Answer
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Hint: Let the speed of stream and boat in still water to be ‘u’ and ‘v’ respectively. Get the downstream speed $=\left( u+v \right)km/hr$ and upstream$=\left( v-u \right)km/hr$. Then use (time taken by boat to 32 km upstream) + (time taken by boat to go 36 km downstream) = 7
Complete step-by-step answer:
Likewise for the next case, put$\dfrac{1}{\left( v+u \right)}=A\ and\ \dfrac{1}{\left( v-u \right)}=B$ in further steps.
We are given that a boat covers 32 km upstream and 36 km downstream in 7 hours. It also covers 40 km upstream and 48 km downstream in 9 hours. We have to find the speed of the boat in still water and that of the stream.
Let us consider the speed of a boat in still water to be $'v'km/hr$. Also, let us consider the speed of the stream to be $'u'km/hr$.
We know that whenever a boat goes downstream, then the direction of stream and direction of boat is in the same direction. Hence the stream supports the boat while going downstream. So, we get,
Downstream speed of boat = speed of boat in still water + speed of stream…………………(i)
We also know that whenever a boat goes upstream then the direction of stream and direction of boat is in the opposite direction.
Hence, the stream opposes the boat while going upstream. So, we get,
Upstream speed of boat = speed of boat in still water – speed of stream …………………..(ii)
As we have assumed the speed of boat in still water as $'v'km/hr$and speed of stream$'u'km/hr$, by putting these in equation (i) and (ii), we get,
Downstream speed of boat $=\left( v+u \right)km/hr$
and upstream speed of boat$=\left( v-u \right)km/hr$
Now let us consider the time taken by boat to go ${{D}_{1}}$ km downstream as td.
Since we know that $time=\dfrac{\text{distance}}{\text{speed}}$, therefore by putting the values of time = td, distance = ${{D}_{1}}$ km and downstream speed $=\left( v+u \right)km/hr$, we get,
$td=\dfrac{{{D}_{1}}km}{\left( v+u \right)km/hr}$
In case 1, as we are given that boat goes 36 km downstream we get,
$t{{d}_{1}}=\dfrac{36km}{\left( v+u \right)km/hr}...............(iii)$
In case 2, as we given that boat goes 48 km downstream,
$t{{d}_{2}}=\dfrac{48km}{\left( v+u \right)km/hr}...............(iv)$
Now, let us consider time taken by boat to go ${{D}_{2}}$km upstream as tu.
Since we know that time = $\dfrac{\text{Distance}}{\text{Speed}}$ , therefore by putting the value of time = tu, Distance = ${{D}_{2}}$km and upstream speed = (v – u)km/hr, we get
$tu=\dfrac{{{D}_{2}}km}{\left( v-u \right)km/hr}$
In case 1, as we are given that boat goes 32 km upstream we get,
$t{{u}_{1}}=\dfrac{32km}{\left( v-u \right)km/hr}.............\left( v \right)$
In case 2, as we are given that boat goes 40 km upstream, we get,
$t{{u}_{2}}=\dfrac{40km}{\left( v-u \right)km/hr}.............\left( vi \right)$
Now, in case 1, we are given that boat takes total 7 hours to go 32 km upstream and 36 downstream, therefore we get,
$t{{d}_{1}}+t{{u}_{1}}=7\ hours$
By putting the values of $t{{d}_{1}}\And t{{u}_{1}}$ from equation (iii) and equation (v) respectively, we get,
$\Rightarrow \dfrac{36}{\left( v+u \right)}+\dfrac{32}{\left( v-u \right)}=7.........\left( vii \right)$
