Francine takes a trip

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Francine takes a trip

by nh8404052006 » Sat Jan 03, 2009 12:02 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 Francine’s average speed for the entire trip?

A. (1800 - x) /2
B. (x + 60) /2
C. (300 - x ) / 5
D. 600 / (115 - x )
E. 12,000 / ( x + 200)

OA : E
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Set 06 Q 37
Source: — Problem Solving |

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Re: Francine takes a trip

by logitech » Sat Jan 03, 2009 12:15 am
IF X = 0% we should have only 60 miles/hr as average

IF X = 100% we should have 40 miles/hr as average


E. 12,000 / ( x + 200)


12000/200 = 60

12000/300 = 40
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by cramya » Sat Jan 03, 2009 12:37 am
Picking numbers would be the easiest way(let me know if u need this also).

Algebric approach would be:

Average speed = Total Distance/Total time taken
Let d be the distance

x% of that xd/100

Remaining distance = (100-x)d/100

Time taken to cover x% of distance = xd/100 / 40 = xd/4000

Time taken to cover 100- x % of distance = (100-x)d/100/ 60
= (100-x)d/6000

Total distance = d

Total time taken = xd/4000+(100-x)d/6000

= 3xd+200d-2xd / 12000
= xd + 200d /12000
= d(200+x)/ 12000


Average speed = Total distance / Total time taken
= d/ d(200+x)/ 12000
= 12000/ (200+x)

Choose E)

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by nh8404052006 » Sat Jan 03, 2009 12:40 am
This is a very quick method to a complicate problem

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by cramya » Sat Jan 03, 2009 12:50 am
Picking numbers(pick smart numbers where computations will be easier when it comes to division/multiplication etc..) strategy.

I have tried to explain in detail but once u get the concept u can avoid all this extra verbage on the test (all of us will and get down to the actual calculations) and the calculations can be done quickly.

Let the distance = 240 (picked 240 since its is a multiple of 60 and 40 and when I take x=50% then 120 will also be a multiple of 60 and 40 which makes calculations easier)


Let x = 50% i.e 50% of 240 = 120
remaining distance = 120

Time taken for first 120 = 120/40 =3
Time taken for next 120 = 120/60 = 2

Total time = 5
Tot distance = 240
Av speed = 240/5 = 48


When u substitute x=50 the answer choices should give u 48 as average speed

A (1800 - x) /2 -> (1800-50) /2 NO

B. (x + 60) /2 -> (50+60)/2 = 55 NO

C. (300 - x ) / 5 -> NO

D. 600 / (115 - x ) -> 600/115-50 = -> NO

E. 12,000 / ( x + 200)-> 12000/ (50+200) = 48 YES

Choose E)

Always go with what you feel most comfortable with (algebric or picking numbers i.e one that can be solved within reasonable amount of time that u can allot to a question 2-2.5 mins max->3 on harder ones if needed)
and practice that skill the best u can before the exam.. The only downside(not really) when picking numbers for questions that have variables in the answer choices is that sometimes u may get 2 choices for which the numbers may work. You may have to pick new set of numbers to eliminate one of the 2 choices left. If we get stuck with algebric approach instead of wasting more time staring at the problem we must resort to picking numbers and solv these questions with variables in the answer choices. GMAC does not care how we solve a problem. Any approach conventional/textbook or unconventional works as long as we can get to to the correct answer within a reasonable time frame.


My bad for writing a textbook for this problem :-) but hope this helps!

Regards,
Cramya

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by rajataga » Sat Jan 03, 2009 2:20 am
A common problem that is often made in these sort of problems, is that we suppose Average speed to be in terms of distance, however, average speed is ALWAYS in terms of time....

Here, the average speed is given for x distance, so it would be tempting to average it out in terms of 'x', however, that would be wrong, since x denotes distance.

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by korkun » Sat Jan 03, 2009 3:21 pm
I believe that "lets say" method is easier than number picking.

now

let's say there is 100 km. of total distance.

than we go x km. with 40km/hour and 100-x km. with 60km/hour

average speed is: total distance/total time

so: 100/[x/40+(100-x)/60] would be the answer.

with a quick calculation you find: 12000/(x+200)

ps: use miles in state of km. :)

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by fruti_yum » Tue Aug 04, 2009 12:49 pm
cramya wrote:Picking numbers(pick smart numbers where computations will be easier when it comes to division/multiplication etc..) strategy.

I have tried to explain in detail but once u get the concept u can avoid all this extra verbage on the test (all of us will and get down to the actual calculations) and the calculations can be done quickly.

Let the distance = 240 (picked 240 since its is a multiple of 60 and 40 and when I take x=50% then 120 will also be a multiple of 60 and 40 which makes calculations easier)


Let x = 50% i.e 50% of 240 = 120
remaining distance = 120

Time taken for first 120 = 120/40 =3
Time taken for next 120 = 120/60 = 2

Total time = 5
Tot distance = 240
Av speed = 240/5 = 48


When u substitute x=50 the answer choices should give u 48 as average speed

A (1800 - x) /2 -> (1800-50) /2 NO

B. (x + 60) /2 -> (50+60)/2 = 55 NO

C. (300 - x ) / 5 -> NO

D. 600 / (115 - x ) -> 600/115-50 = -> NO

E. 12,000 / ( x + 200)-> 12000/ (50+200) = 48 YES

Choose E)

Always go with what you feel most comfortable with (algebric or picking numbers i.e one that can be solved within reasonable amount of time that u can allot to a question 2-2.5 mins max->3 on harder ones if needed)
and practice that skill the best u can before the exam.. The only downside(not really) when picking numbers for questions that have variables in the answer choices is that sometimes u may get 2 choices for which the numbers may work. You may have to pick new set of numbers to eliminate one of the 2 choices left. If we get stuck with algebric approach instead of wasting more time staring at the problem we must resort to picking numbers and solv these questions with variables in the answer choices. GMAC does not care how we solve a problem. Any approach conventional/textbook or unconventional works as long as we can get to to the correct answer within a reasonable time frame.


