Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

can be calculated. It will be equal to that of a solid whose base is the base of the vessel, and whose altitude is the difference in depth above mentioned. If the water in a rectangular box, whose base measures 5 by 3 in., is raised 17 in. by the introduction of a piece of marble, the volume of the latter = 5 × 3 × 17 cu. in.

This method is useful in determining the contents of a solid of irregular shape.

=

13. The radius of the base of (25.1328 yd. ÷ 3.1416) = 4 yd. Volume of cone = (42 x 3.1416 x of 18) cu. yd.

14. See Art. 1141, 22.

15. The slant height of the pyramid = √242 + († of 14)3. See Art. 1283, 13.

1146. 6. [(10¢ × 14) + (37 7 × 13)] × 10840).

=

weight in grams of a 10-mark piece. (1000279) x 15.432349. Dividing gives the number of U. S. gold dollars.

7. 1000 grams ÷ 279 Weight in Troy grains this result by 23

=

1000 x 150 43.2349

279 × 23 22.

NOTE.-23.22 is changed to a whole number by removing the decimal point two places to the right, and a corresponding change is made in one of the numbers in the dividend.

11. Interest on $237453250 @ 3 % = $ 7123597.50

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors]

The interest on the entire amount at 21% would be $30635198.75, the saving being $17256586.75. Ans.

12. $100 worth of stock costs $853. The annual dividend is 5% of $100, or $5. This is (5.8575) per cent on the money invested.

13. $89301.11.

14. See Art. 1122, 15. The following shows the account as it stands on the books of the Interior Department. The items that appear as debits on Mr. Well's books, here appear as credits, and vice versa.

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][subsumed][merged small][subsumed][ocr errors][merged small][merged small][merged small][merged small][merged small][subsumed][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][subsumed][merged small][subsumed][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][subsumed][merged small][merged small][merged small][ocr errors][merged small][merged small]

1148. Multiply the length in feet by the width in feet by the thickness in inches.

16. 3 (per ft.) × 15 × 16 × 3 × 3.

17. The floor contains (36 x 171) sq. ft., or 630 sq. ft. If 1-inch boards were used, 630 board feet would be required. The number of feet of 24-inch planks required = 630 ft. × 24.

[merged small][merged small][merged small][merged small][ocr errors][merged small][ocr errors][ocr errors][merged small][merged small][ocr errors][ocr errors][merged small][ocr errors][ocr errors][ocr errors][merged small][merged small][merged small][merged small][merged small][ocr errors][ocr errors][merged small]

=

M, is $3.56. Ans.

=

480 ft.+360 ft. + 480 ft. + 360

19. The length of the fence ft. 1680 ft. For a fence 4 boards high, (1680 × 4) running feet of boards will be needed, or 6720 running feet. If the boards are ft. wide and 1 in. thick, the number of board feet Cost $18 x 3.36.

= 6720 × × 1=3360 ft.

20. The length of the Surface (250 × 6) sq. ft.

=

=

fence

=

=

(25+100+25+ 100) ft. 1500 sq. ft. As the boards are 1

in. thick, the number of board feet 1500.

=

[blocks in formation]

= $37.50. The number of posts = 250÷640; cost, at 25¢ each, $10. The number of running feet of scantling, two strips, 250 × 2 = 500; the number of board feet = 500 × × 2 = 250; cost, $18 x .25 $4.50. Total cost, $37.50 + $10+ $4.50 = $52. Ans.

=

1149. 1. (a) A note made payable to the order of a certain person or to bearer is negotiable; in the former case, an endorsement is necessary to transfer its ownership. A note payable to bearer does not require endorsement. A note payable to Charles Naumann (without the words, "or order," or the like) is not transferable by endorsement. If Charles Naumann wishes to sell the note, he must assign his interest in it by another document.

NOTE. — The above is the general rule; in some states there are special laws bearing on the subject:

In Alabama and Kentucky, a note to be negotiable must be payable at a fixed place; in Indiana and Virginia, at a bank; in West Virginia, at a bank or public office. In Pennsylvania, it should contain the words "without defalcation"; in New Jersey, "without defalcation or discount"; in Missouri, "negotiable and payable without defalcation or discount."

