Question

A motorboat travels 120 miles in 3 hours going upstream. It travels 150 miles going downstream in the same amount of time. What is the rate of the boat in still water and the rate of the current

Answer 1

Rate of boat in still water is 45 m/h

Rate of current is 5 m/h

Step-by-step explanation

Step 1

let x represents the spped of the boat

let y represents the rate of the current

hence

A) going upstream

the current is opposite to the movement, so we need to subtract the rate of the current form the rate of the motor boat ,so

`\begin{gathered} sped=(x-y) \\ \text{time}\cdot\text{speed}=dis\tan ce \end{gathered}`

replace

`3\mleft(x-y\mright)=120\rightarrow equation(1)`

B) going downstream

we need to add the rates , so the rate is (x+y)

`\begin{gathered} \text{time}\cdot\text{speed}=dis\tan ce \\ 3(x+y)=150\rightarrow equation(2) \end{gathered}`

Step 2

solve the equations

`\begin{gathered} 3\mleft(x-y\mright)=120 \\ 3\mleft(x+y\mright)=150 \end{gathered}`

divide both side of equation(1) by 3

`\begin{gathered} (3(x-y))/(3)=(120)/(3) \\ x-y=40 \\ \text{add y in both sides} \\ x-y+y=40+y \\ x=40+y\rightarrow equation(3) \end{gathered}`

now, divide both sides of the equation (2) by 3 and replace the x value we got previously

`\begin{gathered} 3(x+y)=150 \\ (3(x+y))/(3)=(150)/(3) \\ x+y=50 \\ \text{replace the x value} \\ 40+y+y=50 \\ 2y=50-40 \\ y=(10)/(2) \\ y=5 \end{gathered}`

finally, replace the y value into equation (3) to know x

`\begin{gathered} x=40+y\rightarrow equation(3) \\ x=40+5 \\ x=45 \end{gathered}`

therefore.

Rate of boat in still water is 45 m/h

Rate of current is 5 m/h

I hope this helps you