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

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?

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.

Hey..

The answer is B.

It is given that the distance to be covered is 400kms in the question. Taking speed as s km/hr and time as t hrs, we have the eqn

400 = st ----> (1)

In B, it i given as the same distance would be covered by less than an hour if the car had traveled in a speed greater than 20km/hr that it was.

so this can be equated as,

400 = (s+20)(t-1) ---> (2)

Equating (1) and (2) we have,

st = (s+20)(t-1)
st = st + 20t - s - 20
0 = 20t - s - 20

Again substitute the value of s from eqn (1)
0 = 20t - 400/t - 20
divide it by 20
0 = t -20/t - 1

Multiply by t throughout,
0 = t^2 - 20 - t
so,

t^2 - t - 20 = 0
Solving for t, we get
t=5 , -4
T=-4 is not possible,
hence t = 5 hrs

Hope it helps.

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

400/x - 1 = 400/(x+20)
Simplifying, we get,
400x = 400x - x^2 + 8000 -20x
i.e. x^2 + 20x - 8000 = 0

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!

Using statement 2, if v is the speed of the car then:
400/v = 400/(v+20) + 1

Solve this to get v and then compute 400/v. Sufficient. So (B).

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

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

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:

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/06 ... -of-thumb/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/08 ... nd-points/

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.

Had a question. Should the question not have mentioned that the car is not travelling at a constant speed?

The reason is that if the car is travelling at a constant speed, then even (1) is sufficient to arrive at the answer.

Or on GMAT, are we just supposed to assume that unless otherwise mentioned, we cannot assume constant speed.

We cannot assume constant speed unless otherwise mentioned. The car's average speed while covering the first 200 km could be different from its average speed while covering the second half of the distance.

This is simply beautiful VeritasKarishma. You got my kudos for this reasoning. If it were possible, I would have given you two.

Kind Regards,
Fabio.

\(400\,{\rm{km}}\,\,\,\,\,\left\{ \matrix{\\
\,\,\left( {{\rm{real}}\,\,{\rm{speed}}\,,\,\,{\rm{real}}\,\,{\rm{time}}} \right)\,\,\,{\rm{ = }}\,\,\,\left( {{V_R}\,\,,\,\,{T_R}} \right) \hfill \cr \\
\,\,\left( {{\rm{hypothetical}}\,\,{\rm{speed}}\,,\,\,{\rm{hypothetical}}\,\,{\rm{time}}} \right)\,\,\,{\rm{ = }}\,\,\,\left( {{V_H}\,\,,\,\,{T_H}} \right) \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\,\,\,\left[ {\,{{{\rm{km}}} \over {\rm{h}}}\,} \right]\,\,\,,\,\,\,\,\left[ {\,{\rm{h}}\,} \right]\,\,\,} \right)\)

\(? = {T_R}\,\,\,\left[ {\rm{h}} \right]\)


\(\left( 1 \right)\,\,\left\{ \matrix{\\
\,{\rm{If}}\,\,{\rm{it}}\,\,{\rm{took}}\,\,0.5\,{\rm{h}}\,\,{\rm{in}}\,\,{\rm{the}}\,\,{\rm{last}}\,\,200\,{\rm{km}}\,\,\,\,\, \Rightarrow \,\,\,\,?\,\, = \,\,3\, \hfill \cr \\
\,{\rm{If}}\,\,{\rm{it}}\,\,{\rm{took}}\,\,1\,{\rm{h}}\,\,{\rm{in}}\,\,{\rm{the}}\,\,{\rm{last}}\,\,200\,{\rm{km}}\,\,\,\,\, \Rightarrow \,\,\,\,?\,\, = \,\,3.5\,\, \hfill \cr} \right.\)


\(\left( 2 \right)\,\,{V_H} = {V_R} + 20\,\,\,\left( * \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{T_R} - {T_H} = 1\,\,\,\,\,\,\left[ {\rm{h}} \right]\)

Now it´s time for UNITS CONTROL, one of the most powerful tools of our method:

