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# It takes the high-speed train x hours to travel the z miles

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Re: Manhattan CAT math question [#permalink]

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29 Mar 2013, 02:01
hitman5532 wrote:
Bunuel wrote:
Hi,

Time in which they meet is $$time=\frac{distance}{combined-rate}=\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{xy}{x+y}$$.

I must be making a dumb algebra error somewhere in this step. How did you convert the fraction like this? I keep ending up with somehting slightly different

$$\frac{z}{\frac{z}{x}+\frac{z}{y}}=\frac{z}{\frac{zy+zx}{xy}}=\frac{xyz}{zy+zx}=\frac{xy}{x+y}$$.
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Re: Manhattan CAT math question [#permalink]

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29 Mar 2013, 05:04
Hey Bunuel so something I sort of missed was why we combine the rate (understood this) but then to figure out the time the trains cross each other to simply divide the distance z by the combined rates?

The distance z is the total distance for A to B. Which is different for what we're looking for, no? Aren't we looking for the time (and hence corresponding) distance where A and B cross each other?[/quote]

Two trains are traveling to meet each other.

Distance = 100 miles;
Rate of train A = 20 miles per hour;
Rate of train B = 30 miles per hour.

In how many hours will they meet?

(Time) = (Distance)/(Combined rate) = 100/(20+30) = 2 hours.

Does this make sense?[/quote]

Combining the rates makes sense to me, they are both moving to each other relatively

Here's what doesn't make sense to me.

The distance between the two starting points of the train is 100 miles (z). If we are trying to find what time they will pass each other, that distance MUST BE less than 100 if both trains have a positive velocity.

This distance is less than the starting points of the train from 100 miles (z)?

So I don't see how we can simple plug in z here. Maybe there's a test assumption that simplifies this situation for us.
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Re: Manhattan CAT math question [#permalink]

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29 Mar 2013, 05:43
manimgoindowndown wrote:
Hey Bunuel so something I sort of missed was why we combine the rate (understood this) but then to figure out the time the trains cross each other to simply divide the distance z by the combined rates?

The distance z is the total distance for A to B. Which is different for what we're looking for, no? Aren't we looking for the time (and hence corresponding) distance where A and B cross each other?

Two trains are traveling to meet each other.

Distance = 100 miles;
Rate of train A = 20 miles per hour;
Rate of train B = 30 miles per hour.

In how many hours will they meet?

(Time) = (Distance)/(Combined rate) = 100/(20+30) = 2 hours.

Does this make sense?[/quote]

Combining the rates makes sense to me, they are both moving to each other relatively

Here's what doesn't make sense to me.

The distance between the two starting points of the train is 100 miles (z). If we are trying to find what time they will pass each other, that distance MUST BE less than 100 if both trains have a positive velocity.

This distance is less than the starting points of the train from 100 miles (z)?

So I don't see how we can simple plug in z here. Maybe there's a test assumption that simplifies this situation for us.[/quote]

"Pass" here means "meet".
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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29 Jul 2013, 18:16
speed of high speed train, Vh=z/x
speed of regular train, Vr=z/y
Let p be the distance covered by high speed train when both trains met
Time taken to meet both trains =px/z=(z-p)y/z
p=zy/(x+y)

but required is how much more distance covered by high speed train than regular train
i.e required = 2p-z = zy/(x-y) -z
=z(y-x)/x+y
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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01 Aug 2013, 16:26
It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?

We are given the rate of speed for both trains
We are given the distance both trains travel (which is the same)
We need to find distance traveled by both trains. Distance = rate*time. We need to find time, not distance, so we can multiply the time by the rate to get the distance traveled by each train.

Rate(fast) = (z/x)
Rate(slow) = (z/y)

We have the rate at which each train travels. Now, lets find the time at which they pass one another. If we know what time they pass one another and their rate, we can figure out distance.

