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:

Liam is pulled over for speeding just as he is arriving at work.He [#permalink]

Show Tags

21 Jun 2010, 11:13

2

This post received KUDOS

18

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

56% (02:43) correct
44% (03:38) wrong based on 163 sessions

HideShow timer Statistics

Liam is pulled over for speeding just as he is arriving at work.He explains to the police officer that he could not afford to be late today, and has arrived at work only four minutes before he is to start. The officer explains that if Liam had driven 5mph slower for his whole commute, he would have arrived at work exactly on time. If Liam's commute is 30 miles long,how fast was he actually driving?(Assume that Liam drove at a constant speed for the duration of his commute.)

Of the many ways to solve this problem, two are as follows:

(Method 1) Assume the actual speed of Liam to be r. Distance travelled is 30 miles. So, time taken is 30/r. In the hypothetical case, speed of Liam is (r-5). Distance remains the same. So, time taken is (30/r)+(1/15) because 4 minutes is 1/15th hour. So, translating these values into equations, the hypothetical scenario becomes:

(Method 2) This is the method used in the Manhattan guide. Speed in the actual case is considered to be (r+5). Time taken is therefore 30/(r+5). Speed in the hypothetical case is considered to be r. Time taken is 30/r. Because we know time taken in the hypothetical scenario is 4 minutes more, 30/r = [(30/(r+5))+(1/15)] => 30/r = [((450+r+5)/(15r+75)] => 30(15r+75) = r(455+r) => r^2 +5r-2250 = 0 => (r+50)(r-45) = 0 => r=45.

Can anyone please explain to me why both these methods DON'T yield the same answer? Isn't the first method more appropriate because the hypothetical scenario is the one in which we should assume the speed to be 5mph less than the actual and time taken is 4 minutes more than the actual?

Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink]

Show Tags

21 Jun 2010, 11:35

1

This post received KUDOS

Phoenix9 wrote:

Hi All,

I have a question regarding a problem from Manhattan Strategy Guide: Word Translations (3). Chapter 2. Rates. Page 37.

Problem: Liam is pulled over for speeding just as he is arriving at work.He explains to the police officer that he could not afford to be late today, and has arrived at work only four minutes before he is to start. The officer explains that if Liam had driven 5mph slower for his whole commute, he would have arrived at work exactly on time. If Liam's commute is 30 miles long,how fast was he actually driving?(Assume that Liam drove at a constant speed for the duration of his commute.)

Solution: Of the many ways to solve this problem, two are as follows:

(Method 1) Assume the actual speed of Liam to be r. Distance travelled is 30 miles. So, time taken is 30/r. In the hypothetical case, speed of Liam is (r-5). Distance remains the same. So, time taken is (30/r)+(1/15) because 4 minutes is 1/15th hour. So, translating these values into equations, the hypothetical scenario becomes:

(Method 2) This is the method used in the Manhattan guide. Speed in the actual case is considered to be (r+5). Time taken is therefore 30/(r+5). Speed in the hypothetical case is considered to be r. Time taken is 30/r. Because we know time taken in the hypothetical scenario is 4 minutes more, 30/r = [(30/(r+5))+(1/15)] => 30/r = [((450+r+5)/(15r+75)] => 30(15r+75) = r(455+r) => r^2 +5r-2250 = 0 => (r+50)(r-45) = 0 => r=45.

Can anyone please explain to me why both these methods DON'T yield the same answer? Isn't the first method more appropriate because the hypothetical scenario is the one in which we should assume the speed to be 5mph less than the actual and time taken is 4 minutes more than the actual?

Thanks.

Check the blue quoted. Actual speed = r+5 where r = 45 => r+5 = 50 same answer.
_________________

Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink]

Show Tags

01 Apr 2011, 10:22

Liam is pulled over for speeding just as he is arriving at work. He explains that he could not afford to be late today, and has arrived at work only 5 minutes before he is to start. The officer explains that if he had driven 5mph slower for his whole commute, he would have arrived on time. If his commute is 30 miles, how fast was he actually driving?

What is wrong with the way I'm trying to solve this problem?

Actual: Speed = s time = t - 1/12

Hypothetical: Speed = s-5 time = t

Since distances are equal, equate the two. I cannot seem to get the correct answer...Please help.

Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink]

Show Tags

28 Jun 2011, 11:26

Liam is pulled over for speeding just as he is arriving at work. He explains to the police officer that he could not afford to be late today, and has arrived at work only four minutes before he is to start. The officer explains that if Liam had driven 5 mph slower for his whole commute, he would have arrived at work exactly on time. If Liam's commute is 30 miles long, how fast was he actually driving? (Assume that Liam drove at a constant speed for the duration of his commute.) A. 50 mph B. 45 mph C. 48 mph D. 52 mph E. 60 mph

Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink]

Show Tags

28 Jun 2011, 22:13

Shalom vrk002, Shalom! I like your method the best because of its simplicity. However, could you post the step whereby you solve both equations? The only way I see that I could get the answer by using your method is plugging in all the possible answers until I see the one that is the solution.

Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink]

Show Tags

03 Aug 2011, 08:00

plugging the numbers would be much faster, sometimes, especially if you can't intuitively guess that 2250 = 45*50

Just solve to 30/s + 1/15 = 30(s-5) and plug in the options. Its much much faster. In this case you have to be careful for whether u use s-5 or s+5. The latter can't be used cause you're searching for the faster speed.

Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink]

Show Tags

16 Nov 2014, 22:49

Phoenix9 wrote:

Hi All,

I have a question regarding a problem from Manhattan Strategy Guide: Word Translations (3). Chapter 2. Rates. Page 37.

Problem: Liam is pulled over for speeding just as he is arriving at work.He explains to the police officer that he could not afford to be late today, and has arrived at work only four minutes before he is to start. The officer explains that if Liam had driven 5mph slower for his whole commute, he would have arrived at work exactly on time. If Liam's commute is 30 miles long,how fast was he actually driving?(Assume that Liam drove at a constant speed for the duration of his commute.)

Solution: Of the many ways to solve this problem, two are as follows:

(Method 1) Assume the actual speed of Liam to be r. Distance travelled is 30 miles. So, time taken is 30/r. In the hypothetical case, speed of Liam is (r-5). Distance remains the same. So, time taken is (30/r)+(1/15) because 4 minutes is 1/15th hour. So, translating these values into equations, the hypothetical scenario becomes:

(Method 2) This is the method used in the Manhattan guide. Speed in the actual case is considered to be (r+5). Time taken is therefore 30/(r+5). Speed in the hypothetical case is considered to be r. Time taken is 30/r. Because we know time taken in the hypothetical scenario is 4 minutes more, 30/r = [(30/(r+5))+(1/15)] => 30/r = [((450+r+5)/(15r+75)] => 30(15r+75) = r(455+r) => r^2 +5r-2250 = 0 => (r+50)(r-45) = 0 => r=45.

Can anyone please explain to me why both these methods DON'T yield the same answer? Isn't the first method more appropriate because the hypothetical scenario is the one in which we should assume the speed to be 5mph less than the actual and time taken is 4 minutes more than the actual?

Thanks.

Hi all,

I'm reviewing rates & work and even though I feel I pretty much got it figured out, questions like this one let me doubt myself.

My question for this one is: Phoenix9 introduced 2 approaches to solve the question that use the information given differently, in approach 1 the fact that Liam should go 5mph slower is marked as (r-5) in the hypothetical case. In approach 2 it's (r+5) in the actual case. No questions until here.

I, however, tried like this:

Real case: R: r+5 ; T: (30/r - 1/15); Hypothetical case: R: R ; t: (30/r)

The equation than comes to two negative values for r, which is unsolvable.

My question is: what's the mistake in my approach?

Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink]

Show Tags

25 Aug 2016, 02:09

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.
_________________

Concentration: General Management, Entrepreneurship

GPA: 3.8

WE: Engineering (Energy and Utilities)

Re: Liam is pulled over for speeding just as he is arriving at work.He [#permalink]

Show Tags

02 Aug 2017, 02:35

Phoenix9 wrote:

Liam is pulled over for speeding just as he is arriving at work.He explains to the police officer that he could not afford to be late today, and has arrived at work only four minutes before he is to start. The officer explains that if Liam had driven 5mph slower for his whole commute, he would have arrived at work exactly on time. If Liam's commute is 30 miles long,how fast was he actually driving?(Assume that Liam drove at a constant speed for the duration of his commute.)

Of the many ways to solve this problem, two are as follows:

(Method 1) Assume the actual speed of Liam to be r. Distance travelled is 30 miles. So, time taken is 30/r. In the hypothetical case, speed of Liam is (r-5). Distance remains the same. So, time taken is (30/r)+(1/15) because 4 minutes is 1/15th hour. So, translating these values into equations, the hypothetical scenario becomes:

(Method 2) This is the method used in the Manhattan guide. Speed in the actual case is considered to be (r+5). Time taken is therefore 30/(r+5). Speed in the hypothetical case is considered to be r. Time taken is 30/r. Because we know time taken in the hypothetical scenario is 4 minutes more, 30/r = [(30/(r+5))+(1/15)] => 30/r = [((450+r+5)/(15r+75)] => 30(15r+75) = r(455+r) => r^2 +5r-2250 = 0 => (r+50)(r-45) = 0 => r=45.

Can anyone please explain to me why both these methods DON'T yield the same answer? Isn't the first method more appropriate because the hypothetical scenario is the one in which we should assume the speed to be 5mph less than the actual and time taken is 4 minutes more than the actual?

Thanks.

Let the speed at which Liam drove at office actually be x mph So, the speed which traffic police suggested = x-5

Distance between his home and office = 30 miles

So, difference of time with these 2 speeds = 4 minutes

So, 30/(x-5) - 30/x = 4/60 -> 30*5/x(x-5) = 1/15 -> x(x-5) = 30*5*15 -> x^2 - 5x - 30*5*15 = 0 -> x^2 - 50x + 45x - 30*5*15 = 0 -> (x -50)(x+45) = 0 -> x = 50 (as speed will be +ve) -> So, he was actually driving at a speed of 50 miles per hour