# The speed of a boat in still water is 15km/hr. It goes 30km upstream and returns downstream to the original point in 4 hours 30 minutes. Find the speed of the stream.

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Hint: Let us understand the concept of upstream and downstream first, if we consider a boat to be moving in the direction of the water flow then we call it downstream and when the boat is moving opposite to the direction of flow of water then it is upstream.

We have been given with the time that is taken by the boat in downstream and to calculate the speed of the stream

Now, we have been given the speed of the boat in still water is 15km/hr.

Let us assume the speed of the stream to be x km/hr.

When the boat is moving upstream the speed will be the speed of boat, minus the speed of stream,

Therefore, Speed of boat upstream = 15-x

Similarly, the speed of boat downstream will be = 15+x

We have the distance and the speed for upstream, so we can find the value of time,

Therefore,

Time taken for upstream, ${T_1} = \dfrac{{30}}{{15 - x}}$

Similarly.

Time taken for downstream, ${T_2} = \dfrac{{30}}{{15 + x}}$

But, in the question we have been provided with the value of ${T_1} + {T_2}$ as 4 hours 30 minutes,

Therefore on converting it into fractions we get,

${T_1} + {T_2} = \dfrac{9}{2}$

We will equate the value of ${T_1}$ and${T_2}$,

$ \Rightarrow \dfrac{{30}}{{15 - x}} + \dfrac{{30}}{{15 + x}} = \dfrac{9}{2}$

$ \Rightarrow 30\left( {\dfrac{1}{{15 - x}} + \dfrac{1}{{15 + x}}} \right) = \dfrac{9}{2}$

$ \Rightarrow 30\left( {\dfrac{{15 + x \div 15 - x}}{{\left( {15 - x} \right)\left( {15 + x} \right)}}} \right) = \dfrac{9}{2}$

$ \Rightarrow \dfrac{{30 \times 30}}{{225 - {x^2}}} = \dfrac{9}{2}$

$ \Rightarrow \dfrac{{100}}{{225 - {x^2}}} = \dfrac{9}{2}$

$ \Rightarrow 225 - {x^2} = 200$

$ \Rightarrow {x^2} = 25$

$ \Rightarrow x = 5$Km/hr (As x cannot be negative)

Speed of the stream is 5km/hr

Note: Whenever a boat is in downstream, it takes less effort and time as the boat is moving in the direction of the water but in case of upstream the boat is moving in the opposite direction so takes extra effort and time.

We have been given with the time that is taken by the boat in downstream and to calculate the speed of the stream

Now, we have been given the speed of the boat in still water is 15km/hr.

Let us assume the speed of the stream to be x km/hr.

When the boat is moving upstream the speed will be the speed of boat, minus the speed of stream,

Therefore, Speed of boat upstream = 15-x

Similarly, the speed of boat downstream will be = 15+x

We have the distance and the speed for upstream, so we can find the value of time,

Therefore,

Time taken for upstream, ${T_1} = \dfrac{{30}}{{15 - x}}$

Similarly.

Time taken for downstream, ${T_2} = \dfrac{{30}}{{15 + x}}$

But, in the question we have been provided with the value of ${T_1} + {T_2}$ as 4 hours 30 minutes,

Therefore on converting it into fractions we get,

${T_1} + {T_2} = \dfrac{9}{2}$

We will equate the value of ${T_1}$ and${T_2}$,

$ \Rightarrow \dfrac{{30}}{{15 - x}} + \dfrac{{30}}{{15 + x}} = \dfrac{9}{2}$

$ \Rightarrow 30\left( {\dfrac{1}{{15 - x}} + \dfrac{1}{{15 + x}}} \right) = \dfrac{9}{2}$

$ \Rightarrow 30\left( {\dfrac{{15 + x \div 15 - x}}{{\left( {15 - x} \right)\left( {15 + x} \right)}}} \right) = \dfrac{9}{2}$

$ \Rightarrow \dfrac{{30 \times 30}}{{225 - {x^2}}} = \dfrac{9}{2}$

$ \Rightarrow \dfrac{{100}}{{225 - {x^2}}} = \dfrac{9}{2}$

$ \Rightarrow 225 - {x^2} = 200$

$ \Rightarrow {x^2} = 25$

$ \Rightarrow x = 5$Km/hr (As x cannot be negative)

Speed of the stream is 5km/hr

Note: Whenever a boat is in downstream, it takes less effort and time as the boat is moving in the direction of the water but in case of upstream the boat is moving in the opposite direction so takes extra effort and time.

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