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
OA A
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GMATsid2016
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crackverbal
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Hi GMATsid2016,
The following two points will serve you well when dealing with questions on speed, distance and time where two objects meet each other.
1. Always read the question and figure out the constant out of distance and time (there will always be one).
If the distance is constant then we use the formula S1T1 = S2T2 where S1 and T1 are the speed and time of the first object and S2 and T2 are the speed and time of the second object.
If the time is constant then we use the formula D1/S1 = D2/S2 where D1 and S1 are the distance and speed of the first object and D2 and S2 are the distance and speed of the second object.
2. Once you figure out which of the two above formulae you need to use, always graphically represent the question.
Let us consider z = 100 miles, x = 5 hours and y = 10 hours. Now since both the trains started simultaneously, the time will be constant at the point they meet. So we use the formula D1/S1 = D2/S2.
If O represents the point that the two trains meet,
a/20 = (a - 100)/10 -----> a = 200/3 and 100 - a = 100/3. the difference between the distances traveled by the faster train and the regular train is 200/3 - 100/3 = 100/3 miles.
now substituting z = 100 miles, x = 5 hours and y = 10 hours in the answer choices, the only answer choice which gives us a 100 is A. z(y-x)/x+y
CrackVerbal Academics Team
The following two points will serve you well when dealing with questions on speed, distance and time where two objects meet each other.
1. Always read the question and figure out the constant out of distance and time (there will always be one).
If the distance is constant then we use the formula S1T1 = S2T2 where S1 and T1 are the speed and time of the first object and S2 and T2 are the speed and time of the second object.
If the time is constant then we use the formula D1/S1 = D2/S2 where D1 and S1 are the distance and speed of the first object and D2 and S2 are the distance and speed of the second object.
2. Once you figure out which of the two above formulae you need to use, always graphically represent the question.
Let us consider z = 100 miles, x = 5 hours and y = 10 hours. Now since both the trains started simultaneously, the time will be constant at the point they meet. So we use the formula D1/S1 = D2/S2.
If O represents the point that the two trains meet,
a/20 = (a - 100)/10 -----> a = 200/3 and 100 - a = 100/3. the difference between the distances traveled by the faster train and the regular train is 200/3 - 100/3 = 100/3 miles.
now substituting z = 100 miles, x = 5 hours and y = 10 hours in the answer choices, the only answer choice which gives us a 100 is A. z(y-x)/x+y
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regor60
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Recognize that when the trains meet they will have been moving for the same amount of time and that the sum of their distances traveled equals the distance between the points A and B, which is Z.
Remembering distance = rate x time. Let D equal distance traveled by high speed train, therefore Z-D = distance of regular train.
Rate High = Z/X
Rate Regular = Z/Y
Time = distance/ rate, so
Time High = DX/Z
Time Regular = (Z-D)Y/Z
Equate the two times, since they travel for same amount of time
DX/Z=(Z-D)Y/Z
So, DX = (Z-D)Y
Solve for D
D= ZY/(X+Y). Therefore, distance traveled by regular train = Z-ZY/(X+Y)
The question asks for how much farther faster train travels compared to slower train, so subtract the two above:
ZY/(X+Y) - Z + ZY/(X+Y) = 2ZY/(X+Y) - Z.
Put both on same basis:
(2ZY-Z(X+Y))/(X+Y) = Z(Y-X)/(X+Y) =
A
Remembering distance = rate x time. Let D equal distance traveled by high speed train, therefore Z-D = distance of regular train.
Rate High = Z/X
Rate Regular = Z/Y
Time = distance/ rate, so
Time High = DX/Z
Time Regular = (Z-D)Y/Z
Equate the two times, since they travel for same amount of time
DX/Z=(Z-D)Y/Z
So, DX = (Z-D)Y
Solve for D
D= ZY/(X+Y). Therefore, distance traveled by regular train = Z-ZY/(X+Y)
The question asks for how much farther faster train travels compared to slower train, so subtract the two above:
ZY/(X+Y) - Z + ZY/(X+Y) = 2ZY/(X+Y) - Z.
Put both on same basis:
(2ZY-Z(X+Y))/(X+Y) = Z(Y-X)/(X+Y) =
A
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GMATGuruNY
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Let the high speed rate = 3 miles per hour.GMATsid2016 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 the regular rate = 2 miles per hour.
Combined rate for the trains = 2+3 = 5 miles per hour.
Let z = 30 miles.
Time for the high speed train to travel 30 miles = x = 30/3 = 10 hours.
Time for the regular train to travel 30 miles = y = 30/2 = 15 hours.
Time for the trains to meet = 30/5 = 6 hours.
Distance traveled by the high speed train in 6 hours = r*t = 3*6 = 18 miles.
Distance traveled by the regular train in 6 hours = r*t = 2*6 = 12 miles.
Distance for the high speed train - distance for the regular train = 18-12 = 6. This is our target.
Now plug x=10, y=15, and z=30 into the answers to see which yields our target of 6.
Only answer choice A works:
z(y - x)/(x+y) = 30*(15-10)/(10+15) = 6.
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Here's a cutesy way to guess the answer very quickly.
The high-speed train's rate = z/x
The low-speed train's rate = z/y
So the high-speed train travels (z/x) - (z/y), or z*(y - x) / x*y faster per hour than the low-speed train.
By the Lazy Test Writer Principle, the answer must contain some form of the numerator or the denominator. Only A does, so we're done.
The high-speed train's rate = z/x
The low-speed train's rate = z/y
So the high-speed train travels (z/x) - (z/y), or z*(y - x) / x*y faster per hour than the low-speed train.
By the Lazy Test Writer Principle, the answer must contain some form of the numerator or the denominator. Only A does, so we're done.
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toby001
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Thanks Matt. This is exactly how I arrived at my answer. How do we get from this point to the final answer?
Thanks!
Thanks!
Matt@VeritasPrep wrote:Here's a cutesy way to guess the answer very quickly.
The high-speed train's rate = z/x
The low-speed train's rate = z/y
So the high-speed train travels (z/x) - (z/y), or z*(y - x) / x*y faster per hour than the low-speed train.
By the Lazy Test Writer Principle, the answer must contain some form of the numerator or the denominator. Only A does, so we're done.
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Scott@TargetTestPrep
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We have a converging rate problem in which:GMATsid2016 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
OA A
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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