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
||Cost (in rupees)
|$A – B – C – Z$
||$3 + 5 + 5$ $= 13$
|$A – M – Z$
||$4 + 7$ $= 11$
|$A – M – E – Z$
||$4 + 4 + 2$ $= 10$
|$A – D – E – Z$
||$6 + 7 + 2$ $= 15$
If only one route is followed, all 1200 vehicles will reach all junctions on that route. If two routes are followed, then either 600 (if the junction is part of only one of the routes) or 1200 (if the junction is part of both the routes) vehicles will reach a junction on one of the routes. Neither of this is what we want.
If three routes are followed, then 400, 800 or 1200 vehicles can reach a junction on one of the routes. For 800 vehicles to reach $E$ each day, the traffic has to be distributed equally among three routes and junction $E$ must be a part of two out of those three routes.
The route of least cost is $A – M – E – Z$, with a cost of Rs. 10. We cannot change the cost of travel on any of the roads on this route, since we can only change the costs of travel on $M – Z$ and $D – E$, neither of which is a part of this route. Thus, we need to change the costs of two other routes so that their total cost becomes Rs. 10. The only two routes for which we can change the cost of travel are $A – M – Z$ and $A – D – E – Z$.
For travel on $A – M – Z$ to cost Rs. 10, the cost of $M – Z$ must be Rs. (10 - 4) = Rs. 6
For travel on $A – D – E – Z$ to cost Rs. 10, the cost of $D – E$ must be Rs. (10 - 6 - 2) = Rs. 2
Hence, option E is the correct choice.