Train T leaves station X for station Y at 3 pm. Train S, traveling at

Question

Train T leaves station X for station Y at 3 pm. Train S, traveling at three quarters of the speed of T, leaves Y for X at 4 pm. The two trains pass each other at a station Z, where the distance between X and Z is three-ﬁfths of that between X and Y. How many hours does train T take for its journey from X to Y?

A. 9
B. 12
C. 15
D. 16
E. 18

A. 9
B. 12
C. 15
D. 16
E. 18

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TestPrepUnlimited · Accepted Answer

First, we may set the speed of Train T as 1 per hour since the actual speed is not important and we only need to verify the ratio of speeds is correct. Then Train S would be traveling at a speed of 0.75 per hour. We are given XZ / XY = 3/5, where Z is between X and Y. Therefore XZ/YZ = 3/2. The distance train T and train S traveled has a ratio of 3 to 2.  Set x as the number of hours train S needed to travel from station Y to station Z

Then we have:  \frac{(1*x + 1) }{ (0.75x)} = \frac{3}{2} . Note Train T traveled an extra hour since it left one hour earlier. Solving this gives x = 8 which means train T traveled 9 hours. However, x = 9 hours was needed for train T to cover only 60% of the total distance, to cover the entire distance it needs 9/60% = 15 hours.

Ans: C

firas92 · Answer

Let the speed of Train T be  T mph  and S be  \frac{3}{4}T mph  and the distance XY be  d

We need to find  \frac{d}{T}

1 hour after T starts, the trains are at a distance of  d-T . So the time taken for the trains to meet ( x ) is

(T+\frac{3}{4}T)*x=d-T

x=\frac{4}{7}(\frac{d}{T})-\frac{4}{7}

We also know that  T(1+x)=\frac{3}{5}d  or  \frac{3}{5}(\frac{d}{T})=1+x

So,  \frac{3}{5}(\frac{d}{T})=1+\frac{4}{7}(\frac{d}{T})-\frac{4}{7}

Solving, we get  \frac{d}{T}=15

Answer is (C)

ryansmith123123 · Answer

First lets assume the distance between X and Y is 100 miles. Then, the distance between X and Z becomes 60 miles (given in the question that distance between X and Z =1/3 distance between X and Y)

Let St be the speed of train T
Let Ss be the speed of train S
Let Tt be the time taken by Train T to reach Station Z
Let Ts be the time taken by train T to reach STation Z

Also, given that Ss=1/3St

Let us calculate speed of train T

St=60/Tt

Speed of Train S:

Ss= 40/(Tt-1) [As train S starts one hour after Train T )

As we know Ss=1/2 St

St= 60/Tt -(1)

St=120/(Tt-1) -(2)

Solving, we get Tt=3Hrs

From that we can calculate speed of Train T to be 20mph and Train S to be 20/3 mph

Since we know speed of Train S, Distance between X and Y, we can calculate time taken T = 100/(20/3) = 15 hrs

exc4libur · Answer

d=(1)

3/5(1)/r-1*r=2/5(1)/3r/4
3/5r-8/15r=1
9-8/15r=1
r=1/15

(1)/1/15=15

ans (C)

SiddharthR · Answer

This question is a little tougher than it looks. From the average solve times, it shows that it takes quite sometime to complete. Here is how it is solved --

Let's assume the following :

Distance between X and Y = D --> 1
R(T) --> Rate of Train T --> 2
R(S) --> Rate of Train S -> 3

Given -->
R(S) = 3/4   --> 4
Distance from Station X to Z = 3/5(D) --> 5
Distance from Station Y to Z = 2/5(D) --> 6
Train T leaves Station X at 3:00 PM --> 7
Train S leaves Station Y at 4:00 PM --> 8

From the above given points (7 & 8), we can conclude that Train T travels for 1 hour while Train S is stationary

So, using the Distance = Rate x Time formula, we can conclude that Train T travels a distance (say a). This can be mentioned in the formula as --> a = R(T) --> 9

Using points 5 and 9, we can conclude that after Train T travels for an hour (Lets name this position an arbitrary position O), the remaining distance to Station Z = 3/5(D) - a = 3/5(D) - R(T) --> 10

Now Train T has to travel to Station Z from O and Train S has to travel to Station Z from Y (Two Objects travelling towards each other)

When two objects travel towards each other, you can sum up their rates, the distance they travel becomes the total distance and time traveled by both the objects remains the same. So we have --

R(T) + R(S) =   / T(Time to reach Station Z) --> 11

T(Time to reach station Z) = Time taken by Train T from O to Z (OR) Time taken by Train S from Y to Z   --> 12

From the given rates, we can write the following for equation 12--> T(S) =   /  = 8D/  --> 13

We can substitute equations 4 & 13 in equation 11. After solving we end up with D = 15R(T) --> It takes Train T 15 hours to travel from Station X to Station Y @ rate R(T)

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Train T leaves station X for station Y at 3 pm. Train S, traveling at

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Train T leaves station X for station Y at 3 pm. Train S, traveling at

Prep Toolkit

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