Answer:
The speed of the boat in still water is 12 km/hour.
Explanation:
Given:
Boat goes 50 km downstream and 30 km upstream. The speed of the stream 3 km/hour.
Now, to find the speed of the boat in still water:
Let the speed of boat in still water be
.
The speed of the downstream be
![x+3](https://img.qammunity.org/2020/formulas/mathematics/high-school/woj0rrjm61nev51989rvtudftjl44fz43c.png)
And, the speed of the upstream be
![x-3](https://img.qammunity.org/2020/formulas/mathematics/high-school/tjyfmtu7x6sfy09a0ly5sukhuaamljsadw.png)
And, now we find the time by putting the formula:
![Time = (Distance)/(Rate)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/al77a3w16b55h34ivar6tj7191p8kumbho.png)
So, downstream time is:
![downstream\ time = (50)/(x+3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/saiiwxk9vx3ah6up3kd3spsff8y2r8fiop.png)
So, upstream time is:
![upstream\ time = (30)/(x-3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/mtnvo8ftmrz1fgqh4251hfrx38tw8539ag.png)
According to question:
Time upstream = Time downstream
![(30)/(x-3) = (50)/(x+3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cc53ve15sf03x9i9ox03gca3ywgwm8awba.png)
By cross multiplication:
![30* (x+3)= 50* (x-3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ke4fsknod87qac3tyz1hus3p9g8oonz5nl.png)
![30x+90=50x-150](https://img.qammunity.org/2020/formulas/mathematics/middle-school/rr9ryvawcm061ifqujj4glzn5cn0sivrff.png)
By taking variables in one side and taking numbers on the other side we get:
![90+150=50x-30x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/aym22el6ig97fl3v166spmygo80mqtfkmv.png)
![240=20x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/44z9e7d1keojsxhqoessm2e44ypeyimqlt.png)
Dividing both sides by 20 we get :
![12=x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/mqfl0s9by5tq4dt8qlgh2snergn7smvwsi.png)
Therefore, the speed of the boat in still water is 12 km/hour.