any number of factors each equal to 10 will not enable us to cancel the factor 7 from the denominator. As the same principles apply to any other fraction, it follows that, (d.) No vulgar fraction can be reduced to an exactly equivalent decimal form, if, when reduced to its lowest terms, its denominator contains other factors than such as are found in 10, i. e., other than 2's or 5's; and conversely, that, — (e.) Every vulgar fraction which, when reduced to its lowest terms, contains in its denominator only such factors as are found in 10, can be reduced to an exactly equivalent decimal form, and will contain as many decimal places as there are 2's or 5's to be cancelled from the denominator. Illustrations. 15 15 = and hence can be reduced to an exactly 16 24' equivalent decimal. Moreover, it will contain four decimal places; for 10 must be introduced 4 times as a factor into the numerator, to cancel 24 from the denominator. Again. = 13 and hence can be reduced to an exactly equivalent decimal. Morcover, it will contain 3 decimal places, for 10 must be introduced 3 times as a factor in the numerator, to cancel the factors of the denominator. (f) Vulgar fractions which cannot be exactly reduced give rise to REPEATING or CIRCULATING DECIMALS. Reduce each of the following to a decimal form, carrying the division, when only approximate values can be obtained, to six decimal places. 7.1. 8. 19. 11. 8. 12. 78. 144. Fractional Parts of Denominate Numbers. (a.) What is the value of § of a mile in whole numbers of lower denominations, i. e., in furlongs, rods, yards, &c.? First Solution. This example may be solved by the common pro - cess of compound division, thus: of a mile = of 8 miles = 0 niles, with 8 miles remaining. But 8 miles =64 furlongs, and of 64 furlongs 7 furlongs, with, &c. Second Solution. Since 1 m. = 8 fur., of a mile must equal of 8 fur., which is 7 fur. Since 1 fur. =40 rods, of a fur. must 7 fur., 4 rd., 2 yd., 1 ft., 4 in. fur. fur. fur. rd. = 8 X 8 64 1 X 40 = or = 9' 9 9 rd. yd. in. Hence, 8 m. = 7 fur. 4 rd. 2 yd. 1 ft. 4 in. Third Solution.· Since there are 8 furlongs for every mile, there must be 8 times as large a part of a furlong as of a mile, or, in this instance, 8 times of a furlong, which is 7 furlongs. Since there are 40 rods for every furlong, there must be 40 times as large a part of a rod as of a furlong, or, in this instance, 40 times of a rod, which is 4 rods. Since for every rod there are 5, or yards, there must be of as large a part of a yard as of a rod, or, in this instance, which is, &c. The written work would be the same as before. yard, (b.) The only difference between the processes of this and the previous article is the difference between decimal and denominate numbers. In the former, a unit of any denomination equals 10 of the next lower, while in the latter the number of units of a lower denomination to which any unit is equal, varies with the denomination of the unit considered. What is the value of each of the following in whole num bers of lower denominations? (c.) Reduce .7925 of a £ to whole numbers of lower de nominations. First Solution. Since £1 = 20 s., .7925 of a £ must equal .7925 of 20 s. 20 times .7925 s. 15.8500 s. Since 1 s. = = 12 d., .85 of a s. must equal .85 of 12 d. = 12 times .85 d. = 10.20 d. Since 1 d. = 4 qr., .2 of a penny must equal .2 of 4 qr. = .8 qr. Hence, .7925 of a £= 15 s., 10 d., .8 qr. Second Solution.· Since there are 20 s. for every pound, there must be 20 times as large a part of a shilling as of a pound, or, in this case, 20 times .7925 s. 15.85 s. Since there are 12 d. for every shilling, there must be 12 times as large a part of a penny as of a shilling, or, in this case, 12 times .85 s., &c. The written work is the same as in the last solution. NOTE. The student should be careful to multiply only the fractional part of each number. 9. What part of 1 mile is 7 fur. 4 rd. 2 yd. 1 ft. 4 in.? Solution. Since 1 in. 1 ft. = = a foot, to which adding the 1 foot gives 1 ft. of a foot. Since of a yd., of a foot must equal of of a yd., or of a yd., to which adding the 2 yards gives 2 yd., or 22 yd. Since 1 yd. of a rd., 22 of a yd. must equal 22 off of a rd., or of a rd., to which adding the 4 rd. gives, &c. When the work is written, the following form may be adopted: = Second Solution. - Since there is of a foot for each inch, there must be as many feet as inches, or, in this case, 2 of 4 ft. = 12, or of a ft. Since there is of a yard for each foot, there must be as many yards as feet, or, in this case, of, or yd. =, &c. The written work is the same as in the last solution. NOTE. - The method here given is usually preferable to that of 129, solution of 72d example, inasmuch as it keeps all the fractions reduced to their lowest terms, and enables us to perform very many, if not most, such examples without writing any figures. What part 10. Of 1 A. is 2 R. 36 sq. rd. 8 sq. yd. 2 sq. ft. 36 sq. in.? 11. Of 1 gal. is 3 qt. 1 pt. 1 gi.? 12. Of 1 Cd. ft. is 10 cu. ft. 1382 cu. in.? 13. Of 1 T. is 17 cwt. 3 qt. 2 lb. 12 oz. 74 dr.? 14. Of 1 lb. is 6 oz. 13 dwt. 8 gr.? 15. Of 1 £ is 9 s. 5 d. 11 qr.? 16. Of 1 circumference is 155° 4' 361"? 17. Of 1 lb. is 5 3 6 3 0 9 63 gr.? 18. Of 1 w. is 3 da. 10 h. 17 m. 8 sec.? (d.) Should it be required to give the answers to such questions as the above in a decimal form, it will only be necessary to reduce the vulgar fractions obtained by the preceding process to equivalent decimals. (e.) The following process may also be applied : 19. What part of a pound is 13 s. 7 d. 2 qr.? Solution. Since for every farthing there is of a penny, there must of 2 d. = .5 of a Since for every penny as many shillings as pence, be as many pence as farthings, or, in this case, or, in this case, 1 of 7.5 s. =, &c. The most convenient form of writing the work is to arrange the numbers expressing the various denominations in a vertical column, thus: 4) 2.0 qr. 12) 7.500 d. = 7 d. 2 qr. 20) 13.6250 s. 13 s. 7 d. 2 qr. .68125 £13 s. 7 d. 2 qr. = Ans. In like manner perform the following 20. What part of 1 bu. is 5 pk. 3 qt. 1 pt.? 21. What part of 1 lb. is 6 oz. 13 dwt. 8 gr.? 22. What part of 1 gal. is 3 qt. 1 pt. 3 gi.? 23. What part of 1 Cd. ft. is 10 cu. ft. 13823 cu. in.? 145. To find a Number from a Fractional Part of it. 1. 5861 First Solution. of what number? 5861 = of 7 times 5861, which is 41027. Second Solution. If 5861 = of some number, 7, or the number itself, must equal 7 times 5861, which is 41027. Solution. - If 3476 = - of some number, of that number must be of 3476, which is 3476, and 9, or the number, must be 9 times this result, and may be expressed by multiplying the numerator by 9. Hence we have the following written work:: |