THE SPHERE. 503. A Sphere is a volume bounded by a curved surface, every point of which is equally distant from a point within called the centre. 504. The diameter of a sphere is a line passing through its centre and ending in the surface. The radius is half the diameter. 505. RULE.-To find the surface of a sphere, we multiply the circumference by the diameter, or square the radius and mul tiply it by 4 times 3.1416. 1. Required the surface of a sphere whose diameter is 24 inches. Ans. 1809.5616sq. in. 2. Required the surface of a sphere whose diameter is 96 inches. Ans. 28952.9856sq. ft. 506. RULE. To find the contents of a sphere, we multiply the cube of the diameter by of 3.1416. 1. Required the contents of a sphere whose diameter is 6 inches. Ans. 113.0976cu. in. 2. If the diameter of the earth is 8000 miles, what are its surface and solid contents? Ans. Sur., 201062400sq. mi. 507. RULE.-To find the size of a cube which may be cut from a given sphere, we square the diameter, divide by 3, and extract the square root of the quotient. 1. What is the side of a cube which may be cut from a sphere 21 inches in diameter? MEASURE OF LUMBER. Ans. 12.124in. 508. Boards are measured by the square foot instead of the cubic foot. This measure is called Board Measure. 509. In board measure the board, whatever its thickness, is regarded as an inch thick. Planks, beams, joists, etc., are usually measured by board measure. 510. Round timber is generally measured by the ton, though sometimes it is estimated by board measure. 511. RULE.-I. To find the contents of a board, multiply the length in feet by the width in feet, or, II. Multiply the length in feet, by the width in inches, and divide the product by 12. 1. What are the contents of a board 16 feet long and 11ft wide? Ans 24sq. ft. 2. What are the contents of a board 14 feet long and 9 inches wide? Ans. 10 sq. ft. 3. Required the contents of a board 20 feet long, the ends being 18 and 14 inches respectively. Ans. 263sq. ft. 512. RULE. To find the contents of a plank, beam, joist, etc., take the product of the width and thickness, in inches, multiply by the length, in feet, and divide by 12. 1. How many feet in 14 planks 16 feet long, 18 inches wide, and 4 inches thick? Ans. 1344sq. ft. 2. How many feet in a stick of timber 40 feet long, 14 inches wide, and 9 inches thick? Ans. 420sq. ft. MASONS' AND BRICKLAYERS' WORK. 513. Masons' work is measured either by the cubic foot, or he perch. 514. The Perch is 16ft. long, 11ft. wide, and 1ft. deep, and contains, therefore, 16×14×124 cu. ft. 515. RULE.-To find the number of perches in a piece of masonry, we find the contents of the wall in cubic feet and divide by 241. 1. How many perches in a wall 40ft. long, 12ft. high, 3ft. thick? Ans. 58.18 perches. 2. How many perches in a wall 24ft. 9in. long, 15ft. 6in. high, and 4ft. 6in. thick? Ans. 69 perches. 3. How much was paid for building a wall 60ft. long, 18ft. 9in. high, and 3ft. 8in. thick, at $2.50 a perch? Ans. $416.667. 4. How many bricks each 8 inches long, 4 inches wide, 21 inches thick, will it take to build a wall 240 feet long, 8 feet high, 1 foot thick? Ans. 46080. SECTION XIV. ARITHMETICAL ANALYSIS. Immediately after Fractions we gave some simple cases of arithmetical analysis; we now present some more difficult problems under the same head. Many of these problems are very old, some of them are original with the author, and several are from algebra, first introduced into arithmetic by the author in his Normal Mental Arithmetic. 516. CASE I. 1. A can do a piece of work in 4 days and B can do it in 6 days; in what time can both do it? SOLUTION.-If A can do it in 4 days, in 1 day he can do of it; and if B can do it in 6 days, in 1 day he can do of it, and they together can do plus, which is of it, in one day; hence they will do of it in of a day, and they will do t of it in, or 2 days. 2. A can build a boat in 8 days and B in 12 days; in what time can they together build it? 3. A quantity of flour lasts a man and the wife alone 27 days; how long would alone? Ans. 44 days. wife 9 days, and it last the man Ans. 131 days. 4. A can build a wall in 8 days, B in 12 days, and C in 15 days; Ans. 37 days. in what time can they all build it? 5. M can dig a ditch in 30 days, and M and N in 20 days; how long will it take N to make what remains when M has made Ans. 45 days. of it? 6. A, B, and C can mow a field in 8 days, A and B can do it in 12 days, B and C in 15 days; in what time could each alone do it? Ans. A, 174; B, 40; C, 24. 