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Two trains fast and slow are going from city A to city B

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WABL
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Two trains fast and slow are going from city A to city B [#permalink] New post   14 Feb 2019, 15:10
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Two trains fast and slow are going from city A to city B at the same time. when the fast train has covered 2/3 of the distance. the slow train is 180Km away from city B, when the fast has arrived in city B, the slow train has covered 6/7 of the distance. How long is the distance between A and B?

A. 210 km
B. 315 km
C. 420 Km
D. 490 Km
E. 560 Km
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C
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IanStewart
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Re: Two trains fast and slow are going from city A to city B [#permalink] New post   14 Feb 2019, 15:55
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When the fast train arrives at B, the slow train has covered 6/7 of the distance. So, when they travel for the same amount of time, the slow train travels 6/7 as far as the fast train, which is another way of saying that the slow train's speed is 6/7 of the fast train's speed.

If we call the total distance between the two cities "D", when the fast train has covered 2/3 of that distance, so 2D/3 miles, the slow train will only go 6/7 as far, because its speed is 6/7 of the fast train's, so the slow train will cover (6/7)(2D/3) = 4D/7 miles. But we also know that in this time, the slow train covers D - 180 miles (all but 180 miles of the total distance). So

4D/7 = D - 180
180 = 3D/7
D = 60*7 = 420
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Re: Two trains fast and slow are going from city A to city B [#permalink] New post   14 Feb 2019, 20:27
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Consider Fast Train as FT and Slow train as ST. Total distance as D.
We have information of two instances (consider trains started from time T=0)
1. at T1; (2/3)D covered by FT and
at T2; D covered by FT
since distance covered is proportional to time, we can say, T1/T2= ((2/3)D)/D = 2/3

2. at T1: D-180 covered by ST and
at T2: (6/7)D covered by ST
T1/T2= (D-180)/((6/7)D)

now from both equation you can calculate D and D= 420. Answer is C.
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Re: Two trains fast and slow are going from city A to city B [#permalink] New post   16 Feb 2019, 09:56
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Always try to organize the data into a table to be clearer. now we can deduce two equations:

T1 = \(\frac{\frac{2}{3}D}{Sf}\) = \(\frac{D-180}{Ss}\)
T2 =\(\frac{D}{Sf} = \frac{\frac{6}{7}D}{Ss}\)
from the two equations:
\(\frac{Ss}{Sf}= \frac{D-180}{\frac{2}{3}D} = \frac{\frac{6}{7}D}{D} = \frac{6}{7}\)
\(7D-(180*7) = 4D\)
\(3D = 180*7\)
\(D = 60*7 = 420\)
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Re: Two trains fast and slow are going from city A to city B [#permalink] New post   15 Sep 2023, 11:46
Another interesting solution:
Let, total distance = D
When the Faster train covers D (full) distance, the slower train covers 6/7 of D. It means the slower train always covers 6/7 of the distance covered by the faster train.
In the 1st case,
Faster train covers 2D/3 and the slower train covers 2D/3 * 6/7 = 12D/21
ATQ,
12D/21 = D-180
9D = 3780
D = 420 (Ans) [Option C]
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Re: Two trains fast and slow are going from city A to city B [#permalink]
15 Sep 2023, 11:46
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