Now in case 2, we are given that boat takes total 9 hours to go 40 km upstream and 40 km downstream therefore, we get,
$t{{d}_{2}}+t{{u}_{2}}=9\ hours$
By putting the value of $t{{d}_{2}}\And t{{u}_{2}}$from equation (iv) and equation (vi) respectively, we get,
$\dfrac{48}{\left( v+u \right)}+\dfrac{40}{\left( v-u \right)}=9........\left( viii \right)$
Let us take $\dfrac{1}{\left( v+u \right)}=A\ and\ \dfrac{1}{\left( v-u \right)}=B$. By applying this we get equation (vii) & (viii) as,
36A + 32B = 7…………….(a)
48A + 40B = 9…………….(b)
By dividing by 4 in equation (a), we get,
$\begin{align}
& \dfrac{1}{4}\left( 36A+32B \right)=\dfrac{7}{4} \\
& =9A+8B=\dfrac{7}{4}.............\left( {{a}_{1}} \right) \\
\end{align}$
By dividing by 5 in equation (b) we get,
$\begin{align}
& \dfrac{1}{5}\left( 48A+40B \right)=\dfrac{9}{5} \\
& =\dfrac{48A}{5}+8B=\dfrac{9}{5}.............\left( {{b}_{1}} \right) \\
\end{align}$
By subtracting equation $\left( {{a}_{1}} \right)$ from equation $\left( {{b}_{1}} \right)$, we get,
$\begin{align}
& \left( \dfrac{48A}{5}+8B \right)-\left( 9A+8B \right)=\dfrac{9}{5}-\dfrac{7}{4} \\
& =\dfrac{48A}{5}-9A=\dfrac{\left( 4\times 9 \right)-\left( 5\times 7 \right)}{20} \\
& =\dfrac{48A-45A}{5}=\dfrac{36-35}{20} \\
& =\dfrac{3A}{5}=\dfrac{1}{20} \\
\end{align}$
By cross multiplying above equation we get,
$\begin{align}
& A=\dfrac{5}{20\times 3} \\
& \Rightarrow A=\dfrac{1}{4\times 3}=\dfrac{1}{12} \\
\end{align}$
As we know that $A=\dfrac{1}{\left( v+u \right)}$ therefore we get,
$\begin{align}
& \dfrac{1}{\left( v+u \right)}=\dfrac{1}{12} \\
& or\ v+u=12.........\left( {{a}_{2}} \right) \\
\end{align}$
By putting the value of A in equation $\left( {{a}_{1}} \right)$, we get,
$\begin{align}
& 9\left( \dfrac{1}{12} \right)+8B=\dfrac{7}{4} \\
& =\dfrac{3}{4}+8B=\dfrac{7}{4} \\
& or\ 8B=\dfrac{7}{4}-\dfrac{3}{4} \\
& 8B=\dfrac{4}{4}=1 \\
& B=\dfrac{1}{8} \\
\end{align}$
As we know that $B=\dfrac{1}{\left( v-u \right)}$, therefore we get,
$\begin{align}
& \dfrac{1}{\left( v-u \right)}=\dfrac{1}{8} \\
& or\ \left( v-u \right)=8...........\left( {{b}_{2}} \right) \\
\end{align}$
By adding equation $\left( {{a}_{2}} \right)$ & $\left( {{b}_{2}} \right)$, we get,
$\begin{align}
& \left( v+u \right)+\left( v-u \right)=12+8 \\
& =2v=20 \\
& v=\dfrac{20}{2}=10km/hr \\
\end{align}$
By putting the value of v in equation $\left( {{a}_{2}} \right)$, we get,
$\begin{align}
& 10+u=12 \\
& \Rightarrow u=12-10 \\
& \Rightarrow u=2km/hr \\
\end{align}$
Therefore we get speed of boat in still water (v) = 10km/hr and speed of stream (u) = 2km/hr.
Note: Students must remember this approach of taking $\dfrac{1}{v+4}=A\And \dfrac{1}{v-4}=B$ because if they won’t consider this, the question becomes very confusing and even unsolvable due to term $\left( {{v}^{2}}-{{u}^{2}} \right)$ . Also students should take care that upstream speed is (v – u)km/hr and not (u – v)km/hr where v is speed of boat in still water and u is speed of stream. Students can cross check their answer by putting the values of v and u in various equations and see if it is satisfying them or not.
