Elements of Algebra

CHAP. IX. OF ALGEBRA. 157 step by step. Beginning therefore with the stivers, since $1$ ruble $= 47\frac{1}{2}$ stivers, or $2$ rubles $= 95$ stivers, we shall have Ru. Ru. Stiv. As $2 : 1000 :: 95 : 47500$ stivers; then again, Stiv. Stiv. Flor. As $20 : 47500 :: 1 : 2375$ florins. Also, since $2\frac{1}{2}$ florins $= 1$ Dutch dollar, or $5$ florins $= 2$ Dutch dollars; we shall have Flor. Flor. D.D. As $5 : 2375 :: 2 : 950$ Dutch dollars. Then, taking the dollars of Berlin, according to the exchange, at $142$, we shall have D.D. D.D. Dollars. As $100 : 950 :: 142 : 1349$ dollars of Berlin. And lastly, Dol. Dol. Du. As $3 : 1349 :: 1 : 449\frac{2}{3}$ ducats, which is the number sought. 482. Now, in order to render these calculations still more complete, let us suppose that the Berlin banker refuses, under some pretext or other, to pay this sum, and to accept the bill of exchange without five per cent. discount; that is, paying only $100$ instead of $105$. In that case, we must make use of the following proportion: As $105 : 100 :: 449\frac{2}{3} : 428\frac{16}{21}$ ducats; which is the answer under those conditions. 483. We have shewn that six operations are necessary in making use of the Rule of Three; but we can greatly abridge those calculations by a rule which is called the Rule of Reduction, or Double Rule of Three. To explain which, we shall first consider the two antecedents of each of the six preceding operations: 1st. $2$ rubles : $95$ stivers. 2d. $20$ stivers : $1$ Dutch florin. 3d. $5$ Dutch flor. : $2$ Dutch dollars. 4th. $100$ Dutch doll. : $142$ dollars. 5th. $3$ dollars : $1$ ducat. 6th. $105$ ducats : $100$ ducats. If we now look over the preceding calculations, we shall observe, that we have always multiplied the given sum by the third terms, or second antecedents, and divided the products by the first: it is evident, therefore, that we shall