I hope there is someone out there who is really good with Systems of Equations. I am really stumped with this one. Traveling against the wind, a plane flies 2100 miles from Chicago to San Diego in 4 hours and 40 minutes. The return trip, traveling with a wind that is twice as fast, takes 4 hours. Find the rate of the plane in still air. A) 634.5 mi/h B) 637.5 mi/h C) 650.5 mi/h D) 675 mi/h We have to solve this problem using systems of equations. I am really confused. Usually, I can do these problems, but I cannot figure this one out. Thanks so much! Susan
I thought I figured it out, but I came out with 475 mi/h, so I guess I can't help you either. *shrugs*
Thanks for trying. I wonder if there is a problem with the answer choices. We have found problems before. Blessings, Susan
Susan, Dh and daughter worked and worked on it and couldn't come up with the right answer said something was missing or its a misprint. They came up with the same answer 2littleboys did 475 something was misprinted some where.
A: 2100 mi / 4.67 hr = 449.68 mph (or 450 for argument's sake) B: 2100 mi / 4 hr = 525 mph A: 450 mph + 25 mph winds = 475 mph B: 525 mph - (25 x 2, or 50) mph winds = 475 mph
2 little boys and KrisRV, Thanks for explaining. That is the way that I was trying to do it, but hint said that the formulas were 14/3 (b-c) = 2100 4 (b+ 2c) =2100 I did figure out that 14/3= approximately 4.68 so it is the hours of course the other equation is multiplied by 4 the hours again I still get 475, so that has to be the answer. Thanks for all the help. I feel confident that there was just an error. I guess it was just a typo. Option D should be 475 instead of 675. Bountiful blessings, Susan
