What is the maximum quantity of natural gas than can be transported from M to R?

We know that cities M and P are gas plants.Hence , option A will be correct answer .

From the above given figure ,
The maximum quantity of natural gas that S can receive = 6 + 2 + 9 = 17 units.
Therefore , required answer will be 17 units.

According to question ,
It can be seen that every city is connected to all the other cities (i.e. 3 other cities).
Step 1: Let starting point is A, there are 3 ways in which we could proceed, viz. AB, AD or AC.
Step 2: Once we are at any of these cities (B, D or C), each one of them is connected to 3 other cities. But since we cannot go back to A the originating city, there are only 2 ways in which we could proceed from here.
Step 3: let us assume that we are at B, we can only go to D or C by taking BD or BC respectively. From this point we a choice of either directly going back to A (thus skipping 4th city or go to 4th city and come back to A. )
Step 4:- Now if we are at D, we can either take DA or DCA. So there are 2 more ways to go from here.
So , total number of ways = 3 x 2 x 2 = 12 ways.

As per the given above figure , we can see that
There are four ways to go from A to the first level of nodes. Each of these 4 nodes in turn leads into two more ways to go to the second level nodes.
Each of the second level nodes leads into two more ways to go to the third level nodes.
And from here we have only one way each to go to B.
Hence by fundamental principal of counting, total number of ways = 4 x 2 x 2 x 1 = 16 ways.

From above given route diagram , we can see that
Free capacity at Avanti-Vaishali pipeline is 300, since capacity of each pipeline is 1000 and demand at Vidisha is 400 and 300 flows to Jyotishmati.
Thus, free capacity = {1000 ? (400 + 300)} = 300

On the basis of above given question , we can say that
We can block B to D if A to B means R₁ and B to F means R₂ is blocked.
Therefore minimum 2 ways needed to be block.

Here , number of vertical steps ( v ) = 3
Number of horizontal steps ( h ) = 5
Then in this case total number of ways is given by ^h+vC_h = ^h+vC_v = ⁸C₃ = 6 x 7 x ( 8/6 ) = 7 x 8 = 56.
Hence , 56 distinct routes can a car reach point B from point A .

As per the given figure , we can see that
Let the toll charged at junctions A, B, C and D be a, b, c and d respectively.
Then 1^st we will list down all the routes and corresponding cost of travel.
Since Route BC is under repair hence route S-B-C-T is not in use.
Rest all four have the same toll charges hence 14 + a = 9 + a + b ? b = 14 - 9 = 5
Similarly 10 + c + d = 13 + d ? c = 13 - 10 = 3
Hence Options 4 is ruled out, now if we check option rest 3 options we will find out that option 2 and 3 both are correct. Option (2)/(3) Inconsistent options .

As per the given diagram , we can see that
Let the toll charged at junctions A, B, C and D be a, b, c and d respectively.
Then 1^st we will list down all the routes and corresponding cost of travel.
Here in this case all 5 routes have the same toll charge hence 14 + a = 7 + b + c = 13 + d = 9 + a + b = 10 + c + d
After solving we will get a = 1, b = 5, c = 3 and d = 2

On the basis of above given question , we can say that
There must be one other route other than those involving B.
We must take S - D - C - T as the other route.
S - B - C - T, if toll at B = 3, total cost = 10
S - D - C - T, if toll at D and C is 0, total cost is 10.
Hence ,$ 10 is the least cost.