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

Question

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

A. 210 km 
B. 315 km
C. 420 Km
D. 490 Km
E. 560 Km

IanStewart · Accepted Answer

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

MahmoudFawzy · Answer

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

pankajpaliitkgp · Answer

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.

Abhinandan Dey · Answer

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)

GSV · Answer

Let's assume 
total distance = d
timings required by fast train to cover 2/3d is t1 
timings required by fast train to cover d is t2

S(fast) = (2d/3)/t1 = d/t2 
S(slow) = (d-180)/t1 = (6d/7)/t2

S(fast)/ S(slow) = (2d/3)/ (d-180) = d/ (6d/7) after cancelling t1 and t2 
On solving this we get d = 420

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

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

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