174 De ruptis rotulorum⁴⁵⁶

Item \({1 \over 4}\) \({2 \over 3}\) unius rotuli pro \({1 \over 7}\) \({1 \over 6}\) \({2 \over 5}\) unius bizantii; quantum habuero de rotulis⁴⁵⁷ pro \({1~~4~~\phantom{1}7 \over 8~~9~~10}\) unius

149 b. 11 ℞ \({1 \over 7}\) \({1 \over 6}\) \({2 \over 5}\) \({1 \over 4}\) \({2 \over 3}\) 537   \({1~~4~~\phantom{1}7 \over 8~~9~~10}\) \({3~~\phantom{1}8~~\phantom{1}83~~11 \over 4~~10~~149~~12}\)

⁴⁵⁸ bizantii? Describe questionem sic, et multiplica \({1 \over 4}\) \({2 \over 3}\) per \({1~~4~~\phantom{1}7 \over 8~~9~~10}\) et divide per \({1 \over 7}\) \({1 \over 6}\) \({2 \over 5}\), quod sic fit: accipe prius \({1 \over 4}\) \({2 \over 3}\), et multiplicabis 2 que sunt super 3 in 4 et 1 quod est super 4 in 3 et adde insimul; erunt 11, que pone super \({1 \over 4}\) \({2 \over 3}\). 175 Et invenies numerum de \({1 \over 7}\) \({1 \over 6}\) \({2 \over 5}\); erunt 149, que pone super \({1 \over 7}\) \({1 \over 6}\) \({2 \over 5}\) et invenies numerum de \({1~~4~~\phantom{1}7 \over 8~~9~~10}\); erunt 537, et multiplicabis 11 per 537, que per ruptos [F:42r] qui sunt sub virgulis sub 149, scilicet per 5 et per 6 et per 7, et divides summam per 149 et per ruptos qui sunt sub 11 et sub 537, hoc est per \({1~~0~~0~~0~~0 \over 3~~4~~8~~9~~10}\) , qui insimul coaptati sunt \({1~~0~~0~~0~~0 \over 8~~9~~10~~149~~12}\), et⁴⁵⁹ evitabis hoc quod evitare poteris; exibunt \({3~~\phantom{1}8~~\phantom{1}83~~11 \over 4~~10~~149~~12}\).