf\red{ are \ 8 \ km/hr \ and \ 3 \ km/hr \ Respectively. }}}}$

Answer 2

$\Large{\textsf{\textbf{\underline{\underline{Given\::}}}}}$

• Time taken for 30 km upstream and 44 km downstream is 10 hours.

• Time taken for 40 km upstream and 55 km downstream is 13 hours.

$\Large{\textsf{\textbf{\underline{\underline{To\:do\::}}}}}$

⠀⠀❶ The speed of the stream.

⠀⠀❷ The speed of the boat in still water.

$\Large{\textsf{\textbf{\underline{\underline{Solution\::}}}}}$

Let,

• x be the speed of the stream.

• y be the speed of the boat in still water.

As we know that,

✯ Upstream means the boat moves against the stream of water.

Thus,

$\red\star\:\bf{Upstream \:speed\: =\: speed \:of \:the \:boat \:- \:speed\: of\: the\: stream\:}$

$:\implies\:\bf{Upstream \:speed\: =\: y \:- \:x\:}$

And,

✯ Downstream means a boat moves along the stream of water.

Thus,

$\red\star\:\bf{Downstream ~speed~ =~ speed~ of~ the ~boat ~+ ~speed~ of~ the ~stream~}$

$:\implies\:\bf{Downstream ~speed~ =~ y~+ ~x~}$

As we know that,

$\orange\bigstar\:{\Large\mid}\:\bf\purple{Speed\:=\:\dfrac{Distance}{Time}\:}\:{\Large\mid}\:\green\bigstar$

$:\implies\:\bf{Time\:=\:\dfrac{Distance}{Speed}\:}$

Now,

Time taken for 30 km upstream is,

Here,

• Distance = 30 km

• Speed (due to upstream) = (y - x) km/hr

$\longrightarrow\:\bf{Time\:=\:\left(\dfrac{30}{y\:-\:x}\right)\:hours\:}$

And,

Time taken for 44 km downstream is,

Here,

• Distance = 44 km

• Speed (due to downstream) = (y + x) km/hr

$\longrightarrow\:\bf{Time\:=\:\left(\dfrac{44}{y\:+\:x}\right)\: hours\:}$

According to the question,

☛ Time taken for 30 km upstream and 44 km downstream is 10 hours.

$:\implies\:\tt{\dfrac{30}{y\:-\:x}\:+\:\dfrac{44}{y\:+\:x}\:=\:10\:}---(1)$

Similarly,

☛ Time taken for 40 km upstream and 55 km downstream is 13 hours.

$:\implies\:\tt{\dfrac{40}{y\:-\:x}\:+\:\dfrac{55}{y\:+\:x}\:=\:13\:}---(2)$

Multiplying equation(1) with 4 & equation(2) with 3, we get

$\dashrightarrow\:{\tt{\begin{cases} \dfrac{120}{y\:-\:x}\:+\:\dfrac{176}{y\:+\:x}\:=\:40\:---(a) \\ \\ \dfrac{120}{y\:-\:x}\:+\:\dfrac{165}{y\:+\:x}\:=\:39\:---(b) \end{cases}\:}}$

Substracting equation (b) from equation (a), we get

$\dashrightarrow\:\tt{\dfrac{120}{y\:-\:x}\:+\:\dfrac{176}{y\:+\:x}\:-\:\dfrac{120}{y\:-\:x}\:-\:\dfrac{165}{y\:+\:x}\:=\:40\:-\:39\:}$

$\dashrightarrow\:\tt{\dfrac{176}{y\:+\:x}\:-\:\dfrac{165}{y\:+\:x}\:=\:1\:}$

$\dashrightarrow\:\tt{\dfrac{11}{y\:+\:x}\:=\:1\:}$

$\dashrightarrow\:\tt{y\:+\:x\:=\:11\:}$

Putting the value of '(y + x)' in the equation (2), we get

$\dashrightarrow\:\tt{\dfrac{40}{y\:-\:x}\:+\:\dfrac{55}{11}\:=\:13\:}$

$\dashrightarrow\:\tt{\dfrac{40}{y\:-\:x}\:+\:5\:=\:13\:}$

$\dashrightarrow\:\tt{\dfrac{40}{y\:-\:x}\:=\:13\:-\:5\:}$

$\dashrightarrow\:\tt{\dfrac{40}{y\:-\:x}\:=\:8\:}$

$\dashrightarrow\:\tt{y\:-\:x\:=\:\dfrac{40}{8}\:}$

$\dashrightarrow\:\tt{y\:-\:x\:=\:5\:}$

From the above we get,

$\dashrightarrow\:{\tt{\begin{cases} y\:+\:x\:=\:11\:---(i) \\ \\ y\:-\:x\:=\:5\:---(ii) \end{cases}\:}}$

Adding both equation, we get

$\dashrightarrow\:\tt{y\:+\:x\:+\:y\:-\:x\:=\:11\:+\:5\:}$

$\dashrightarrow\:\tt{2y\:=\:16\:}$

$\dashrightarrow\:\tt{y\:=\:\dfrac{16}{2}\:}$

$\dashrightarrow\:{\textsf{\textbf{\pink{y\:=\:8\:km/hr\:}}}}$

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

$\dashrightarrow\:\tt{8\:+\:x\:=\:11\:}$

$\dashrightarrow\:\tt{x\:=\:11\:-\:8\:}$

$\dashrightarrow\:{\textsf{\textbf{\pink{x\:=\:3\:km/hr\:}}}}$

Therefore,

❶ The speed of the stream is 3 km/hr.

❷ The speed of the boat in still water is 8 km/hr.