# Section-1: Critical Path Solved QuestionLogical Reasoning Discussion

Common Information

1200 vehicles travel every day from Point $A$ to Point $Z$ on a network of one-way roads as shown in the diagram below.

Points $B$, $C$, $M$, $D$ and $E$ are junctions in this network. The number adjacent to the ray depicting each road stands for the cost (in rupees) of travelling on that road. Each vehicle takes the path of least cost from $A$ to $Z$.

If two or more paths have the same cost, then the vehicles are distributed equally on those paths.

 Q. Common Information Question: 3/4 If the cost of travel on the road from C to Z was reduced by 3 rupees, how many vehicles would travel to junction B every day?
 ✖ A. 300 ✖ B. 400 ✔ C. 600 ✖ D. 1000 ✖ E. 1200

Solution:
Option(C) is correct

Let us analyze the given information.

Here the roads have associated costs, but the nodes do not have associated costs.

From the diagram, observe that the possible routes from $A$ (the initial point) to $Z$ (the final point) are:

$A – B – C – Z$, $A – M – Z$, $A – M – E – Z$ and $A – D – E – Z$.

Let us tabulate the total costs incurred in travelling along each path.

Table below can be scrolled horizontally

Route Cost (in rupees)
$A – B – C – Z$ $3 + 5 + 5$ $= 13$
$A – M – Z$ $4 + 7$ $= 11$
$A – M – E – Z$ $4 + 4 + 2$ $= 10$
$A – D – E – Z$ $6 + 7 + 2$ $= 15$

Since the cost of the routes involving path $CZ$ changes, one can either draw a new table with recalculated values or change the value in only one cell. The path $CZ$ appears only in one route. Hence, redrawing the entire table may be unnecessary.

The new cost of travelling from $C$ to $Z$ is Rs. 5 – 3 = Rs. 2. Hence, the redrawn table is as shown below.

Table below can be scrolled horizontally

Route Cost (in rupees)
$A – B – C – Z$ $3 + 5 + 2$ $= 10$
$A – M – Z$ $4 + 7$ $= 11$
$A – M – E – Z$ $4 + 4 + 2$ $= 10$
$A – D – E – Z$ $6 + 7 + 2$ $= 15$

The two paths, $A – B – C – Z$ and $A – M – E – Z$ have the same cost (Rs.10) which is the least among the calculated values. Thus, the traffic divides itself equally among these two paths.

Hence, 600 vehicles travel along route $A – B – C – Z$ and 600 vehicles travel along route $A – M – E – Z$. All the cars travelling through $A – B – C – Z$ pass through junction $B$. Hence, 600 vehicles pass through junction $B$ on a daily basis.

Hence, option D is he correct choice.