cents, to make a mixture worth 74 c. a bushel. To make a mixture worth 69 c. a bushel. Last ans. 1 bu. at 80 c., 1 at 75 c., 1 at 70 c., 18 at 68 c. 2. B has contracted to furnish 5000 lb. of wool worth 50 c. a pound. He desires to make it up of three different qualities, worth respectively 45, 48, and 60 cents. How many pounds must he take of each ? 3. A trader wishes to mix 60 lb. of coffee, which cost him $11.40, with other qualities that cost 15 c. and 16 c. a pound, so that he can sell the mixture for 18 c. and make a profit of 1 c. a pound. How much of the two latter kinds must he take? Ans. 40 lb. of each. 4. How much water must be put with 18 gal. of rum, half of which cost $2 a gallon and the rest $3 a gallon, that the mixture may be worth $2 a gallon? 5. A confectioner, having 72 lb. of candy worth 16 c. a pound, wishes to mix it with other qualities worth respectively 20 c., 25 c., and 30 c. a pound, in such proportions that the mean value of the whole may be 22 c. What quantities must he take? 6. A fruit-dealer sold 208 oranges for $5.20, charging 1, 2, 3, 4, and 5 cents apiece, according to their size. In what relative quantities might he have sold the different sizes, to receive the sum named? 7. In what relative quantities must pure silver and silver.950, .800, and .850 fine, be melted together, to make a compound .900 fine? 8. A farmer purchased 36 animals for $216; sheep, at $7 a head, lambs at $3, calves at $5, and 12 pigs at $10. How many of each kind did he buy? Ans. 4 sheep, 16 lambs, and 4 calves. 9. A man bought 10 cows, at $45 each, some sheep at $6, horses at $125, and young cattle at $20. The whole averaged $49 apiece; how many horses, etc., did he buy? CHAPTER XXXI. EVOLUTION. [It may be well to review Involution, Arts. 98-104.] 628. Evolution is the process of resolving a number into two or more equal factors. One of these equal factors is called a Root of the number resolved. Resolving 9 into two equal factors, we get 3 and 3. This process is Evolution, and 3 is a root of 9. Evolution is the opposite of Involution, Art. 98. In the latter, the power of a root is required; in the former, the root of a power. 629. Roots, how named.-Roots take their names, Square Root, Cube Root, Fourth Root, Fifth Root, etc., from those of the corresponding powers. 9 is the square of 3, and 3 is the square root of 9; 27 is the cube of 3, and 3 is the cube root of 27. 630. Roots, how indicated.-Roots are indicated by a character called the Radical Sign, √, placed before the number whose root is to be extracted. The Index of a root is a figure placed above the radical sign at the left, to denote what root is to be taken,—that is, into how many equal factors the number is to be resolved. To express the square root, the radical sign is used without any index. 9, read square root of 9, = 3; for 3 X 3 = 9. 27, read cube root of 27, = 3; for 3 X 3 × 3 = 27. 81, read fourth root of 81, = 3; for 3 × 3 × 3 × 3 = 81. Roots are also indicated by means of a fractional index placed after the number. Thus, 9 = √/ 9. 12: = √12. 631. Surds. Some numbers are incapable of being resolved into equal factors; their roots can only be indi cated. Such roots are called Surds, or Irrational Quantities. 1/3 is a surd. 632. The most important operations in Evolution are the extraction of the Square and the Cube Root. Square Root. 633. Extracting the square root of a number is resolving it into two equal factors; as, 4 = 2 x 2. 634. General Principles.—Taking the smallest and the greatest number that can be expressed by one figure, by two, three, and four figures, let us see how the number of figures they contain compares with the number of figures in their squares : Squares, 1 81 1'00 98'01 1'00'00 99'80'01 1'00'00'00 99'98'00'01 We find from these examples, which might be extended indefinitely, that, if we separate a square into periods of two figures each, commencing at the right, there will be as many figures in the square root as there are periods in the square,—counting the left-hand figure, if there is but one, as a period. 635. Let us square 15, regarding it as composed of 1 ten (10) and 5 units, and merely indicating the operations: Adding partial products, 102+2 (10 × 5) +52 225 225 From the result on the left, which is shown on the right to equal in value the square of 15 as obtained in the ordinary way, we deduce the following law : The square of a number composed of tens and units, equals the square of the tens, plus twice the product of the tens and units, plus the square of the units. The diagram in the margin will illustrate this law of numbers, by means of surfaces. Let AC be 15 feet, AB being 10, and BC 5. Let CK be 15 feet, CJ being 10, and JK 5. It will be seen that A K (the square of 15 ft.) is made up of A H (the square of 10), the two rectangles C H, H L (twice the product of 10 and 5), and HK (the square of 5). Fifteen feet. A Fifteen feet. 10% 10X5 B 10X5 H J 52 K 636. Extracting the Square Root.-The method of extracting the square root is derived from the opposite operation of squaring, as performed in Art. 635. Ex. 1.-Required, the square root of 225. Separating 225 into periods of two figures, beginning at the right (2'25), according to Art. 634 we find that the root will contain two fig ures. According to the law laid down in Art. 635, 225 must equal the square of the tens in its root, plus twice the product of the tens and units, plus the square of the units. The square of the tens must be found in the left-hand period 2 (hundreds). The greatest number whose square is contained 2'25 (15 1 10×220) 125 (20 × 5)+52 = 125 the 2 hundreds. Bringing down the remaining period, we have 125, which must equal twice the product of the tens and units, plus the square of the units. Hence, to find the units' figure of the root, we divide 125 by twice 1 ten, or 20. The quotient is 6; but, if we multiply 20 by 6, and to the product add the square of 6, we shall have 156, which exceeds 125. Accordingly, 6 being too great, we write 5 in the root as its units' figure. Twice the product of the tens and units (that is, twice 10 × 5), plus the square of the units (52), is now 125. Placing this under the dividend 125, and subtracting, we have no remainder. 2'25 (15 1 25 125 In practice, we write twice the tens' figure (2) on the left as a trial divisor, and annex to it the units' figure (5) when found. Multiplying the divisor thus completed by the units' figure (which is equivalent to adding twice the product of the tens and units to the square of the units), we have the same result as before, 125. 637. RULE.-1. Separate the given number into periods of two figures each, beginning at the units' place. 2. Find the greatest number whose square is contained in the left-hand period, and place it on the right as the first root figure. Subtract its square from the first period, and to the remainder annex the second period for a dividend. 3. Double the root already found, and, placing it on the left as a trial divisor, find how many times it is contained in the dividend with its last figure omitted. Annex the quotient to the root already found and to the trial divisor: Multiply the divisor thus completed by the last root figure, subtract, and bring down the next period as before. 4. To the last complete divisor add the last root figure for a new trial divisor, and proceed as before till the periods are exhausted. If any trial divisor is not contained in the dividend with its last figure omitted, annex 0 to the root already found and to the trial divisor, bring down the next period, and find how many times it is then contained. In finding the root figures after the first, allowance must be made for the completion of the trial divisor. If, on multiplying a complete divisor by the last root figure, the product is greater than the dividend, |