Byron S. answered • 10/27/14

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To solve this I drew a triangle to represent the situation. I labeled the vertices A, B, and C. The length of (number of routes between) BC I labeled x, AB y, and AC z.

You can calculate the total number of routes by adding the direct routes plus the product of the two indirect routes. The number of routes between A and B is y + xz, between A and C is z + xy, and between B and C is x + yz.

You know that

y+xz = 89

z+xy = 82

x+yz > 99

If you subtract the first two equations you get

(y-z) + x(z-y) = 7

-1 + x = 7/(z-y) (divide by (z-y). remember that y-z = -(z-y) )

x-1 = 7/(z-y)

Since x has to be a whole number, (z-y) must be a factor of 7. The only choices are 1 and 7.

IF z-y = 1

z = y+1

x-1 = 7

x = 8

y+xz = 89

y+8(y+1) = 89

9y + 8 = 81

y=9

z=9+1=10

The number of routes between B&C is

x+yz = 8 + 9*10 = 98.

This is too small! This answer is incorrect.

IF z-y = 7

z=y+7

x-1=1

x=2

y+xz = 89

y+2(y+7) = 89

3y + 14 = 89

y = 25

z = 25+7 = 32

The number of routes between B&C is

x+yz = 2 + 25*32 = 802

x+yz = 2 + 25*32 = 802

The answer is (e)