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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: Difficult DS Problem- Need Help! [#permalink]  08 Sep 2010, 09:09
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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
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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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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
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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.
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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?
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Re: Data Sufficiency - Rate & Time problem [#permalink]  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)

[20 *(t-1)] km/hr * t hr = 400 km

Upon solving we get t= 5hrs.

Thanks Karishma.
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Re: Difficult DS Problem- Need Help! [#permalink]  12 Oct 2013, 10:27
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
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Re: How much time did it take a certain car to travel 400 [#permalink]  31 Dec 2013, 06:59
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
Re: How much time did it take a certain car to travel 400   [#permalink] 31 Dec 2013, 06:59
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