By calculating the cross-sectional area of the shapes (counting squares), the follow volumes can be obtained:
| Letter | Volume $/\textrm{m}^3$ |
| I | 14 |
| L | 14 |
| O | 14 |
| V | 13 |
| E | 14 |
| M | 14 |
| A | 13 |
| T | 14 |
| H | 14 |
| S | 14.5 |
Therefore {I, L, O, E, M, T, H} all take the same time to fill. S takes the longest and A the shortest time.