OG 13 ,Q no 162, couldn't understand it's explanation

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sohailmbaprep: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Thu Feb 23, 2012 11:18 pm

OG 13 ,Q no 162, couldn't understand it's explanation

by sohailmbaprep » Fri May 25, 2012 2:01 am

1)Can Somebody please solve this for me.
2) this question is of what difficulty level(600 or 650 or like 700 or like 550)
During a trip,Francine traveled x percentage of the total distance at an average speed of 40 miles/hr and rest of the distance at an average speed of 60 miles/hr.In terms of x,what was Francine's average speed for the entire trip ?

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Source: Beat The GMAT — Problem Solving | Fri May 25, 2012 2:01 am

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by GMATGuruNY » Fri May 25, 2012 2:18 am

sohailmbaprep 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)

To make the math easy:
Let x=50%, implying that 50% of the trip is traveled at 40mph, with the other half traveled at 60mph.
Let the distance for each half of the trip = the LCM of 40 and 60 = 120.

Time for the first 1/2 of the trip = d/r = 120/40 = 3.
Time for the second 1/2 of the trip = d/r = 120/60 = 2.
Average speed for the entire trip = (total distance)/(total time) = 240/5 = 48. This is our target.

Now we plug x=50 into the answer choices to see which yields our target of 48.

Only answer choice E works:
(12,000)/(x+200) = 12000/(50+200) = 48.

The correct answer is E.

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by Anurag@Gurome » Fri May 25, 2012 3:06 am

sohailmbaprep wrote:1)Can Somebody please solve this for me.
2) this question is of what difficulty level(600 or 650 or like 700 or like 550)
During a trip,Francine traveled x percentage of the total distance at an average speed of 40 miles/hr and rest of the distance at an average speed of 60 miles/hr.In terms of x,what was Francine's average speed for the entire trip ?

Let us take the total distance = 1 mile
Then x% of 1 = x/100
So, Francine traveled x/100 mile at 40 mph. So, time taken = distance/speed = (x/100)/40 = x/4000
Remaining distance = 1 - x/100 = (100 - x)/100
Time taken to travel (100 - x)/100 mile at 60 mph = distance/speed = [(100 - x)/100]/60 = (100 - x)/6,000

Total time = x/4000 + (100 - x)/6000 = [3x + 2(100 - x)]/12000 = (x + 200)/12000

Therefore, total average speed = distance/time = 1/[(x + 200)/12000] = [spoiler]12000/(x + 200)[/spoiler]

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ronnie1985: Legendary Member; Posts: 626; Joined: Fri Dec 23, 2011 2:50 am; Location: Ahmedabad; Thanked: 31 times; Followed by:10 members

by ronnie1985 » Fri May 25, 2012 10:33 am

Assume that the total distance of the trip is 100 miles

Distance x miles at speed 40 miles / hr , time = x/40
Similarly 100-x miles at 60 miles / hr , time = (100-x)/60

Avg Speed = (Total Distance)/(Total Time)

= 100/[x/40+(100-x)/60] = 12000 / [200+x]

I think this is 650 level question

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