A few days ago
anonymous

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
Anonymous

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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