What is the difference between maximum amount and minimum amount that Mr. Ricky can spend on a day if he wants to travel from his home (X) to office (Y) if toll tax at each toll booth is Rs 50?

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 .

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

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 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

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.

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 .

According to question ,
Here ,we will eliminate the options one by one. Option (A) cannot be true as there are many routes that satisfy the given condition. Option (C) is also not true as we can have a route starting from D (e.g. DEBDCBAC).
The route need not necessarily end at E, which is apparent. Option B satisfies all the conditions.
Hence , A route can either start at C or end at C, but not both.

On the basis of above given figure , we can see that
City A is connected by 2 roads, B by 4 roads, C by 3 roads, D by 3 roads and E by 2 roads.
For a city to be starting city for such a route, it has to be connected by odd number of roads.
Hence the required answer is 2, i.e. C and D.

As per the above result ,
Route -1 :- 5 + AB = 13 ? AB = 13 - 5 = 8 km
Hence , required distance between A and B is 8 km .