Amy has to visit towns B and C in any order. The roads conne

Thanks Paresh and Karishma .Kudos to u both

Clockwise:
Lets assume Weight of A is 1, So the weight of B is 1 as only one route is there to reach.
Weight of C = 1+1 (2 ways to reach C from B) = 2
Weight of A (the last point in the journey ) = 2 + 2 (2 ways to reach A from C) = 4

Similarly for Anti-clokwise :
Lets assume weight of A is 1, so the weight of C = 1+1 (2 ways to reach to C from A) = 2
Weight of B = 2+2 (as there are 2 ways to reach B from C) = 4
Weight to reach A = 4 (as only one way to reach A from C) = 4

hence total is 4+4 = 8 ways (option B))

For any questions on route, this is the best way to solve.

To make the concept a little more clear, I have just added a example which is attach.

How many ways a person can reach from A to I if only direction allowed to traverse is from left to right?


Weight of A = 1
Weight of B = 1
weight of C = 1
Weight of D = 1
weight of E = 1+1 = 2
weight of F =1
weight of G = 1
weight of H = 1+ 2 + 1 = 4
weight of I = 1 + 4 + 1 = 6

so 6 ways are possible.

The questions can be made a little more complex as there might be a circular path to a node or in one of the path back traversal is allowed (only once for example), then the weights can be calculated again. This method takes care of double counting and with practice, it can be done in secs.
One of my Quant teacher used to say "Clicking the option takes more time than to solve the question".

Hi Bunuel,

Can you shed some light on this concept of weights? Is this type of question frequent in gmat?

Thanks Karishma for a simple explanation. +1 to you!

[quote="kumar83"]

Given: Amy has to visit towns B and C in any order. The roads connecting these towns with her home are shown on the diagram.
Asked: How many different routes can she take starting from A and returning to A, going through both B and C (but not more than once through each) and not travelling any road twice on the same trip?

Number of ways to visit B from A = 2
Number of ways to visit C from B = 2
Number of ways to visit A from C = 2 (There are 3 ways out of which 1 is already used for B-C movement)

Total number of ways = 2*2*2 = 8

IMO B

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