with Berlin is at 142, that is to say, for 100 Dutch dollars, 142 dollars are paid at Berlin. Lastly, the ducat is worth 3 dollars at Berlin.

425. To resolve the questions proposed, let us proceed step by step. Beginning therefore with the stivers, since 1 ruble = 47 stivers, or 2 rubles = 95 stivers, we shall have 2 rubles : 95 stivers 1000:.... Answer, 47500 stivers. If we go further and say 20 stivers: 1 florin 47500 stivers: .... we shall have 2375 florins. Further, 2 florins = 1 Dutch dollar, or 5 florins =2 Dutch dollars; we shall therefore have 5 florins: 2 Dutch dollars=2375 florins:.... Answer, 950 Dutch dollars.

Then taking the dollars of Berlin, according to the exchange at 142, we shall have 100 Dutch dollars: 142 dollars = 950: the fourth term, 1349 dollars of Berlin. Let us, lastly, pass to the ducats, and say S dollars: 1 ducat = 1349 dollars: Answer, 4493 ducats.

426. 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; 105: 100 449 a fourth term, which is 42816 ducats.

427. 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. To explain this rule, we shall first consider the two antecedents of each of the six operations.

If we now look over the preceding calculations, we shall observe, that we have always multiplied the given sum by the second terms, and that we have divided the products by the first; it is evident therefore, that we shall arrive at the same

results, by multiplying, at once, the sum proposed by the product of all the second terms, and dividing by the product of all the first terms. Or, which amounts to the same thing, that we have only to make the following proportion; as the product of all the first terms is to the product of all the second terms, so is the given number of rubles to the number of ducats payable at Berlin.

428. This calculation is abridged still more, when amongst the first terms some are found that have common divisors with some of the second terms; for, in this case, we destroy those terms, and substitute the quotient arising from the division by that common divisor. The preceding example will, in this manner, assume the following form.*

429. The method, which must be observed, in using the rule of reduction, is this; we begin with the kind of money in question, and compare it with another, which is to begin the next relation, in which we compare this second kind with a third, and so on. Each relation, therefore, begins with the same kind, as the preceding relation ended with. This operation is continued, tiil we arrive at the kind of money which the answer requires; and, at the end, we reckon the fractional remainders.

* Divide the 1st and 9th by 2, the 3d and 12th by 20, the 5th and 12th (which is now 5) by 5, also the 2d and 11th by 5.

430. Other examples are added to facilitate the practice of this calculation.

If ducats gain at Hamburg 1 per cent. on two dollars banco; that is to say, if 50 ducats are worth, not 100, but 101 dollars banco; and if the exchange between Hamburg and Konigsberg is 119 drachms of Poland; that is, if 1 dollar banco gives 119 Polish drachms, how many Polish florins will 1000 ducats give?

30 Polish drachms make 1 Polish florin.

431. We may abridge a little further, by writing the number, which forms the third term, above the second row; for then the product of the second row, divided by the product of the first row, will give the answer sought. ́

Question. Ducats of Amsterdam are brought to Leipsick, having in the former city the value of 5 flor. 4 stivers current; that is to say, 1 ducat is worth 104 stivers, and 5 ducats are worth 26 Dutch florins. If, therefore, the agio of the bank* at Amsterdam is 5 per cent., that is, if 105 currency are equal to 100 banco, and if the exchange from Leipsick to Amsterdam, in bank money, is 38 per cent. that is, if for 100 dollars we pay at Leipsick 1334 dollars; lastly, 2 Dutch dollars making 5 Dutch florins; it is required to find how many dollars we must pay at Leipsick, according to these exchanges, for 1000 ducats?

*The difference of value between bank money and current money.

Answer, 26391 dollars, or 2639 dollars and 15 drachms.

CHAPTER IX.

Of Compound Relations.

432. COMPOUND RELATIONS are obtained, by multiplying the terms of two or more relations, the antecedents by the antecedents, and the consequents by the consequents; we say then, that the relation between those two products is compounded of the relations given.

Thus, the relations a: b, c : d, e: f, give the compound relation a ce: bdf.*

433. A relation continuing always the same, when we divide both its terms by the same number, in order to abridge it, we may greatly facilitate the above composition by comparing the antecedents and the consequents, for the purpose of making such reductions as we performed in the last chapter.

For example, we find the compound relation of the following given relations, thus ;

* Each of these three ratios is said to be one of the roots of the compound ratia.

So that 2: 5 is the compound relation required.

434. The same operation is to be performed, when it is required to calculate generally by letters; and the most remarkable case is that, in which each antecedent is equal to the consequent of the preceding relation. If the given relations are a: b

b: c

c: d

d: e

e: a

the compound relation is 1 : 1.

485. The utility of these principles will be perceived, when it is observed, that the relation between two square fields is compounded of the relations of the lengths and the breadths.

Let the two fields, for example, be A and B ; let A have 500 feet in length by 60 feet in breadth, and let the length of B be 360 feet, and its breadth 100 feet; the relation of the lengths will be 500: 360, and that of the breadths 60: 100. So that we have

Wherefore the field A is to the field B, as 5 to 6.

436. Another example. Let the field A be 720 feet long, 88 feet broad; and let the field B be 660 feet long, and 90 feet broad; the relations will be compounded in the following man