Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

During a 40-mile trip, Marla traveled at an average speed of [#permalink]

Show Tags

27 Apr 2010, 09:55

38

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

62% (02:19) correct
38% (01:30) wrong based on 988 sessions

HideShow timer Statistics

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x miles per hour for the last 40 - y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she had traveled at an average speed of x miles per hour for the entire trip?

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?

Q: \(\frac{t_1}{t_2}=\frac{0.25y+40}{1.25x}*\frac{x}{40}=\frac{0.25y+40}{1.25}*\frac{1}{40}\). So we see that the value of this fraction does not depend on \(x\), only on \(y\).

First read the stem and derive the appropriate equations.

We get,

[(y/x) + (40-y)/1.25x] is total time of the journey done by Marla. By simplifying, we get (y+160)/5x. Now, lets process this, 'what percent of the time it would have taken her if she had traveled at an average speed of x miles per hour for the entire trip' i.e. total time '40/x'

(40/x) P% = (y+160)/5x On simplifying, the x is removed from the equation and we are left with only y. i.e. P% = (y+160)/(5*40)

So we need only y value to obtain value of P. Therefore, we need only B.

Re: During a 40-mile trip, Marla traveled at an average speed of [#permalink]

Show Tags

29 Jul 2013, 11:14

1

This post received KUDOS

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?

(1) x = 48.

Answering the stem requires that we know how many miles she covered at what speed. For example, if someone covers 99 miles at 1 mile/hour and 1 mile at 1000 miles/hour, their average speed (and the time it takes for them to cover point A to B will be far more than if someone covers 99 miles at 1000 miles/hour and 1 mile at 1 mile/hour. This problem is no different. We know the speed she covered for the first portion of the trip and the speed she covered for the second portion of the trip (i.e. 1.25*48) but we don't know how many miles she covered for each speed. INSUFFICIENT

(2) y = 20.

If y = 20 that means she traveled 1/2 of the trip at speed x and 1/2 of the trip at 1.25x. Regardless of the speed of x the ratio of x to 1.25x will be the same because the distance covered by speed x and the distance covered by speed 1.25 x is the same. SUFFICIENT

I understand this problem conceptually (and in my post above this, got it right) but I have difficultly setting up the algebra to solve the problem as I do with many word and rate/distance problems. Here are my questions:

I.) I know why we have to add up y and (40-y) but I don't understand why it's set up as y/x + (40-y)/1.25x. Is it because distance/speed = time and we are looking for time in this question?

II.) Time 2 is x/40 which = speed/distance = time so we are multiplying the two averages (y/x + (40-y)/1.25x) combined by the single average (x/40) to get the percentage difference between the two averages and the one average, but why in your solution do you flip x/40 to 40/x?

Thanks!!!

Bunuel wrote:

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?

Q: \(\frac{t_1}{t_2}=\frac{0.25y+40}{1.25x}*\frac{x}{40}=\frac{0.25y+40}{1.25}*\frac{1}{40}\). So we see that the value of this fraction does not depend on \(x\), only on \(y\).

I understand this problem conceptually (and in my post above this, got it right) but I have difficultly setting up the algebra to solve the problem as I do with many word and rate/distance problems. Here are my questions:

I.) I know why we have to add up y and (40-y) but I don't understand why it's set up as y/x + (40-y)/1.25x. Is it because distance/speed = time and we are looking for time in this question?

II.) Time 2 is x/40 which = speed/distance = time so we are multiplying the two averages (y/x + (40-y)/1.25x) combined by the single average (x/40) to get the percentage difference between the two averages and the one average, but why in your solution do you flip x/40 to 40/x?

Thanks!!!

Bunuel wrote:

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?

Q: \(\frac{t_1}{t_2}=\frac{0.25y+40}{1.25x}*\frac{x}{40}=\frac{0.25y+40}{1.25}*\frac{1}{40}\). So we see that the value of this fraction does not depend on \(x\), only on \(y\).

