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Re: Train-A and train-B crosses each other in 8 seconds, while running in [#permalink]
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TheNightKing wrote:
Train-A and train-B crosses each other in 8 seconds, while running in opposite direction. Train-B crosses a pole in 8.4 seconds and train-A crosses a 90 meters long tunnel in 12 seconds. If speed of train-A is 15 km/hr more than the speed of train-B, then find the ratio of length of train-A to length of train-B.
(a) 8 : 7
(b) 11 : 7
(c) 5 : 4
(d) 3 : 2
(e) None of the above.



In such questions, we should form equations. Here you would have to form >3 equations and then remember that speed is given in km/hr while time spoken about is in seconds. So, not likely in GMAT in present form. But can help you in clearance of concepts in SPEED AND DISTANCE.

So let us form equations first..
Quote:
Train-A and train-B crosses each other in 8 seconds, while running in opposite direction.

Let the length be x & y and speeds be A and B respectively. 'Running in opposite directions' means we have to add the speeds as they are moving towards each other.
so, speed = A+B and distance traveled=x+y.......Time = \(8=\frac{x+y}{A+B}......x+y=8A+8B\)....(i)

Quote:
Train-B crosses a pole in 8.4 seconds

So speed = B and distance traveled=y.......Time = \(8.4=\frac{y}{B}......y=8.4B\) ....(ii)

Quote:
train-A crosses a 90 meters long tunnel in 12 seconds

So speed = A and distance traveled=x+90.......Time = \(12=\frac{x+90}{A}......x+90=12A.....x=12A-90\) ....(iii)


Substitute values of x and y from (ii) and (iii) in (i).....
\(x+y=8A+8B........8.4B+12A-90=8A+8B.........4A+0.4B=90\)......(iv)


Finally,
Quote:
speed of train-A is 15 km/hr more than the speed of train-B

As we have taken speeds in m/s, convert 15km/h in m/s......\(\frac{15km}{h}=\frac{15*1000m}{60*60sec}=\frac{15*5}{18}=\frac{25}{6}\)
So \(A=B+\frac{25}{6}\)...(v)


Substitute value of A in (iv)..
\(4A+0.4B=90.......0.4B+4(B+\frac{25}{6})=90.....2.4B+24B+100=540.....26.4B=540-100=440.....B=\frac{44*2*5}{44*0.6}=\frac{50}{3}\),
so from (ii) \(y=8.4B=8.4*\frac{50}{3}=2.8*50=140\)

and \(A=B+\frac{25}{6}=\frac{50}{3}+\frac{25}{6}=\frac{125}{6}\)
So from (iii) \(x=12A-90=12*\frac{125}{6}-90=250-90=160\)

We are looking for \(\frac{x}{y}=\frac{160}{140}=\frac{8}{7}\)

A
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Re: Train-A and train-B crosses each other in 8 seconds, while running in [#permalink]
I think after making equations, plugging in the choice is a better option.
Let the length of train A and B are a mts and b mts. Let the speed of train B is x mts/sec and speed of train B is x+ 25/6 mts/sec.
Now, (a+b)/2x+ 25/6 = 8
b = 8.4x
(a +90)/ x+ 25/6 =12
Now trying first option let a=8c and b=7c
Now we will calculate value of c from b= 8.4 x that is c= 1.2 x
From first equation
15c = 16x + 8×25/6
Putting c =1.2x in this equation
X=50/3
Putting this value of x in last equation that is
(a+90)/ x+ 26 =12 , if it comes true then this option is correct otherwise we have to try another option.
a=8c and c= 1.2x and x = 50/3 in the equation gives 12 hence option A is the answer.

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Re: Train-A and train-B crosses each other in 8 seconds, while running in [#permalink]
25/6 is m/sec conversion of 15 km/hr.
Your mistake is length you have taken in meters and speed in km/ hour.

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Re: Train-A and train-B crosses each other in 8 seconds, while running in [#permalink]
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Re: Train-A and train-B crosses each other in 8 seconds, while running in [#permalink]
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