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03 Apr 2020, 07:59
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C
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Difficulty:
75%
(hard)
Question Stats:
71% (03:29) correct
29%
(03:16)
wrong
based on 202
sessions
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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
Are You Up For the Challenge: 700 Level Questions
A. 9
B. 12
C. 15
D. 16
E. 18
Are You Up For the Challenge: 700 Level Questions
03 Apr 2020, 08:15
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Bunuel
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
General Discussion
firas92
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03 Apr 2020, 11:20
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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 \(\frac{d}{T}\) [since T*Time=d]
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)
We need to find \(\frac{d}{T}\) [since T*Time=d]
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)
