2 Would be the answer.
Hope that helps!!
Answer:
2 haha
Step-by-step explanation:
The speed of wind is 656 kmph and speed of plane is 70 kmph.
<u>SOLUTION:
</u>
Given that, an air-plane travels 4688 kilometers against the wind in 8 hours
And 5808 kilometers with the wind in the same amount of time.
We have to find the rate of the plane in still air and the rate of wind
Now, let the speed of wind be a kmph and speed of plane be b kmph.
And we know that, 

Substituting (1) in (2) we get,


On substituting (3) in (1) we get,

Hence, the speed of wind is 656 kmph and speed of plane is 70 kmph.
We don't know what WCLM means or have any contacts clues