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

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How much time did it take a certain car to travel 400 [#permalink]  06 Sep 2010, 11:09
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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.
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Re: Difficult DS Problem- Need Help! [#permalink]  06 Sep 2010, 11:41
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jjewkes wrote:
Can you help me with this problem??

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

(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.

Hope it helps.
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Re: Data Sufficiency - Rate & Time problem [#permalink]  03 Dec 2010, 18:02
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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
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Current Student Status: What's your raashee? Joined: 12 Jun 2009 Posts: 1847 Location: United States (NC) Concentration: Strategy, Finance Schools: UNC (Kenan-Flagler) - Class of 2013 GMAT 1: 720 Q49 V39 WE: Programming (Computer Software) Followers: 22 Kudos [?]: 225 [2] , given: 52 Re: Difficult DS Problem- Need Help! [#permalink] 08 Sep 2010, 09:09 2 This post received KUDOS anticipation wrote: It can be solved in an easier way. 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 _________________ If you like my answers please +1 kudos! Current Student Joined: 06 Sep 2013 Posts: 2036 Concentration: Finance GMAT 1: 770 Q0 V Followers: 39 Kudos [?]: 432 [1] , given: 355 Re: Difficult DS Problem- Need Help! [#permalink] 12 Oct 2013, 10:27 1 This post received KUDOS hemanthp wrote: 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. Hope it helps Current Student Joined: 06 Sep 2013 Posts: 2036 Concentration: Finance GMAT 1: 770 Q0 V Followers: 39 Kudos [?]: 432 [1] , given: 355 Re: How much time did it take a certain car to travel 400 [#permalink] 31 Dec 2013, 06:59 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 Is this all clear? Cheers! J Manager Status: Last few days....Have pressed the throttle Joined: 20 Jun 2010 Posts: 71 WE 1: 6 years - Consulting Followers: 3 Kudos [?]: 38 [0], given: 27 Re: Difficult DS Problem- Need Help! [#permalink] 07 Sep 2010, 01:51 Agree ..I get +5 and -4 on solving the quadratic equation...so B is sufficient _________________ Consider giving Kudos if my post helps in some way Intern Joined: 15 Oct 2009 Posts: 12 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: Difficult DS Problem- Need Help! [#permalink] 07 Sep 2010, 21:11 It can be solved in an easier way. 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. Manager Status: Keep fighting! Affiliations: IIT Madras Joined: 31 Jul 2010 Posts: 236 WE 1: 2+ years - Programming WE 2: 3+ years - Product developement, WE 3: 2+ years - Program management Followers: 4 Kudos [?]: 309 [0], given: 104 Re: Difficult DS Problem- Need Help! [#permalink] 20 Oct 2010, 05:46 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? Intern Joined: 02 Jun 2011 Posts: 4 Followers: 0 Kudos [?]: 1 [0], given: 0 Re: Data Sufficiency - Rate & Time problem [#permalink] 19 Jun 2011, 06:40 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) [20 *(t-1)] km/hr * t hr = 400 km Upon solving we get t= 5hrs. Thanks Karishma. GMAT Club Legend Joined: 09 Sep 2013 Posts: 8160 Followers: 416 Kudos [?]: 111 [0], given: 0 Re: How much time did it take a certain car to travel 400 [#permalink] 12 Jun 2015, 10:51 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. _________________ Senior Manager Joined: 10 Mar 2013 Posts: 285 GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39 Followers: 1 Kudos [?]: 42 [0], given: 2403 Re: How much time did it take a certain car to travel 400 [#permalink] 31 Oct 2015, 18:21 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. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6216 Location: Pune, India Followers: 1674 Kudos [?]: 9587 [0], given: 196 Re: How much time did it take a certain car to travel 400 [#permalink] 02 Nov 2015, 20:26 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. Check out these two posts for more on this topic: http://www.veritasprep.com/blog/2011/06 ... -of-thumb/ http://www.veritasprep.com/blog/2011/08 ... nd-points/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: How much time did it take a certain car to travel 400 [#permalink]  08 Nov 2015, 04:57
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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.
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Re: How much time did it take a certain car to travel 400 [#permalink]  17 Dec 2015, 11:03
Statement 1: What about the remaining 200 kms. Did it take 10,00 hours or 10 hours or 2.356 hours? Insufficient.

Statement 2: Let initial speed be x and time taken be y. Distance(i.e. 400) = xy. According to information in Statement 2, (x+20)(y-1) = 400.

Eliminating laziness and solving the equations we would get xy = xy + 20y - x - 20

20y - x = 20

Hence B
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Re: How much time did it take a certain car to travel 400   [#permalink] 17 Dec 2015, 11:03
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