A few days ago
I need help on an equation. (Algebra)?
An airplane whose speed in still air is 530 mi/h carries enough air gas for 10 hours of flight. On a certain flight it flies against a wind of 30 mi/h. On the return flight it travels with a wind of 30 mi/h. How far can the plane fly without refueling?
Please show work! Thanks!
Top 1 Answers
A few days ago
Favorite Answer
Let t be the time on the outward-bound leg and t’ the return time. You know first off that
t + t’ = 10
You also know that the distance out equals the distance in. Rate times time equals distance. Going out, the rate is (530 – 30) = 500. Coming back, the rate is (530 + 30) = 560. So
500t = 560t’
Two equations, two unknowns – you should be able to solve for the time. Given that, the distance is just one multiplication away.
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