cant figure out this OG question!

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cant figure out this OG question!

by hongwang9703 » Sat Nov 21, 2009 7:16 pm
The question reads: A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? -assume that the driver did not make any stops during the 40 mile trip-

a. 65mph
b. 68mph
c. 70mph
d. 75mph
e. 80mph

[a: d][/spoiler]
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by papgust » Sun Nov 22, 2009 2:31 am
Given:
Total distance = 40 miles
Speed of first 20 miles = 50 mph

Now,
You need to compute remaining 20 miles speed.

Average speed of trip must be 60 mph. So time taken should be Total distance/Average Speed = 40/60 = 40 min.
Already first 20 miles is covered. Time taken for this is 20 miles / 50 mph = 24 min.

Remaining Time is 40 - 24 = 16 min (Remaining 20 miles should be covered in 16 min in order to clock average of 60 mph for the entire trip)

Therefore, Speed for second half of the trip ==> remaining distance / remaining time = (20 / 16) * 60 = 75 mph.

Choose D.

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by hongwang9703 » Sun Nov 22, 2009 2:09 pm
your explaination is much more clear than the one OG gave me. THank you so much!
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by heshamelaziry » Sun Nov 22, 2009 5:28 pm
hongwang9703 wrote:The question reads: A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? -assume that the driver did not make any stops during the 40 mile trip-

a. 65mph
b. 68mph
c. 70mph
d. 75mph
e. 80mph

[a: d][/spoiler]

1st portion of trip:

d=20
r= 50
t= 2/5 of an hour

2nd portion:

d=20
r= x
t= 20/x

- to acheive 60 mph for all trip, total time will be 2/3 of an hour for entire trip
- time for second portion will be 2/3 - 2/5 = 4/15 of an hour
- 20/x = 4/15 -------> x = 75 mph for second portion

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by adamsmith2009 » Mon Nov 23, 2009 3:25 pm
Another way to do these types of questions is the formula: 2*S1*S2/S1+S2 = Average speed where S1 is the speed of the first trip and S2 is the speed of the second trip. This only works when the distance is split equally between the trips and there are no stops. I find this way to be much quicker.

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by bblast » Sun Jul 24, 2011 11:01 pm
hi ,
The question is pretty straight and can be solved by total distance/ total time = avg speed


my whacky brain posted an interesting query on this question. y cant this be treated as a weighted average question ?


20 miles at 50mph------------mean = (20*50)+(20*70) / 40 = 60-----------------20 miles at 70mph



I understand weighted average does not work here as the answer would be clearly 70 as shown above. But why?



sorry if my query is silly.
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by Geva@EconomistGMAT » Sun Jul 24, 2011 11:21 pm
bblast wrote:hi ,
The question is pretty straight and can be solved by total distance/ total time = avg speed


my whacky brain posted an interesting query on this question. y cant this be treated as a weighted average question ?


20 miles at 50mph------------mean = (20*50)+(20*70) / 40 = 60-----------------20 miles at 70mph



I understand weighted average does not work here as the answer would be clearly 70 as shown above. But why?



sorry if my query is silly.
not a silly question - in fact, it's the kind of common mistake whoever wrote the question wanted you to make. If I drive half the distance at a speed of 50, and I want the average speed to be 60, why don't I just drive the other half at 70? 70+50/2=60, right?

What you're doing is not a weighted average - it's a straight down the middle average between 50 and 70. A true weighted average should bring into account the different weights of the speeds (50 mph and the other speed). The problem is that where speed is involved, the weight is actually measured in TIME, not in DISTANCE. The car travels the same distance at each speed (20 miles each), but it travels a different amount of TIME at each speed, and that TIME is the relative weight of each speed.

A classic example here would be if you were to travel one hour at 100 mph, then spend the next 100 hours walking at 1 mph. Sure, the distance covered is the same (100 miles each), but is your average speed simply 100+1/2 = 50.5 miles per hour? It took you four days to cover 200 miles!

Back to our question: Since we cover the distance at two speeds (one slower at 50 mph, one faster), we spend more time at the slower speed: more time at 50 mph than at the other speed. So the final average speed needs to be closer to 50 than to the other end. Another way of looking at the same: Since we're going to be spending less time at the other speed, it has to be faster than simply 70 mph to achieve the same 60 mph average. Hence 75 mph.
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by GMATGuruNY » Mon Jul 25, 2011 2:38 am
bblast wrote:hi ,
The question is pretty straight and can be solved by total distance/ total time = avg speed


my whacky brain posted an interesting query on this question. y cant this be treated as a weighted average question ?


20 miles at 50mph------------mean = (20*50)+(20*70) / 40 = 60-----------------20 miles at 70mph



I understand weighted average does not work here as the answer would be clearly 70 as shown above. But why?



sorry if my query is silly.
The problem above requires almost no math.

If you drive x miles at one speed and another x miles at another speed, the average speed for the entire trip will be just a bit less than the average of the two speeds.

If you go to work at 30 miles per hour and return along the same route at 50 miles per hour, the average speed for the whole trip will be just a bit less than (30+50)/2 = 40 miles per hour.

The reason: more time will be spent traveling at the slower speed, so the average speed for the whole trip will be weighted toward the slower speed.

To illustrate:
Let the distance between home and work = 150 miles.
Time at 30 miles per hour = 150/30 = 5 hours.
Time at 50 miles per hour = 150/50 = 3 hours.
Average speed for the whole trip = 300/8 = 37.5, just a bit less than the average of the two speeds.
A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? -assume that the driver did not make any stops during the 40 mile trip-

a. 65mph
b. 68mph
c. 70mph
d. 75mph
e. 80mph


We can plug in the answers, which represent the speed for the remaining x miles.

Answer choice C: 70 miles per hour.
The average speed for the whole trip will be just a bit less than (50+70)/2 = 60 miles per hour.
Too slow.
Eliminate A, B and C.

Answer choice E: 80 miles per hour
The average speed for the whole trip will be just a bit less than (50+80)/2 = 65 miles per hour.
Too fast.
Eliminate E.

The correct answer is D.
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by navami » Mon Jul 25, 2011 6:52 am
ans is 75 mph
This time no looking back!!!
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by bblast » Mon Jul 25, 2011 10:08 am
Thank you Geva and Thank you Mitch, there was an important takeaway in this relatively straight forward problem. I am happy that I posted the query.
Cheers !!

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by ECMoyano » Mon Jul 25, 2011 12:13 pm
First time poster :)

I did the old Box method that I picked up in my review book.

R * T = D

50mph * unknown minutes = 20 miles
unknown rate * unknown rate = 20 miles

50 miles traveled/60 minutes. I simplify to 5miles every 6 minutes. 5 miles goes into 20 a total of 4 times. 6 minutes times 4 = 24

So I know
In 24 minutes going 50mph I'll go 20 miles. (Plug in 24 for first unknown)

going 60 mph or 1 mile a minute means the journey must take 40 minutes, since its 40 miles away.
24 + x = 40
X=16

20 Miles/16 minutes * 60 minutes = 75 mph

I complete my box
50MPH * 24 minutes = 20miles
75MPH * 16 minutes = 20miles

Took me 75 seconds, 15-20 seconds too long I think.