# A motor boat, whose speed is 20 km/hr in still water takes 1 hour more to go 48 km upstream than to return downstream to the same spot. Find the speed of the stream.

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Hint: Assume a variable x which will represent the speed of the stream in km/hr. Since it is given that the speed of the motorboat is 20 km/hr in still water, the speed of the motorboat while going downstream will be equal to (x+20) km/hr and the speed of the motorboat while going upstream will be equal to (20-x) km/hr. Using this information, we can solve this question.

Complete step-by-step answer:

Before proceeding with the question, we must know the formulas that will be required to solve this question.

We have a formula, time = distance/speed $......................\left( 1 \right)$

In the question, it is given that the motorboat takes 1 hour more to go 48 km upstream than to return downstream to the same spot. Also, it is given that the speed of the motor boat in still water is equal to 20 km/hr. We are required to find the speed of the stream.

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

It is given that the speed of the motor boat in still water is equal to 20 km/hr.

While going upstream, the direction of the stream will be against the direction of the motorboat. So, the speed of the motorboat while going upstream will be equal to (20-x) km/hr. It is given that the distance to be travelled is equal to 48 km. So, using formula $\left( 1 \right)$, the time taken by the motorboat to go upstream is equal to $\dfrac{48}{20-x}...........\left( 2 \right)$

While going downstream, the direction of the stream will be along with the direction of the motorboat. So, the speed of the motorboat while going downstream will be equal to (20+x) km/hr. It is given that the distance to be travelled is equal to 48 km. So, using formula $\left( 1 \right)$, the time taken by the motorboat to go downstream is equal to $\dfrac{48}{20+x}...........\left( 3 \right)$

In the question, it is given that the motor boat takes 1 hour more to go 48 km upstream than to return downstream. So, we can say that,

(Time taken while going downstream) + 1 = Time taken while going upstream

Substituting these times from equation $\left( 1 \right)$ and equation $\left( 3 \right)$, we get,

\[\begin{align}

& \dfrac{48}{20+x}+1=\dfrac{48}{20-x} \\

& \Rightarrow \dfrac{48}{20-x}-\dfrac{48}{20+x}=1 \\

& \Rightarrow \left( 48 \right)\left( \dfrac{\left( 20+x \right)-\left( 20-x \right)}{\left( 20-x \right)\left( 20+x \right)} \right)=1 \\

& \Rightarrow \left( 48 \right)\left( 2x \right)=\left( 20-x \right)\left( 20+x \right) \\

& \Rightarrow 96x=400+20x-20x-{{x}^{2}} \\

& \Rightarrow {{x}^{2}}+96x-400=0 \\

& \Rightarrow {{x}^{2}}-4x+100x-400=0 \\

& \Rightarrow x\left( x-4 \right)+100\left( x-4 \right)=0 \\

& \Rightarrow \left( x+100 \right)\left( x-4 \right)=0 \\

& \Rightarrow x=4,x=-100 \\

\end{align}\]

Since speed cannot be negative, hence, the speed of the stream is 4 km/hr.

Note: There is a possibility that one may commit a mistake while writing the equation that is relating the time taken while going upstream and going downstream. There is a possibility that one may write it ha \[\dfrac{48}{20+x}=\dfrac{48}{20-x}+1\] instead of \[\dfrac{48}{20+x}+1=\dfrac{48}{20-x}\]. So, in order to avoid this mistake, one has to read the question carefully.

Complete step-by-step answer:

Before proceeding with the question, we must know the formulas that will be required to solve this question.

We have a formula, time = distance/speed $......................\left( 1 \right)$

In the question, it is given that the motorboat takes 1 hour more to go 48 km upstream than to return downstream to the same spot. Also, it is given that the speed of the motor boat in still water is equal to 20 km/hr. We are required to find the speed of the stream.

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

It is given that the speed of the motor boat in still water is equal to 20 km/hr.

While going upstream, the direction of the stream will be against the direction of the motorboat. So, the speed of the motorboat while going upstream will be equal to (20-x) km/hr. It is given that the distance to be travelled is equal to 48 km. So, using formula $\left( 1 \right)$, the time taken by the motorboat to go upstream is equal to $\dfrac{48}{20-x}...........\left( 2 \right)$

While going downstream, the direction of the stream will be along with the direction of the motorboat. So, the speed of the motorboat while going downstream will be equal to (20+x) km/hr. It is given that the distance to be travelled is equal to 48 km. So, using formula $\left( 1 \right)$, the time taken by the motorboat to go downstream is equal to $\dfrac{48}{20+x}...........\left( 3 \right)$

In the question, it is given that the motor boat takes 1 hour more to go 48 km upstream than to return downstream. So, we can say that,

(Time taken while going downstream) + 1 = Time taken while going upstream

Substituting these times from equation $\left( 1 \right)$ and equation $\left( 3 \right)$, we get,

\[\begin{align}

& \dfrac{48}{20+x}+1=\dfrac{48}{20-x} \\

& \Rightarrow \dfrac{48}{20-x}-\dfrac{48}{20+x}=1 \\

& \Rightarrow \left( 48 \right)\left( \dfrac{\left( 20+x \right)-\left( 20-x \right)}{\left( 20-x \right)\left( 20+x \right)} \right)=1 \\

& \Rightarrow \left( 48 \right)\left( 2x \right)=\left( 20-x \right)\left( 20+x \right) \\

& \Rightarrow 96x=400+20x-20x-{{x}^{2}} \\

& \Rightarrow {{x}^{2}}+96x-400=0 \\

& \Rightarrow {{x}^{2}}-4x+100x-400=0 \\

& \Rightarrow x\left( x-4 \right)+100\left( x-4 \right)=0 \\

& \Rightarrow \left( x+100 \right)\left( x-4 \right)=0 \\

& \Rightarrow x=4,x=-100 \\

\end{align}\]

Since speed cannot be negative, hence, the speed of the stream is 4 km/hr.

Note: There is a possibility that one may commit a mistake while writing the equation that is relating the time taken while going upstream and going downstream. There is a possibility that one may write it ha \[\dfrac{48}{20+x}=\dfrac{48}{20-x}+1\] instead of \[\dfrac{48}{20+x}+1=\dfrac{48}{20-x}\]. So, in order to avoid this mistake, one has to read the question carefully.

Last updated date: 27th Sep 2023

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