Endorsed, Nov. 15th, 1845, $250.375. March 1st, 1844, $65.625. How much was due July 4th, 1845 ? $1000. PHILADELPHIA, June 20th, 1841. 46. Six months after date, I promise to pay Messrs. Carey, Hart & Co., or order, one thousand dollars, with interest at 5 per cent., value received. HORACE PRESTON. Endorsed, Jan. 10th, 1844, $125. June 16th, 1844, $93. Feb. 20th, 1845, $200. CONNECTICUT RULE. AF 249. “Compute the interest on the principal to the time of the first payment; if that be one year or more from the time the interest commenced, add it to the principal, and deduct the payment from the sum total. If there be after payments made, compute the interest on the balance due to the next payment, and then deduct the payment as above; and, in like manner, from one payment to another, till all the payments are absorbed; provided the time between one payment and another be one year or more. But if any payments be made before one year's interest hath accrued, then compute the interest on the principal sum due on the obligation, for one year, add it to the principal, and coinpute the interest on the sum paid, from the time it was paid up to the end of the year; add it to the sum paid, and deduct that sum from the principal and interest added as above.* “ If any payments be made of a less sum than the interest arisen at the time of such payment, no interest is to be computed, but only on the principal sum for any period.”—Kirby's Reports. VERMONT RULE. 249.a. First find the amount of the given principal for the whole time; then find the amount of the several payments from the time they were endorsed to the time of settlement. Finally, subtract the amount of the several payments from the amount of the principal, and the remainder will be the sum due. * If a year does not extend beyond the time of payment; but if it does, then nd the amount of the principal remaining unpaid, up to the time of settlement, likewise the amount of the endorsements from the time they were paid to the time of settlement, and deduct the sum of these several amounts from the amount of the principal. 12 Note.-It will be an excellent exercise for the pupil to cast the interest on each of the preceding notes by each of the above rules. 47. What is the interest of £175, 10s. 6d. for 1 year, at 5 per cent. ? Operation. We first reduce the 10s. 6d. £175.525 Prin. to the decimal of a pound, .05 Rate. (Art. 200,) then multiply the £8.77625 Int. for 1 yr. off the product as in Art. 241. principal by the rate and point 20 The figure 8 on the left of the s. 15.52500 decimal point is pounds, and those on the right are decid. 6.30000 mals of a pound, and must be 4 reduced to shillings, pence, far. 1.20000 and farthings. (Art. 201.) Ans. £8, 15s. 64d. Hence, 250. To compute the interest on pounds, shillings, &c. Reduce the given shillings, pence, and farthings, to the decimal of a pound ; (Art 200 ;) then find the interest as in Federal Money; finally, reduce the decimal figures in the answer to shillings, pence, and farthings. (Art. 201.) 48. What is the interest of £56, 153. for one year and 6 months, at 6 per cent. ? Ans. £5, 2s. 14d. 49. What is the interest of £75, 12s. 6d. for 1 year and 3 months, at 7 per cent. ? 50. What is the interest of £96, 18s. for 2 years and 6 months, at 45 per cent. ? 51. What is the amount of £100 for 2 years and 4 months, at 5 per cent. ? 52. What is the amount of £430, 16s. 10d. sor 1 year and 5 months, at 6 per cent. ? QUEST.-250. How is interest computed on pounds, shillings, &c. ? PROBLEMS IN INTEREST. 251. It will be observed that there are four parts or terms connected with each of the preceding operations, viz: the principal, the rate per cent., the time, and the interest, or the amount. These parts or terms have such a relation to each other, that if any three of them are given, the other may be found. The processes of finding the fourth from the three given parts are usually denominated Problems. Obs. 1. A Problem is a question proposed which requires a solution. 2. A number or quantity is said to be given, when its value is stated or may be easily inferred from the conditions of the problem under consideration. Thus when the principal and interest are known, the amount may be said to be given, because it is merely the sum of the principal and interest. So if the principal and the amount are known, the interest may be said to be given, because it is the difference between the amount and the principal. 