POSITION.*

POSITION is a rule by which, from the assumption of one or more false answers to a problem, the true one is obtained. It admits of two varieties, Single Position, and Double Position.

In SINGLE POSITION the answer is obtained by one assumption in DOUBLE POSITION it is obtained by two.

Single Position may be employed in resolving problems, in which the required number is any how increased or diminished in any given ratio; such as when it is increased or diminished by any part of itself, or when it is multiplied or divided by any number.

Double Position is used, when the result obtained by increasing or diminishing the required number in a given ratio, is increased or diminished by some number which is no known part or multiple of the required number; or when any root or power of the required number, is either directly or indirectly contained in the result given in the question.

SINGLE POSITION.†

RULE. (1.) Assume any number, and perform on it the operations mentioned in the question as being performed on the required number: (2.) Then, as the result thus obtained, is to the assumed number, so is the result given in the question, to the number required.

Exam. 1. Required a number to which if one half, one third, one fourth, and one fifth of itself be added, the sum may be 1644.

Suppose the number to be 60: then, if to 60 one half, one third, one fourth, and one fifth of itself be added, the sum is 137. Hence, according to the rule, as 137: 1644:: 60: 720, the. number required. The truth of the result is proved by adding to 720, one half, one third, &c. of itself, and the sum is found to be 1644. The number 60 was here assumed, not as being near the truth, but as being a multiple of 2, 3, 4, and 5; and in this way the operation was kept free from fractions. By the assumption of any other number, however, the answer would have been found correctly, but often not so easily. The reason of the operation is obvious from the principles of proportion.

Ex. 1. Divide £2000 between A, B, and C, giving A as much as B and a fifth part more, and C as much as both together. Answ. A's part £545 9 1, B's £454 10 11, C's £1000.

This rule is sometimes called the Rule of False, or the Rule of False Position, or the Rule of Trial and Error. It might properly be called the Rule of Supposition.

+ Every question that can be resolved by this rule, may also be resolved by the rule for Double Position, or without Position, by some of the preceding rules; and hence this rule is of little importance.

2. One third of a ship belongs to A, and one fifth to B, and A's part is worth £1000 more than B's: required the value of the ship. Answ. £1500.

3. It is required to divide 252 into three parts, such that one third of the first, one fourth of the second, and one fifth of the third, shall all be equal to one another. Answ. 63, 84, and 105. 4. To find a number such that if it be multiplied by 10, and the product be divided by 13, the quotient, increased by the number itself, and by 80, will amount to 1000. Answ. 520.

5. Required a number to which if one half of itself, one third of that half, and one fourth of that third, be added, the sum will be 500. Answ. 292.

6. A father bequeaths to his three sons £7000 in such a manner, that if the share of the eldest be multiplied by 5, that of the second by 6, and that of the third by 7, the products are all equal. What are their shares? Answ, £2747 13 34, £2289 14 42, and £1962 12 4.

7. The number of a gentleman's horses is two fifths of the num ber of his black cattle, and for every four of the latter he has eleven sheep. Required the number of each, the number of the sheep exceeding that of the horses by 141. Answ. 24 horses, 60 black cattle, 165 sheep.

DOUBLE POSITION.

RULE I. (1.) Assume two different numbers, and perform on them separately the operations indicated in the question: (2.) Then, as the difference of the results thus obtained, is to the difference of the assumed numbers, so is the difference between the true result and either of the others, to the correction to be applied, by addition or subtraction, as the case may require, to the assumed number which gave this result.

This rule, which was first published in substance by Mr. Bonnycastle in his larger work on Arithmetic, in 1810, is the simplest and easiest that has yet appeared for the resolution of questions in which the given result is a known number, independent on the required number: and these questions are generally the most useful. Mr. Bonnycastle appears, however, not to have been aware, that this rule fails in relation to the whole class of questions, in which the result of the operations to be performed, according to the question, on the required number, is not a known, determi nate number, but the required number, or one depending on it, such as some multiple or part of it. In that case the following rule will be necessary. This rule has also the advantage of being applicable in every case whatever.

RULE II. (1.) Having assumed two different numbers, perform on them separately the operations indicated in the question, and find the errors of the results. (2.) Then, as the difference of the errors, if both results be too great or both too little, or as the sum of the errors, if one result be too

great and the other too small, is to the difference of the assumed numbers, so is either error to the correction to be applied to the number that produced that error.

Exam. 1. Required a number, from which if 2 be subtracted, one third of the remainder will be 5 less than half the required number.

Here, suppose the required number to be 8, from which take 2, and one third of the remainder is 2. This being taken from one half of 8, the remainder is 2, the first result. Suppose again, the number to be 32, and from it take 2: one third of the remainder is 10, which being taken from the half of 32, the remainder is 6, the second result. Then, the difference of the results being 4, the difference of the assumed numbers 24, and the difference between 5, the true result, and 6 the result nearer it, being 1; as 4: 24:: 1:6, the correction to be subtracted from 32, since the result 6 was too great. Hence, the required number is 26.

Exam. 2. If one person's age be now only four times as great as another person's, though 7 years ago it was six times as great: what is the age of each ?

Here, suppose the age of the younger to be 12 years; then would the age of the older be 48. Take 7 from each of these, and there wil remain 5 and 41, their ages 7 years ago. Now, 6 times 5 is 30, which taken from 41, leaves an error of 11 years. By supposing the age of the younger to be 15, and proceeding in a similar manner, the error is found to be 5 years. Hence, as 6, the difference of the errors, (both results being too small,) is to 3, the difference of the assumed numbers, so is 5, the less error, to 2, the correction; which being added to 15, the sum, 17, is the age of the younger, and consequently that of the older must be 70.

