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).

From the above given diagram , we can see that
Maximum distance is when path taken is-
A ? D ? C ? B ? G ? E ? F ? H = 1300 + 500 + 200 + 400 + 400 + 300 + 600 = 3700 km
Minimum distance is when path taken is-
A ? C ? G ? H = 600 + 400 + 400 = 1400 km
Hence , Required ratio is 37 : 14.

Number of paths is as follows:
Starting from HGB: HGBA, HGBCA, HGBCDA, HGBCEDA, HGBCEFDA (Total 5 paths)
Starting from HGC: HGCA, HGCBA, HGCDA, HGCEDA, HGCEFDA (Total 5 paths)
Starting from HGE: HGECA, HGECBA, HGECDA, HGEDCA, HGEDCBA, HGEDA, HGEFDCBA, HGEFDCA, HGEFDA (Total 9 paths)
Starting from HEG: HEGBA, HEGCBA, HEGCA, HEGCDA (Total 4 paths)
Starting from HEC: HECGBA, HECBA, HECA, HECDA (Total 4 paths) Starting from HED: HEDCBGA, HEDCBA, HEDCA, HEDA (Total 4 paths)
Starting from HEF: HEFDCGBA, HEFDCBA, HEFDCBA, HEFDCA, HEFDA (Total 5 paths)
Starting from HFE: HFEGBA, HFECGBA, HFECBA, HFECA, HFEDCGBA, HFEDCBA, HFEDCA, HFEDA, (Total 8 paths)
Starting from HFD: HFDEGBA, HFDEGCA, HFDEGBCA, HFDECGBA, HFDECBA, HFDECA, HFDCEBGA, HFDCGBA, HFDCBA, HFDCA, HFDA (Total 11 paths)
Total number of paths is: 5 + 5 + 9 + 4 + 4 + 4 + 5 + 8 + 11 = 55
Hence , the total number of paths from city H to city A is 55 .

As per the given above diagram , we can see that
For minimum distance path is HGCA = 400 + 400 + 600 = 1400
Therefore , the minimum distance between H to A without visiting city E is 1400 km .

On the basis of above given diagram , we can say that
The minimum distance between A and H through D is HEDCA and distance is 500 + 500 + 500 + 600 = 2100.
Hence , required answer will be 2100 km.

From solution of question number 26 ,
Starting from HEC: HECGBA, HECBA, HECA, HECDA (Total 4 paths)
Hence , the total number of such paths is 4 .