Aptitude - Problems on Trains - Discussion

Lets assume that:

1. They meet at time t.

2. total distance between trains before they start is d.

3. when they meet, one of the train having speed "v1" has traveled "d1" distance. Hence other train having speed "v2" must have "d-d1" distance traveled.

Explanation:

Using the formula , distance=speed * time.

d1=v1*t ......(1) for train 1.

d-d1=v2*t ......(2) for train 2.

By using (1)and (2).
v1/v2=d1/(d-d1) .......(3).

Now , second train takes t+16 hours to cover "d" distance and other train takes t+9 hours for same distance.

So,

d=v1*(t+9) ...........(4).

d=v2*(t+16) ...........(5).

For result (4),

d=t*v1+9*v1.
Replacing t*v1 by result (1).

Then we get
d=d1+9v1 ................(6).

For result (5),
d=t*v2+16*v2.

Replacing t*v2 by result (2).
Then we get,
d=d-d1+16*v2 ................(7).

From (6) and (7) we can write,
9v1/16v2 = (d-d1)/(d1) ...............(8).

From (8) and (3) we write that,
9v1/16v2=v2/v1.

Hence:
v1*v1/v2*v2=16/9.
v1/v2=sqrt(16/9).
v1/v2=4/3.

Lets assume that:

1. They meet at time t.
2. total distance between trains before they start is d.
3. when they meet, one of the trains having speed "v1" has travelled "d1" distance. Hence other train having speed "v2" must have "d-d1" distance traveled.

Explanation:

Using the formula, distance=speed * time.
d1=v1*t -----> (1) for train 1.
d-d1=v2*t -----> (2) for train 2.

By using (1)and (2).
v1/v2=d1/(d-d1) -----> (3).

Now, the second train takes t+16 hours to cover "d" distance and other train take t+9 hours for the same distance.

So,

d=v1*(t+9)-----> (4).
d=v2*(t+16)-----> (5).

For result (4),

d=t*v1+9*v1.
Replacing t*v1 by the result (1).

Then we get,
d=d1+9v1-----> (6).

For result (5),
d=t*v2+16*v2.

Replacing t*v2 by the result (2).
Then we get,
d=d-d1+16*v2-----> (7).

From (6) and (7) we can write,
9v1/16v2 = (d-d1)/(d1)-----> (8).

From (8) and (3) we write that,
9v1/16v2=v2/v1.

Hence:
v1*v1/v2*v2=16/9.
v1/v2=sqrt(16/9).
v1/v2=4/3.

Howrah(A) ------------------------- Patna(B).

Let, distance between Howrah and Patna be x; S1 = speed of 1st train from A to B; S2 = speed of 2nd train from B to A.

Let, after t hours both trains meet each other.

So, distance covered by 1st train in t hours = S1 * t = distance covered by 2nd train after he meets the 1st train and goes towards station A.

So, S1 * t = S2 * 16 Or, S1/S2 = 16/t--> (I)

Again, distance covered by 2nd train in t hours = S2 * t = distance covered by 1st train after he meets the 2nd train and goes towards station B.

So, S2 * t = S1 * 9 Or, S1/S2 = t/9--> (II)
From equations (I) and (II), we get,
16/t = t/9
=> t^2 = 9*16 = 144
=>t = 12 hours ( as t is time hence -12 is neglected).

Hence, ratio of speeds,
S1/S2 = 12/9 = 4/3 (putting t=12 in equation(II)).
Or, S1 : S2 = 4 : 3.

It is a completely wrong answer.

<Case-1> Speed = Distance/Time

(As distance is same, so Speed(S1)= Distance(D)/Time(T1).
Similarly, (S2)= (D)/(T2), Now, (S1)/(S2)=[(D)/(T1)]/[(D)/(T2)]
=[(D)/(T1)]*[(T2)/(D)]
=(T2)/(T1)
So,ratio of speeds is inversely proportional to ratio of times.

<Case-2>

How can (16/9)=(4/3) or (9/16)=(3/4)??? Please Explain???

In Ratio,Proportion and percentage,(9/16)*100 = 56.25%;
But,(3/4)*100 = 75%.

Now, PLEASE EXPLAIN,IS 56.25% = 75% ???

