From the solution of question number 25 ,
Maximum distance is when path taken is:
A ? D ? C ? B ? G ? E ? F ? H = 1300 + 500 + 200 + 400 + 400 + 300 + 600 = 3700 km.
Therefore , the maximum distance between A and H is 3700 km.

Map of the town is given as below:
It is given that Neelam took the shortest path hence she has to pass through path EF.
Number of ways to reach E from A is ²⁺²C₂ = ⁴C₂ = 6.
Number of ways to reach B from F is ⁴⁺²C₂ = ⁶C₂ = 15.
Hence total number of ways to reach B from A is 6 x 15 = 90.

From the answer of previous question,
Total number of ways from A to B is 90.
Now we need to calculate total number of ways from B to C that is 13 (on the same logic)
Hence, total number of paths from A to C is 90 x 13 = 1170.

As per the given conditions,
we require surveillance cameras that would cover all roads converging at an intersection. The cost for covering all the roads has to be minimum.
If we put a camera on '1', roads from '7' and '6' to 1 will be covered.
If we put a camera on intersection '2', roads converging from '3' to '2' will be covered.
If we put a camera on '3', roads from '3' to '5', '6' to '5' and '7' to '5' will be covered.
Now we are left with roads from 7 to 4 and from 6 to 4 both of them converge from 4 hence these will be covered by putting a camera on '4'. And then all the roads will be covered and the minimum cost is 2 + 5 + 1 + 4 = 12 lakhs

On the basis of above given question ,we can say that
Similar to the above answer cameras are put on 2, 5, 1 and 4 to minimize the cost.

As per the given above diagram , we can see that
Total number of paths from X to Y is 7.

From the solution of previous question ,
we have following results:
T₃ = T₇ = 0,
T₁ = 20 + T₂ + T₄, T₅ = 30 + T₆ + T₈.
From the given options option B satisfies the condition.
Hence , required answer will be 50, 20, 0, 10, 60, 20, 0, 10 .

On the basis of above given diagram , we can see that
Maximum expenditure is in path 3 and that is 260 + 50 x 6 = 260 + 300 = 560
Minimum expenditure is in path 5 and that is 230 + 50 x 3 = 230 + 150 = 380
? Required difference = Maximum expenditure is in path 3 - Minimum expenditure is in path 5 = 560 ? 380 = 180
Hence , Required difference is 180.

From the given condition ,
cost incurred in path 2 is equal to that in path 3 then T₃ = 0
Similarly ,equating cost of path 3 and path 4 T₇ = 0
Equating cost of path 4 and path 5 we will get
260 + T₆ + T₈ = 230 + T₅
? T₅ = T₆ + T₈ + 30
Equating cost of path 1 and path 2 we will get ,
240 +T₁ = 260 + T₂+ T₄
or T₁ = T₂ + T₄ + 20
For minimum total cost T₁ = 20 and T₅ = 30 then cost incurred is Rs 260.
Therefore , the minimum cost of travel from X to Y is Rs 260 .

If toll tax at all of the toll booths is same and let it be k then expense in each path is as follows:
Path 1: 240 + 3k
Path 2: 260 + 5k
Path 3: 260 + 6k
Path 4: 260 + 5k
Path 5: 230 + 3k
So from the given condition On solving we will get k = 50
Tool booth 3 will come in Path 3 and Path 4.
So cost incurred is:
Path 3: 260 + 6k = 260 + 300 = 560
Path 4: 260 + 5k = 260 + 250 = 510
Therefore , 560 or 510 is the cost of travel from X to Y through toll Both T3 .