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:

How much time did it take a certain car to travel 400 [#permalink]

Show Tags

06 Sep 2010, 12:09

1

This post received KUDOS

8

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

67% (02:03) correct
33% (01:16) wrong based on 475 sessions

HideShow timer Statistics

How much time did it take a certain car to travel 400 kilometers?

(1) The car traveled the first 200 kilometers in 2.5 hours. (2) If the car's average speed had been 20 kilometers per hour greater than it was, it would have traveled the 400 kilometers in 1 hour less time than it did.

(1) The car traveled the first 200 km in 2.5 hours.

(2) If the car's average speed had been 20km per hour greater than it was, it would have traveled the 400km in 1 hour less than it did.

I know that statement 2 is sufficient, but I cannot figure out why. I am getting a difficult quadratic equation when I try to put it into formulas. Obviously statement 1 is insufficient. Please help me understand this logic/algebra. Thanks!

Jeremiah

Let the average speed of the car be \(s\) km/h and the time it spent to cover 400 km be \(t\) hours, then \(st=400\) (\(s=\frac{400}{t}\)). Question: \(t=?\)

(1) The car traveled the first 200 km in 2.5 hours --> we don't know the time the car needed to travel the second 200 km, so this statement is clearly insufficient.

(2) If the car's average speed had been 20km per hour greater than it was, it would have traveled the 400km in 1 hour less than it did --> \((s+20)(t-1)=400\) --> \(st-s+20t-20=400\) --> as \(st=400\) and \(s=\frac{400}{t}\) --> \(400-\frac{400}{t}+20t-20=400\) --> 400 cancels out and after simplification we'll get: \(20t^2-20t-400=0\) --> \(t^2-t-20=0\) --> \((t-5)(t+4)=0\) --> \(t=5\) or \(t=-4\) (not a valid solution as time can not be negative), so \(t=5\). Sufficient.

Say it traveled initially with 80km per hour. So it covered 400 km in 5 hours. Now adding 20km per hour . i.e 100 km per hour, it would have covered 400 km in 4 hours . Which is 1 hour less as per statement 2.

Say it traveled initially with 80km per hour. So it covered 400 km in 5 hours. Now adding 20km per hour . i.e 100 km per hour, it would have covered 400 km in 4 hours . Which is 1 hour less as per statement 2.

Makes life easy when u plugin numbers.

totally agree. while algebra is nice sometimes number plugin is easier to understand _________________

Do we always have to solve the quadratic in such cases? You have 1 equation and you will obviously get 1 right answer as these are real speeds and we know for sure a solution exists. Given that this is a DS problem, is it safe to arrive at the equation and move on?

Re: Data Sufficiency - Rate & Time problem [#permalink]

Show Tags

03 Dec 2010, 19:02

4

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

Yellow22 wrote:

How much time did it take a certain car to travel 400 km?

1) The car travelled the first 200 KM in 2.5 hrs? 2) If the Car's average speed was 20 KM per hour greater than it was, it would have travelled the 400 KM in 1 hour less time than it did.

I guess the solution is already clear to you.

Just for intellectual purposes, look at an alternative method:

If the car actually took t hours to cover 400 km, in the last 1 hour, the car travels a distance which is equal to 20*(t - 1) km This must be the distance it covered in each hour since we are considering average speed. 400 = 20*(t - 1)*t t(t - 1) = 20 t = 5 hrs _________________

Re: Data Sufficiency - Rate & Time problem [#permalink]

Show Tags

19 Jun 2011, 07:40

1

This post was BOOKMARKED

VeritasPrepKarishma wrote:

Yellow22 wrote:

How much time did it take a certain car to travel 400 km?

1) The car travelled the first 200 KM in 2.5 hrs? 2) If the Car's average speed was 20 KM per hour greater than it was, it would have travelled the 400 KM in 1 hour less time than it did.

I guess the solution is already clear to you.

