land, and 6 pieces of Holland to 14 yards of satin, and 12 yards of satin to 8 pieces of lace, and 9 pieces of lace to 8 £. 2 s., how many pairs of hose may be bought for 2 £. 16 s.? 24 pairs. Ans. 74. A man was hired 50 days, on condition that for every day he worked, he should receive 6 s., and for every day he was idle, he should forfeit 2 s. At the expiration of the time, he received $27.50. How many days did he work, and how many was he idle? Ans. He labored 40 days, and was idle 10 days. Note.-The York. currency of the preceding sum is that of New 75. A hare starts 40 yards in advance of a greyhound, and is not perceived by him till she has been up 40 seconds. She scuds away at the rate of 10 miles per hour, and the hound pursues after her at the rate of 18 milés per hour. In what time will the hound overtake the hare, and how far will he have run? Ans. The required time is 605 sec.; the distance run, 530 yards. 76. If 8 men can build a wall 15 rods long in 10 days, how many men will be required to build 45 rods of wall in 5 days? Ans. 48 men. 77. A man when he married was three times as old as his wife; after they had been married 15 years, his age was only double that of his wife's. How old were they when they married? Ans. The man was 45 years, his wife 15 years old. 78. There is in a pasture a certain number of sheep, cows, and oxen; there are twice as many sheep as cows, and three times as many cows as oxen, and the whole number is 80.. How many are there of each kind? Ans. 8 oxen, 24 cows, and 48 sheep. 79. Sold coffee at 15 cents per pound, and thereby lost 10 per cent. on the first cost. Afterwards sold a quantity of the same for $525, by which I gained 40 per cent. quantity sold, and what the price per pound? 20 cwt. 10 lb.; price, 23 cents. What was the Ans. Quantity, 80. Two sons, one 11 and the other 16 years of age, received a bequest of $10000, to be so divided between them that the shares being put on interest at 5 per cent. should amount to equal sums, when they became respectively 21 years age. What were the shares of each? Ans. The elder received $54541, and the younger, $4545 of APPENDIX. Prob. 1. To find the greatest common measure of two or more numbers. Note 1st.-The greatest common measure of two or more Lumbers, is the greatest number that will divide them separately without remainders. RULE.-If two numbers only are given, divide the greater of them by the less, and if nothing remain, that divisor is the common measure; but if there be a remainder, divide the preceding divisor by it; and so continue to divide each preceding divisor by the last remainder, till the division is effected without remainder; the last divisor will be the common measure required. When more than two numbers are given, find the common measure of any two of them first, and then of that common measure, and either of the remaining numbers. This process carried through all the numbers will give their greatest common measure. Ex. 1. What is the greatest common measure of 72 and 108? OPERATION. 72) 108(1 36) 72(2 72 00 36 is, therefore, the common measure required. Proof, 108 -36=3, and no remainder. 72÷36=2, and no remainder. 2. What is the greatest common measure of 27 and 99? Ans. 9. 3. What is the greatest common measure of 25, 45, and 90? Ans. 5. 4. What is the greatest common measure of 16, 32, 48, and 96? Ans. 16. Prob. 2. To determine how many different positions any given number of objects may assume with regard to each other. RULE.-Represent the number of objects by the figures 1, 2, 3, 4, 5, &c. making the numbers of figures equal to the number of objects. The product of these figures will determine the number of changes. Ex. 1. How many changes may be made by the first three letters of the alphabet ? Operation: 1×2×3=6, Ans. These changes are as follows: 1st, a, b, c; 2d, a, c, b; 3d, b, a, c; 4th, b, c, a; 5th, c, b, a; 6th, c, a, b. 2. How many changes may be made by the first 6 letters of the alphabet? Ans. 720. 3. At a certain boarding house there are 12 boarders. How many different positions may they occupy at the table? Ans. 479001600. 4. Five men engaged board at a tavern for as many days as the landlord might be able to seat them in different positions. How long did they remain? Ans. 120 days. Prob. 3. To find the area of a square. RULE.-Multiply the length of one of the sides by itself; or, its linear measure. (See fig. 2d, Square Root.) square Ex. 1. There is a room just 8 feet square, what is the area? Operation, 8 x8-64 square feet, Ans. 2. What is the area of a floor 19 feet square? Ans. 361 square feet. 3. What is the area in square inches of a board 3 feet square? Ans. 1296 square inches. Prob. 4. To find the area of a parallelogram. RULE.-Multiply the length of the parallelogram by its breadth. (See fig. 6th, Square Root.) Ex. 4. How many square feet are there in a floor 16 feet long and 12 broad? 16×12 192 square feet, Ans. Note 1st.-If the dimensions of a field in the form of a square or parallelogram are given in rods, the area is reduced to acres by dividing the square rods by 160. 2. How many acres are there in a piece of land 80 rods in length, and 40 in breadth? 80 × 40=3200, and 3200÷160= 20 acres, Ans. 3. What is the number of acres in a piece of land 63 rods long and 49 rods wide? Ans. 19160 47 Note 2d. If the parallelogram be not right-angled, the length must be multiplied into the perpendicular distance between the sides. (See fig. 1st.) The dotted line in the center is the perpendicular distance required; therefore, 45 × 15 = 675, the area of the figure.. FIG. 1. 45. ft. Prob. 5. To find the area of triangles. The area of any triangle is just equal to one half the area of a square or parallelogram of the same height. This is obvious with regard to right-angled triangles, and true of all others. The diagonal of a square or parallelogram divides it into two equal right-angled triangles, and since the area of the whole figure is found by multiplying FIG. 2. the length into the breadth, the area of its half, that is, of either of the triangles into which it is divided by the diagonal, is found by multiplying the same length or base into one half the breadth. We have then the following rule : RULE.-If the triangle be right-angled, multiply its base into half its perpendicular height. But if it be not right-angled, drop a perpendicular from one of the angles to the opposite side or base; then multiply the base or side upon ular falls, by one half the perpendicular. which the perpendic Ex. 1. What is the area of a triangular piece of land whose base is 40 rods, and whose perpendicular is 30 rods? 30÷2 =15, and 40 × 15=600 square rods, Ans. Note. The same result is obtained by multiplying the perpendicular by one half the base. Thus, 40÷2=20, and 30× 20=600. 2. There is a room 26 feet long and 18 feet high. If a line be drawn from one of the upper corners to its opposite lower corner, the side of the room will be divided into two equal triangles. What is the area of each triangle, and also of the side of the room? Ans. Each triangle contains 234 square feet, and the side, 468. 3. How many acres are there in a triangular piece of land, whose base is 42 rods and whose perpendicular height is 36 rods? Ans. 428 acres. Prob. 6. Given the diameter of a circle, to find its circumference. The diameter of a circle is to its circumference, as 7 to 22; or, as 1 to 3.141592. Therefore, RULE.-Multiply the diameter of a circle by 3.141592, and the product will be the circumference; or, multiply the diameter by 22, and divide the product by 7. Ex. 1. What is the circumference of a wheel whose diameter is 8 feet. 8×22÷7=25.14+ feet, Ans. Or, 8×3.141592 =25.132736, nearly the same as before. 2. What is the circumference of a circle whose diameter is 6 feet? Ans. 18.85 feet. 3. What is the circumference of a circle whose diameter is 42 feet? Ans. 132 feet. Prob. 7. Given the circumference of a circle, to find the diameter. pro RULE.-Multiply the circumference by 7 and divide the duct by 22; or, (what will produce nearly the same result,) divide the circumference by 3.141592. |