NOTE. 245. THE" six-per-cent method." If the teacher so prefers, the problems on the three preceding pages, as well as those that follow, may be solved by the "six per cent method." EXPLANATORY. The interest at 6% for 1 yr. = .06 of the principal. The interest at 6% for 1 mo. = of .06, or .005 of the principal. = Find the interest of $243.25 for 2 yr. 5 mo. 18 da. at 6%. (3) How find the interest at 8%? at 9%? at 4% ? at 3% ? Interest. 246. PROBLEMS TO BE SOLVED BY THE "SIX PER CENT METHOD." 13 da. at 6 %.* 1. Find the interest of $265 for 1 yr. 3 mo. 2. Find the interest of $346 for 2 yr. 5 mo. 20 da. at 6 %.† 3. Find the interest of $537 for 1 yr. 7 mo. 10 da. at 6 %.‡ 4. Find the interest of $428 for 3 yr. 3 mo. 14 da. at 6 %. 5. Find the interest of $150 for 1 yr. 6 mo. 15 da. at 6 %. (a) Find the sum of the five results. 6. Find the interest of $245.30 for 6 mo. 18 da. at 7 %. § 7. Find the interest of $136.25 for 8 mo. 10 da. at 7 %. 8. Find the interest of $321.42 for 5 mo. 15 da. at 7 %. 9. Find the interest of $108.00 for 10 mo. 8 da, at 7 %. 10. Find the interest of $210.80 for 7 mo. 21 da. at 7 %. (b) Find the sum of the five results. 11. Find the amount of $56.25 for 2 yr. 4 mo. 2 da. at 6%. 12. Find the amount of $31.48 for 1 yr. 5 mo. 11 da. at 6%. 13. Find the amount of $55.36 for 2 yr. 8 mo. 12 da. at 6 %. 14. Find the amount of $82.75 for 2 yr. 10 mo. 8 da, at 6 %. 15. Find the amount of $27.35 for 1 yr. 1 mo. 1 da. at 6%. (c) Find the sum of the five results. 16. Find the amount of $875 for 3 yr. 8 mo. 15 da. at 5 %. 17. Find the amount of $346 for 2 yr. 6 mo. 12 da. at 4 %. 18. Find the amount of $500 for 1 yr. 7 mo. 18 da. at 3 %. 19. Find the amount of $600 for 2 yr. 5 mo. 21 da. at 8 %. 20. Find the amount of $825 for 1 yr. 9 mo. 24 da. at 9 %. (d) Find the sum of the five results. First find the interest at 6%; then add of that sum to itself. 247. TO FIND THE TIME BETWEEN TWO DATES. EXAMPLE. How many years, months, and days from Sept. 25, 1892, to June 10, 1896 ? 1896-6-10 3-8-15 THE USUAL METHOD. From the 1896th yr., the 6th mo., and the 10th day, subtract the 1892nd yr., the 9th mo., and the 25th day. Regard a month as 30 days. A BETTER METHOD. From Sept. 25, 1892 to Sept. 25, 1895, is 3 years. NOTE.-The two methods will not always produce the same results. The greatest possible variation is two days. Find the time from Jan. 22, 1895, to March, 10, 1897, by each method, and compare results. The difference arises from the fact that the month as a measure of time is a variable unit-sometimes 28 days, sometimes 31. The "usual method" regards each month as 30 days; the "better method" counts first the whole years, then the whole months, then the days remaining. By the "usual method," the time from Feb. 28, 1897 to March 1, 1897, is 3 days; by the "better method," it is 1 day. PROBLEMS. Find the time by both methods and compare the results. 1. From March 15, 1894 to Sept. 10, 1897. 7. James has $54.20, and James's money equals 40 per cent of Henry's money. How much money has Henry? 8. Mr. Williams's annual expenses are $791.20; this is 92 per cent of his annual income. How much is his annual income? 9. Mr. Randall has 450 sheep; these equal 125 per cent of Mr. Evans's sheep. How many sheep has Mr. Evans? *TO THE TEACHER.-Encourage pupils to solve the 7th, 8th, and 9th problems in three ways: 1st, by the method given above; 2nd, arithmetically (Prob. 7— Henry's money equals 100 40ths of James's money); 3rd, by the application of the formula, 100 n TO THE TEACHER.-Require the pupils to solve the first five prob. lems in four ways: ((1) Let x = the per cent and solve as the "example" is solved. (2) Using one hundredth of each base as a divisor and the other number mentioned in the problem as a dividend, find the quotient. (3) Find what part the first number mentioned in each problem 13 of the base, and change the fraction thus obtained to hundredths. 100 n b (4) Apply the formula, = x. * The b in problem 6, and in the formula may be thought of as standing for the base. |