we are given
A bee flies at 20 feet per second directly to a flower bed from its hive
so, forward speed is
![s_f=20](https://tex.z-dn.net/?f=%20s_f%3D20%20)
flies directly back to the hive at 12 feet per second
so, backward speed is
![s_b=12](https://tex.z-dn.net/?f=%20s_b%3D12%20)
It is away from the hive fore 20 minutes total
so,
total time = forward time + backward time + stay time
The bee stays at the flower bed for 15 minutes,
20 =tf+tb+15
![t_f+t_b=5min](https://tex.z-dn.net/?f=%20t_f%2Bt_b%3D5min%20)
now, we can change it into seconds
![t_f+t_b=5*60sec](https://tex.z-dn.net/?f=%20t_f%2Bt_b%3D5%2A60sec%20)
![t_f+t_b=300sec](https://tex.z-dn.net/?f=%20t_f%2Bt_b%3D300sec%20)
(a)
we know that
distance between hive and flower bed will remain same
Let's assume
forward time =x
so,
![t_f=x](https://tex.z-dn.net/?f=%20t_f%3Dx%20)
![t_b=300-x](https://tex.z-dn.net/?f=%20t_b%3D300-x%20)
we know that
distance = speed*time
![d=s_f*t_f](https://tex.z-dn.net/?f=%20d%3Ds_f%2At_f%20)
now, we can plug values
![d=20*x](https://tex.z-dn.net/?f=%20d%3D20%2Ax%20)
![d=s_b*t_b](https://tex.z-dn.net/?f=%20d%3Ds_b%2At_b%20)
now, we can plug values
![d=12*(300-x)](https://tex.z-dn.net/?f=%20d%3D12%2A%28300-x%29%20)
both distances are same
![20x=12*(300-x)](https://tex.z-dn.net/?f=%2020x%3D12%2A%28300-x%29%20)
(b)
now, we can solve above equations and find x
![20x=12*(300-x)](https://tex.z-dn.net/?f=%2020x%3D12%2A%28300-x%29%20)
![20x=3600-12x](https://tex.z-dn.net/?f=%2020x%3D3600-12x%20)
add both sides 12x
![20x+12x=3600-12x+12x](https://tex.z-dn.net/?f=%2020x%2B12x%3D3600-12x%2B12x%20)
![32x=3600](https://tex.z-dn.net/?f=%2032x%3D3600%20)
![x=112.5](https://tex.z-dn.net/?f=%20x%3D112.5%20)
now, we can find distance
![d=20*112.5](https://tex.z-dn.net/?f=%20d%3D20%2A112.5%20)
![d=2250](https://tex.z-dn.net/?f=%20d%3D2250%20)
so, distance between hive and flower bed is 2250 feet.........Answer