# The boat goes \[24\] km upstream and \[28\] km downstream in \[6\] hours. It goes \[30\] km upstream and \[21\] km downstream in $6\dfrac{1}{2}$ hours. Find the speed of the boat in still water and also speed of the stream.

Answer

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Hint: Assume speed of boat in still water $=u$ km/h and speed of stream $=v$ km/h. Then calculate the speed of the boat while moving upstream and downstream. Use $\text{time taken=}\dfrac{\text{distance travelled}}{\text{speed}}$ for the two separate parts of the journey to calculate speed.

Complete step-by-step answer:

The main concept behind this type of question is that when an object moves in the direction of water current (downstream) then the net speed of object is the sum of its speed in still water and that of stream and when it flows in the direction opposite to that of water current (upstream) then the net speed of object is the difference of speed of object and that of stream. This concept is known as relative velocity.

Let speed of boat in still water be ‘\[u\]’ km/h

And, speed of stream be ‘\[v\]’ km/h.

Therefore, speed of boat travelling upstream = \[\left( u-v \right)\]km/h

And, speed of boat travelling downstream = \[\left( u+v \right)\]km/h

We know that; \[\text{distance travelled = speed}\times \text{time taken}\]

Hence, \[\text{time taken = }\dfrac{\text{distance travelled}}{\text{speed}}.\]

$\therefore \text{Total time taken}=\dfrac{\text{distance travelled upstream}}{\text{speed of boat upstream}}\text{+}\dfrac{\text{distance travelled downstream}}{\text{speed of boat downstream}}..................\text{(A)}$

Now, considering the journey of boat \[24\] km upstream and \[28\] km downstream, from equation $(\text{A})$ we have;

$\dfrac{24}{u-v}+\dfrac{28}{u+v}=6................................(1)$

Now, considering the journey of boat \[30\]km upstream and \[21\]km downstream, from equation $(\text{A})$ we have;

$\dfrac{30}{u-v}+\dfrac{21}{u+v}=6\cdot 5...........................(2)$

Substituting $\dfrac{1}{u-v}=x$ and $\dfrac{1}{u+v}=y$ in equations $(1)$ and $(2)$, we get;

$\begin{align}

& 24x+28y=6...................................(3) \\

& 30x+21y=6\cdot 5...............................(4) \\

\end{align}$

Multiplying equation \[\left( 3 \right)\] by \[5\] and equation \[\left( 4 \right)\] by \[4\], we get;

$\begin{align}

& 120x+140y=30..............................(5) \\

& 120x+84y=26................................(6) \\

\end{align}$

Subtracting equation \[\left( 6 \right)\] from\[\left( 5 \right)\], we get;

\[\begin{align}

& 56y=4 \\

& \therefore y=\dfrac{4}{56} \\

& \therefore y=\dfrac{1}{14}............................................(7) \\

\end{align}\]

Putting this value of $y$ in equation\[\left( 5 \right)\], we get;

\[\begin{align}

& 120x+140\times \dfrac{1}{14}=30 \\

& \therefore 120x+10=30 \\

& \therefore 120x=30-10 \\

& \therefore 120x=20 \\

& \therefore x=\dfrac{20}{120} \\

& \therefore x=\dfrac{1}{6}...............................................(8) \\

& \because \dfrac{1}{u-v}=x=\dfrac{1}{6} \\

& \therefore u-v=6..........................................(9) \\

& \because \dfrac{1}{u+v}=y=\dfrac{1}{14} \\

& \therefore u+v=14........................................(10) \\

\end{align}\]

Adding equations \[\left( 9 \right)\] and \[\left( 10 \right)\], we get;

$\begin{align}

& 2u=20 \\

& \therefore u=\dfrac{20}{2}=10. \\

\end{align}$

Subtracting equation \[\left( 9 \right)\] from \[\left( 10 \right)\], we get;

$\begin{align}

& 2v=8 \\

& \therefore v=\dfrac{8}{2}=4. \\

\end{align}$

Hence the speed of the boat is \[u\] km/h $=10$ km/h and the speed of stream is \[v\] km/h$=4$ km/h.

