SOLUTION: An airplane takes 1 hour longer to go a distance of 375 miles flying against the headwind then on the return trip with a tail wind. if the speed of the wind is 50 miles per hour fi

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Question 1053493: An airplane takes 1 hour longer to go a distance of 375 miles flying against the headwind then on the return trip with a tail wind. if the speed of the wind is 50 miles per hour find the speed of the plane in still air.
I know the answer is 200mph because my test gave me the right answer after getting it wrong I just need help on how to get that answer and solve it right for next time.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39613) (Show Source):
You can put this solution on YOUR website!
This is a certain kind of constant travel rates problem that is asked OFTEN as an exercise. OFTEN!!! You may want to solve this generally, and that is what I will ~~try~~ start to do here.

SPEED TIME DISTANCE AGAINSTWIND r-50 t+1 375 WITHWIND r+50 t 375

Some two variables are implied in the data table there.

Instead of using those other given values, the tabulated data can be done in variables only, and again you can figure what is being assigned:

SPEED TIME DISTANCE AGAINSTWIND r-w t+h d WITHWIND r+w t d

The UNKNOWN variables are r and t. All of the units are miles,hours,milesPerHour.

You already understand and can use ;

CAN YOU SOLVE THIS SYSTEM FOR t AND FOR r ?

Answer by MathTherapy(10549) (Show Source):
You can put this solution on YOUR website!
An airplane takes 1 hour longer to go a distance of 375 miles flying against the headwind then on the return trip with a tail wind. if the speed of the wind is 50 miles per hour find the speed of the plane in still air.
I know the answer is 200mph because my test gave me the right answer after getting it wrong I just need help on how to get that answer and solve it right for next time.

Let speed in still air be S
Total speed against the wind: S - 50, and total speed with the wind: S + 50
We then get the following TIME equation:
375(S + 50) = 375(S - 50) + (S - 50)(S + 50) ------- Multiplying by LCD, (S - 50)(S + 50)

(S - 200)(S + 200) = 0
S, or speed in still air = OR S = - 200 (ignore)
It's that simple.
You just need to solve for what is asked for, and that's speed in still air. Solving for time is TOTALLY UNNECESSARY. Don't do it! It makes no sense.