4. Its weight 4 + √15 − (4 −√15) = 2√15 = √60. = A cubic centimeter.39373 cu. in.⇒(.39373÷1728) cu. ft. weight of 1 gram = (.3937 x 1000 ÷ 1728) oz. = [(.3937 × 1000) (1728 x 16)] lb. A kilogram = [(1000 x .39373 × 1000) ÷ (1728 × 16)] lb. The weight of the anchor in kilograms = 6500 lb. ÷ the number of pounds in a kilogram; or, - 6500 x 1728 × 16 1000 x .3937 × .3937 × .3937 × 1000 63 360 5. x-(x XT80 X 8) = 1500; 493x= 750000; x= 1521.30. Ans. $1521.30. The proceeds of the new note + $200 must pay the note of $2000; the proceeds must therefore be $1800. 6. 272 liquid quarts = 231 cu. in. x 272÷4. A dry quart 7. The tub holds 12 qt. x 454 qt. Both pipes discharge 12 qt. + 83 qt. = 951 qt. The time required to fill it =(54951) min. 8. The number of hours that must elapse before all will again be together at the starting point, is the least common multiple of,,35. The least common multiple of the numerators is 70. The smallest fraction that will contain the above fractions an exact number of times must have 70 for its numerator, and for the denominator the largest number that will divide 36, 9, and 99 without a remainder; i.e. the greatest common divisor of these numbers. The G. C. D. is 9, and the fraction is 3o. In 30 hr., therefore, A, B, and C will be at the starting point. A will have walked around the circle (3+) 56 times; B, (703) 35 times; C, (7035) 22 times. NOTE. — The scholars will readily understand that the fraction which is the least common multiple of,, and, should have 70 for its numerator. The following may make clear to them why 9 should be the denominator: An examination of the second line, in which the divisors are inverted, will show that 70 contains the three denominators, 5, 2, and 35, an exact number of times; the numerators, 36, 9, and 99, should contain x an exact number of times; x, therefore, must be a divisor of these numbers, etc. 1172. This work may be slightly abbreviated by combining the interest on the annual interest into one item of 6 years' interest instead of the three separate ones of 3 years' interest, 2 years' interest, and 1 year's interest. In beginning a new topic, however, pupils should not be confused by short methods. 2. $1200+ $300+ (4+3+2+1) years' interest on $60. For partial payments on notes bearing annual interest, see Art. 1308. The special rules for New Hampshire and Vermont will be found in Arts. 1309 and 1310. 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 this topic. 2. of section measures 80 rd. by 160 rd. 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 16 × 480 7920. A fence 4 boards high requires 7920 ft. x 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 × 80 × 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 × [(68 ÷ 10) × 36]; i.e. 14 marks × 6.8 × 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. = (6400 ÷ 231) gal. 15. A liter of water, 64 cu. in., weighs a kilo. 1 cu. in. of water + 1998 oz.; 64 cu. in. = [(64000 ÷ 1728) ÷ 16] lb. = 4000 1000 lb. ÷ 1728. = 16. 400000000 in. circumference. See Arithmetic, Art. 1177. 17. A square meter = (40 × 40) sq. in. 21. 1000 grams weigh (4000 lb. ÷ 1728); 1 gram weighs 4 lb. ÷ 1728 = 28000 grains ÷ 1728. 22. A kilometer =(6.3364) Km. = 40000 in.; a mile = 63360 in.; a mile 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. 2. x 22+ 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 X of original sum. The sum lost is 1-, or of original As this is $22, the original sum = $22 × 1 = $30. Ans. sum. 3. Let x = time past noon; x+12= time past midnight. x+12; 5x=x+12; 4 x = 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 (10 -11) minutes 120 min. 11 time is 101 min. past 3. 5. == = 1010 min. The A= B; A=4 B; 5 B=30. B's age: age 24 yr. Ans. 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 525, or 5, of the profits; he owns, therefore, of the capital. If § of the capital +$30 = 4, or 38, of the capital, of the capital = $30; etc. 7. Let x= the number of sheep; х 35 80 cost of each; x 5 160x 800 120x; 40x Or, if he received $40 for 800; x= = 20. Ans. 20 sheep. of the remainder, he would have received $60 for the remaining sheep. $60 being of $80, 4 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; |