Complete step-by-step answer:
Likewise for the next case, put$\dfrac{1}{\left( v+u \right)}=A\ and\ \dfrac{1}{\left( v-u \right)}=B$ in further steps.
We are given that a boat covers 32 km upstream and 36 km downstream in 7 hours. It also covers 40 km upstream and 48 km downstream in 9 hours. We have to find the speed of the boat in still water and that of the stream.
Let us consider the speed of a boat in still water to be $'v'km/hr$. Also, let us consider the speed of the stream to be $'u'km/hr$.
We know that whenever a boat goes downstream, then the direction of stream and direction of boat is in the same direction. Hence the stream supports the boat while going downstream. So, we get,
Downstream speed of boat = speed of boat in still water + speed of stream…………………(i)
We also know that whenever a boat goes upstream then the direction of stream and direction of boat is in the opposite direction.
Hence, the stream opposes the boat while going upstream. So, we get,
Upstream speed of boat = speed of boat in still water – speed of stream …………………..(ii)
As we have assumed the speed of boat in still water as $'v'km/hr$and speed of stream$'u'km/hr$, by putting these in equation (i) and (ii), we get,
Downstream speed of boat $=\left( v+u \right)km/hr$
and upstream speed of boat$=\left( v-u \right)km/hr$
Now let us consider the time taken by boat to go ${{D}_{1}}$ km downstream as td.
Since we know that $time=\dfrac{\text{distance}}{\text{speed}}$, therefore by putting the values of time = td, distance = ${{D}_{1}}$ km and downstream speed $=\left( v+u \right)km/hr$, we get,
$td=\dfrac{{{D}_{1}}km}{\left( v+u \right)km/hr}$
In case 1, as we are given that boat goes 36 km downstream we get,
$t{{d}_{1}}=\dfrac{36km}{\left( v+u \right)km/hr}...............(iii)$
In case 2, as we given that boat goes 48 km downstream,
$t{{d}_{2}}=\dfrac{48km}{\left( v+u \right)km/hr}...............(iv)$
Now, let us consider time taken by boat to go ${{D}_{2}}$km upstream as tu.
Since we know that time = $\dfrac{\text{Distance}}{\text{Speed}}$ , therefore by putting the value of time = tu, Distance = ${{D}_{2}}$km and upstream speed = (v – u)km/hr, we get
$tu=\dfrac{{{D}_{2}}km}{\left( v-u \right)km/hr}$
In case 1, as we are given that boat goes 32 km upstream we get,
$t{{u}_{1}}=\dfrac{32km}{\left( v-u \right)km/hr}.............\left( v \right)$
In case 2, as we are given that boat goes 40 km upstream, we get,
$t{{u}_{2}}=\dfrac{40km}{\left( v-u \right)km/hr}.............\left( vi \right)$
Now, in case 1, we are given that boat takes total 7 hours to go 32 km upstream and 36 downstream, therefore we get,
$t{{d}_{1}}+t{{u}_{1}}=7\ hours$
By putting the values of $t{{d}_{1}}\And t{{u}_{1}}$ from equation (iii) and equation (v) respectively, we get,
$\Rightarrow \dfrac{36}{\left( v+u \right)}+\dfrac{32}{\left( v-u \right)}=7.........\left( vii \right)$
Now in case 2, we are given that boat takes total 9 hours to go 40 km upstream and 40 km downstream therefore, we get,
$t{{d}_{2}}+t{{u}_{2}}=9\ hours$