Note: We have to understand the concept of relative velocity, quadratic equation and speed-distance formula. For an object moving downstream the net speed of object is the sum of object in still water and that of stream and when the object moves upstream the net speed of object is the difference of speed of object in still water and that of stream.

Complete step-by-step answer:

The main concept behind this type of question is that when an object moves in the direction of water current (downstream) then the net speed of object is the sum of its speed in still water and that of stream and when it flows in the direction opposite to that of water current (upstream) then the net speed of object is the difference of speed of object and that of stream. This concept is known as relative velocity.

Let speed of boat in still water be ‘\[u\]’ km/h

And, speed of stream be ‘\[v\]’ km/h.

Therefore, speed of boat travelling upstream = \[\left( u-v \right)\]km/h

And, speed of boat travelling downstream = \[\left( u+v \right)\]km/h

We know that; \[\text{distance travelled = speed}\times \text{time taken}\]

Hence, \[\text{time taken = }\dfrac{\text{distance travelled}}{\text{speed}}.\]

$\therefore \text{Total time taken}=\dfrac{\text{distance travelled upstream}}{\text{speed of boat upstream}}\text{+}\dfrac{\text{distance travelled downstream}}{\text{speed of boat downstream}}..................\text{(A)}$

Now, considering the journey of boat \[24\] km upstream and \[28\] km downstream, from equation $(\text{A})$ we have;

$\dfrac{24}{u-v}+\dfrac{28}{u+v}=6................................(1)$

Now, considering the journey of boat \[30\]km upstream and \[21\]km downstream, from equation $(\text{A})$ we have;

$\dfrac{30}{u-v}+\dfrac{21}{u+v}=6\cdot 5...........................(2)$

Substituting $\dfrac{1}{u-v}=x$ and $\dfrac{1}{u+v}=y$ in equations $(1)$ and $(2)$, we get;

$\begin{align}

& 24x+28y=6...................................(3) \\

& 30x+21y=6\cdot 5...............................(4) \\

\end{align}$

Multiplying equation \[\left( 3 \right)\] by \[5\] and equation \[\left( 4 \right)\] by \[4\], we get;

$\begin{align}

& 120x+140y=30..............................(5) \\

& 120x+84y=26................................(6) \\

\end{align}$

Subtracting equation \[\left( 6 \right)\] from\[\left( 5 \right)\], we get;

\[\begin{align}

& 56y=4 \\

& \therefore y=\dfrac{4}{56} \\

& \therefore y=\dfrac{1}{14}............................................(7) \\

\end{align}\]

Putting this value of $y$ in equation\[\left( 5 \right)\], we get;

\[\begin{align}

& 120x+140\times \dfrac{1}{14}=30 \\

& \therefore 120x+10=30 \\

& \therefore 120x=30-10 \\

& \therefore 120x=20 \\

& \therefore x=\dfrac{20}{120} \\

& \therefore x=\dfrac{1}{6}...............................................(8) \\

& \because \dfrac{1}{u-v}=x=\dfrac{1}{6} \\

& \therefore u-v=6..........................................(9) \\

& \because \dfrac{1}{u+v}=y=\dfrac{1}{14} \\

& \therefore u+v=14........................................(10) \\

\end{align}\]

Adding equations \[\left( 9 \right)\] and \[\left( 10 \right)\], we get;

$\begin{align}

& 2u=20 \\

& \therefore u=\dfrac{20}{2}=10. \\

\end{align}$

Subtracting equation \[\left( 9 \right)\] from \[\left( 10 \right)\], we get;

$\begin{align}

& 2v=8 \\

& \therefore v=\dfrac{8}{2}=4. \\

\end{align}$

Hence the speed of the boat is \[u\] km/h $=10$ km/h and the speed of stream is \[v\] km/h$=4$ km/h.

Note: We have to understand the concept of relative velocity, quadratic equation and speed-distance formula. For an object moving downstream the net speed of object is the sum of object in still water and that of stream and when the object moves upstream the net speed of object is the difference of speed of object in still water and that of stream.

Last updated date: 06th Jun 2023

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