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garuhape: Senior | Next Rank: 100 Posts; Posts: 74; Joined: Fri Dec 17, 2010 12:37 pm; Location: Munich; Thanked: 3 times

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by garuhape » Tue Feb 15, 2011 6:33 am

During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was the average speed for the entire trip.

Official Answer: 12000/(x+200)

I don't understand the solution.

I did it like that:

x*40+(1-x)*60=average speed of total trip.

Let's say that she traveled 30% of the distance at an average speed of 40 and 70% at an average speed of 60. Thus the average speed of the total trip is:

0.3*40+0.7*60=54

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Source: Beat The GMAT — Problem Solving | Tue Feb 15, 2011 6:33 am

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pm; Thanked: 127 times; Followed by:10 members

by Night reader » Tue Feb 15, 2011 6:55 am

OA may skip important details

see my solution below

D=distance, Speed=S, T=time; xD/100 at average S(1)=40; (100-x)D/100 at average S(2)=60; find D/(T1+T2)-?
T1=xD/100 : 40 and T2=(100-x)D/100 :60; T1+T2=[3xD+2D(100-x)]/(100*120)= (xD+200D)/(100*120)

D: (xD+200D)/(100*120)= 12000/(x+200)

garuhape wrote:During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was the average speed for the entire trip.

Official Answer: 12000/(x+200)

I don't understand the solution.

I did it like that:

x*40+(1-x)*60=average speed of total trip. <--- it says precisely x% but we don't know the value of x, e.g. x=20% you take 1-20 which is -19, but should be x%=x/100 and (100-x)/100

Let's say that she traveled 30% of the distance at an average speed of 40 and 70% at an average speed of 60. Thus the average speed of the total trip is:

0.3*40+0.7*60=54

:

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Tue Feb 15, 2011 7:14 am

GMATGuruNY wrote:During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?

a. (180-X)/(2)
b. (x+60)/(4)
c. (300-x)/(5)
d. (600)/(115-x)
e. (12,000)/(x+200)

Plug in a value for the distance and a value for x.

Let distance = 100 miles.
Let x = 40.
Distance for 40% of the trip = .4*100 = 40 miles.
Since the rate for this portion is 40mph, time = d/r = 40/40 = 1 hour.
Distance for remainder of the trip = 100-40 = 60 miles.
Since the rate for this portion is 60mph, time = d/r = 60/60 = 1 hour.
Total time = 1+1 = 2 hours.
Average speed for the whole trip = (total distance)/(total time) = 100/2 = 50. This is our target.

Now we plug x = 40 into all the answer choices to see which yields our target of 50.

Only answer choice E works:
(12,000)/(x+200) = 12000/(40+200) = 50.

The correct answer is E.[/quote]

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garuhape: Senior | Next Rank: 100 Posts; Posts: 74; Joined: Fri Dec 17, 2010 12:37 pm; Location: Munich; Thanked: 3 times

by garuhape » Tue Feb 15, 2011 7:21 am

Using what you have written in blue

x/100*40 + (100-x)/100*60 = 40x/100+60(100-x)/100 = 40x/100+(6000-60x)/100 = (6000-20x)/100 = (300-x)/5,

which is one of the answer choices but which apparently is wrong. However, if you try it out the formula works perfectly fine.