A train starts from a Jaipur towards Delhi at 10 am. .. A train starts from a Jaipur towards Delhi at 10 am. Another train starts from Delhi towards Jaipur at 11 am. Both reach their destinations at same time. At 1 pm, the sum of distances that had been traveled by them together is 90% of distance between Delhi and Jaipur. At what time did they reach their destinations? 
I could only make equation like this :
Train(Jaipur's speed) = x
Train(Delhi's speed) = x+1 (since it starts one hour later and and both reaches at same time)
so equation : 3x + 2(x+1) = 90% of distance
now I'm out of any clue where to move... Thanks in advance :) 
 A: If  $d$ is the distance & $t$ is the time taken by the first train
$\dfrac dt\cdot3+\dfrac d{t-1}\cdot2=\dfrac{90d}{100}$
A: If the speed of the first train is $x$ and $t$ is the time it takes it to get to Delhi, then the distance between Delhi and Jaipur would simply be this: $s=xt$. If the speed of the second train is $y$, that same distance can be expressed with this expression: $s=y(t-1)$. It takes the second train one hour less to get to Delhi. Thus:
$$
x=\frac{s}{t},\ y=\frac{s}{t-1}.
$$
1 PM means that it's been 3 hours for the first train since it made its departure from Jaipur. For the second train, it means that it's been only 2 hours since it made its departure from Jaipur. With that information, we can obtain the following relation:
$$
3x+2y=0.9s.
$$
Substitute $x$ and $y$ with the expressions we got a few lines earlier and divide the whole thing by the common factor $s$:
$$
3\frac{s}{t}+2\frac{s}{t-1}=0.9s\implies\\
\frac{3}{t}+\frac{2}{t-1}=0.9\implies\\
9t^2-59t+30=0.
$$
All you have to do now is solve that quadratic equation for $t$. One of the solutions is going to be extraneous. Add the one that is the solution ($t=6$ hours) to 10 PM and that is going to be the time the trains reach Delhi (4 PM).
