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the remainder with peach-trees; how much of the lot was reserved for peaches ?

Ans. 10 A. 146 P. 9. Mr. Albert having purchased a section of land from the Government at $1.25 an acre, sold Mr. Hull a halfquarter section at $2.75 an acre; Mr. Snyder a quarterquarter section at $3.00 an acre; Mr. Landis a half-quartersection at $2 an acre; and Mr. Anderson a quarter-quartersection at $2.50 an acre; how many acres bas he remaining, and what is his entire profit, if he disposes of the remainder at $2.25 an acre ?

Ans. 400 acres ; $700. COST OF ARTIFICERS' WORK. 480. By Artificers' Work we mean plastering, painting, papering, paving, stone-cutting, etc.

481. Plastering, painting, papering, paving, and ceiling are estimated by the square foot or square yard. Roofing, flooring, partitioning, slating, etc., generally by the square, which consists of 100 square feet, but sometimes by the square foot or yard.

482. Shingles, which commonly measure 18 in. by 4 in., are estimated by the thousand or bundle. 1000 are generally allowed to a square of 100 sq. ft.

EXAMPLES FOR PRACTICE. 1. What will be the cost of plastering a room 40 ft. long, 25 ft. 3 in. wide, and 9 ft. 9 in. bigb, at $.38 a square yard ?

SOLUTION.—The surface equals 40 X 25/+(40 X 92 X2)+(254 X 9 X 2), or 203625 sq. ft., which equals 226-325 sq. yd., hence the cost is $.38X 226-39, or $85.971).

2. What will it cost to pave a street 36 st. wide and 2240 ft. long, with Nicholson pavement, at $.30 per square foot ?

Ans. $24192. 3. What will be the cost of slating the roof of my barn 65 ft. long, and 24 ft. 6 in. from eaves to ridge @ $8.75 per square ?

Ans. $278.68% 4. How many slates 12x 24, { exposed to the weather, will be required to cover a roof 112 ft. long and 40 ft. from eaves to ridge ?

Ans. 13440.

5. What will be the expense of shingling a roof 85 ft. long and measuring 25 ft. from the caves to the ridge, the shingles being worth $14.25 p M. ? Ans. $605.624.

6. How many shingles will it take to cover a roof 65 ft. long, and 30 ft. from the eaves to the ridge, each shingle being exposed one-third to the weather, and the first course being double?

Ans. 23790. 7. What cost the painting of a wainscot 4 ft. 6 in, high, in a room 25 ft. by 16 ft. 8 in. at $.45 a sq. yd.? Ans. $18.75.

8. What will it cost to plaster a room 40 ft. long, 22 ft. wide, and 9 ft. 6 in. high, at $3.15 per square of 100 ft., deducting 96 sq. ft. for doors and windows? Ans. $61.80.

9. What will it cost to plaster a house of 10 rooms, 4 of 20 x 16 ft. and 8.ft. high ; 4 of 20 x 16 ft. and 92 ft. high; 2 of 20 x 15, one being 8 ft. and the other 9 ft. high, 2 halls 32X8 ft., one being 8. ft. and the other 9 ft. high, allowance being made for 20 doors, 71x3, and 24 windows 61x3, at 18€ per sq. yd.?

Ans. $211.36.

CARPETING, PAPERING, ETC. 483. In Carpeting, Papering, etc., it is frequently necessary to find the quantity of material of a given width required to cover or line a given surface. We do this by the following

Rule.- Divide the surface we wish to cover by the area contained in a yard of the material.

EXAMPLES FOR PRACTICE. 1. How many yards of carpeting of a yard wide, are required to cover a floor 24 ft. 9 in. by 18 ft. 3 in.?

SOLUTION.—24 ft. 9 in. equals 247 ft. ; 18 ft. 3 in. equals 184 ft. ; 247 x 184 equals 45111 sq. ft., or 50, sq. yd.; the area of 1 yard of the carpet is of a square yard, and dividing 50, by, we have 80.3, the number of yards of carpet required.

2. What will it cost to carpet a room 33 ft. long and 24 ft. wide, with Brussels carpeting of a yard wide, at $2.25 a yard ?

Ans. $316.80. 3. A lady wishes to carpet a parlor 33 ft. long and 16 ft. 4 in. wide with Brussels carpet i yd. wide ; how many yards must she buy, allowing nothing for waste ? Ans. 792yd.

4. A lady made a patchwork bedquilt 6 ft. long, and 54 ft. wide; how much yard-wide muslin must she buy to line it, allowing 1 inch in length for waste ? Ans. 4 yd.

5. I have a flower-bed 16 ft. long by 12 ft. 8 in. wide, around which I wish to make a sod border 8 in. wide ; how many sods 1 foot square will be required ? Ans. 40 sods.

6. I have a table 6 ft. long and 3 ft. 4 in. wide, which I wish to cover with a baize cloth hanging down 10 inches on each side ; how many square yards do I require, and how many yards in length ? Ans. 427 sq. yd. ; 25 yd.

