
A motorboat covers a given distance in $6hours$ moving downstream on a river. It covers the same distance in \[10hours\] moving upstream. The time it takes to cover the same distance in still water is:
(A) $6.5hours$
(B) $8hours$
(C) $9hours$
(D) $7.5hours$
Answer
137.1k+ views
Hint: We are given a situation where a motorboat covers a distance upstream at a given amount of time and covers a distance downstream at a given amount of time and are asked to find the time it takes to cover the same distance in steady water. Thus, we will form the equation for all the given situations.
Complete Step By Step Solution:
Let$v$ be the speed of the boat in steady water and ${v_s}$ is the speed of stream.
Also,
Let the distance covered be $d$ .
Thus,
For downstream, we get the equation to be
$ \Rightarrow \dfrac{d}{{v + {v_s}}} = 6$ , and we will name it equation $1$
For upstream, we get the equation to be
$ \Rightarrow \dfrac{d}{{v - {v_s}}} = 10$ , and we will name it equation $2$
Further, simplifying the equations $1$ and $2$ , we get
$ \Rightarrow d = 6\left( {v + {v_s}} \right)$
And
$ \Rightarrow d = 10\left( {v - {v_s}} \right)$
Now,
Equating the values, we get
$ \Rightarrow 3\left( {v + {v_s}} \right) = 5\left( {v - {v_s}} \right)$
Further, we get
$ \Rightarrow 3v + 3{v_s} = 5v - 5{v_s}$
Then, we get
$ \Rightarrow v = 4{v_s}$
Now,
Substituting this value in one of the equations, we get
$ \Rightarrow d = 30{v_s}$
Thus,
The time required to cover the same distance in steady water, we get
$ \Rightarrow t = \dfrac{{30{v_s}}}{{4{v_s}}}$
Further, we get
$ \Rightarrow t = 7.5hours$
Hence, the correct option is (D).
Note: So to better understand the question, we should also know what is upstream and what is downstream. So, upstream is when you swim or float against the flow of water. Downstream is defined as when a person swims or floats in the direction of the flow of the water.
Complete Step By Step Solution:
Let$v$ be the speed of the boat in steady water and ${v_s}$ is the speed of stream.
Also,
Let the distance covered be $d$ .
Thus,
For downstream, we get the equation to be
$ \Rightarrow \dfrac{d}{{v + {v_s}}} = 6$ , and we will name it equation $1$
For upstream, we get the equation to be
$ \Rightarrow \dfrac{d}{{v - {v_s}}} = 10$ , and we will name it equation $2$
Further, simplifying the equations $1$ and $2$ , we get
$ \Rightarrow d = 6\left( {v + {v_s}} \right)$
And
$ \Rightarrow d = 10\left( {v - {v_s}} \right)$
Now,
Equating the values, we get
$ \Rightarrow 3\left( {v + {v_s}} \right) = 5\left( {v - {v_s}} \right)$
Further, we get
$ \Rightarrow 3v + 3{v_s} = 5v - 5{v_s}$
Then, we get
$ \Rightarrow v = 4{v_s}$
Now,
Substituting this value in one of the equations, we get
$ \Rightarrow d = 30{v_s}$
Thus,
The time required to cover the same distance in steady water, we get
$ \Rightarrow t = \dfrac{{30{v_s}}}{{4{v_s}}}$
Further, we get
$ \Rightarrow t = 7.5hours$
Hence, the correct option is (D).
Note: So to better understand the question, we should also know what is upstream and what is downstream. So, upstream is when you swim or float against the flow of water. Downstream is defined as when a person swims or floats in the direction of the flow of the water.
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