what legacy did the father give to each, supposing their ages were 12, 13, 14, 15 years, and allowing compound interest at 5 per cent. per annum? 1. Put r1.05 the amount of 1 dollar for one year, m= = 100,000 dollars; and let the four shares be denoted by x, rx, r2x, r3x; then x + rx + r2x + r3x=m, hence x = m ÷ (1 + r — p2 + p3) = 23201.18 dolls., the first share; then the other three shares are 24361.242, and 25579.304, and 26858.27. For, by table 3, the amount of each of these sums will be 35992.65 dollars. =a. 2. Put p=100000, and r = 1.05; then will the ratios of the shares be denoted by 1, r, r2, r3, the sum of which is 4.310125 Then as a : 73:: p: 26858.27 dollars, the eldest son's share. a: r2::p:25579.30 ar::p:24361.24 a:1::p:23201.18 66 the second " 66 the third "< 66 66 the fourth 66 3. The present worths of 1 dollar, forborne during their respective minorities, are .74622, and .71068, and .67684, and .64462 dolls., the sum of which is 2.77836; consequently 2.77836: 100000 :: .74622 26858.27 dolls., the sum left to the eldest son. In like manner the sums bequeathed to the others are found to be 25579.30, and 24361.24, and 23201.18 dollars, Ans. as before. 587. A sum of money, p, is to be divided among A, B, C, in such a manner, that at the end of a, b, c years, when they respectively come of age, they are to possess equal sums: required the share of each, allowing compound interest at a given rate. Ans. A's share B's share= and C's share= 588. A lends B 1000 dollars, for which B repays him as folFows, viz. at the end of three months 180 dollars, of 5 months 150 dollars, of six months 140 dollars, of eight months 100 dollars, of nine months 90 dollars, of ten months 120 dollars, and at the year's end 250 dollars. The rate of interest is required. Let 2 rate per annum; then x (according to art. 137) will be the rate for one month, x2 for two months, &c., whence the present value of 180 dolls. to be received at the end of 3 months will 180 150 140 100 90 120. 250 be 180 &c.; .. + + + + + + = 23 212 1000 (per quest.); and therefore x12 — .18xo — .15x21 — .14xo .09.x3 .12.25: Solved, x = 1.003852; and 12 = 1.047216; and consequently 4.72 the rate per cent. required. 589. A gentleman bought an estate in houses for 1500 dollars, which, being let, brought him in 120 dollars per annum, clear of all expenses and deductions. At the end of 10 years, (most of the houses being out of repair, and he not choosing to be at the expense of fitting them up,) he sold the whole estate again for 800 dollars. The question is, to find what interest he made of his money. Let a 1500 dollars, b = 800 dollars, c = 120 dollars, and x the required rate of interest; Then will ax = principal and interest, and ax c = amount after the first payment is deducted. And in the same manner we have ax1o — cao —— cx3 — cx2 - cx6. Cx5 сха cx3 схов CX cb (per question), or ax1o - c(x1o — 1) ÷ (x — 1) = b; therefore ar11 (a—c)x10. bx+b+ c=0. Solved, x=1.04142, &c., and the rate of interest required $4.14 per cent. See Index. 590. How long must a capital a, $3600, remain at the interest p, 5, to become as much as a capital a', $5000, at the interest p' =4, for n, 12 years? log. a'n log. p'-log. a) years Ans. log. p 16 nearly. 591. A capital a is put out at the interest p, and at the expiration of each year the interest is added to the principal; but at the same time it increases or diminishes yearly by the same sum b. What will this capital amount to n years hence? Ans. It is ap" ± b(p" —1) p-1 where the upper sign obtains for the capital in creased by b, and the lower the capital diminished by b. 592. A capital a is lent out at the rate p. In what time will it become a', if the capital, with the interest and compound interest added to it, be augmented or diminished yearly by the sum b? log. (p-1)a'b}-log. {(p-1)±b}, Ans. Log. n= number of years.) The upper sign of b, the lower the deduction b. log. p ; (n the denotes the addition of 593. If, in the preceding problem, b be yearly taken away, and b be greater than the interest of the capital a; in this case, in how many years will the capital be spent? Ans. Log. n= log.blog.b-(p-1)a} log. p gives the number of yrs. 594. What is the present value w of an annuity r at p per cent. which a person has to enjoy n years ? If an annuity r, having n years to run, be worth in cash w, what 596. How long has an annuity to run, if its present value be considered as equal to w dollars, the interest being 5 per cent.? Ramount of 1 dollar for 1 year at the given rate. 597. An annuity r is required for the given space of n' years, whose present value is equal to another annuity r for n years; if both be calculated at p per cent., what is the annuity?* 598. But how great is the required annuity, if it be payable for m years; consequently, if computing after a lapse of m years, it be payable n' years after this period? * Ans. r' = (p"-1)p"-" -r. Ans. r' = p" — 1 (ph ·1)p 599. The amount of an annuity of $1 to continue n years = $P, and the amount of the same annuity to continue 2 years = $Q; what is the rate per cent. allowing compound interest ? Ans. 100. 600. Given (s) the sum of an arithmetic progression, (n), the number of terms, and (p) the product of the first and last terms, to determine the progression. com. dif. 2 n-1 { where the upper or lower sign must be used, according as the progression is a decreasing or an increasing one. 601. Seven years ago, I was just three times as old as my eldest son; but seven years hence, if we should both live, I shall be only twice as old. What are our present ages? Ans. 49 and 21. 