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.
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 .
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 .
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 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 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 .
From the above result length of route -2 = 11 km
Therefore , the length of route -2 is 11 km .
As per the given above figure , we can see that
The number of routes from P to Q = 2 + 4 = 6
Hence , the number of routes from P to Q is 6 .
Clearly 1st, 8th, 15th, 22nd and 29th October are Sundays.
So, 31st October is Tuesday.
? 1st November will be Wednesday.
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