In the adjoining figure, the lines represent one-way roads allowing travel only northwards or only westwards. Along how many distinct routes can a car reach point B from point A?

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.

From the above given figure and the conditions R can receive only 6 units (5 + 1) of natural gas if utilization is 100%.
Therefore , required answer will be 6 units.

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 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.

According to question ,
If all the five routes have the same cost, then there will be an equal flow in all the five routes, i.e. 20% in each route.
But then the percentage of traffic in S - A = 20% (Only one route involving S - A)
S - B = 40% (As there are two routes involving S - B)
S - D = 40% (As there are two routes involving S - D)
But here the given condition that traffic in S-A is equal to that in S - B, which in turn is equal to S - D is not satisfied.
Of the routes, that can be used the number of routes involving S - A must be the same as S - B, which in turn is same as that as S - D.
That is possible only when we block the junction C and that can be done by taking higher toll charge at C to achieve this goal c > 3.
Hence , required answer will be option A .

From the given diagram ,
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 the cost of travel including toll on routes S-A-T, S-B-C-T, S-B-A-T and S-D-C-T is the same. And D-T has no traffic due to high toll charge at D.
From the last solutions we will get b = 5, 14 + a = 7 + b + c = 12 + c, or a + 2 = c 7 + b + c = 10 + c + d = 12 + c or d = 2, hence the result is B = 5, d = 2 and c-a = 2 that is satisfied by option (E).