Then
| Number of cutters in set. | Number of cutter. | |||
| 8 | - | 7 | = | 1 |
| Number of teeth in smallest wheel. | The number of the cutter. | The number of the teeth in largest wheel. | ||
| 12 | × | 8 | ÷ | 300 |
| 12 | |||||
| 8 | |||||
| 300 | ) | 96 | 0 | ( | 0.32 |
| 90 | 0 | ||||
| 6 | 00 | ||||
| 6 | 00 | ||||
Now add the 1 to the .32 and we have 1.32, which we must divide into the 96 first obtained.
Thus
| 1.32 | ) | 96 | .00 | ( | 72 | |
| 92 | 4 | |||||
| 3 | 60 | |||||
| 2 | 64 | |||||
| 96 | ||||||
Hence No. 8 cutter may be used for all wheels that have between 72 teeth and 300 teeth.
To find the range of wheels to be cut by the next cutter, which we will call No. 7, proceed again as before, but using 7 instead of 8 as the number of the cutter.