Time = distance/combined rate
Time = z / (z/x)+(z/y)
Time = z/(z/x)(y/y) + (z/y)(x/x)
Time = z/(zy/xy) + (zx/xy)
Time = z/ zy+zx/xy
Time = (zxy)/(zy+zx)
Time = xy/y + x

Now we have the time at which they pass one another. Distance = Rate * Time. Now that we have the distance each train travels plus the time at which they pass one another (which represents the time each train has been traveling for) we can solve. When going through the problem, we don't solve for t because doing so would require that we use distance (z) which only tells us the distance between points a and b. We need to find the distance traveled by each train which adds up in total to distance z. That means we need to find the rate each train traveled at and how long it traveled for (which is when they pass one another) Remember, we aren't looking for how many miles the fast train traveled. we are looking for how many more miles it traveled than the slow train.

distance(fast) - distance(slow):
(z/x)*xy/(y + x) - (z/y)*xy/(y + x)
zy/y+x - zx/y+x
(zy-zx)/(y+x)
z(y-x)/(y+x)

(A) z(y – x)/x + y

I would love to know someone's explanation as to how they knew what steps to take to solve this problem. Though the actual algebra wasn't too bad, knowing what steps to take and when made it extremely tough!
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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01 Aug 2013, 19:43
WholeLottaLove wrote:

I would love to know someone's explanation as to how they knew what steps to take to solve this problem. Though the actual algebra wasn't too bad, knowing what steps to take and when made it extremely tough![/color]

1. First formulate what is to be found in precise terms. That is let the distance traveled by the high speed train till both the trains meet be "a". The distance traveled by the regular train is z-a. So what we need to find is a-(z-a) = 2a-z. ---(1)
2. To find "a" we need to know the speed of the high speed train which we know as z/x. --(2) We also need to know the time elapsed till the two trains meet. This we can find out since we know the time taken for each train to travel the whole distance as x and y. Thus the time elapsed when they meet is equal to xy/(x+y) -- (3)
3. So a= xy/(x+y) * z/x
4. Substitute this in (1) and you get the answer.
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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15 Aug 2013, 11:12
It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?

We know that when both trains meet, they will have been traveling for the same amount of time. We are given the distance both trains travel (the same distance) and the time in which they do it (x and y respectively) We need to figure out the difference in distance between the fast train and the slow train. first, we need to figure out what rate each train travels at (x/z and x/y for the high speed and slow speed train) To figure out distance (d=r*t) we also need to find the time the two trains traveled for. The distance (z) is equal to the distance traveled by the fast train plus the distance traveled by the slow train.

Lets call the fast train HS and the slow train LS.

speed = distance/time

Speed (HS) = z/x
Speed (LS) = z/y

Now that we have the speed, we need to find the distance each train traveled. The distance the high speed train travels - the distance the slow speed train travels will get us the amount of distance more the high speed train travels.

Distance = rate*time
We know the time it takes each train to reach their respective destinations. For HS it will be less than LS. However, in this problem, we know that both will travel for the same amount of time when they meet each other.

We know the speed of each train, now we must figure out the time each train traveled. Distance = speed*time

z = (z/x)*t + (z/y)*t
z = zt/x + zt/y
z = (zty/xy) + (ztx/xy)
z = (zty+ztx)/xy
zxy = (zty+ztx)
xy = ty + tx
xy = t(y+x)
t = xy/(x+y)

The time each train traveled was: xy/(x+y)
Now that we know the value for T, we can solve for their respective distances by plugging in for distance=rate*time

Distance (HS) = z/x*xy/(x+y)
zy/x+y

Distance (LS) = z/y*xy/(x+y)
zx/x+y

(zy/x+y) - (zx/x+y) = z(y – x)/x + y

ANSWER: (A) z(y – x)/x + y
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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15 Aug 2013, 12:45
joyseychow wrote:
It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?

(A) z(y – x)/x + y

(B) z(x – y)/x + y

(C) z(x + y)/y – x

(D) xy(x – y)/x + y

(E) xy(y – x)/x + y

........
s = z/x t
z-s = z/y t
..................
(+), z = zt (x+y/ xy)
or, t = xy/x+y

Now, z/x . t - z/y . t = zt (y-x/xy) = z . xy/(x+y) . (y-x)/xy = z (y-x/y+x)
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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15 Sep 2013, 20:00
All the methods looks like it will take more than 2 minutes to complete. There's got to be a faster way.
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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16 Sep 2013, 00:51
haotian87 wrote:
All the methods looks like it will take more than 2 minutes to complete. There's got to be a faster way.

