The Rule of Discount teaches us to find the amount of the allowance, when we know the rate of discount, which ought to be the same as the interest fixed by the law. The Rule of Discount amounts to this question of interest; find the interest of a principal, which, joined with this same interest, formsa given sum. In effect, let us suppose a bill of exchange of 3000£. at 6 months sight; since the merchant who has in hand this bill of exchange, is to pay it only at the end of 6 months, he is in the same state as if a certain sum had been lent to him, which, joined to its interest, amounts to 3000; and as if the terms of the loan were such, that he was to remit the sum and its interest, at the expiration of 6 months; but if the proprietor of the bill of exchange requires to be paid before the time has expired, the merchant then ought only to remit him the sum supposed to be lent, and to retain for himself the surplus, which we call Discount. The method we are to follow, to find the discount, is general; we must always begin by seeking the interest of 100. for the time that the bill of exchange was to run; then adding this interest to the 100£. we shall make this proportion; the principal and interest of 100. is to the interest, as the amount of the bill of exchange, is to the discount required.

Question. What is the discount of a bill of er change of 6484£. for 4 months and 18 days, at 6 per cent?

We must first seek the interest of 100£. for 4 months and 18 days; we find it by reasoning thus; if in 12 months we have 6 pounds interest, how many shall we have in 4m. 18 days? that is to say, 12m:6£:: 4m..18d: X; this proportion becomes, by dividing the two terms of the first ratio by 6, and afterwards taking the half of the two antecedents,

1m:1£::2m. 9d :X2£;* the interest of 100£. for 4m. 18d.

Now to find the discount, we shall make this proportion 102+:+:: 6484: X; making the denominators disappear, we shall have 1023:23::6484;X= 145£.--15s. --6d., which is the discount demanded. Subtracting the discount from the amount of the bill of exchange, the difference, which is 6338£. - 4s.-5d., wili be the sum due to the owner of the bill of exchange.

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For the proof of the rule we must make this proportion, 100:26338 £.-4s.-5d.57:145 £.-15s.-6d.311, which will give the product of the extremes equal to that of the means, if the operation has been well performed.

RULE OF FALSE POSITION.

The Rule of False Position, consists in dividing a number into parts proportional to numbers, which we determine relatively to the state of the question.

* Here, beginners are apt to be puzzled; they know not what to conclude from this proportion, considering that 1 neither multiplies nor divides; but they must take notice, that X, being, by the state of the question, a number of pounds, and equal (as the 4th number of the proportion) to 1.X(2m. 9d.)

1m.

they must, in this expression, take for

multiplicand 1. and for multiplier 2 months 9 days, which, according to the rules of Complex Multiplication, gives X=2£••

3

We cannot better explain this rule, than by examples.

Question First. Divide 658£. among three persons, in such a manner, that the second may have three times as much as the 1st, and the third as much as the other two together.

We plainly see, that if we knew the part of the first, we should easily obtain that of the two others; let us suppose this part to be 1. then the part of the second will be 3£.and that of the third 1£+3=4£; the whole sum of these three parts will be 8£; thus the supposition is false, for the whole amount of the parts required, ought to be 658; nevertheless, as the supposed parts are evidently proportional to the true. parts, this false supposition will serve us to solve the question; for, we may say, the whole of the false parts is to the first false part, as the whole of the true parts is to the first true part. Thus we have 8£:1:: 658: the first true part

Taking three times this part, we shall have the 2d true part

Adding these two 1st parts, we shall have the 3d true part

5s.

=82£.

}

=246

15

For the proof, it is evident, that the whole sum of the parts must be equal to the number to be divided. Question Second. Find a number, whereof the half, the fourth, and the fifth, make together 60.

Let us represent the number sought by 1; its half will be, its fourth, its fifth; I add together these three fractions, which give me the fraction for the whole sum of the supposed parts; the supposition is false, but it will give the true number sought, by making this proportion; if the fraction con

tains the half, the fourth, and the fifth of 1, of what number does 60 contain the half, the fourth, and the fifth parts? Thus we have 48:1::60: the number sought=63.

Question Third. Divide the number 720 into 3 parts, in such a manner, that the 1st may be to the 2d. as 3 is to 4; and the 2d to the 3d, as 5 is to 6.

If we knew the first part, we should have the others; let us suppose it to be 1; then making this proportion 3:4::1; a fourth term; this fourth term

will evidently represent the second part; and then makingthis other proportion 5:6::: a fourth term, this fourth term will represent the third part; the whole sum of these three parts is 1+1+%, which added together, make ; whence we see that the supposition is false; but we shall have the first true part by making this proportion, :1::720: the first true part

Now, according to the state of the questi

we have 3:4::1835, the second

on,

1835

24455

292

720

Question Fourth. Divide 300£. among 3 persons, in such a manner, that the 2d may have twice as much as the 1st, and 6£. more; and the 3d as much as both the others, and 10£. more.

We must observe here, that since the second person is allowed 6£. and the 3d 16£. whatever may be the part of the 1st, we plainly see, that we must first subtract these two numbers from the sum to be

divided, which reduces it to 278.; then the question becomes just like the first example; now the question is to divide 278. among 3 persons, in such a manner, that the 2d may have twice as much as the 1st, and the 3d as much as both the others together.

Thus, if we suppose 1. to be the part of the 1st. 2. will be the part of the 2d, and 3. that of the 3d; the whole sum of the false parts will then be 6£. which gives the proportion.

6:1::278: the first true part Doubling this 1st part. and adding 6£, we have the 2d part

Adding to the sum of these two parts, 10. we have the 3d part

Question Fifth. Divide 300. among 3 persons, in such a manner, that the 2d may have twice as much as the 1st, less 6£. and the 3d as much as the two others, less 10£.

Since, in the preceding example, we were obliged to subtract 22. from the sum to be divided, it is plain that we must, on the contrary, add them in this instance, and to find the first true part, make this proportion.

6:1:322: the first true part Doubling this 1st part, and subtracting 6. we have the 3d part

Subtracting 10. from the sum of these two parts, we have the 5th part