Both the rules above given for Double Position depend on the principle, that the differences between the true and the assumed numbers, are proportional to the differences between the result given in the question and the results arising from the assumed numbers. This principle is quite correct in relation to all questions which in Algebra would be resolved by simple equations, but not in relation to any others; and hence, when applied to others, it gives only approximations to the true results. In this case the assumed numbers should be taken as near the true answer as possible. Then, to approximate the required number still more nearly, assume for a second operation the number found by the first, and that one of the two first assumptions which was nearer the true answer, or any other number that may appear to be nearer it still. In this way, by repeating the operation as often as may be necessary, the true result may be approximated to any assigned degree of accuracy. When applied in this way, Double Position is of considerable use in Algebra, affording in many cases a very convenient mode of approximating the roots of equations, and finding the values of unknown quantities in very com. plicated expressions, without the usual reductions.

Exam. 3. Required a number to which if twice its square be added, the sum will be 100.

It is easy to see that this number must be between 6 and 7. These numbers being assumed, therefore, the sum of 6 and twice its square is 78, and the sum of 7 and twice its square 105. Then, as 105-78; 7 — 6 :: 105 — 100:18; which being taken from 7, the remainder, 6.82, is the required number nearly. To this let twice its square be added, and the result is 99-8148. Then, as 105-99-8448: 7 — 6·82 :: 105-100: 1746; which being taken from 7, the remainder is 6·8254, the required number still more nearly and if the operation were repeated with this and the former approximate answer, the required number would, be found true for seven or eight figures.

Ex. 1. A merchant increased his capital each year by a fourth of itself, except an expenditure of £300 annum, and at the end of four years found himself worth £5000. What was his original capital? Answ. £2756 9 7z.

2. Suppose every thing to be as in the last exercise, except that at the end of four years the merchant found himself possessed of twice his original stock: how much had he to begin with? Answ. £3918 11 8.

3. Required as in the two preceding exercises, supposing every thing as before, except that at the end of the time the merchant finds himself possessed of only half his original capital. Answ.

£890 18 11.

4. Required a number from which if 84 be taken, three times the remainder will exceed the required number by a fourth of itself. Answ. 144.

5. Required a number such that if it be multiplied by 11, and 320 be taken from the product, the tenth part of the remainder will be 20 less than the number itself. Answ. 120.

6. Required a number whose half is as much less than 1000, as its double is greater than 999. Answ. 7998.

7. A farmer engaged a labourer on condition of paying him 1/4 a day for every day he should work, and of charging him 9d. for his boarding every day he should be idle. Now, at the end of a Year (313 days) the man was entitled to £12: how many days then did he work? Answ. 22733 days.

8. How many guineas of £1 2 9, and moidores of £1 9 3 each, will pay a bill of £130, the number of pieces of both kinds being 100? Answ. 50 of each.

9. If I be added to a number, and 100 divided by the sum, the quotient is 3 less than if I had been subtracted from the number, and 100 divided by the remainder: required the number. Answ. 8.2259751.

10. Required a number which exceeds 3 times its square root by 11. Answ. 26.4201648.

11. Required a number to which if twice its square and 3 times its cube be added, the sum will be 2000. Answ. 8.506744.

12. Given the sum of two numbers = 20, and the sum of their squares 324: required the numbers. Answ. 17.8740079, &c.

COMPOUND INTEREST.

The method that naturally presents itself for finding the amount of a sum at compound interest, is to find its amount at simple interest at the end of the first year; then to take this amount as a new principal, and find its amount in like manner, which would be the amount at compound interest at the end of the second year, and the principal for the third year, the amount of which must be found in like manner. Continuing the process, we should thus find the amount at the end of the proposed time. be illustrated in the following example.

This will

Exam. 1. Required the amount of £2500 at the end of 4 years, at 6 cent. annum, compound interest.

Here, the amount for 1 year is £2650; the amount of which for 1 year also is £2809, the amount at compound interest for two years. The amount of this again for 1 year, or the amount of the given sum at the end of the third year, is £2977 10 91: the amount of which for the same time is £3156 3 10, the amount of £2500 for four years.-The amount at simple interest would have been £3100, which is less than the amount at compound interest by £56 3 10.

When the time is short, this method may be practised without much trouble; but when the time is long, the labour would become very great. In this case, the methods that follow should be employed.

RULE I. To find the amount of one pound sterling for any number of years, at compound interest: (1.) Divide the amount of £100 for 1 year by 100, and the quotient will be the amount of one pound for 1 year: (2.) This amount involved to a power denoted by the number of years, will be the amount of 1 pound for that number of years.

The contracted mode of multiplication of decimals is peculiarly useful in this rule, and in computations in compound interest and annuities in general. So also is the contracted method of dividing decimals. (See pages 118, 121, and 203.)

Exam 2. Required the amount of one pound sterling for 20 years at 4 cent. annum, compound interest.

Here, the amount of £100 for 1 year is £104-5, the hundredth part of which is £1.045, the amount of £1 for a year. The second power of this is £1.092025, the amount of £1 at the end of the second year. The product of this by itself, by the contracted method of multiplication, is £1·192518, the amount at the end of the fourth year. The square of this again is £1.422099, the amount for eight years; the square of which is £2.022366, the amount at the end of the sixteenth year. Finally, the product of