Aptitude - Problems on Trains - Discussion

Let distance between the cities = x.

Let speeds be s1 and s2.

The 2 trains will meet after time = x/(s1+s2)....(1) [using relative velocities].

Distance covered by T1 in the above time = s1*(1) = s1*x/(s1+s2)....(2).

Therefore dist remaining for T1 to destination = x-(2) = s2*x/(s1+s2).

Therefore time required to cover this dist by T1 = s2*x/s1*(s1+s2)....(3).

Similarly time required by T2 to cover its remaining distance = s1*x/s2*(s1+s2)....(4).

Now these times are given to be 9 and 16 respectively.

Dividing (3) and (4) eliminates x to give:

(s2/s1)^2 = 9/16.

Hence, s2/s1 = 3:4.

Let the trains be named as A & B.

And the total distance b/w Howrah to Pune be d.
WHEN BOTH THE TRAINS MEET,

Distance covered by train A, d1 = s1*t.........1.
Distance covered by train B, d2 = s2*t.........2.

Divide equation 1 by 2,
d1/d2 = s1/s2.........3.

WHEN BOTH TRAINS HAVE CROSSED AND ARE MOVING TO THEIR RESPECTIVE STATIONS,

Distance covered by train A, d2=s1*9..........4.
Distance covered by train B, d1=s2*16..........5.

Divide equation 5 by 4.
d1/d2=s2*16/s1*9.........6.

Solving equation 3 & 6.
We get,

s1/s2 = s2*16/s1*9.
s1*s1/s2*s2 = 16*9.

s1/s2 = 4*3. Answer.

Divide distance into two parts based on the point where they meet.

Let first half of journey took x hrs which is same for both trains.

Total time for train 1 = 9+x.
Total time for train 2 = 16+ x.

Let the speed of the first train(took 9 hrs to complete the second half of its journey) be a

Speed of the second train( took 16 hrs to complete the second half of its journey) be b

The total distance between the cities is 9a+16b.

Now,

Speed = distance/ time.

a = 9a+16b/9+x.
b= 9a+16b/16+x.

eliminate x.
ax= 16 b,
bx = 9a.

16b/a =9a/b.

a/b = 4/3.

Let's suppose the train meets after x hours.
speed of train1 = s1 and speed of train2 =s2,
So, distance covered by train1 in x hours = xs1 and distance covered by train2 in x hours = xs2.
So total distance from Howrah to Patna(d)=xs1+xs2.

After they meet, they reach their destination after 9 and 16 hours respectively.
So, speed of train1 =d/x+9 and speed of train2=d/x+16
So, s1=xs1+xs2/x+9 and s1x+9s1=xs1+xs2, x=9s1/s2,
speed of train2=d/x+16.
So s2=xs1+xs2/x+16, xs2+16s2=xs1+xs2.
x=16s2/s1, 9s1/s2 = 16s2/s1,(s1/s2)^2 = 16/9.
So, s1/s2 = 4/3.

Total distance = x.
Time is taken to cross = t.
Speed of both trains = U and V.
Time taken after t seconds to reach Patna = t+9.
Time taken after t seconds to reach Howrah = t+16.

x=(U+V)t-----------(1)
after time t total distance covered by train A.

x=U(t+9)------------(2)
After time t total distance covered by train B.

x=V(t+16)-----------(3)
By solving eq(1) and eq(2).

we get Vt=9U-----------(4)
by solving eq(1) and eq(3).

we get Ut=16V----------(5)
by solving eq(4) and eq(5).

We get U/V=4/3
therefore the ratio is 4:3.

Let say the distance between them is d, and it takes time t when both the trains meet each other.

So relative speed = v1+v2.

So t = d/ (v1+v2) -----> (1).

Now for the train 1 distance remains = d - t*v1.
For the train 2 distance remains = d-t*v2.

According to the question, it takes 9 hrs for the first train and 16 hrs for the second train.
So, 9 = (d-t*v1) /v1 and 16 = (d-t*v2) /v2.

Now from the equation 1, d = t*v1+t*v2 substitue above.
So 9 = t*v2/v1 and 16 = t*v1/v2.

Equating t, 9v1/v2 = 16v2/v1 so v1/v2 = 4/3.

Two trains start simultaneously,one traveling at vh the other at VP.

These trains must travel a distance of length l.

They meet at a point x from Howrah that is l-x from Patna. Since they both started at the same time it means they took equal amount of time t to cover the distances.

x = vh*t.
l-x = vp*t.

x/(l-x) = vh/vp.

Now it takes the Patna bound train 9 hours to cover l-x.
It takes the Howrah bound train 16 hours to cover x.

x = 16vp.
l-x = 9vh.

16vp/9vh = vh/vp.
vh2/vp2 = 16/9.

Perform the square root.

Let's say.

Train A travels 'x' is the distance from Patna with speed 'V1' till the point of meet and
train B travels 'y' is the distance from Patna with speed 'V2' till the point of the meet.

the time taken for them to meet 't' = x/V1= y/V2 ---- (1)

Now Train A will travel 'y' distance that was previously travelled by train 'B' and vice-versa.

As per given info;
y/V1 = 9 and x/V2 = 16.

Taking ratio:
y/V1*V2/x = 9/16.

but from (1) -----> y/x = V1/V2,

=> ( V1/V2)^2 = 9/16.
=> V1:V2 = 3:4.

Let the total distance b/w two station = x.

Let after t time they will meet and their speed is a and b respectively, then
distance travel by trains,

at +bt = x ... (eq1).

After meeting they reach their destinations after 9hr and 16 hrs,

So, a(t+9) = x ... (eq2).
And b(t+16) = x ... (eq3).

Now, by equating the eq1 and eq2.
at+bt = a(t+9).

a/b = t/9... (eq4).

By, eq1 and eq3.
a/b = 16/t ... (eq5).


by, eq4 and eq5.

t*t = 9*16.
t = 3*4.
Put the value of t in eq5.

a/b = 4/3 answer.

Consider first train speed v' & other v"

After meeting point v'=(meeting point to patna)/9 = b/9.
v"=(meeting point to howrah)/16 =a/16
v'/v"=[b/a][16/9] ------> (1)
when train start and meet each other
time taken is same ie. T.

v'= (howrah to meeting point)/T = a/T.
v"=(patna to meeting point)/T = b/T.
then from above eq. v'/v"=a/b -----> (2)put in eq. 1.
From eq. 1;
v'/v" = (v"/v')[16/9].
v'^2/v"^2 = 16/9.
v'/v" = 4/3.