7. How many rolls of wall paper, each containing 4 sq. yards, are required to paper the walls and ceiling of a room 25 ft. long, 15 ft. 8 in. wide, and 9i ft. high, deducting 51 sq. ft. for doors and windows ?

Ans. 3027 8. What will be the cost of papering the above room at $2.374 a roll, putting also a gilt moulding around the top of the walls, at 9 cents a foot ?

Ans. $80.941. 9. My parlor contained 4 windows curtained with damask of a yard wide; for each window 8 yards @ $1.75 are required, the curtains being lined with silk of a yard wide at $1; also 54 yards of trimming @$1.25, and a cornice @$4.50; required the number of yards of silk, and the whole cost of the curtains.

Ans. 26yd.; $128.163.

MEASURES OF VOLUME. 484. A Volume is that which has length, breadth, and thickness or height. These three elements are called dimensions. A volume is also called a solid.

485. A Rectangular Volume or Solid is a volume bounded by six rectangles. The bounding rectangles are

CUBIC YARD called faces. Cellars, boxes, rooms, etc., are examples of rectangular volumes.

3 feet wide.

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3 feet high.

486. A Cube is a volume bounded by six equal squares. Or, a cube is a rectangular volume whose faces are all equal.

487. By the Contents or Solidity of a volume we mean the amount of space it contains. The contents are expressed by the number of times it contains a cube as a unit of measure.

Rule I.— To find the contents of a cube or rectangular volume, take the product of its length, breadih, and height.

For, in the volume above, the number of cubic units on the base equals the length multiplied by the breadth, or 3X3= 9, and the whole number of cubic units equals the number on the base multiplied by the number of layers of these cubes, or 9X3= 27; hence the whole number of cubes, or the contents, equals the product of the length, breadth, and height.

Rule II.- To find either dimension, divide the contents by the product of the other two dimensions.

EXAMPLES FOR PRACTICE. 1. What are the contents of a box 2 ft. long, 1 ft. 6 in. wide, and 1 ft. 9 in. high, measured on the inside ?

SOLUTION.—To find the contents, we multiply the length, breadth, and height together, and we have 2X11 X1$=54 cubic feet. 2. What are the contents of a cube whose edge is 15 ft.?

Ans. 3375 cu. ft. 3. How many cubic feet in a block of granite 6 ft. long, 4 ft. wide, and 3 ft. high?

Ans. 72 cu. ft. 4. How thin must a cubic inch of gold be beaten to cover a floor 36 ft. long and 20 ft. wide ? Ans. .000009+in.

5. How much air is in a room 24 ft. 6 in. long, 14 ft. 8 in. wide, and 10 ft. 6 in. high? Ans. 139 cu. yd. 20 cu. ft.

6. A piece of masonry is 17 yd. 2 ft. 7 in. long, 1 yd. 2 ft. 3 in. thick, and its contents are 301 cu. yd. 7 cu. ft. 1071 cu. in.; what is its height?

Ans. 9 yd. 1 ft. 11 in. 7. How much earth will be taken out of a cellar 32 ft. 7 in. long, 16 ft. 9 in. broad, and 7 ft. 6 in. deep?

Ans. 151 cu. yd. 16 cu. ft. 486 cu. in. 8. A hall 60 ft. long, 40 ft. wide, and 20 ft. high, is capable of seating 600 persons; how long before the air contained in it becomes unfit for respiration, allowing 10 cu. ft. a minute to each person ?

Ans. 8 minutes.

THE CYLINDER. 488. A Cylinder is a round body of uniform size, with equal and parallel circles for its ends. The two circular ends are called bases.

489. The Altitude of a cylinder is the dis- eu tance from the centre of one base to the centre of the other.

490. The Convex Surface of a cylinder is the surface of the curved part.

Rule I.To find the convex surface of a cylinder, multiply the circumference of the base by the altitude.

Rule II.To find the contents of a cylinder, multiply the area of the base by the altitude.

EXAMPLES FOR PRACTICE. 1. What is the convex surface of a cylinder, the diameter of whose base is 10 inches, and whose altitude is 18 inches?

SOLUTION.--The circumference of the base equals 10 in.X 3.1416, which is 31.416 inches; multiplying by the altitude, 18, we have 565.488 square inches, the convex surface.

2. What is the surface of a marble column 20 ft. high, and 24 inches in diameter ?

Ans. 125.664 sq. ft. 3. What is the length of a log of wood 18 inches in diameter, whose convex surface is 47.124 square feet ?

Ans. 10 ft. 4. How many cubic feet of water will a cistern hold whose depth is 71 ft. and diameter 5] ft. ? Ans. 178.1876.

5. A cistern is to be dug in a place where its diameter can only be 6 ft., but is to contain 420 cu. ft. of water ; what must be the depth ?

Ans. 14.85+ft. 6. What will it cost to line a cylindrical cistern with tin, at 50 cents a square foot, the diameter being 6 ft. and depth 8 ft.?

Ans. $89.54.

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WOOD MEASURE. 491. The Measure of Wood is the cord, which is divided into cord feet, etc.

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