Aptitude - Problems on Trains - Discussion

Before Meet these trains;

x = Sa * t.
D-x = Sb * t.

After meet the trains.
D-x = 9 * Sa,
x = 16Sb.

Then D-x, x put on the other equation;
get Sa * t = 16b , Sb * t = 9Sa.
and make it equal 16Sb/Sa = 9Sa/Sb.

Sa/Sb = 4/3.
And the Answers is 4:3.

Good questions, Thanks all for giving explanations.

Very good to practise. Thanks all.

Good. Thanks everyone for explaining the answer.

Nice, Thanks all.

Here we can use the ratio method:
Train 1 ( Howrah to Patna) = A.
Train 2 ( Patna to Howrah) = B/

Ratio of speed = √speed of A/ √speed of B
= √16/√9.
= 4/3.

That is the answer is 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.

Very useful. Thanks for the explanation.

Thank you for giving useful questions.

Thanks everyone for explaining the answer.