times 10; it makes 7901⁄181⁄271⁄541⁄81 cubed cubits. Add 1⁄2 of it to it; it makes 11851⁄61⁄54, its contents in khar. 1⁄20 of this is 591⁄41⁄108. 591⁄41⁄108 times 100 hekat of grain will go into it.
Method of working out:
| 2 | 172⁄31⁄9 | |
| 4 | 351⁄21⁄18 | |
| \ | 8 | 711⁄9 |
| \ | 2⁄3 | 52⁄31⁄61⁄181⁄27 |
| 1⁄3 | 22⁄31⁄61⁄121⁄361⁄54 | |
| \ | 1⁄6 | 11⁄31⁄121⁄241⁄721⁄108 |
| \ | 1⁄18 | 1⁄31⁄91⁄271⁄1081⁄324 |
| Total | 791⁄1081⁄324 | |
| 1 | 791⁄1081⁄324 | |
| 10 | 7901⁄181⁄271⁄541⁄81 | |
| 1⁄2 | 3951⁄361⁄541⁄1081⁄162 | |
| Total | 11851⁄61⁄54 | |
| 1⁄10 | 1181⁄21⁄54 | |
| \ | 1⁄29 | 591⁄41⁄108 |
This problem is exactly like 41, but with 10 instead of 9 for diameter there are many fractions. The 1⁄108 of 100 hekat should be reduced to "Horus eye" fractions and ro, but in the papyrus this fraction is omitted.
Problem 43
A cylindrical granary of diameter 9 and height 6. What is the amount of grain that goes into it?
Do it thus: Take away 1⁄9 of it, namely, 1, from 9; the remainder is 8. Add to 8 its 1⁄3; it makes 102⁄3. Multiply 102⁄3 times 102⁄3; it makes 1132⁄31⁄9. Multiply 1132⁄31⁄9 times 4, 4 cubits being 2⁄3 of the height; it makes 455 1⁄9, its contents in khar. Find 1⁄20 of this, namely, 221⁄21⁄41⁄80. The amount of grain that will go into it is 221⁄21⁄4 times 100 hekat 1⁄21⁄321⁄64 hekat 21⁄21⁄41⁄16 ro.
Method of working out:
| \ | 1 | 8 |
| 2⁄3 | 51⁄3 | |
| \ | 1⁄3 | 22⁄3 |
| Total | 102⁄3. | |