(1) x = 48. Not sufficient.

(2) y = 20. Sufficient.

Answer: B.

x mph for the first y miles --> \(time=\frac{distance}{speed}=\frac{y}{x}\) 1.25x mph for the last 40-y miles of the trip --> \(time=\frac{distance}{speed}=\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\) So, \(t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\);

x miles per hour for the entire trip --> \(t_2=\frac{40}{x}\).

But why when you multiply by x/40 do you flip it to 40/x?

Bunuel wrote:

WholeLottaLove wrote:

Hi Bunuel,

I understand this problem conceptually (and in my post above this, got it right) but I have difficultly setting up the algebra to solve the problem as I do with many word and rate/distance problems. Here are my questions:

I.) I know why we have to add up y and (40-y) but I don't understand why it's set up as y/x + (40-y)/1.25x. Is it because distance/speed = time and we are looking for time in this question?

II.) Time 2 is x/40 which = speed/distance = time so we are multiplying the two averages (y/x + (40-y)/1.25x) combined by the single average (x/40) to get the percentage difference between the two averages and the one average, but why in your solution do you flip x/40 to 40/x?

Thanks!!!

Bunuel wrote:

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?

Q: \(\frac{t_1}{t_2}=\frac{0.25y+40}{1.25x}*\frac{x}{40}=\frac{0.25y+40}{1.25}*\frac{1}{40}\). So we see that the value of this fraction does not depend on \(x\), only on \(y\).

(1) x = 48. Not sufficient.

(2) y = 20. Sufficient.

Answer: B.

x mph for the first y miles --> \(time=\frac{distance}{speed}=\frac{y}{x}\) 1.25x mph for the last 40-y miles of the trip --> \(time=\frac{distance}{speed}=\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\) So, \(t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\);

x miles per hour for the entire trip --> \(t_2=\frac{40}{x}\).

But why when you multiply by x/40 do you flip it to 40/x?

Bunuel wrote:

WholeLottaLove wrote:

Hi Bunuel,

I understand this problem conceptually (and in my post above this, got it right) but I have difficultly setting up the algebra to solve the problem as I do with many word and rate/distance problems. Here are my questions:

I.) I know why we have to add up y and (40-y) but I don't understand why it's set up as y/x + (40-y)/1.25x. Is it because distance/speed = time and we are looking for time in this question?

II.) Time 2 is x/40 which = speed/distance = time so we are multiplying the two averages (y/x + (40-y)/1.25x) combined by the single average (x/40) to get the percentage difference between the two averages and the one average, but why in your solution do you flip x/40 to 40/x?

Thanks!!!

x mph for the first y miles --> \(time=\frac{distance}{speed}=\frac{y}{x}\) 1.25x mph for the last 40-y miles of the trip --> \(time=\frac{distance}{speed}=\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\) So, \(t_1=\frac{y}{x}+\frac{40-y}{1.25x}=\frac{0.25y+40}{1.25x}\);

x miles per hour for the entire trip --> \(t_2=\frac{40}{x}\).

Ahh...okay. You divide t1/t2 and you flip the denominator in the process. The problem therefore is looking for the difference between the average of the two times and the average if she traveled for the one figure of time?

Ahh...okay. You divide t1/t2 and you flip the denominator in the process. The problem therefore is looking for the difference between the average of the two times and the average if she traveled for the one figure of time?

Read the question: The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?
_________________

Re: During a 40-mile trip, Marla traveled at an average speed of [#permalink]

Show Tags

08 Oct 2013, 07:03

Is it possible to solve the problem by taking the weighted averages of the distance and speed, and equating the same to find the average speed of the trip.

My thinking was that if there was a total line, it would have been total miles/total rate --> 40/(x + 1.25x), which does not equal the above T1 + T2

Why is this the case?

