A transcontinental jet travels at a rate of x-100 mph with a headwind and x+100 mph...

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BTGmoderatorLU: Moderator; Posts: 2558; Joined: Sun Oct 15, 2017 1:50 pmFollowed by: 6 members

by BTGmoderatorLU » Tue Jul 28, 2020 6:24 pm

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Source: Magoosh

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 hours 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

The OA is C

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Source: Beat The GMAT — Problem Solving | Tue Jul 28, 2020 6:24 pm

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Re: A transcontinental jet travels at a rate of x-100 mph with a headwind and x+100 mph...

by Brent@GMATPrepNow » Wed Jul 29, 2020 7:44 am

BTGmoderatorLU wrote: ↑
Tue Jul 28, 2020 6:24 pm
Source: Magoosh

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 hours 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

The OA is C

Let's start with a "word equation"

(time with headwind) = (time with tailwind) + 2 hours 40 minutes

2 hours 40 minutes = 2 2/3 hours = 8/3 hours
Time = distance/speed

Plug the given values into the word equation to get: 3200/(x – 100) = 3200/(x + 100) + 8/3
Multiply 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

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8156; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

Re: A transcontinental jet travels at a rate of x-100 mph with a headwind and x+100 mph...

by Scott@TargetTestPrep » Fri Jul 31, 2020 6:43 am

BTGmoderatorLU wrote: ↑
Tue Jul 28, 2020 6:24 pm
Source: Magoosh

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 hours 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

The OA is C

Solution:
We can create the equation:
Time traveling headwind = Time traveling tailwind + 2 hours 40 minutes
3,200/(x - 100) = 3,200/(x + 100) + 2 + 40/60
3,200/(x - 100) = 3,200/(x + 100) + 8/3
Multiplying the equation by 3(x - 100)(x + 100), we have:
9,600(x + 100) = 9,600(x - 100) + 8(x - 100)(x + 100)
1,200(x + 100) = 1,200(x - 100) + x^2 - 10,000
1,200x + 120,000 = 1,200x - 120,000 + x^2 - 10,000
x^2 - 250,000 = 0
(x - 500)(x + 500) = 0
x = 500 or x = -500
Since x can’t be negative, x = 500, and the tailwind speed of the jet is x + 100 = 600 mph.
Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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