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by goyalsau » Thu Dec 02, 2010 9:25 am
There are three cities a,b,c .Each of these cities is connected with the other two
cities by at least one direct road.If a traveller wants to go from one city to another city,she can do so either by tranversing a road connecting the two cities directly or by transversing two roads,the first connecting the origin to the third city and the second connecting the third city to destination .In all there are 33 routes from a to b ,including through c.similarly there are 23 routes from b tp c ,including a.How many roads are there from a to c directly?

A)6 B)3 C) 5 D)10 E)8
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by Rahul@gurome » Thu Dec 02, 2010 11:03 am
Here is the solution.
But in my opinion it's definitely not a GMAT type question.
It will take 3-4 minutes to solve.

Image

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.
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