Just for intellectual purposes, look at an alternative method:

If the car actually took t hours to cover 400 km, in the last 1 hour, the car travels a distance which is equal to 20*(t - 1) km This must be the distance it covered in each hour since we are considering average speed. 400 = 20*(t - 1)*t t(t - 1) = 20 t = 5 hrs

Nice one. Could't understand completely at first glance until I broke the equation down :

( Let original Avg Speed = X km/hr, Time taken = t hr., Distance = 400 km)

From statement 2,

=> (X+20) km/hr * (t-1) hr = 400 km => [ X km/hr * (t-1)hr ]+ [20 km/hr * (t-1)hr] = 400 km => [ X km/hr * (t-1)hr ]+ [20 *(t-1) km/hr * 1hr] = 400 km ---> Equation (1)

From the original problem statement, the car travels 400 km at X km/hr in t hrs

So if the car travels at X km/ hr , the distance covered in the first (t-1hrs) is given by [ X km/hr * (t-1)hr ] and the distance covered in the last 1 hr is given by [20 *(t-1) km/hr * 1hr]

Hence speed in the last 1 hr = [20 *(t-1)] km/hr ( and this would be the avg speed for the entire distance of 400km)

Do we always have to solve the quadratic in such cases? You have 1 equation and you will obviously get 1 right answer as these are real speeds and we know for sure a solution exists. Given that this is a DS problem, is it safe to arrive at the equation and move on?

I guess something i've noticed in this kind of questions is the following. Once you have that a increase in rate will obviously lead to a decrease in time, say in the form (r+x)(t-y), where r and t are rate and time respectively, and asuming you know the value for rt as well, or the distance, then you know for sure you are going to have a quadratic equation with two roots one positive and one negative. Therefore, I usually stop solving right there and move on.

Would like to here the opinion of the Math experts in regards to this. It could save us a nice 40 seconds at least if it is usually correct. I know we must be careful though cause there might be some exceptions on problems where this might not work, although I haven't seen one where it has not worked yet.

Re: How much time did it take a certain car to travel 400 [#permalink]

Show Tags

31 Dec 2013, 07:59

1

This post received KUDOS

jjewkes wrote:

How much time did it take a certain car to travel 400 kilometers?

(1) The car traveled the first 200 kilometers in 2.5 hours. (2) If the car's average speed had been 20 kilometers per hour greater than it was, it would have traveled the 400 kilometers in 1 hour less time than it did.

Statement 1

Clearly insuff

Statement 2

Now here's a trick, when I get something like this (r+20)(t-1) = 400, where rt=400

Then I know two things:

First, I will be able to eliminate the first 'rt' with the 400 at the other side and I will also multiply all the expression again by 'r' to get 400 again

Second, I will end up with a quadratic with negative sign and two different values that will give me a positive solution and a negative solution

Since time can only be positive then I know that this statement is going to be sufficient without even solving

With not much more to add, this answer is a clear B

Re: How much time did it take a certain car to travel 400 [#permalink]

Show Tags

12 Jun 2015, 11:51

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: How much time did it take a certain car to travel 400 [#permalink]

Show Tags

02 Nov 2015, 21:26

Expert's post

TooLong150 wrote:

I'm surprised that no one realized that we (2) allows us to solve for both s and t, since we now have a system of two equations with two unknowns.

Note that it is not necessary that 2 equations in 2 variables will give you a unique solution. The lines depicted by the equations might be parallel or the same line. Similarly, it is not necessary that a quadratic will give two solutions - it might give a unique solution.

Hence, these situations warrant further inspection if you go the algebra way.

Re: How much time did it take a certain car to travel 400 [#permalink]

Show Tags

08 Nov 2015, 05:57

Expert's post

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

How much time did it take a certain car to travel 400 kilometers?

(1) The car traveled the first 200 kilometers in 2.5 hours. (2) If the car's average speed had been 20 kilometers per hour greater than it was, it would have traveled the 400 kilometers in 1 hour less time than it did.

There are 2 variables (v=velocity, t=time), one equation(vt=400) and 2 further equations from the 2 conditions, so there is high chance (D) will be our answer. From condition 1, the fact that the car traveled the first 200km in 2.5hrs is not helpful; there is no explanation that the velocity is constant, so we do not know anything about the next 200km, so this is insufficient. From condition 2, (v+20)(t-1)=400, vt=400. This is sufficient, so the answer becomes (B).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E. _________________

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...

This highly influential bestseller was first published over 25 years ago. I had wanted to read this book for a long time and I finally got around to it...