There are three pillars X, Y and Z of different heights. Three spiders A, B and C start to climb on these pillars simultaneously. In one chance, A climbs on X by 6 cm but slips down 1 cm. B climbs on Y by 7 cm but slips down 3 cm. C climbs on Z by 6.5 cm but slips down 2 cm. If each of them requires 40 chances to reach the top of the pillars, what is the height of the shortest pillar?

1. First, let's find the 'Net Gain' per chance for each spider to see who is the slowest. This is a concept of Net Displacement [1].\n - Spider A: Climbs 6, Slips 1. Net = 5 cm.\n - Spider B: Climbs 7, Slips 3. Net = 4 cm.\n - Spider C: Climbs 6.5, Slips 2. Net = 4.5 cm.\n\n2. The question asks for the *shortest* pillar. Logic dictates that for the same number of chances (40), the spider with the lowest Net Gain will cover the shortest distance. Spider B has the lowest Net Gain (4 cm). Therefore, Pillar Y is the shortest.\n\n3. Now, we only need to calculate the height for Spider B (Pillar Y). We use the Boundary Condition logic [8] where the last step is purely a climb.\n - Total Chances = 40.\n - First 39 chances: Spider climbs and slips (Net Gain of 4 cm).\n - 40th chance: Spider climbs 7 cm and reaches the top (No slip).\n\n4. Calculation:\n - Height = (39 chances * 4 cm) + 7 cm\n - Height = 156 + 7 = 163 cm.\n\n5. Checking options: 163 cm is Option (B). No need to calculate others.