Train A traveling at 60 m/hr leaves New York for Dallas at 6

Thank you Sir. This was tough but relative speed as I can see, often, is the key

Train A and Train B meets somewhere between NY and Dallas. This means the miles covered by Train A and Train B were equal during that point.
Also, we know that Train A travelled for 3 hours before Train B started.

Distance of Train A + Distance of Train A before Train B starts = Distance of Train B

\(60t + 60(3) = 90t\)
\(t = 6 hrs\)

Train B has covered 90 x 6 = 540 miles when it was met by Train C (which covered the rest of the tracks between Dallas and NY)
1260 - 540 = 720 miles

Train C also started the same time as B.
\(R = 720/6 = 120 mph\)

AnsweR: C

..
(A) s=60t
(B) s = 90(t-3)
(C) 1260-s=V(t-3)
from these t=9, s=540 and velocity of C = 120

Train A traveling at 60 m/hr leaves New York for Dallas at 6 P.M. Train B traveling at 90 m/hr also leaves New York for Dallas at 9 P.M. Train C leaves Dallas for New York at 9 P.M. If all three trains meet at the same time between New York and Dallas, what is the speed of Train C if the distance between Dallas and New York is 1260 miles?

All three trains meet at the same time. If we can find how many miles it takes for the faster train B to catch up to the slower train A, we can figure out at what time all three trains meet.

Train B gains 30miles per hour on train B every hour it travels (90mi/hr - 60 mi/hr = 30 mi/hr) Therefore, if B started when A was 180 mi ahead, it will need 180/30 = 6 hours to catch up to A.

In the 6 hours it takes for train B to catch up to A, it travels 6+90 = 540 miles. Therefore, when A,B and C meet C has traveled 1260-540 = 720 miles. If train C left at 9PM and it took 6 hours for A to catch up to B (and meet C) then the speed of C is d/t ===> 720/6 = 120 mi/hour.

ANSWER: C. 120 m/hr

In 3 hours A travels 60*3=180km
Train B needs 6 hours to catch A 180/(90-60)=6hours

Trains b and c will meet (90+x)*6=1260.
X=120

Got it

First since A starts 3 hours before it travels 180 miles in that time span

Now, B needs to catch up first so it will need (30)(t) = 180 ---> 6 hours to do so

In 6 hours, B will travel 540 miles

Now, C needs to meet them at the same time

So in 6 hours it will need to cover 1260 - 540 = 720 miles

720 / 6= 120 miles/hour

Hence C is the answer

Provide kudos if you think that this post was helpful

Cheers!
J

I understand how the problem is solved but my doubt is: why do they meet at that determinate point with C (720 miles). Why not later? there is no equation that relates the first two trains with the C.

Thank you in advance,

It is given to you in the question, we are not assuming it: "If all three trains meet at the same time between New York and Dallas,"
This is important data that the question stem gives us. It tells us that their speeds are related in a way that all three trains meet together - just when B crosses A, C also passes A and B.

A a question for all,

Do RTD PS questions such as this often take you well over 2 minutes? I find it often takes me a minute to read the question - then re-read to write down all the pertinent information. I often need another 2 minutes to complete all of the calculations. Just wondering if it's just me, or if others are in the same boat.

Well I can only speak for myself but short answer is no, it doesn't at least its a very complex RTD question such as the one with roses for example. But like everything its just practice and its good that you read carefully cause thats the most important thing in RTD.

Cheers
J

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Can you please explain : To catch up A (to meet A), B will need 180/30=6 hours

Indeed, relative speed is the best way. I tried it without using it; arrived at the answer, but it's obviously not as elegant.

Assuming trains A, B and C travel for a, (a-3) and (a-3) hours.

Assume speed of C as x m/hr

So, 60a + x*(a-3) = 1260
Also, 90(a-3) + x*(a-3) = 1260

x*(a-3) = 1260 - 60a

90(a-3) + (1260 - 60a) = 1260

30a=270
a = 9

So, they are travelling for 9, 6 and 6 hours.

x*(a-3) = 1260 - 60a
Substituting a = 9

x*(9-3) = 1260 - 60*9
6x = 720
-> x = 120

Train A has already taken 3hrs when train B started , lets make a equation of distance .