602. Find the ratio of the length and breadth of a rectangular field, consisting of 2 acres of ground, that shall have the same perimeter as a square field consisting of 4 acres. Ans.. 2+v2 603. A person being asked how many horses he kept, said, "For want of room in my stable, I am obliged to put 8 of them out to hire; but I am now building a new one, twice as large as the former, which, when finished, will enable me to accommodate 8 horses more than I now have. How many did he keep? Ans. 24. 604. A company of 18 persons, consisting of men and women, having dined together at an inn, their reckoning came to £9 18s., in the settling of which each of them paid as many shillings as there were men in company. How many men and women were there? Ans. 11 men and 7 women. 605. Find two geometrical mean proportionals between 3 and 24, and four geometrical means between 3 and 96. Ans. 6 and 12; and 6, 12, 24, and 48. 606. Supposing that 19 pounds of gold weigh 18 pounds in water, and 10 pounds of silver 9 pounds in water, and that a mass of gold and silver of 106 pounds weighs 99 pounds in water; it is required to find the quantity of gold and silver contained in the Ans. 76 pounds of gold, and 30 pounds of silver. mass. Let x and y, denote the weight of gold and silver in each mass then x+y =100, and As 19: 18:: x: 18x; 10:9::y: Poy; and 18x+y=99, fx+1by=11, Ans x=76 and y = 30. 607. The dimensions of a rectangular floor are such, that if it had been 2 feet broader and 3 feet longer, it would have been 64 square feet larger, and if it had been 3 feet broader and 2 feet longer, it would have been 68 square feet larger. What is the length and breadth of the floor? Ans. 14 feet long and 10 broad. Let (x+2)(y+3)=64—xy, or 2x+3y=58 and (x+3y) (y+2) 68+xy or 3x + 2y = = 62, and x = 14, y 10 Ans. 608. Find three numbers in arithmetical progression, such that the square of each, added to the product of the other two, shall make 576, 612, and 792 respectively. Ans. 12, 18, and 24. 609. Find two such whole numbers, x and y, that the sum of the aliquot parts of x shall be two thirds of y, and the sum of the aliquot parts of y three fourths of x. Ans. x = 20, y = 33. 610. Find three numbers in harmonical proportion, such that their three differences shall be in geometrical proportion. * = 611. The sum of two numbers is 2, and the sum of their ninth powers 32; required the numbers by a quadratic equation. Ans. *5-1, 6+2/5, and 2. Ans. 1±√(6√34—33). 612. The sum of the two extremes of six numbers, in geomet❤ rical progression, is 165, and the sum of the four means 150; required the terms of the progression, by a quadratic equation. Ans. 5, 10, 20, 40, 80, and 160. 613. Given xy(2+ y2)=3, and x2y2(x* + y1)=7, to find the values of x and y by a quadratic equation. x= Ans. (5+ 1), y = {(√√5—1). 614. Given x+y+z= 23, xy+xz+yz 167, and xyz: 385, to find the values of x, y, and z. Ans. x=5, y=7,z= =11. 615. Given (y + z) = 207, and y(x+2)=87, z.(x + y) 240, to find the values of x, y, and z. Ans. x=9, y=3, z=20. 616. Given x2+xy+y=5, and x + x'y' + y'= 11, to find the values of x and y by a quadratic equation. x= Ans. 10+√5, y=√10—No5. 617. Given x2+yz=920, y2+xz-980, and 22+xy=1000, to find the values of x, y, and z by a biquadratic equation, this being one of the least dimensions to which the question can be reduced. Ans. x= 19.5991, y = 22.7788, and z = 23.5276. 518. Required the sum of n terms of the series 12+22x +32x2 +42x2+52x2+ &c. 1 Ans. (1 —x)3° 519. Find the sum of the series +4.6.8 +6.8.108.10.12 &c. continued ad infinitum. 2.4.6 Ans. 1 620. Required the sum of the series + + + 2 1.3.5 3.5.7 5.7.9 3 621. Find the sum of the series + &c. continued to infinity. 1 + 2.3.4.5 5.6.7.8+ 8.9.10.11 Ans. x= 1. Ans. 280 622. Given 1+3x+6x2+10x3+15x+ &c. ad infinitum = 10, to find the value of x. 100. 623. Given + fx2+18x2+382 + 384025+ &c. ad infinitum, to find x. Ans. x=-16+36-512+2560-&c. 624. Find the sum of 100 terms of the series (1X2)+(2×3) +(3×4)+(4×5) + (5×6) + &c. Ans. 343400. 625. Required the sum of 25 terms of the series 1 x 35+ 2 x 36+3x37+4x 38+5 x 39+ &c. which gives the number of shot in a complete oblong pile, consisting of 25 tiers, the number of shot in the uppermost row being 35. Ans. 16575. 626. Required the approximate value of the series 1-1+ -16+25- &c. continued ad infinitum. Ans. .822467. 627. Find the approximate value of the hypergeometrical series 1.2-2.32.3.4-2.3.4.5+ 2.3.4.5.6-&c. continued ad infinitum. See Index. Ans. .59634736. 628. There are two columns, in the ruins of Persepolis, left standing upright; one is 64 feet above the plane, the other 50: Between these, in a right line, stands an ancient statue, the head whereof is 97 feet from the summit of the higher, and 86 feet from the top of the lower column; the base whereof measures just 76 feet to the centre of the figure's base. Required the distance of the tops of the columns. CD 169.96. E 629. Wanting to know the height of an inaccessible tower; at the least distance from it, on the same horizontal plane, I took its angle of elevation equal to 58°; then going 300 ft = AC directly from it, found the angle there to be, only 320: Find its height, and my distance from it at the first station. S height 307.53 Ans. distance 192.15 |