Time and distance: Buses traveling in opposite directions Two buses starting from two different places A and B simultaneously, travel towards each other and meet after a specified time. If the bus starting from A is delayed by 20 minutes, they meet 12 minutes later than the usual time. If the speed of the bus starting from A is 60km/hr, what is the speed of the second bus?
My attempt:
Let t be the usual time of meeting and s(A) and s(B) be the speeds of Bus A and Bus B respectively, then
Applying relative velocity concept (stopping bus B i.e. taking it's speed =0 and adding it to speed of A), i got:
s(A)= 60 (given)
(60 + s(B))t = d (distance between the buses)
Now if bus A is delayed by 20 mins, then B's travelling time is 20 mins more than A's to cover the same distance.
t(B) = t(A) + 20/60
hence, t(B)= t(A) + 1/5
distance covered by A = t(A) * 60
distance covered by B = t(B) * s(B) = (t(A) +1/5)*s(B)
The problem: How do i relate the delay of 12 mins in their meeting time?
 A: When the buses meet, Bus A has travelled $8$ fewer minutes than usual, and therefore $8$ fewer km. 
These $8$ km were travelled by B in the $12$ extra minutes. So the speed of B is $\dfrac{8}{12}$ km per minute. 
A: So, the speed of $A$ is $1 KM/minute$
Let the speed of $B$ is $u KM/minute$
If they normally meet in $t$ minutes, 
the distance covered by $A$  is $t$ KM and the distance covered by $B$  is $u\cdot t$ KM 
So, total distance $u\cdot t+t$ KM
If $A$ delays 20 minute, $A$ travels for $t-20+12=t-8$ minutes,
the distance covered by $A$  is $t-8$ KM and the distance covered by $B$  is $u(t+12)$ KM
So, total distance becomes $t-8+u(t+12)$ KM
So, $t-8+u(t+12)=ut+t\implies 12u=8,$ or $u=\frac23$
So, the speed of $B$ bus is $\frac23  KM/minute=60\cdot \frac23  KM/Hour=40 KM/Hour$
A: ADDED. Solution using your equations and making the necessary corrections and additions. 
(60 + s(B))t = d 
distance covered by A = t(A) * 60
distance covered by B = t(B) * s(B) = (t(A) +1/5)*s(B)
(distance between buses = d = distance covered by A + distance covered by B)

t(B) = t(A) + 20/60
hence, t(B) = t(A) + 1/5

Correction: 20/60 = 1/3; hence, t(B) = t(A) + 1/3.

How do i relate the delay of 12 mins in their meeting time?

t(B) = t + 1/5.

Let $d$ be the distance from $A$ to $B$, $v_{A}=60$ km/hr be the speed of
the bus departing from $A$ and $v_{B}$ the speed of the bus departing from $B$. If both buses start simultaneously, $v_{A}t$ is the distance traveled by bus $A$ and $v_{B}t$ is the the distance traveled by bus $B$, where $t$ is the usual time they spend to meet each other. We have
$$v_{A}t+v_{B}t=d.$$
(The units have to be consistent: e.g. speed in km/hr, time in hours and distance in km). If the bus starting from $A$ is delayed by 20 minutes =$\frac{1}{3}$ hour,
then bus $A$ travels the distance $v_{A}(t-1/3+1/5)$ and bus $B$ travels $v_{B}(t+1/5)$, because they meet $12$ minutes = $\frac{1}{5}$ hour  later than the usual time $t$. 
$$v_{A}(t-1/3+1/5)+v_{B}(t+1/5)=d.$$
Equate both equations 
$$v_{A}t+v_{B}t= v_{A}(t-1/3+1/5)+v_{B}(t+1/5),$$
and simplify 
$$\frac{2}{15}v_{A}-\frac{1}{5}v_{B}=0.$$
Since $v_{A}$ is 60km/hr, we have
$$\frac{2}{15}\left( 60\right) -\frac{1}{5}v_{B}=0,$$ 
whose solution is $v_{B}=40$ km/hr.
A: let their initial meeting point was C.As bus from A starts 20 min late , so both buses would have met at the same place if bus from B also have been late by 20 minutes , as the bus from B reached C at its usual time, now if their 2nd meeting point is D then from C to D bus from B takes 12 minutes while the bus from A will take 8 minutes from D to see, so ratio of their speed = 12/8 = 3/2
so speed of bus from B = 60 *2/3 = 40 kmph
for more : https://www.handakafunda.com/motion-of-two-bodies-in-a-straight-line/ 
