Answer:
Rate of boat in still water = 70 km/hr
Rate of current = 15 km/hr
Explanation:
Let the speed of motorboat in still water =
![u\ km/hr](https://img.qammunity.org/2021/formulas/mathematics/high-school/o27e6oqkmx8x6wpglxdqp3zp4qruwlvvsb.png)
Let the speed of current =
![v\ km/hr](https://img.qammunity.org/2021/formulas/mathematics/high-school/sp3841vvw3jhilzoam0bdsugsdo5f68yt2.png)
Equivalent Speed upstream =
![(u-v)\ km/hr](https://img.qammunity.org/2021/formulas/mathematics/high-school/jw1jj302w2db8rlix1yva2vq8wrluhy5ki.png)
Equivalent Speed downstream =
![(u+v)\ km/hr](https://img.qammunity.org/2021/formulas/mathematics/high-school/vywoq2hqt2pjbd678dych7x4bzak3qhuvs.png)
Distance traveled upstream = 440 km
Time taken upstream = 8 hours
Using the formula:
![Speed = (Distance)/(Time)](https://img.qammunity.org/2021/formulas/mathematics/high-school/jdl1y54prqag7xwcb644jzaygmxghn0sdk.png)
![u-v = (440)/(8)\\\Rightarrow u -v=55 ..... (1)](https://img.qammunity.org/2021/formulas/mathematics/high-school/fnxrvvri9kypmm6st2cwwnx5a6kejhxy7i.png)
Distance traveled downstream = 510 km
Time taken downstream = 6 hours
![u +v = (510)/(6)\\\Rightarrow u+v=85 ...... (2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/rtugidjbcxodi42re720hz3yewdojl1wth.png)
Adding (1) and (2):
![2u = 140\\\Rightarrow u = 70\ km/hr](https://img.qammunity.org/2021/formulas/mathematics/high-school/22x0jvx05mxet8f9mhh73almj1bb1uu8fo.png)
By equation (1):
![v = 85-70\\\Rightarrow v =15\ km/hr](https://img.qammunity.org/2021/formulas/mathematics/high-school/pkbo3gaxm28hdjbrnd7bf61acjq2fz1q7s.png)
Therefore, the answer is:
Rate of boat in still water = 70 km/hr
Rate of current = 15 km/hr