(b) A person unable to write his name, makes his mark, as shown below, in the presence of a witness:

Witness:

THEODORE H. FICKLIN.

his

WILLIAM X DEVERS.

mark

3. In old deeds, the contents of a farm are given in acres (A.), roods (R.), and poles (P.), the rood being acre, and containing 40 poles, or square rods. In long measure, the word pole is occasionally employed instead of rod.

[ocr errors][merged small][merged small]

7. The distance between the ships is the hypotenuse of a right-angled triangle, whose other sides are 72 mi. and 128 mi., respectively.

=

8. The first capital = $3500+1.40 $2500; C put in 388,

or of $2500 = $1500; etc.

9. See Art. 1026, 10.

10. See Supplement.

1150. 3. x2+x2= hypotenuse2 = 100. x2=50, the area of the inscribed square. Ans.

4. Area of circle = (5 x 5 x 3.1416) sq. in.

5. Arc of 90° =

= √50 in.

=

78.54 sq. in.

[blocks in formation]

6. Arc of 90° in a circle whose radius is 10 in.

[blocks in formation]
[ocr errors]

- 15.708 in.

€78.54 sq. in. Ans.

50 sq. in.; area of

- 50 sq. in. = 28.54 sq. in. Ans.

9. Area of outer circle in square yards = (152 × 3.1416) ÷ 9; of inner circle = (102 × 3.1416) ÷ 9.

10. (125 × 3.1416) sq. ft.

11. [(62 — 32) × 3.1416] sq. in. The radius of the outer circle

is 6 in.; of the inner circle, 3 in.

12. 36 x 3.1416: 9 x 3.1416 =4:1.

13. [(30 × 30) — (20 × 20)]

sq. ft.

[blocks in formation]

X

30

[ocr errors]

20

[blocks in formation]

of the center lines is 100 ft. The area of the walk is (100 × 5)

sq. ft.

The area of the circular frame in 11 can be ascertained in the same way. The center of the frame is 44 in. from the center of the glass. The middle line of the frame × 2) in. = 28.2744 in. The area

[blocks in formation]

=

(3.1416 × 4

15. The area of the first = 240 sq. ft.; of the second, 960 sq. ft. 1153. 18. The surface of the sphere = (4x) sq. ft. = 3.1416 sq. ft.; the convex surface of the cylinder = 3.1416 sq. ft.

=

19. The entire surface = 3.1416 sq. ft.+ (2 x x 3.1416) sq. ft. 11⁄2 times 3.1416 sq. ft.

20. 4 times (circumference) 10 × ( diameter)

[blocks in formation]
[ocr errors]

10

3.1416

=

1154. 1. Rate per cent = 18750 ÷ 12500 = 1. 1% of $6000 $90.

[ocr errors]

3. The price of silver is now given by the ounce. 4. The interest on $200 at 41%

$112.50. Ans.

=

Total liabilities

=

$9.

[blocks in formation]
[ocr errors]

=

=

5. $2100 + $4400 + $13000 + $7200 (90% of $8000) $26700 total assets. $1625 +$5625 $7250. $26700 - $7250 = $19450. To this, add the amounts withdrawn, $850 + $1075 = $1925, making the sum of the capital and profits $21375. Of this, H is entitled to, $7125, less the amount withdrawn by him, $1075, or $6050.

1157. 9-11. Find the cube root of the numerator and of the denominator separately.

12-15. Reduce to an improper fraction before extracting the cube root; then reduce the root to a mixed number.

1159. 2. [× 3.1416 × (3 × 3 × 3)] cu. in. Cancel.

3. The volume of the first in cubic inches = {π×}=}π. The volume of the second TX1=&T.

4. The volume of the sphere = .5236 cu. in.; the volume of the cube 1 cu. in.

« ΠροηγούμενηΣυνέχεια »