You would have to divide because 30/5= 6 and 6/2= 3
The first one is d and the second one is b
Distance between Kalamazoo and Ann Arbor is 152 kilometers (94 miles).
The speed of wind and plane are 105 kmph and 15 kmph respectively.
<u>Solution:</u>
Given, it takes 6 hours for a plane to travel 720 km with a tail wind and 8 hours to make the return trip with a head wind.
We have to find the air speed of the plane and speed of the wind.
Now, let the speed wind be "a" and speed of aeroplane be "b"
And, we know that, distance = speed x time.

Now at head wind → 
So, solve (1) and (2) by addition
2a = 210
a = 105
substitute a value in (1) ⇒ 105 + b = 120
⇒ b = 120 – 105 ⇒ b = 15.
Here, relative speed of plane during tail wind is 120 kmph and during head wind is 90 kmph.
Hence, speed of wind and plane are 105 kmph and 15 kmph respectively.
It is not factorable. Factors of 21 are. 21 and 1
7 and 3. None of those add up to 12