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mbaprep2016
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It takes the high-speed train x hours to travel the z miles 23 Apr 2017, 07:34
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Bunuel
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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

Please help me with this question.

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} --> \(\frac{xy}{x+y}*\frac{z}{x}-\frac{xy}{x+y}*\frac{z}{y}=\frac{z(y-x)}{x+y}\).

Answer: A.
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Bunuel VeritasPrepKarishma
Can anybody please tell me what's wrong with my approach

High speed train x miles/hour
slow train at y miles/hour
combined speed x+y
total time when they meed would be z/x+y

distance covered by FAST train will be x(z/x+y)
by slow train would be y(z/x+y)

Difference will be x(z/x+y) - y(z/x+y)

I am getting B as answer
help me please
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It takes the high-speed train x hours to travel the z miles 24 Apr 2017, 01:14
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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

Please help me with this question.

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} --> \(\frac{xy}{x+y}*\frac{z}{x}-\frac{xy}{x+y}*\frac{z}{y}=\frac{z(y-x)}{x+y}\).

Answer: A.

Bunuel VeritasPrepKarishma
Can anybody please tell me what's wrong with my approach

High speed train x miles/hour
slow train at y miles/hour
combined speed x+y
total time when they meed would be z/x+y

distance covered by FAST train will be x(z/x+y)
by slow train would be y(z/x+y)

Difference will be x(z/x+y) - y(z/x+y)

I am getting B as answer
help me please
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Note that x hours is the time taken by the high speed train, not its speed. Similarly, y hours is the time taken by the regular train, not its speed.
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It takes the high-speed train x hours to travel the z miles 07 Dec 2017, 17:50
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joyseychow
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
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We have a converging rate problem in which:

Distance(1) + Distance(2) = Total Distance

We are given that 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. Thus, we know the following:

rate of the high-speed train = z/x

rate of the regular speed = z/y

We are also given that they leave at the same time, so we can let the time when both trains pass each other be t. We can substitute our values into the total distance formula.

Distance(1) + Distance(2) = Total Distance

(z/x)t + (z/y)t = z

zt/x + zt/y = z

We can divide the entire equation by z and we have:

t/x + t/y = 1

Multiplying the entire equation by xy gives us:

ty + tx = xy

t(y + x) = xy

t = xy/(y + x)

Now we can calculate the distance traveled by both trains for time t, using the formula

distance = rate x time

distance of high speed train = (z/x)[xy/(y + x)]

distance of regular speed train = (z/y)[xy/(y + x)]

Now we need to calculate the difference between the distance of the high speed train and the distance of the regular speed train:

difference of distance traveled = distance of high speed train - distance of regular speed train

difference of distance traveled = (z/x)[xy/(y + x)] - (z/y)[xy/(y + x)]

We can factor out [xy/(y + x)]:

difference of distance traveled = [xy/(y + x)](z/x - z/y)

difference of distance traveled = [xy/(y + x)][(yz - xz)/(xy)]

The xy terms cancel and we are left with:

difference of distance traveled = (yz - xz)/(y + x) = z(y-x)/(x + y)

Answer: A
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It takes the high-speed train x hours to travel the z miles 08 Jun 2020, 00:44
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joyseychow
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
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Variables in the answer. Lets use plug in

z=100 miles
x=1 hr
y=2 hrs
high speed = 100 miles/hr
regular speed = 50 miles /hr
combined speed 150 miles/hr
time taken to meet= d/s =100/150 = 2/3 hrs

in 2/3 hrs, high speed would have traveled 66 2/3 miles
regular speed would have traveled 33 1/3 miles

difference is 33 1/3 miles

Choice A gives this value when plugged in with the above values
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It takes the high-speed train x hours to travel the z miles 24 Jun 2022, 19:21
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joyseychow
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
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x, y, and z? No thank you!!! Let's Plug In.

High-speed travels 4 mph. Regular travels 2mph. Total distance is 6 miles.

If they both start at the same time heading toward each other, in one hour, high-speed will have gone 4 miles and regular will have gone 2 miles, so they meet.
High-speed has traveled 2 miles farther than regular.

High-speed can cover the 6 miles in 6/4 hours, so x = 1.5.
Regular can cover the 6 miles in 6/2 hours, so y = 3.
z = 6

(A) z(y – x)/x + y = \(\frac{6(3-1.5)}{1.5+3} = \frac{6(1.5)}{4.5} = 1 \)Keep it.

(B) z(x – y)/x + y = Different numerator from (A) but same denominator, so it can't also be right. Eliminate.

(C) z(x + y)/y – x = \(\frac{6(3+1.5)}{3-1.5} = \frac{6(4.5)}{1.5}\) Eliminate.

(D) xy(x – y)/x + y = \(\frac{(1.5)(3)(1.5-3)}{1.5+3}\) Numerator is negative and denominator is positive. Eliminate.

(E) xy(y – x)/x + y = \(\frac{(1.5)(3)(3-1.5)}{1.5+3} = \frac{4.5*1.5}{4.5} = 1.5\) Eliminate.

Answer choice A.


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