\({{{\rm{km}}} \over {\,\,\,{{{\rm{km}}} \over {\rm{h}}}\,\,}} = {\rm{h}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\{ \matrix{\\
{T_R} = {{400} \over {{V_R}}} \hfill \cr \\
{T_H}\,\mathop = \limits^{\left( * \right)} \,\,{{400} \over {{V_R} + 20}} \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,{{400} \over {{V_R}}} - {{400} \over {{V_R} + 20}} = 1\)

\({{400\left( {{V_R} + 20} \right)} \over {{V_R}\,\,\left( {{V_R} + 20} \right)}} - {{400\,\,{V_R}} \over {\left( {{V_R} + 20} \right)\,\,{V_R}}} = 1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,400 \cdot 20 = {V_R}\,\,\left( {{V_R} + 20} \right)\)

\({V_R}^2 + 20{V_R} - 400 \cdot 20 = 0\,\,\,\,\,\,\mathop \Rightarrow \limits_{{\rm{of}}\,\,{\rm{roots}}}^{{\rm{product}}} \,\,\,\,\,\,\,\left( {{c \over a} = } \right)\,\,\,{{ - 400 \cdot 20} \over 1} < 0\,\,\,\, \Rightarrow \,\,\,\,\,{V_R}\, > 0\,\,\,\,{\rm{unique}}\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\)

\(\left( {\,\,\,\left. \matrix{\\
{V_R}\,\,{\rm{unique}}\,\, \hfill \cr \\
{\rm{400}}\,{\rm{km}} \hfill \cr} \right\}\,\,\,\, \Rightarrow \,\,\,\,\,? = \,\,{T_R}\,\,\,{\rm{unique}}\,\,\,} \right)\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Target question: How long did it take to travel 400km
To find the travel time, we need to know the average speed traveled.
Let x = the average speed traveled.

REPHRASED target question: What is the value of x?

Statement 1: The car traveled the first 200 km in 2.5 hrs
No info about the 2nd half of the trip, so we can't determine the overall average speed (aka x).
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: If the car’s average speed had been 20 km/h faster, it would have traveled the 400 km in 1 hour less time.
Let's start with a word equation:
(travel time at x km/h) - 1 = (travel time at x+20 km/h)
Since time = distance/speed, we can now write:
(400/x) - 1 = 400/(x+20)

IMPORTANT: At this point, we need only determine whether this equation will yield 1 or 2 valid values of x. If it yields only 1 valid value, then statement 2 is sufficient. If it yields 2 valid values, then statement 2 is not sufficient.

Rewrite as: (400-x)/x = 400/(x+20)
Cross multiply: (400)(x) = (400-x)(x+20)
Simplify: 400x = 8000 + 380x - x²
Rewrite: x² + 20x - 8000 = 0
Factor: (x + 100)(x - 80) = 0
So, x = -100 or 80
Since x cannot be negative, it must be the case that x = 80
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent

Hi VeritasKarishma
The solution appears to be way shorter, but I don't understand a certain aspect.

"in the last 1 hour, the car travels a distance which is equal to 20*(t - 1) km"
THis is the part I am confused about. The speed has been increased by 20. SO the time taken has been reduced by 1 hour. So the distance that is covered from this increase in speed is:20(t-1).
This is the total extra distance. so then what do you mean when you say:
1." IN the last 1 hour"
2.400 = 20*(t - 1)*t (why are we then multiplying by t again)

Regards

Current case: Car travels 400 km in t hrs

Hypothetical case: Car travels 20 km extra for every (t - 1) hrs and covers the distance in (t - 1) hrs. Why did the car take 1 hr less here? Because it travelled some extra distance for (t - 1) hrs. What was that extra distance? 20*(t - 1) km. This is the distance it travels in its current case in the last hour (because the current case speed is lower).

If it travels 20*(t - 1) km in its last hour in our current case, it travels the same 20*(t - 1) in each previous hour too because average speed is considered.

So speed of the car = 20*(t - 1) km/hr (in current case)

20*(t - 1)*t = 400 (speed * time = distance in current case)