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4 Bumps

# Need help with a math problem..

Flying with the wind, a plane flew 1220 mi in 2 hrs. Against the wind, the plane could fly only 1080 mi in the same amount of time. Find the rate of the plane in calm air.

I have notes, and the text, but it's asking to find the rate of the plane in CALM air, so it's throwing me off..

Asked by careph at 12:40 PM on Mar. 22, 2011 in Money & Work

Level 10 (411 Credits)
• Answer by AnuMeha at 12:42 PM on Mar. 22, 2011

• I believe it is 1150 miles in 2 hours.

Answer by JeremysMom at 12:48 PM on Mar. 22, 2011

• How did you get that? I'm so overloaded with tables, formulas, etc. it's insane

Comment by careph (original poster) at 12:50 PM on Mar. 22, 2011

• Don't look at me- I'm a math dunce! lol But here's a bump anyway!

Answer by lovingmy4babies at 12:51 PM on Mar. 22, 2011

• I am on mobile so I will try to explain as best as I can. Since they gave you both scenarios ( from the wind hindering to helping) you need to find the difference. The difference is 70 so you can either add 70 to the 1080 speed or subtract it from 1220. Either way, you get 1150.

Answer by JeremysMom at 12:56 PM on Mar. 22, 2011

• With the wind, the plane flies 1220 miles in 2 hours. Divide 1220 by 2, and the plane flies 610 miles in one hour.

Against the wind, the plane flies 1080 miles in 2 hours. Divide 1080 by 2, and the plane flies 540 miles in one hour.

Add these two speeds together, then divide by two, to find the average.

610 + 540 = 1150

1150 divided by 2 = 575

Answer by FelipesMom at 12:56 PM on Mar. 22, 2011

• 575 miles per hour. When it is flying with the wind the rate is 610 mph. Against the wind the rate is 540 mph. The wind is acting equally on the plane no matter which direction it is flying so the wind has its own rate. The difference between these rates (610-540) is 70. So the wind rate must be 35 mph. Thus, the rate of the plane is 575 mph no wind. Flying with the wind the plane now flies at 610 mph. In two hours...1220 miles.

Does that make sense?

Answer by heat631 at 12:56 PM on Mar. 22, 2011

• Thank you all so much! I spent idk how long on this problem last night (after my daughter's bedtime routine, and 4 hours of tests and homework, I was just braindead).

Comment by careph (original poster) at 12:58 PM on Mar. 22, 2011

• You may want to add a comment along the lines of, "Assuming the same wind speed in both scenarios".

Answer by anng.atlanta at 1:02 PM on Mar. 22, 2011

• You can opt for an online math tutor. They provide 1 to 1 tutoring at much lower cost. As a mother, sometimes I also become blank, when it comes to help my daughter with her math homework. I use tutorteddy.com. You can try it.

Answer by AAfterSearch at 2:33 AM on Mar. 23, 2011

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