Question

A boat covers 24 km upstream and 36 km downstream in 6 hours while it covers 36 km upstream and 24 km downstream in $6\dfrac{1}{2}$ hours. What is the speed of the current?
(a) 1 km/hr
(b) 1.5 km/hr
(c) 2 km/hr
(d) 2.5 km/hr

Answer 1

Hint: Focus on the point that while a boat covers downstream, the speed of the current is added to the boat while in case of the upstream, speed of the current is subtracted from that of the boat.

Complete step-by-step answer:
Let’s start with what is speed. Speed is a scalar quantity defined as the distance travelled by a particle or object per unit time.
Generally, we deal with two kinds of speeds. One is instantaneous, and the other is the average speed. For uniform motion, both are identical.
Average speed is defined as the total distance covered by a body divided by the time taken by the body to cover it.
$\therefore {{v}_{avg}}=\dfrac{\text{distance covered}}{\text{time taken}}$
$\Rightarrow \text{time taken}=\dfrac{\text{distance covered}}{{{v}_{avg}}}$

Now, starting with the solution to the above question. Let the original average speed of the boat be x km/hr, and the speed of the current be y km/hr.
It is given in the question that when a boat covers 24 km upstream and 36 km downstream it takes a total of 6 hours to complete the journey. We know $\text{time taken}=\dfrac{\text{distance covered}}{{{v}_{avg}}}$ , so, we get
$\text{time taken}=\dfrac{\text{distance covered downstream }}{{{v}_{avg}}+{{v}_{current}}}+\dfrac{\text{distance covered upstream }}{{{v}_{avg}}-{{v}_{current}}}$
$\Rightarrow 6=\dfrac{\text{36 }}{x+y}+\dfrac{\text{24 }}{x-y}$
$\Rightarrow 6=\dfrac{\text{36}\left( x-y \right)\text{+24}\left( x+y \right)\text{ }}{{{x}^{2}}+{{y}^{2}}}$
$\Rightarrow 6\left( {{x}^{2}}-{{y}^{2}} \right)=60x-12y..........(i)$
It is also given in the question that when a boat covers 36 km upstream and 24 km downstream it takes a total of $6\dfrac{1}{2}$ hours to complete the journey.
$\text{time taken}=\dfrac{\text{distance covered downstream }}{{{v}_{avg}}+{{v}_{current}}}+\dfrac{\text{distance covered upstream }}{{{v}_{avg}}-{{v}_{current}}}$
$\Rightarrow 6\dfrac{1}{2}=\dfrac{24}{x+y}+\dfrac{\text{36 }}{x-y}$
$\Rightarrow \dfrac{13}{2}=\dfrac{\text{36}\left( x+y \right)\text{+24}\left( x-y \right)\text{ }}{{{x}^{2}}+{{y}^{2}}}$
$\Rightarrow \dfrac{13}{2}\left( {{x}^{2}}-{{y}^{2}} \right)=60x+12y..........(ii)$
Now if we divide equation (ii) by equation (i), we get
$\dfrac{13}{12}=\dfrac{60x+12y}{60x-12y}$
$\Rightarrow \dfrac{13}{12}=\dfrac{5x+y}{5x-y}$
On cross-multiplication, we get
$65x-13y=60x+12y$
$\Rightarrow 5x=25y$
$\Rightarrow x=5y$
Now we will substitute the value of x in equation (i). So, we get
$6\left( {{\left( 5y \right)}^{2}}-{{y}^{2}} \right)=60\times 5y-12y$
\[\Rightarrow 144{{y}^{2}}=288y\]
And we know the current speed cannot be zero. So, we have
\[\Rightarrow y=\dfrac{288}{144}=2\text{ }Km/hr\]
Therefore, the speed of the current is 2 km/hr, and the answer to the above question is option (c).

Note: Always try to keep the quantities according to a standardised unit system, this helps us to solve the question in an error-free manner. Also, it is prescribed to write each and every statement given in the question in mathematical form as it ensures that we are not missing any information given in the question.