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

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Re: It takes the high-speed train x hours to travel the z miles [#permalink]
I solved it this way:

Speed of high speed train =z/x
Speed of normal train = z/y

Let d be the distance covered by high speed train when they meet.
So distance covered by normal train is z-d

When they meet, the time taken by both is same. So,

d/x = (z-d)/y
d*y = z*x - d*x
d(y+x) = z*x
d = z*x / x+y

the difference between the distance would be d - (z-d) which is equal to 2d - z
= 2z*x / (x+y) - z
= 2z*x - z*x - z*y / (x+y)
= z(x-y)/x+y

which is option b

please let me know where am I going wrong.
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]
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Prax wrote:
I solved it this way:

Speed of high speed train =z/x
Speed of normal train = z/y

Let d be the distance covered by high speed train when they meet.
So distance covered by normal train is z-d

When they meet, the time taken by both is same. So,

d/x = (z-d)/y
d*y = z*x - d*x
d(y+x) = z*x
d = z*x / x+y

the difference between the distance would be d - (z-d) which is equal to 2d - z
= 2z*x / (x+y) - z
= 2z*x - z*x - z*y / (x+y)
= z(x-y)/x+y

which is option b

please let me know where am I going wrong.

You can solve this way too, but you made a mistake in calculation: $$time=\frac{distance}{rate}=\frac{d}{\frac{z}{x}}=\frac{z-d}{\frac{z}{y}}$$ --> $$dx=zy-dy$$ (not d/x = (z-d)/y) --> $$d=\frac{zy}{x+y}$$.

Difference $$2d-z=\frac{2zy}{x+y}-z=\frac{z(y-x)}{x+y}$$.

You could spot that answer B cannot be the correct choice as it's negative (numerator x-y<0) (high speed train needs less time to cover the distance than regular train, so x<y) but the difference in distances cannot be negative as high speed train would cover greater distance than regular train when they meet (for exact same reason choice D can be eliminated as well).

Hope it's clear.
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]
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The problem can be solved very easily with the concepts of Relative velocity..

if we assume that slow moving tarin is at stand still and high speed train is moving with the speed : x+y

time taken to travel the distance : z/(x+y)

diffrence in the distance tarvelled = distance traveled by high speed tarin - distance traveled by slow moving train
= x[z/(x+y)] - y[z/(x+y)]
= z(x-y)/(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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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

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.
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]
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Bunuel wrote:
ishanand 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)

z represents distance, x and y represent time.
The answer requires an expression with units of distance.
We can immediately eliminate answers D and E, both have units of time squared and not distance.

The regular train is slower than the high-speed train, so necessarily y > x. We can eliminate choice B, being negative.

Since (x + y)/(y - x) > 1, we can eliminate choice C, as neither of the two trains could have traveled a distance greater than z until they passed each other.

We are left with the only choice A.
The above posts confirm that it 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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Rate of high speed train: $$\frac{z}{x}$$
Rate of regular train: $$\frac{z}{y}$$

t: time it takes for two trains to pass each other from opposite direction

$$\frac{z}{x}(t)+\frac{z}{y}(t)=z$$

Calculate t: t = $$\frac{x+y}{xy}$$

Distance travelled by high-speed train minus regular train:

$$(\frac{z}{x})(\frac{xy}{x+y})-(\frac{z}{y})(\frac{xy}{x+y})$$

Answer: $$z(\frac{y-x}{x+y})$$
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]
Bunuel wrote:
Prax wrote:
Hi,

I have another doubt:

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?

z(y – x)/x + y

z(x – y)/x + y

z(x + y)/y – x

xy(x – y)/x + y

xy(y – x)/x + y

It takes the high-speed train x hours to travel the z miles --> rate of high-speed train is $$rate_{high-speed}=\frac{distance}{time}=\frac{z}{x}$$;

It takes the regular train y hours to travel the same distance --> rate of regular train is $$rate_{regular}=\frac{distance}{time}=\frac{z}{y}$$;

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

Difference in distances covered: {Time}*{Rate of high-speed train} - {Time}{Rate of regular train} --> $$\frac{xy}{x+y}*\frac{z}{x}-\frac{xy}{x+y}*\frac{z}{y}=\frac{z(y-x)}{x+y}$$.

Hey can someone help me. 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? That is not the same as z (to me) but should be shorter than z
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Re: It takes the high-speed train x hours to travel the z miles [#permalink]
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manimgoindowndown wrote:
Bunuel wrote:
Prax wrote:
Hi,

I have another doubt:

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?

z(y – x)/x + y

z(x – y)/x + y

z(x + y)/y – x

xy(x – y)/x + y

xy(y – x)/x + y

It takes the high-speed train x hours to travel the z miles --> rate of high-speed train is $$rate_{high-speed}=\frac{distance}{time}=\frac{z}{x}$$;

It takes the regular train y hours to travel the same distance --> rate of regular train is $$rate_{regular}=\frac{distance}{time}=\frac{z}{y}$$;

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

Difference in distances covered: {Time}*{Rate of high-speed train} - {Time}{Rate of regular train} --> $$\frac{xy}{x+y}*\frac{z}{x}-\frac{xy}{x+y}*\frac{z}{y}=\frac{z(y-x)}{x+y}$$.

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

Let us call the trains as H and R resp.

Given:

Distance = z
Time taken by H = x
Time taken by R = y

Question: When the trains meet how much more distance has H traveled than R? For that we need to calculate the speed of both the trains and the time taken for them to meet.

Deductions:

Speed of H = z/x
Speed of R= z/y

To calculate the time taken for them to meet we need to use the total distance between the two towns and the combined speed as they are moving towards each other.

Time taken for the trains to meet= z/(z/x +z/y) = xy/(x+y)

Distance traveled by H when the trains meet= time taken to meet* speed of H

=xy/(x+y) * z/x

=zy/(x+y)

Similarly for R , we have the distance = zx/(x+y)

The difference is z(y-x)/(x+y)

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

Originally posted by KarishmaB on 23 Jun 2014, 19:50.
Last edited by KarishmaB on 22 Jun 2022, 20:53, edited 1 time in total.
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