Aptitude - Problems on Trains - Discussion

Speed = Distance/Time; Then how will you write root of time?

How can you calculate speed as the square-root of time?

It is formula, see below for important formulas.

Problems on Trains - Important Formulas

Can someone prove that?

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

Here in this question, one train had reached in 9 hrs and another one in 16 hrs.The first train reached faster than the second one ,clearly it should have more speed than the second one, as we know that time and speed are inversely proportional and comparing all answers, 4:3 is most optimal.

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).