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2 x 12 x 12 =

3024 x 66 x 10 x 8

924
3024 X 66 X 10 X8_100
924 x 12 x 12

Ans. 120 in. = 10 ft. 9. The weight of the provisions = 3 lb. x 32 x 45. Dividing by 40 gives the daily allowance for the increased crew; dividing this by their number, 32 + 16, gives the allowance of each.

3 lb. x 32 x 45

40 X 48 12. The difference in deposits = $450, which sum produces $18 interest. The rate is 4%.

14. $7500 at 3% for 37 years produces $ 1125 interest.

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200

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19. A receives 37% of selling price; he invested, therefore,

of $600; etc. 20. F receives 75. E's share is is more than D's. If is = $90, is =$45; F's share, 15, = $315.

21. The bank receives at the end of 63 days, the sum loaned, $ 593.70, + $6.30 interest. The problem is: At what rate will $593.70 produce, in 63 days, $6.30 interest?

593.70 x 100 * 360 = 6.30,
_630 x 100 x 360_

P=61,2%

59370 x 63 22. The two supply pipes fill [ + , or é, of tank in 1 hour. If all the pipes are set to work when the tank is full, the exhaust pipe takes off each hour of the tank more than the others supply. To empty the tank would, therefore, require 6 hours. 1103. 7. The diagonal = V13742 + 1374?.

9. Let 100x = value. Buying price = 90 x; selling price= 110x; gain = 20x, which is of the cost, 90x, or 22%.

10. 420¢ = (&$ - *)= number.

1106. 1. $500 = 5= cost of one portion ; 500 -- = cost of other portion.

4. If of farm is sold for cost, the selling price of the whole farm at the same rate would be cost = = f cost, making the gain } cost = 127%.

7. 10$ per bu. of 60 lb. = 163& per 100 lb., or $ higher than 16% per 100 lb., or 24 of 16¢ higher, or 41%.

8. See Art. 1084, 15.

10. If of the selling price is profit, the cost must be of the selling price; the latter is therefore of the cost. This is a profit of of the cost, or 663%. Ans.

Or, a profit of 2-fifths on a cost of 3-fifths is of the cost.
11. x +41x = 601
14. 14% of 2x = 150.
17. See Arithmetic, Art. 915, 6-8.
22. (100 = 54) years. The $200 is unnecessary.
23. (96 x 48 x 45) = (231 x 311). Cancel.

24. 20 acres = 3200 sq. rd. Each side measures V3200 rd. The diagonal = the square root of the sum of the squares of two equal sides. The square of each = 3200.

The diagonal= V3200 + 3200 rd. = V6400 rd. = 80 rd.

NOTE. — By constructing a square on the diagonal of a square, the pupils will see that the former will be twice as large as the latter ; that is, that a square on the diagonal of the above will contain 6400 sq. rd., making each side 80 rd.

In the above example, the square root of 3200 should not be extracted.

1107. 6. Find the proceeds of $1572.50 for 81 da. (Omitting days of grace, the proceeds for 78 da. = $ 1552.06.)

9. One costs 857% of $ 1500; the other costs 1024% of $1300. 10. ($ 2562.50 = 1.025) -- $62.50 = number of acres.

1. It is frequently difficult, for various reasons, to measure the altitude of a triangular field. On this account, a method of determining the area when the lengths of the sides are given, is useful, even though the underlying principles be not understood. A pupil can satisfy himself as to its accuracy, by calculating the area of the right-angled triangle in 3, and of the isosceles triangle in 5.

2. The half sum= 1 of (35 + 84 +91) = 105. The remainders are: (105 – 35) 70, (105 - 84) 21, and (105 – 91) 14. Area = V105 x 70 x 21 x 14 sq. ft. = 1470 sq. ft.

3. This is a right-angled triangle, since 212 +282 = 352; its area, therefore, is į of (21 x 28) sq. rd., or 294 sq. rd.

By the other method, the area = V42 21 x 14 x 7 sq. rd., or 294 sq. rd.

