and these with an additional 1⁄2 make 1; for applied to 45 these fractions are equal to
111⁄451⁄21⁄8541⁄211⁄211⁄81and 15.
The author does not explain how he gets the fractions 1⁄91⁄40. Following the method of the first two problems of this section he would have to multiply 45 so as to get 651⁄8. This he might have done as follows:
| 1 | 45 | |
| 1⁄10 | 41⁄2 | |
| 1⁄5 | 9 | |
| \ | 1⁄9 | 5 |
| 1⁄20 | 21⁄4 | |
| \ | 1⁄40 | 11⁄8 |
| Total | 1⁄91⁄40 | |
In the proof of this problem, where the number to be obtained is 2⁄3, he adds in also 1⁄3 and so gets 1, which is itself the number to be obtained in Problems 21 and 22.
SECTION V
Problems 24-29. 'AHA' or Quantity Problems
Problem 24
A quantity and its 1⁄7 added together éecome 19. What is the quantity?
Assume 7.
| \ | 1 | 7 |
| \ | 1⁄7 | 1 |
| Total | 8. | |
As many times as 8 must be multiplied to give 19, so many times 7 must be multiplied to give the required number.
| 1 | 8 | |
| \ | 2 | 16 |
| 1⁄2 | 4 | |
| \ | 1⁄4 | 2 |
| \ | 1⁄8 | 1 |
| Total | 21⁄41⁄8. | |
| \ | 1 | 21⁄41⁄8 |
| \ | 2 | 41⁄21⁄4 |
| \ | 4 | 91⁄2 |
Do it thus:
| The quantity is | 161⁄21⁄8 | |
| 1⁄7 | 21⁄41⁄8 | |
| Total | 19. |