By putting the value of $t{{d}_{2}}\And t{{u}_{2}}$from equation (iv) and equation (vi) respectively, we get,
$\dfrac{48}{\left( v+u \right)}+\dfrac{40}{\left( v-u \right)}=9........\left( viii \right)$
Let us take $\dfrac{1}{\left( v+u \right)}=A\ and\ \dfrac{1}{\left( v-u \right)}=B$. By applying this we get equation (vii) & (viii) as,
36A + 32B = 7…………….(a)
48A + 40B = 9…………….(b)
By dividing by 4 in equation (a), we get,
$\begin{align}
& \dfrac{1}{4}\left( 36A+32B \right)=\dfrac{7}{4} \\
& =9A+8B=\dfrac{7}{4}.............\left( {{a}_{1}} \right) \\
\end{align}$
By dividing by 5 in equation (b) we get,
$\begin{align}
& \dfrac{1}{5}\left( 48A+40B \right)=\dfrac{9}{5} \\
& =\dfrac{48A}{5}+8B=\dfrac{9}{5}.............\left( {{b}_{1}} \right) \\
\end{align}$
By subtracting equation $\left( {{a}_{1}} \right)$ from equation $\left( {{b}_{1}} \right)$, we get,
$\begin{align}
& \left( \dfrac{48A}{5}+8B \right)-\left( 9A+8B \right)=\dfrac{9}{5}-\dfrac{7}{4} \\
& =\dfrac{48A}{5}-9A=\dfrac{\left( 4\times 9 \right)-\left( 5\times 7 \right)}{20} \\
& =\dfrac{48A-45A}{5}=\dfrac{36-35}{20} \\
& =\dfrac{3A}{5}=\dfrac{1}{20} \\
\end{align}$
By cross multiplying above equation we get,
$\begin{align}
& A=\dfrac{5}{20\times 3} \\
& \Rightarrow A=\dfrac{1}{4\times 3}=\dfrac{1}{12} \\
\end{align}$
As we know that $A=\dfrac{1}{\left( v+u \right)}$ therefore we get,
$\begin{align}
& \dfrac{1}{\left( v+u \right)}=\dfrac{1}{12} \\
& or\ v+u=12.........\left( {{a}_{2}} \right) \\
\end{align}$
By putting the value of A in equation $\left( {{a}_{1}} \right)$, we get,
$\begin{align}
& 9\left( \dfrac{1}{12} \right)+8B=\dfrac{7}{4} \\
& =\dfrac{3}{4}+8B=\dfrac{7}{4} \\
& or\ 8B=\dfrac{7}{4}-\dfrac{3}{4} \\
& 8B=\dfrac{4}{4}=1 \\
& B=\dfrac{1}{8} \\
\end{align}$
As we know that $B=\dfrac{1}{\left( v-u \right)}$, therefore we get,
$\begin{align}
& \dfrac{1}{\left( v-u \right)}=\dfrac{1}{8} \\
& or\ \left( v-u \right)=8...........\left( {{b}_{2}} \right) \\
\end{align}$
By adding equation $\left( {{a}_{2}} \right)$ & $\left( {{b}_{2}} \right)$, we get,
$\begin{align}
& \left( v+u \right)+\left( v-u \right)=12+8 \\
& =2v=20 \\
& v=\dfrac{20}{2}=10km/hr \\
\end{align}$
By putting the value of v in equation $\left( {{a}_{2}} \right)$, we get,
$\begin{align}
& 10+u=12 \\
& \Rightarrow u=12-10 \\
& \Rightarrow u=2km/hr \\
\end{align}$
Therefore we get speed of boat in still water (v) = 10km/hr and speed of stream (u) = 2km/hr.
Note: Students must remember this approach of taking $\dfrac{1}{v+4}=A\And \dfrac{1}{v-4}=B$ because if they won’t consider this, the question becomes very confusing and even unsolvable due to term $\left( {{v}^{2}}-{{u}^{2}} \right)$ . Also students should take care that upstream speed is (v – u)km/hr and not (u – v)km/hr where v is speed of boat in still water and u is speed of stream. Students can cross check their answer by putting the values of v and u in various equations and see if it is satisfying them or not.
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