Also, as a broader question: why is it that in distance problems, such as this one, we cannot add the speeds but in work problems, we commonly add, for example, the rates of two machines working individually to find their combined rate?

Re: During a 40-mile trip, Marla traveled at an average speed of [#permalink]

Show Tags

27 Feb 2015, 18:18

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: During a 40-mile trip, Marla traveled at an average speed of [#permalink]

Show Tags

19 Nov 2015, 11:50

Bunuel,

My approach was:

The question stem deals with a ratio (total time in cenario 1 / total time in cenario 2).

We know by definition that ratios are not affected if both a/b are increased or decreased by the same percentage. Statement 1 stablishes a steady 25% difference between average speeds, so that alone could not give us a unique ratio. Eliminate Statement 1.

Statement 2 gives a value for distance, which is not subject to any proportion rule. X could be anything, as long as the 25% increase in speed still holds. Bingo! That alone could affect our final ratio. Statement 2 is enough.

Re: During a 40-mile trip, Marla traveled at an average speed of [#permalink]

Show Tags

24 Nov 2016, 19:23

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: During a 40-mile trip, Marla traveled at an average speed of [#permalink]

Show Tags

29 Jan 2017, 01:30

From the question, I see that there are two situations we are comparing: (1) The time for traveling x mph for the first y miles and 1.25x mph for the last 40-y miles. We'll call this T1. (2) The time for traveling x mph the entire 40 miles. We'll call this T2.

We want to find T1 / T2 * 100%

Let's rephrase T2 first. Time = Distance / Rate, so T2 = 40 miles / x mph = 40/x hours.

Now let's take a look at T1. Using an RTD Chart: ----------Rate * Time = Distance----- 1st Part x mph * ?? = y miles 2nd Part 1.25x mph * ?? = 40-y miles

The time for the first part is y/x hours, and the time for the second part is (40-y)/(1.25x) hours.

Now let's plug these values for T1 and T2 into the the formula. T1 / T2 * 100% = [(40-y)/(1.25x)] / (40/x) * 100% = [(40-y)/(1.25x)][x / 40] * 100% = [(100)(40-y)]/[(40)(1.25)]

Our rephrased question is "what is y?"

Clearly (2) alone is clearly sufficient, so (B) is the answer. Hope that helps.
_________________

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x miles per hour for the last 40 - y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she had traveled at an average speed of x miles per hour for the entire trip?

(1) x = 48. (2) y = 20.

We are given that during a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and at an average speed of 1.25x miles per hour for the last 40 - y miles of the trip. We are asked to determine what percentage the travel time for the 40-mile trip was of the time it would have taken her, had she traveled the entire trip at an average speed of x miles per hour.

Since time = distance/rate, the time it would have taken her to travel the entire trip at an average speed of x miles per hour is 40/x, and the actual time it took her to travel the entire trip at an average speed of x miles per hour for the first y miles of the trip and at an average speed of 1.25x miles per hour for the last 40 - y miles of the trip is y/x + (40 – y)/(1.25x). So the question becomes:

y/x + (40 – y)/(1.25x) is what percent of 40/x ?

[y/x + (40 – y)/(1.25x)]/(40/x) * 100 = ?

[y/x + (40 – y)/(1.25x)]/(40/x) * 100 = ?

[y/x + (40 – y)/(1.25x)] * (x/40) * 100 = ?

[y/40 + (40 – y)/50] * 100 = ?

2.5y + 2(40 – y) = ?

Thus, knowing the value of y will allow us to determine the above percentage.

Statement One Alone:

x = 48

Since we do not have any information about y, statement one alone is not sufficient. We can eliminate answer choices A and D.

Statement Two Alone:

y = 20

Since we have a value for y, statement two alone is sufficient. For practice, we can calculate the percentage as follows:

2.5(20) + 2(40 – 20) = 50 + 40 = 90

Answer: B
_________________

Jeffery Miller Head of GMAT Instruction

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...