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

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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.

(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?

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]

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19 Jun 2011, 06:40

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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)

Re: Car and 400 Kms Distance - Ivy 26 [#permalink]

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13 Nov 2011, 02:21

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The only information given in the question is that the total distance travelled by the car is 400kms. We have to find the time taken by the car to do so.

Now, let's take a look at the given options.

Statement 1 says that the car traveled the first 200 kilometers in 2.5 hours. This does not give us any information about the remaining 200 km that the car travelled. Hence this is clearly Insufficient.

Statement 2 says that 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 hours less time than it did. This seems to be quite a lengthy statement, so let's break this down into simpler sentences to make the given information clear.

Let's assume that the car's average speed for the 400 km travel was 'x' km per hour. Also, let the time taken to travel be 't' hours. So, we have t = 400/x -----> (A)

Statement 2 says that if the car's average speed had been 20 km per hour greater than it was (earlier). This indicates that the new speed is 'x+20' km per hour The car would have travelled 400 km in 1 hour less time than it did (earlier). This indicates that the time taken later (in the new scenario) is 't-1' hours We know that the car still travelled 400 km, so the new equation can be written as t-1 = 400/(x+20) -----> (B)

But we know from 'A' that t = 400/x Substituting this value in the equation 'B', we get

Solving the above quadratic equation yields the solutions x = 80 OR x = -100 As the speed of the car cannot be negative, the required speed is 80 km per hour

Using this, we can determine the time taken by the car to travel 400 km. This comes to 5 hours. Hence Statement 2 alone is sufficient to answer the question.

Hope this helps.

Cheers!
_________________

MBA Candidate 2015 | Georgetown University McDonough School of Business

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]

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31 Dec 2013, 06:59

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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]

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12 Jun 2015, 10:51

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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.

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.
_________________

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

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01 Aug 2016, 09:39

jlgdr wrote:

jjewkes wrote:

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

J

I have one questions to your approach. How do you know that there won't be two positive solutions? If I only think of quadratic equations with negative sign there also can be two positive solutions (y=x^2-3x+2).

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

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01 Aug 2016, 09:59

22gmat wrote:

jlgdr wrote:

jjewkes wrote:

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

J

I have one questions to your approach. How do you know that there won't be two positive solutions? If I only think of quadratic equations with negative sign there also can be two positive solutions (y=x^2-3x+2).

Thank you for your help

The equation above is of the form (x-a)(x+b) which will always give one positive and one negative solution.
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