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 .
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.
From the given diagram ,
Let the toll charged at junctions A, B, C and D be a, b, c and d respectively.
Then 1st 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).
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 .
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.
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 1st 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
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 .
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.
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