Roadies.

Here is the solution.
But in my opinion it's definitely not a GMAT type question.
It will take 3-4 minutes to solve.



In the above image, the direct paths connecting a, b and c are marked with x, y and z, which represents the number of direct paths between them. Thus between a and b the number of direct paths is x etc...

From a to b :

• The number of direct paths = x.
  There are y direct paths from a to c. For each of the direct paths from a to c, there are z direct paths between b and c. Thus number of paths from a to b through c = y*z = yz
  
  Total number of paths from a to b = (x + yz) = 33

From a to b :

• By the same logic, total number of paths from b to c = (z + xy) = 23

Addition of the equations results, (x + z + xy + yz) = 56
=> [x + z + y(x + z)] = 56
=> [(x + z)(y + 1)] = 56

The possible two factor factorizations of 56 are (2*28), (4*14) and (7*8). Let's analyze each of the cases individually.

Take (2*28)

• (1) If (x + z) = 2 and (y + 1) = 28, then y = 27 => 2nd equation is not possible.
  (2) If (x + z) = 28 and (y + 1) = 2, then y = 1 => (x + z) = 33 from 1st equation and (x + z) = 23 from 2nd equation. Impossible.

Take (4*14)

• (1) If (x + z) = 4 and (y + 1) = 14, then y = 13 => Form 2nd equation x cannot be greater than 1, which means x = 1 => z = 3. But these values don't satisfy the equations. Impossible.
  (2) If (x + z) = 14 and (y + 1) = 4, then y = 3 => Solving the two equations with y = 3 result fractional values of x and z. Impossible.

Take (7*8)

• (1) If (x + z) = 7 and (y + 1) = 8, then y = 7 => Solving the two equations with y = 3 result fractional values of x and z. Impossible.
  (2) If (x + z) = 8 and (y + 1) = 7, then y = 6 => Solving the two equations with y = 6 results x = 3 and z = 5. Possible

Thus y = 6

The correct answer is A.