This is not an easy question, so it's OK if it takes a bit more than 2 minutes to solve. Though if you understand the logic used here: it-takes-the-high-speed-train-x-hours-to-travel-the-z-miles-94564.html#p727726, here: it-takes-the-high-speed-train-x-hours-to-travel-the-z-miles-94564.html#p727737 and here: it-takes-the-high-speed-train-x-hours-to-travel-the-z-miles-94564.html#p1040729 it won't take mcut time.
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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29 Oct 2013, 08:48
Is there a resource to really hammer on this type of problem? I've gotten much better over the last couple months at every type of problem, but so far in 5 practice tests I've gotten every single one of these problems wrong.
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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29 Oct 2013, 23:50
AccipiterQ wrote:
Is there a resource to really hammer on this type of problem? I've gotten much better over the last couple months at every type of problem, but so far in 5 practice tests I've gotten every single one of these problems wrong.

All DS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=44
All PS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=64

Hope this helps.
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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07 Nov 2013, 11:49
It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?

I solved this problem a bit unusually....rather than do out the math, I picked variables and drew out a diagram. I said that the length was 60 and that the high speed train traveled 60 miles/hour and the regular train traveled 30 miles/hour. In one half hour, the fast train traveled 30 miles while the slow train traveled 15 miles. This meant that there was a 15 mile gap to close. Seeing as each train started out at the same time, and my variables had the faster train traveling twice as fast as the slow train for every to miles the fast train traveled, the slow train traveled just one. I basically drew out my diagram and counted off the miles until the two trains reached each other. I found that the slow train traveled 20 miles to the fast trains 40. From there, I simply plugged in the numbers until I got the right answer:z

(after 1/2 hour)
0 60mi
___________________F__________S__________
30mi 45mi

(After 2/3 hour)
0 60mi
_________________________FS______________
40mi 20mi

(A) z(y – x)/x + y
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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09 Nov 2013, 06:07
How would I do this with a RTD chart?
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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16 Jun 2014, 18:26
Since there are variables in the choices, plug in approach would work very well.

Assume x=10, z=100 and y=20. When travelling in the opposite directions and the distance is 100 miles, they would meet when the high speed train had traveled 66 2/3 miles and the regular train 33 1/3 miles. So the high speed train would have traveled 33 1/3 miles more.

Substitute the values of x, y and z in the choices. Choice A gives the value of 33 1/3 and is the correct answer.
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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23 Jun 2014, 16:41
I really had to work on this several times to get the equation right. Is there anyway to avoid silly mistakes?
for example even finding difference in the distance has given totally wrong answer.

Are there any tips like plugging in values in these so many variables question?

Another question is , in distance rate problems, usually based on what variable equations are easier . is it time or distance?

Thanks
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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23 Jun 2014, 18:50
GMatAspirerCA wrote:
I really had to work on this several times to get the equation right. Is there anyway to avoid silly mistakes?
for example even finding difference in the distance has given totally wrong answer.

Are there any tips like plugging in values in these so many variables question?

Another question is , in distance rate problems, usually based on what variable equations are easier . is it time or distance?

Thanks

Note that most of these TSD questions can be done without using equations.

You can also use ratios here.

Ratio of time taken by high speed:regular = x:y
Ratio of distance covered in same time by high speed:regular = y:x (inverse of ratio of speed)
So distance covered by high speed train will be y/(x+y) * z
and distance covered by regular train will be x(x+y) * z
High speed train will travel yz/(x+y) - xz/(x+y) = z(y-x)/(x+y) more than regular train.

Plugging numbers when there are variables works well but it gets confusing if there are too many variables. I am good with number plugging when there are one or two variables - usually not more.

Whether you should make the equation with "total time" or "total distance" will totally depend on the question - sometimes one will be easier, sometimes the other.
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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11 Aug 2015, 14:17
Pick numbers. In this case you can easily eliminate B, C, D and E.

Using: total distance of 200 miles.
High-Speed Train: 200mph; 1h travel time
Regular-Train: 100mph; 2h travel time

A) 200*(2-1)/(2+1) = 200/3, can be this one
B) Negative, not possible
C) 200*(2+1)/(2-1) = 600, not possible
D) Negative, not possible
E) 2*(2-1)/(2+1) = 2/3, too small

Therefore, choose A.
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]

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18 Aug 2016, 03:02
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