A well of diameter 6 m is dug 14 m deep. All the earth taken out is spread evenly over a rectangular platform 18 m long and 11 m wide. Find the height of the platform. (Take π = 22/7.)

Earth dug out = volume of the cylindrical well, radius 3 $m:\pi r^{2}h=\frac{22}{7}\times 9\times 14=396m^3$. Spread over the platform it becomes a cuboid of base $18\times 11=198{\text{m}}^2$, so height = 396 ÷ 198 = 2 $m.8m$ comes from using the diameter 6 $m$ as the radius; 6.83 $m$ divides by the platform's perimeter 58 $m$ instead of its area; 1 $m$ halves the correct height.