Aptitude - Problems on Trains - Discussion

Let distance = x, a as speed of train a & b as speed of train b

9 x a = x
16 x b= x

9a = 16b

a/b = 16/9, simplify this with square root both up & down we get 4/3 & ratio 4:3.

Suppose the total distance is x,

Time taken by trains to meet = t

if speeds be s1 and s2, and they meet after time t,
=> s1*t + s2*t =x (eqn 1)

Train A takes total t+9 hours to travel x with speed s1,

=> s1(t+9)=x (eqn 2)

for train B, total time t+16

=> s2(t+16)=x (eqn 3)


9*s1= t*s2 (equate eqn 1 and 2)

t*s1= 16*s2 (equate eqn 1 and 3)

divide,

9/t= t/16 => t*t= 9*16

substitute t,

=> s1/s2= sqre-root (16/9)

Explain formula used.

SIMPLY THEY ARE USING DISTANCE FORMULA DISTANCE=TIME*SPEED

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).

Ishita Narula z correct....

How can we calculate the time to cover the total distance, for this sum?

@abhisek.

Please be reasonable yaar. Otherwise you will be a laughing stuff to everyone.

Ishita narula is correct. Thank you.

One of the smart way of doing the question is by seeing the options itself. Since the first train reach its destination earliest after meeting the second train, then its speed would be greater than the second train and so as the ratio. In the options given above, only in B the ratio of the first train greater than the second.