252. To find the interest on any given sum, as in the foregoing examples, the principal, the rate per cent., and the time are always given. This process constitutes the First and most important Problem in interest. The other Problems will now be illustrated. PROBLEM II.* 1. A man loaned $75 to one of his neighbors for 4 years, and received $24 interest : what was the rate per cent. ? Note.- In this example it will be observed that the principal, the interest, and the time, are given, to find the rate per cent. Quest.–25). How many terms are connected with the preceding examples? What are they? Are they all given? When three are given, can the fourth be found? What is the process of finding the fourth term usually called? Obs. What is a problem? When is a number or quantity said to be given ? 252. What terms are given when it is required to find the interest? * Should this and the following Problems be deemed too difficult for beginners, they can be omitted till review. Analysis.—The interest of $75 at 1 per cent. for 1 year, is $.75, and for 4 years it is $.75 X4=$3. (Art. 238.) Now since $3 is 1 per cent. interest on the given principal for the given time, $24 must be 34 of 1 per cent., which is equal to 8 per cent. (Art. 121.) Or, we may reason thus: If $3 is 1 per cent. on the principal for the given time, $24 must be as many per cent. as $3 is contained times in $24 ; and $24---$338. Ans. 8 per cent. PROOF.-$75 X.08=$6.00, the interest for 1 year at 8 per cent., and $6 X4=$24, the interest of $75 for 4 years at 8 per cent. Hence, 253. To find the rate per cent. when the principal, interest, and time, are given. Make the given interest the numerator, and the interest of the given principal at 1 per cent. for the given time, the denominator of a common fraction, which reduced to a whole number will give the per cent. required. (Art. 121.) Or, simply divide the given interest by the interest of the given principal at 1 per cent. for the given time, and the quotient will be the per cent. and $42 interest, what rate per cent. do I pay 7? Operation. The interest of $300 for 2y., at $6)$42 1 per ct., is $6. (Art. 238.) 7 Ans. 7 per ct. PROOF. +$300 X.07 X 2=$42. 3. If I borrow $460 for 3 years, and pay $82.80 interest, what is the rate per cent. ? 4. A man loaned $500 for 8 months, and received $40 interest: what was the rate per cent. ? 5. At what rate per cent. must $450 be loaned, to give $56.50 interest in 1 year and 6 months ? Quest.—253. When the principal, interest, and time, are given, how is the rate per cent. found ? 6. At what per cent. must $750 be loaned, to give $225 in 4 years ? 7. A man has $8000 which he wishes to loan for $600. per annum for his support: at what per cent. must he loan it ? 8. A gentleman deposited $1250 in a savings bank, for which he received $31.25 every 6 months : what per cent. interest did he receive on his money? 9. A capitalist invested $9260 in Railroad stock, and drew a semi-annual dividend of $416.70: what rate per cent. interest did he receive on his money ? 10. A man built a Hotel at an expense of $175000, and rented it for $8750 per annum: what per cent. interest did his money yield him ? PROBLEM III. 11. What sum must be put at interest, at 6 per cent., to gain $30 in two years ? Note. In this example it will be observed that the interest, the rate per cent., and the time, are given, to find the principal. Analysis.—'The interest of $1 for 2 years at 6 per cent., (the given time and rate,) is 12 cents. Now 12 cents is 10 of its principal $1. Consequently $30, the given interest, must be 10% of the principal required. The question therefore resolves itself into this : $30 is 10% of what number of dollars? If $30 is 1o, do is $21; and 188 =$2} x 100, which is $250, the principal required. Or, we may reason thus : Since 12 cents is the interest of 1 dollar for the given time and rate, 30 dollars must be the interest of as many dollars for the same time and rate, as 12 cents is contained times in 30 dollars. And $30= .12=250. Ans. $250. PROOF.—$250 X.06=$15.00, the interest for 1 year at the given per cent., and $15 X2=$30, the given interest. Hence, |