517. CASE II. 1. A receives $3 a day for his labor and pays $1 a day for his board, and at the expiration of 50 days has saved $70; how many days was he idle? SOLUTION. Had he labored the 50 days he could have saved 50 times $2, or $100, hence he lost by his idleness $100-$70, or $30. Each day he was idle he lost $3, hence to lose $30 he must have been idle as many days as $3 is contained times in $30 which is 10 days. 2. A agrees to labor for $2.50 on condition that he should forfeit 50ct. every day he is idle; at the end of 100 days he receives $190; how many days was he idle? Ans. 20 days 3. A boy agreed to carry 150 oranges to market for 1ct each on condition that he should forfeit 4ct. for each one he ate; he received $1.70; how many did he cat? Ans. 10. 4. A girl agreed to carry 80 glasses to the depot for 2 cents each, on condition that she should forfeit 12ct. for every one she broke; she received 20ct.; required the number she carried safely. Ans. 68. 518. CASE III. 1. If 80lb. of sea water contain 2lb. of salt, how much fresh water must be added to these 80lb. so that 10lb. of the new mixture may contain of a pound of salt? SOLUTION.-If 10lb. of the mixture contain lb. of salt, to contain 1lb. of salt will require 6 times 10lb. or 60lb. of the mixture, and to contain 21b. of salt will require 2 times 601b., or 120lb.; hence there must be added 1201b. minus 80lb. or 40lb. 2. If 100lb. of sea water contain 3lb. of salt, how much salt must be added to these 100lb. so that 201b. of the new mixture may contain lb. of salt? Ans. 201b. 3. In a mixture of silver and copper consisting of 60oz. there are 4oz. of copper; how much silver must be added that there may be oz. of copper in 12oz. of the mixture? Ans. 12oz. 4. In a mixture of gold and silver consisting of 100oz. there are 6oz. of silver; how much gold must be added that there may be oz. of silver to 10oz. of gold? Ans. 56oz. 519. CASE IV. 1. Suppose that for every 4 cows a farmer has he should plough an acre of land, and allow one acre of pasture for every 2 cows; how many cows could he keep on 30 acres? SOLUTION.-If for 4 cows he ploughs 1 acre, for 1 cow he ploughs of an acre; and if for 2 cows he pastures an acre, for 1 cow he pastures of an acre; hence 1 cow requires or of an acre, and on 30 acres he could keep as many cows as is contained times in 30, or 40 COWS. 2. Suppose that for every 5 cows a farmer has he ploughs 1 acre of land, and allows 1 acre of pasture for every 4 cows: how many cows, could he keep on 54 acres? Ans. 120 cows. 3. Suppose that for every 6 cows a farmer has he ploughs 2 acres, and pastures 2 acres for every 10 cows; how could he keep on 64 acres? many cows Ans. 120 cows. 4. A farmer has 110 acres; he ploughs 3 acres for 7 cows and pastures 4 acres for 9 cows; how many cows did he keep and how many sheep, if of the number of cows equals of the number of sheep? Ans. 126 cows; 210 sheep. 520. CASE V. 1. A has $80 in gold and silver, and for every $3 of gold he has $2 of silver; how much gold must be added that for every $4 of gold there may be $2 of silver? SOLUTION. Since the gold is to the silver as 3 to 2, we find there are $48 of gold and $32 of silver. After the addition to the gold, there will be twice as much gold as silver, hence there will be 2 times $32 or $64 of gold, and hence there was added $64 — $48, or $16 of gold. 2. A boy has 150 apples and pears, and he has twice as many apples as pears; how many apples must he buy that he may have three times as many apples as pears? Ans. 50. 3. A farmer has 100 ducks and geese; for every 2 ducks he has 3 geese; how many ducks must he buy that he may have 2 ducks to 1 goose? Ans. 80. 4. A drover has 280 animals, consisting of horses and cows, and for every 3 horses there are 4 Cows; how many cows must he sell that there may be 4 horses to 3 cows? 521. CASE VI. Ans. 70. 1. The head of a fish is 20 inches long, the tail is as long as the head and of the body, and the body is as long as the head and tail both; required the length of the fish. SOLUTION. By the conditions, of the length of the body plus 20 inches equals the length of the tail, and this, plus 20 inches, the length of the head, equals of the length of the body plus 40 inches, which equals of the length of the body; then of the length of the body minus of the length of the body, which is of the length of the body, equals 40 inches, etc. 2. The head of a fish is 24 inches long, the tail is as long as |