Bunuel wrote:
A transcontinental jet travels at a rate of x – 100 mph with a headwind and x + 100 mph with a tailwind between Wavetown and Urbanio, two cities 3,200 miles apart. If it takes the jet 2 hr 40 minutes longer to complete the trip with a headwind, then what is the jet’s rate flying with a tailwind?
(A) 500
(B) 540
(C) 600
(D) 720
(E) Cannot be determined by the information given.
Let's start with a
"word equation"(time with headwind) = (time with tailwind) + 2 hours 40 minutes2 hours 40 minutes = 2 2/3 hours = 8/3 hours
Time = distance/speedPlug the given values into the word equation to get:
3200/(x – 100) = 3200/(x + 100) + 8/3Multiply both sides of the equation by 3 to get: 9600/(x – 100) = 9600/(x + 100) + 8
Multiply both sides of the equation by (x – 100) to get: 9600 = 9600(x – 100)/(x + 100) + 8(x – 100)
Multiply both sides of the equation by (x + 100) to get: 9600(x + 100) = 9600(x – 100) + 8(x – 100)(x + 100)
Expand and simplify: 9,600x + 960,000 = 9,600x – 960,000 + 8x² – 80,000
Subtract 9,600x from both sides: 960,000 = –960,000 + 8x² – 80,000
Divide both sides of the equation by 8 to get: 120,000 = –120,000 + x² – 10,000
Simplify right side: 120,000 = x² – 130,000
Add 130,000 to both sides: 250,000 = x²
Solve: x = 500 or x = -500
Since the speed cannot be negative, we know that x = 500
What is the jet’s rate flying with a tailwind?x + 100 mph = speed with a tailwind
So, the speed with a tailwind = 500 + 100 = 600
Answer: C
Cheers,
Brent
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Brent Hanneson – Creator of gmatprepnow.com
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