\(180 + 60*T_A = 90*T_B\)

\(6 + 2*T_A = 3*T_B\)

since \(T_A = T_B\)
so , \(T_A=T_B=6hrs.\)
Distance travelled by B in 6hrs = 90*6 = 540
Net time taken by A = 6+3= 9 hrs. (just for information)

Let C be speed of train C , then

\(\frac{1260-90*6}{C} = 6\)

C= 120 Answer.

A gets a head start of 60*3= 180 miles.
B closes in on A with 30 miles/hour.
He will catch up with A in 180/30=6 hours.
At that time A has travelled 60*(6+3)=540 miles.
This means that C has travelled 1260-540=720 miles in 6 hours. 720/60=120.
Hence, C should be correct.

Hi Anu26,

The question uses the concept of relative speed. Let me explain it you its working.

Refer the following diagram for this question:



The diagram shows the relative positioning of trains at 9 PM. Trains B & C are starting from NY & Dallas respectively while train A is at a distance of 180 miles from NY ( as train A started from NY at a speed of 60 miles/hr 3 hours before at 6 PM. SO distance traveled by train A = 60 * 3 = 180 miles)

Since all the trains meet at the same point, it would be enough if we found out meeting point of trains A & B. Train A is 180 miles ahead of train B at 9 PM and the difference between the speeds of trains A & B is 30 miles/hr (Train B travels at a speed of 90 miles/hr and train A travels at a speed of 60 miles/hr).

Using the formula for Distance = Speed * Time, we can say that 180 = 30 * t i.e. t= 6 hours. This means that train B and A will meet 6 hours after 9 PM. In other words, train B will catch up with train A after 6 hours.

So, distance travelled by train B in 6 hours = 90 * 6 = 540 miles. Thus train C will need to travel 1260 - 540 = 720 miles in 6 hours. So, speed of train C would be \(\frac{720}{6} = 120\) miles/hour

In Distance & Speed question which involves concept of relative speed, it is always helpful to draw a diagram and visualize the solution.

Hope its clear!

Regards
Harsh

I was wondering if my method would also be correct.

Speed of A: 60mph
Speed of B: 90mph
Speed of C: C

A started at 6 PM but B and C started at 9 PM. So by the time B and C started, A already traveled 180 miles.

Now the distance between A and C is 1260 - 180 = 1080 miles and the distance between B and C is 1260 miles.

We are told that they all meet together at the same time meaning A and B will be at the same point when C meets them.

Time it takes for A and C to meet = Time it takes for B and C to meet = Distance between two trained/combined speed

(Distance between A & C)/Combined speed of A and C = (Distance between B & C)/Combined Speed of B and C

--> 1080/(60+C) = 1260/(90+C)
--> 1080(90+C) = 1260(60+C)
--> 6(90+C)=7(60+C)
--> 540 + 6C = 420 + 7C
--> C=120mph

Kudos if this helped you!

We can calculate the solution in this manner too. Lets say Train A starts at 6pm at 60miles/hr and train B at 90miles/hr.
Now by 9 pm train A would have travelled 180 miles.
lets say train A and train B meet after x hours after 9 pm hence if we equate the distance traveled by two train in X hours after 9 PM it will be:
60(x+3) = 90x (x+3 for train because of time from 6pm to 9 pm)
solving above equation we get x = 6 hours
So the distance traveled in 6 hours by train B is 90 * 6 = 540 miles/hour
Question says train C also meets train A and B at same time. So distance traveled by train C is 1260 - 540 = 720 miles.
Again we are told train C started at 9 pm. So hours traveled by train C before meeting train A and B is 6 hours.
So speed of train C = 720/6 = 120 miles/hour


Please give kudos if you like the explaination