Aptitude - Problems on Trains - Discussion

Total distance of the 2 trains are 220 (108+112) :first don't change anything. Just change the speed 50 km/hr into m/sec.

If we change speed 50 ino m/sec then it will be 125/9. The equation becomes.

(125/9) +x=220/6.

We get x=410/18 m/sec.

Now convert to km/hr we get x= (410/18) (18/5).

X=82.

108+112=220.

220/6 is the relative speed per sec.

(220/6)*(18/5)=132 is the relative speed per hour.

Speed of the 2nd train is 132-50 = 82kmph.

Easiest method is,

220*18 = 6(250+5x).

Multiply by 6(250+5x) you get 1500+30x.

3960 = 1500+30x.

Subtract the value of 3960-1500 = 2460.

Then divide by this form,
2460/30 = 82 kmph

Distance = 108+112 = 220m.

Time = 6s.

Effective speed = x+y (+ because in opp direction).

=> let x = 50 kmph in terms of m/s = 50*(5/18) = 250/18.

eff.sp = 250/18+y (since v need to find y).

So distance = speed*time=>220 = (250/18+y)6.

220 = (250+18y)/18*6.

On simplification,

we get, 410/18.

Converting m/s to kmph, (410/18)*(18/5) = 82kmph.

Hey shyam sharma if I do in your method what you said the problem formula not getting answer yar.

t = 220/ (v+50)*(5/18).

t = 6.

v = 82 got directly.

And hence no need 2 convert again 2 m/sec why?

Why should we convert to m/sec as in above problem (x+50)*5/18.

But in answer the required ans in km/hr again why we are taking 5/18.

Can anyone explain me please?

220/250+5x/18 = 6.

step 1: 18*220 = 3960.
step 2: 250+5x * 6 = 1500+30x
step 3: 3960 = 1500+30x.
step 4: 3960-1500 = 30x
step 5: 2460 = 30x.
step 6: x=2460/30.
step 7: x=82.

Total distance = 108+112 = 220 m.

Time = 6s.

Relative speed = distance/time = 220/6 m/s = 110/3 m/s.

= (110/3)x(18/5) km/hr = 132 km/hr.

=> 50 + speed of second train = 132 km/hr.

=> Speed of second train = 132-50 = 82 km/hr.

Given,

Speed of first train = 50km/hr (after convert it into m/s it becomes 125/9m/s).

Speed of second train = ?

Length of first train = 108m.

Length of second train = 112m.

Now, lets take the speed of second train be x m/s.

So by speed formula i.e.

Speed = distance/time keeping the values in this formula we'll get,

125/9+x = 108+112/6 (here we are adding these value b'coz they are moving in opposite direction).

By taking lcm of 9 on L.H.S it becomes,

(125+9x)/9 = 220/6.

By cross multiplication.

6 (125+9x) =220*9.

750+54x = 1980.

54x = 1980-750.

54x = 1230.

x = 1230/54.

x = 205/9 m/s.

But now we've to convert it into km/hr so.

(205/9*18/5) km/hr.

After cancellation it become.

Answer: 82 km/hr.

Formula : T=(x+y)/(u+v).

Where,
x,y -- Distance.
u,v -- Speed.

6 = (108+112)/((50*5/8)+V).

6v+250/3 = 220.

Therefore, v = 205/9 m/sec.

v = 205/9*(18/5) = 82km/hr.