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With a tailwind, a plane flew the 
3000km from Calgary to Montreal in 5 hours. The return flight, against the wind, took 6 hours. Find the wind speed and the speed of the plane.

With a tailwind, a plane flew the 3000 km 3000 \mathrm{~km} from Calgary to Montreal in 55 hours. The return flight, against the wind, took 66 hours. Find the wind speed and the speed of the plane.

Full solution

Q. With a tailwind, a plane flew the 3000 km 3000 \mathrm{~km} from Calgary to Montreal in 55 hours. The return flight, against the wind, took 66 hours. Find the wind speed and the speed of the plane.
  1. Define speeds: Let's define the speed of the plane as pp km/h and the speed of the wind as ww km/h. The effective speed of the plane with the tailwind is (p+w)(p + w) km/h, and against the wind, it is (pw)(p - w) km/h.
  2. Set up equations: Using the formula distance=speed×time\text{distance} = \text{speed} \times \text{time}, we can set up the equations for the flights. For the flight from Calgary to Montreal, the equation is 3000=(p+w)×53000 = (p + w) \times 5.
  3. Flight to Montreal: Simplifying the equation from the previous step, we get p+w=600p + w = 600.
  4. Return flight: For the return flight against the wind, the equation is 3000=(pw)×63000 = (p - w) \times 6.
  5. Solve equations: Simplifying this equation, we get pw=500p - w = 500.
  6. Find speed: Now, we have two equations:\newline11. p+w=600p + w = 600\newline22. pw=500p - w = 500\newlineWe can solve these equations by adding them together.
  7. Substitute and solve: Adding the equations, we get 2p=11002p = 1100. Solving for pp, we find p=550km/hp = 550 \, \text{km/h}.
  8. Substitute and solve: Adding the equations, we get 2p=11002p = 1100. Solving for pp, we find p=550km/hp = 550 \, \text{km/h}. Substituting p=550km/hp = 550 \, \text{km/h} back into the equation p+w=600p + w = 600, we solve for ww: \newline550+w=600550 + w = 600\newlinew=50km/h.w = 50 \, \text{km/h}.

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