In states in which the collection of annual interest is not allowed, the teacher should omit this topic. 1173. In the older states, time should not be spent upon topic. 1176. 1. of a section of 640 A. 2. of = section measures 80 rd. by 160 rd. this 3. A line from the southwest corner of Sec. 1, to the northeast corner of Sec. 30 (see township diagram on the opposite page), is the hypotenuse of a right-angled triangle whose perpendicular, the eastern boundaries of Secs. 11, 14, and 23, is 3 mi. long; and whose base, the southern boundaries of Secs. 20, 21, 22, and 23, is 4 mi. long. 5. The number of rods of fence 80 + 160 + 80+ 160 480. The number of feet = 161⁄2 × 480 · 7920. A fence 4 boards high requires 7920 ft. × 4, or 31680, running feet of boards. If the latter are ft. wide, the number of board feet = 31680 X = 15840. 1184. 2. 26.50 x .85. 4. Multiply 135 by 69, and point off two places in the product. 5. Find the base. 6. 8.50 francs × (10 × 1 × 3.25). 7. Each dimension can be expressed in decimeters, 105 x 80 x 65, whose product is the number of liters; or the product of the dimensions in meters -10.5 × 8 × 6.5 may be multiplied by 1000. 8. 07.75 means .757., the denomination in France being generally written before the decimal. 9. 1.25 marks x [(68÷10) × 36]; i.e. 11 marks x 6.8 x 36. 10. The number of liters 50 x 40 x 3060000; 92% of this number gives the weight in kilograms (kilos). 1186. These problems are given for practice in obtaining the approximate values of the metric units in terms of our weights and measures. The use of 39.37 in. makes the work too tedious. 13. 4 in. by 4 in. by 4 in. A quart=231 cu. in, = 14. A hectoliter = 100 liters = 6400 cu. in. (6400+2150.4) bu. 6400 cu. in. (6400231) gal. = 15. A liter of water, 64 cu. in., weighs a kilo. 1 cu. in. of water=1928 oz.; 64 cu. in. = [(64000 ÷ 1728) ÷ 16] lb. 1000 lb. ÷ 1728. 16. 400000000 in. 1177. 4000 circumference. See Arithmetic, Art. 17. A square meter = (40 × 40) sq. in. 18. An are = (400 × 400) sq. in. A hectare = [(400 × 400) × 100] sq. in. 21. 1000 grams weigh (4000 lb. ÷ 1728); 1 gram weighs 4 lb. ÷ 1728 = 28000 grains ÷ 1728. 22. A kilometer =(6.3364) Km. 1197. The average pupil should be permitted to use a pencil for his first solution of these problems. 1. Let x the value of the second suit. Since $12 and the overcoat 2x, the overcoat = 2x - 12. The second suit (x) and the overcoat (2x — 12) = three times the first suit (36). x+2x-12=36; etc. The second suit is worth $16; the overcoat, $20. Ans. The arithmetical analysis might assume some such form as this: The remainder of the remainder, or of the remainder of original sum. The remainder of original sum × The sum lost is 1-, or of original of original sum. 3. Let x = time past noon; x+12= time past midnight. =x+12; 5x=x+12; 4x = 12; x = 3. The time is 3 hr. x= 5 past noon, or 3 P.M. Ans. 4. At 3 o'clock, the hour hand is 15 minute spaces in advance. To be only 5 spaces behind, the minute hand must gain 10 spaces. While the minute hand goes 1 space, the hour hand goes space; so that each minute, the minute hand gains space. To gain the 10 spaces necessary, the minute hand must travel (1011) minutes 120 min. ÷ 11 = 1010 min. The time is 101 min. past 3. 5. = A= B; A=4 B; 5 B=30. B's age= 6 yr.; A's 6. A takes $15 less than of the profits. If his capital is $30 less than of the whole, the latter must be double the profits, or $1440. A's capital = $525 × 2 = $1050; B's = $390. Ans. Or, A takes $25, or 5, of the profits; he owns, therefore, of the capital. If ğ of the capital +$30 = 4, or 3 §, of the capital, of the capital = $30; etc. 7. Let x the number of sheep; = 80 cost of each; x − 5 160 x 800 = 120x; 40x=800; x=20. Ans. 20 sheep. Or, if he received $40 for of the remainder, he would have received $60 for the remaining sheep. $60 being of $80, of the sheep remained, and of them died, or 5 sheep. The whole number was, therefore, 20 sheep. 8. Let x= A's age; then x + 10 = B's age; 1198. 3. 7) 19 mi. 180 rd. 2 yd. 0 ft. 9 in. 2 mi. 254 rd. 1 yd. 2 ft. 8 in. × 12 33 mi. 172 rd. 0 yd. 2 ft. 11⁄2 in. 4. 18 hr. 24 min. 12 sec. ÷ 15 = 1 hr. 13 min. 36 sec. 1 hr. 45 min. time difference in longitude. As the place has erly. = 1° 45' × 15 = 26° 15' difference the later time, it is more east = 7. Calling it 1 in. thick, the number of board feet 16 x $40 × (16 × × 2) 1000-$1.20. Ans. ÷ 9. [96 (in.) × 90 (in.) × 48 (in.)] ÷ 2150.4. 10. If the strips run lengthwise, their number will be 8 yd. ÷ yd. = 10. The number purchased must be 11, each 9 yd. long, or 99 running yards of carpet. XVIII NOTES ON CHAPTER FIFTEEN While the work contained in this chapter is intended more particularly for use in such schools as extend their instruction beyond the eighth year of the elementary course, it can profitably replace some of the less useful arithmetical topics taught during the eighth school year. 1199. These exercises should be taken up without any preliminary explanations. Their previous work in simple equations has so familiarized the pupils with the use of letters to express numbers, etc., that they need no assistance in the first ten examples. The necessary technical terms should be employed as occasion requires, and their meanings should be made clear; but exhaustive treatment of the different operations should be left for the study of the science of algebra in the high school. 1200. The explanation of the meaning xy, abc, etc., may be deferred until Art. 1238. For the present, the use of the word coefficient may be limited to simple numerical ones, as given in the text-book. The teacher should not yet explain that in the expression 5 xy, 5x may be considered the coefficient of y; nor that in 9abc, 9a may be considered the coefficient of bc, and 9 ab the coefficient of c. 1204. So far, the pupils have been required to add only single columns containing the same letters. When the signs are alike throughout, as in Art. 1199, they have found the sum of the coefficients, annexed the letter or letters, and prefixed the |