LESSON 2. 1817 at 3 qrs. Here as in lesson first we say 1817d. which sum, divided by 12 and by 20, will make £7 11s. 5d. as before; then the half of this sum, and the half of that half, will be equal to the amount at 3q.: because 2 q. are equal to half a penny, and Iq. is equal to half of 2q. 1691 at 1d. 2q. In this lesson we say 1691 at 1s. makes £84 11s. Then 1d. is of a shilling and 24. of a penny; that is, the of £ 84 11 and the half of that will give the answer, viz. £ 10 11 4 2 LESSON 6 and 7. What will 568 pairs of mittens come to at 2s. 6d. a pair, and at 3s. 4d. a pair ? Ans. £71 at 2s. 6d. and £94 13s. 4d. at 3s. 4d. EXPLANATION. In the first place we consider 568 pairs at 20s. equal to £ 568. Then as 2s. 6d. is of a pound, we divide by 8, the quotient is the answer at 2s. 6d. Secondly, we say 3s. 4d. is of a pound, and divide by 6, the quotient will be the answer at 3s. 4d. Say 678 at 20s.£678. Then take parts of that sum thus: 4s.} of a pound; 2s. of 4s. ; 18.={ of 2s. ; 6d. of 1s. and 2q=12 of 6d. OPERATION. 4s. of 1£.£678 at 7s. 6d. 2q. PROOF. LESSON 11. 1 8s. 3d. Answer, £255 13s. 3d. 459 at 15s. 8d. 3q. Here we will suppose that pound; 5s. of 10s.; 6d. and 1q. of 2q. OPERATION. PROOF. 6d.of 1s. 459 X 3 642 at 37s. 6d. 2q. LESSON 12. STATEMENT. Say 642 pounds; this will include the amount of the integers at 20%. each: then for 17s. 6d. 2q. say 10s. of a pound; 5s.the half of 10s; 2s. 6d. the half of 5s. and 2q.=3 of 2s. 6d. What cost 15 tons 9 cwt. 3 qrs. 25 lb. of iron, at £2 16s. a hundred ? DIRECTIONS. Answer, £867 18s. 6d. 1. Multiply the price of 1 cwt. by 20, the product will be the price of 1 ton. 2. Multiply the price of 1 ton by 15, the product will be the price of 15 tons. 3. Multiply the price of 1 cwt. by 9, the product will be the price of 9 cwt. 2 qrs. of 1 cwt.187. $ 840 0 Proof in dollars and cents, at the rate of 8s. to the dollar. cwt. 187 price of 1 cwt. 20 cwt. 1 ton. How to find the time a Note or Bond has been on Interest. LESSON 1. Suppose a note was given on interest April 2, 1813, and paid March 4, 1816; how many years, months, and days does it draw interest? Answer, 2 years 11 mo. and 2 days. Allow 30 days to be 1 month and 12 months 1 year. OBSERVATIONS. 1. The Note was paid on the 4th of March 1816, which time we call 1816, years on New-year's day last before the payment, and the 4th of March we say is 2 months and 4 days after New-year's day. 2. So again for the time of signing the note, we say that the 2nd of April was 3 months and 2 days after New-year's day 1813. Subtract and carry at 30 and at 12. APPLICATION. What is the interest of a Note of $ 450.75 for 2 years 11 mo. and 2 days at 7 per cent. per annum? Answer, $92.20+ DIRECTIONS. First, find the interest for three years; then take one twelfth of one year's interest from that sum, the difference will be 2 years 11 mo. interest, Secondly, take the 15th part of one month's interest for 2 days. OPERATION. 45075 cents. 3155.125 one year's interest. 9465.175 three years' interest. 262.93 one month's interest subtracted. 9202.82 two years' and 11 months' interest. 9220.1348 Answer, $ 92.20 cents, and .348+ thousandths of a cent. LESSON 2. If a note were given on interest, 5th May, 1812, and paid March 2, 1817, what time should be allowed for interest? Ans. 4 years, 9 mo. 27 days. 4 APPLICATION. What is the amount of $16.50, for 4 years, 9 months and 27 days, at 7 per cent. per annum ? OBSERVATIONS. Ans. $22.07+ 1. In this case obtain the interest for 4 years; then take half the interest for 1 year, and the half of that half for 9 months. 2. For 27 days, divide the interest of 1 year by 12, the quotient will be the interest of 1 month, viz. 9.6250 cents; subtract of a month's interest, that is 3 days, from that sum, the remainder will be 27 days' interest, viz. 8.6625 cents. OPERATION. 1650 cents, principal. 6 mo.a yr.)115150 interest for 1 year. 12)115.50 cts. 1 year. 10)9.6250 1 month. 4 8.6625 27 days. RULE. Add the rate per cent. to 100; find the Logarithm of that sum, and subtract from it the Logarithm of 100; call the difference, Logratio; multiply the Logratio by the number of years, and to that product add the Logarithm of the principal; the natural number of the last Logarithm will be the amount in Compound Interest. EXAMPLE 1. What is the amount of $500 for three years, at 7 per cent. per annum, Ans. $612.50 compound interest ? OPERATION. Rate per cent. 7+100107. Logarithm of 107 is Logarithm of 100 is 2.02938 2.00000 0.02938 3 0.08814 |