THE CORRECT ANSWER TO THE ABOVE QUESTION IS (16/9) or 16:9.
OR IF WE CONSIDER THE REVERSE RATIO(RECIPROCAL)OF SPEEDS, THEN (9/16) or (9:16).

d1+d2= D;(total distance)

v1=d1/t1; =>v1=(D-d2)/t1----> EQ1

v2=d2/t2; => v2=(D-d1)/t2---->EQ2

d1 is distance traveled by first train to meet the 2nd, similarly d2 is distance traveled by second train to meet first train;

t1 is time taken by train 1 to meet the other
t2 is time taken by train 2 to meet first train

now d1+d2= D;(total distance)

according to given data

v1=(D-d1)/9;
v2=(D-d2)/16;


v1/v2=(16/9) *((D-d1)/(D-d2));

=(16/9) *((v2/t2)/(v1/t1))----->from EQ1 and EQ2

(v1/v2)^2 =(16/9)(t1/t2)

t1 and t2 are equal as trains start at same time and both meet after equal amounts of time so, memorize the above formula

For this question the direct formulas are:
(A's speed) : (B's speed) = root(b) : root(a)
Time at which they meet= root(a*b)

Derivation to the formulas:

Let X be the total distance, t be the time at which they meet
a,b are times at which they reach the destination after they meet.
s1, s2 are speeds of train1, train2 respectively.

then X = t(s1+s2) ...(1)
X = (t+a)s1 ...(2)
X = (t+b)s2 ...(3)

after equating (1) and (2)
t.s2 = a.s1 ...(4)
after equating (1) and (3)
t.s1 = b.s2 ...(5)

multiply (4) and (5)
t.t.a.b = s1.s2.a.b

" t = root(a*b) "

substitute t value in either (4) or (5)

then " s1/s2 = root(b)/root(a) "

Let's assume the speed of train A = x
Speed of train B = y
Trains A& B meet after 't' hours.

(Distance traveled by train A in 't' hours(distance travelled=speed*time=x*t) will be the distance train B going to travel in 16 hours after they meet(speed*time=y*16), in the same way, the distance traveled by train B in t hours(y*t) will be the distance train A going to travel in 9 hours after the meet(9*x).)
from the above statement.

xt=16y ---> equation1
yt=9x ---> equation2

From eq 2,
t=9x/y ----> put this in eq 1
x*9x/y = 16y.
9x^2 = 16y^2.
x^2 / y^2 = 16/9.
Taking the root in both sides of the equation.
x/y = 4/3.
x:y = 4:3.

Suppose total distance is x, and trains are T1 and T2;
Suppose speeds are S1 and S2;
T1 and T2 meet at time t;
.'. total distance is S1.t+S2.t=x;--->(eqn 1)

But according to question T1 and T2 covers the distance 'x' at (t+9) and (t+16) hr respectively.
.'. for T1, S1.(t+9)=x;--->(eqn 2)
And for T2, S2.(t+16)=x;--->(eqn 3)

Now from eqn1 and eqn2;
S2.t=S1.9;
=>S1/S2=t/9;--->(eqn 4)
And from eqn1 and eqn3;
S1.t=S2.16;
=>S1/S2=16/t;--->(eqn 5)

Now multiply eqn4 by S1/S2 from both sides we get
(S1/S2)^2=(t/9).(S1/S2);
=>(S1/S2)^2=(t/9).(16/t); [using eqn5]
=>S1/S2=sq root of (16/9);
=>S1/S2=4/3;

Let the time taken before they meet be 't' hours.
After meeting 1st train takes 9 hours and 2nd takes 16 hours.

dist= speed*time, but here speed is relative before they meet,

Therefore, d=(s1+s2)t -----(1).
d=s1.t+s2.t.

After they meet,

d= s1(t+9) ----------(2) for train-1.
d= s1.t+s1.9.
d= s2(t+16)----------(3) for train-2.
d= s2.t+s2.16.

Equate (1) and (2),
s2.t = s1.9.
s1/s2= t/9 ---------------(4).

Equate (1) and (3),
s1.t = s2.16.
s1/s2=16/t----------------(5).

Divide (4) by (5),
1=(t/9)/(16/t).
16/t = t/9.
t*t=16/9.
Taking square root on both side,
t=4/3.

Therefore, speeds are in ratio 4:3.

@Pawan :

Detailed explanation:
---------------------
Formula: Distance = (Speed x Time).

Speed = 90 km/hr

Convert km/hr into m/sec.

Therefore, Speed = (90 x 1000/3600) = 90000/3600 = 25 m/s

Now, Time = 4 hr 40 min

Convert it into secs.

Therefore = (4 x 3600) + (40 x 60 )= 14400+2400 = 16800 secs.

Now, Distance = 25 m/sec x 16800 sec = 25 x 16800 m

= 4,20,000 m [or]

= 420 km. [1000 m = 1 km]

Some short-cut tricks :
-----------------------

Distance = (Speed x Time)

We can write 4 hr 40 mins = (4 + 40/60) hrs.

So, Distance = 90 x (4 + 2/3)

= (90 x 4) + (90 x 2/3)

= 360 + 60

= 420 km. [or] 420000 m.