My bad for writing a textbook for this problem :-) but hope this helps!

Regards,
Cramya



If I pick D =100 and x = 10..

t1= 1/4
t2 = 3/2
total time = 7/4
total distance = 100

I get avg speed = 400/7 = 57.1

The closest answer is C does anyone know what I did wrong?

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by fruti_yum » Tue Aug 04, 2009 12:52 pm
fruti_yum wrote:
cramya wrote:Picking numbers(pick smart numbers where computations will be easier when it comes to division/multiplication etc..) strategy.

I have tried to explain in detail but once u get the concept u can avoid all this extra verbage on the test (all of us will and get down to the actual calculations) and the calculations can be done quickly.

Let the distance = 240 (picked 240 since its is a multiple of 60 and 40 and when I take x=50% then 120 will also be a multiple of 60 and 40 which makes calculations easier)


Let x = 50% i.e 50% of 240 = 120
remaining distance = 120

Time taken for first 120 = 120/40 =3
Time taken for next 120 = 120/60 = 2

Total time = 5
Tot distance = 240
Av speed = 240/5 = 48


When u substitute x=50 the answer choices should give u 48 as average speed

A (1800 - x) /2 -> (1800-50) /2 NO

B. (x + 60) /2 -> (50+60)/2 = 55 NO

C. (300 - x ) / 5 -> NO

D. 600 / (115 - x ) -> 600/115-50 = -> NO

E. 12,000 / ( x + 200)-> 12000/ (50+200) = 48 YES

Choose E)

Always go with what you feel most comfortable with (algebric or picking numbers i.e one that can be solved within reasonable amount of time that u can allot to a question 2-2.5 mins max->3 on harder ones if needed)
and practice that skill the best u can before the exam.. The only downside(not really) when picking numbers for questions that have variables in the answer choices is that sometimes u may get 2 choices for which the numbers may work. You may have to pick new set of numbers to eliminate one of the 2 choices left. If we get stuck with algebric approach instead of wasting more time staring at the problem we must resort to picking numbers and solv these questions with variables in the answer choices. GMAC does not care how we solve a problem. Any approach conventional/textbook or unconventional works as long as we can get to to the correct answer within a reasonable time frame.


My bad for writing a textbook for this problem :-) but hope this helps!

Regards,
Cramya



If I pick D =100 and x = 10..

t1= 1/4
t2 = 3/2
total time = 7/4
total distance = 100

I get avg speed = 400/7 = 57.1

The closest answer is C does anyone know what I did wrong? I wud have picked C in the exam!!!!!!!

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by linfongyu » Fri Dec 11, 2009 5:26 pm
This is a toughie. I agree that under the circumstances, picking numbers might be the best approach. But under the time constraint of the test, this would be a fairly difficult problem.

First, we should assume x = 50, both distances are the same. To find the average speed over the same distance, the equation is: 2*s1*s2/(s1+s2). In this case, that's 2*40*60/100, which is 48.

So, plug 50 back into the choices for x, and look for 48... E works.

I initially did what Logitech did, looking for extremes of X = 100 and 0 %. But C and E both satisfies that -

C (300-x)/5, when x = 0, result is also 60, when x = 100, result is also 40
E 12000/(x+200), when x = 0, result is 60, when x = 100, result is 40

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by Testluv » Sat Dec 12, 2009 1:05 am
linfongyu's method was excellent: treating it as a round-trip problem in order to use the special round trip average speed formula: whenever an object's journey is divided into two legs of equal distance, you can always find the average speed using the formula that lingfonyu posted. I call it the "round trip" formula b/c the most common situation on the GMAT where you will have one object's journey divided into two legs of equal distance is in round trip problems where an object goes from A to B, and then back to A again.

In case you didn't know that formula, you could have also picked numbers as follows.

Let total distance be 100 miles. Let x be 40; then, x percent of total distance is: 40% of 100 = 40 miles. Then, she went for 40 miles at 40 mph, which, therefore, took 1 hour.

Leaving 100-40 = 60 miles, which she travelled at 60 miles per hour, so this also took 1 hour. So, with these numbers, her total time was 2 hours.

Average speed = total distance/total time = 100/2 = 50. So, when x is 40, her average speed is 50. Go to the answer choices, plug in 40 for x, and eliminate those answer choices that don't result in 50; only E works.

_____________

When picking numbers, we want to make the problem as manageable as possible!
With percent problems, picking 100 is often the best way to go!

___________

When you got to the point where you're ready to plug in to the answer choices, in this question, you should definitely start from the bottom...can anyone tell me why?! (think about it!..think like the test-maker!)
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by Stuart@KaplanGMAT » Sat Dec 12, 2009 1:41 am
linfongyu wrote:To find the average speed over the same distance, the equation is: 2*s1*s2/(s1+s2).
Great way to solve!

Does this formula look familiar? It should, since it's simply the combined work formula.

Remember, to find the time it takes two people to complete a job together, we simply calculate:

a*b/(a+b)

in which a and b are the times it takes each individual to do the job alone.

You'll note that in linfongyu's variation of the equation, there's an extra multiplier of 2. Why is that? Simply because the two people aren't just doing one job - it's a round trip problem, so they're doing two jobs.

For example, the following two questions are identical:

Bob drives to work at 40mph and home from work at 60mph. If the distance is the same in both directions, what's Bob's average speed for his daily commute?

and

Working alone, Lisa can paint a room in 40 hours and Bob can paint an identical room in 60 hours. How long would it take them, working individually but together, to paint both rooms?
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