4. The sides of one triangle measure 39, 52, and 65 rd. respectively. Its area = V78 39 x 26 x 13 sq. rd. = 1014 sq. rd. The sides of the other are 25, 60, and 65 rd., respectively ; and its area = V75 x 50 x 15 x 10 sq. rd. = 750 sq. rd. The area of the quadrilateral =1014 sq. rd. + 750 sq. rd. = 1764 sq. rd. Each of these triangles is right-angled, AC being their common hypotenuse. Their areas are of (39 x 52) sq. rd., and 1 of (25 x 60) sq. rd., respectively.

5. Since the altitude AC, Fig. 2, Arithmetic, Art. 1263, of an isosceles triangle divides the base into two equal parts, BC = 15 yd. AC2 = ABBC? = 625 - 225 = 400; AC= 20 yd. The area = 1 of 600 sq. rd. = 300 sq. rd.

The area by the other method = 140 x 10 x 15 x 15 sq. yd.

6. In the right-angled triangle ACB (Art. 1263, Fig. 2), BC = 1 of 96 ft. = 48 ft; AB= 64 ft.; AC= V642 – 482 ft. = V4096 — 2304 = 1792 ft. The area = 1 of (96 x 42.332) sq. ft.

7. sum=9 ft. Area = 19 x 3 x 3 x 3 sq. ft.

8. The base = V 702 — 422 ft. = 56 ft. Area = £ (42 x 56) sq. ft.

9. V50? — 482 = 14, the number of feet in one-half the base. See Fig. 2, Arithmetic, Art. 1263.

10. The common base of the two triangles will be one diagonal, 2 in. The altitude of a triangle will be one-half the other diagonal = V22 - 1’in. = V3 in. = 1.732 in. The diagonal = 3.464 in. The area of each triangle = { (2 x 1.732) sq. in.= 1.732 sq. in. The area of the rhombus = 3.464 sq. in.

11. The scholars should be led to ascertain for themselves the approximate relation between the diameter of a circle and its circumference. Place a point marked on the circumference of a spool, or something similar, on a given point on the surface of a sheet of paper. Roll the spool until the point on the circumference again touches the surface of the paper. The distance between the two points on the paper will be equal to the circumference of the circle. Measure this distance very carefully, also the diameter of the circle, and ascertain the ratio.

12. See Art. 1274, 6-14, for the reason for the rule for determining the area of a circle.

13. The circumference = 1 (2x x 3.1416) = 3.1416 x; the diameter = x; the area = 3.1416 x x x = 3.1416 x.

14. 3.1416x = area = 314.16. .

? = 100; x = 10. 15. Diameter = 15.708 ft. = 3.1416=5 ft.

Area =[(1 of 15.708) x (1 of 5)] sq. ft. 1108. The balance, $851.72, is found by adding the credits, and subtracting from $2535.35 in one operation. See Art. 384. 1110. 1. Amount of $500, July 25 to April 1, 250 da.,

$520.83 Amount of $100, Sept. 18 to April 1, 195 da., $103.25 “ “ $200, Feb. 5, " " " 55 " 201.83 305.08 Due April 1, 1894,

$215.75 3. Amount of $ 600 for 354 da.,

$635.40 " $300 “ 152 "

$307.60 " " $ 200 “ 57 "

201.90 509.50 Due at settlement,

$125.90 5. In the debit column, place the interest for 329 da., $ 46.06. The total of this column is written on the same line as the total of the credit column, and is $ 886.06. The amount is also written as the total of credits. The total interest ($28.38) on $500 for 297 da., $24.75, and on $ 200 for 109 da., $3.63, is written among the credits; and the cash payment is ascertained by writing in its place the sum necessary ($157.68) to make the total, $ 886.06. See Art. 384.

6. From the amount of $ 725 for 234 da. + the amount of $ 603 for 174 da., take the amount of $600 for 183 da. + the amount of $300 for 37 da.

1113. 3. The field contains 160 sq. rd. x 71. Its length is (160 x 71 - 30) rd. = 40 rd. The diagonal= V40' + 30 rd. = 50 rd.

4. The rate per cent that will produce $36 interest in 1s yr. is 73. The interest on $212.50 at 71% for the given time= $52.02. Ans.

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