69. My chaise having been injured by a very bad boy, I am obliged to sell it for $68.75, which is 40 per cent. less than its original value, what was the cost? Ans. $114.581. 70. Charles Webster's horse is valued at $ 120, but he will not sell him for less than $134.40; what per cent. does he intend to make? Ans. 12 per cent. 71. Three merchants, L. Emerson, E. Bailey, and S. Curtiss engaged in a cotton speculation. Emerson advanced $3600, Bailey $4200 and Curtiss $2200. They invested their whole capital in cotton, for which they received $ 15000 in bills on a bank in New Orleans. These bills were sold to a Boston broker at 15 per cent. below par, what is each man's net gain?

Ans. Emerson $990.00. Bailey $1155.00. $ 605.00.

Curtiss

72. Bought a box made of a plank 3 inches thick. Its length is 4ft. 9in., its breadth 3ft. 7in., and its height 2ft. 1lin. How many square feet did it require to make the box, and how many cubic feet does it contain?

Ans. 70 square feet, 294 cubic feet. 73. How many bricks will it require to construct the walls of a house, 64 feet long and 32 feet wide, and 28 feet high; the walls are to be 1ft. 4in. thick, and there are also three doors 7ft. 4in. high, and 3ft. 8in. wide; also 14 windows 3 feet wide and 6 feet high, and 16 windows 2ft. 8in. wide and 5ft. 8in. high. Each brick is to be 8 inches long, 4 inches wide, and 2 inches thick.

Ans. 167,480 bricks. 74. John Brown gave to his three sons, Benjamin, Samuel, and William, $1000 to be divided in the proportion of ,, and respectively; but William, having received a fortune by his wife, resigns his share to his brothers. It is required to divide the whole sum between Benjamin and Samuel.

Ans. Benjamin $571.429. Samuel $428.574. 75. Peter Webster rented a house for one year to Thomas Bailey for $100; at the end of four months, Bailey rented one half of the house to John Bricket, and at the end of eight months, it was agreed by Webster and Bailey to rent one third of the house to John Dana What share of the rent must each pay ?

Ans. Webster $61, Bailey $273, and Dana $11}.

76. Bought 365 yards of broadcloth, for which I paid £576. 17s. 9d. ; for how much must the cloth be sold per yard to gain 25 per cent. Ans. £1. 19s. 645d. 77. John Brown's house is 40 feet square; the roof comes to a point over the centre of the house, and this point is 12 feet above the garret floor. Required the length of a rafter, which extends from one of the corners of the house to the highest part of the roof.

Ans. 30.72+ feet. 78. Minot Thayer sold broadcloth at $4.40 per yard, and by so doing he lost 12 per cent.; whereas he ought to have gained 10 per cent. For what should the cloth have been sold per yard? Ans. $5.50. 79. John Crowell sold cloth at $5.50 per yard, and gained 10 per cent.; whereas, the cloth having been damaged, he should have sold it 12 per cent. less than the cost. What in justice should he have charged per yard ? Ans. $4.40. 80. Jacob How has cloth, which he purchased for 12 per cent. less than its value; but he sells it at 10 per cent. more than it is worth, and by so doing he gains $1.10 on each yard. What per cent. did he make on his purAns. 25 per cent.

chase? 81. A gentleman has five daughters, Emily, Jane, Betsey, Abigail, and Nancy, whose fortunes are as follows. The first two and the last two have $ 19,000; the first four $19,200; the last four $20,000; the first and the last three $20,500; the first three and the last $21,300. What was the fortune of each?

Ans. Emily has $5,000; Jane $4,500; Betsey $6,000; Abigail $3,700; and Nancy $5,800.

APPENDIX.

CANCELLING METHOD.

By the Cancelling Method the scholar is enabled to solve many questions with less than half the labor, that would be required by the usual process. It cannot, however, be applied to all the rules of arithmetic, nor to all the questions under any one rule; but it is generally used in the operations of those questions which require Multiplication and Division. The system is not new. It has been before the public in some form or other for centuries. John Birks, who published the second edition of his most excellent system of “Arithmetical Collections" in London, 1764, has made many improvements in the system. Since that period, but little advance has been made in it. Whether the author has made his system more plain and intelligible than has been done by others, the candid public must judge. He has spared no pains to exhibit its applicability and utility to those departments of arithmetical science where it can be advantageously employed. He believes the system can be of but little use to the pupil, until he can perform the questions by the common method. Hence the propriety of deferring attention to this method, until the common rules of arithmetic are thoroughly understood.

GENERAL RULE.

1. Equal divisors and dividends cancel each other.

2. When the product of two divisors is equal to the product of two dividends, they cancel each other.

*

I. Cancelling applied to Compound Fractions.

RULE 1.—If there be numbers in the numerators and denominators, that be alike, an equal number of the same value may be cancelled.

23

1. Reduce of of of of to a simple fraction.

STATEMENT.

CANCELLED.

2×3×4×7×8

=

= Ans. 3x4x5x8x9 3×4×5×8×9 45

In this question, we find a 3, 4, and 8 among the numerators, and also the same numbers among the denominators. These we cancel before we commence the operation.

4. Reduce of of of of 43 to a simple fraction.

5×11×12×17×19 5

11×12×17×19× 4

- -= 1 Ans.
4

5. Required the value of of of 19 of 1 of 40.

7×9×10×13×40

9X10X13X24

280
24

=

113 Ans.

6. Reduce of of 23 to its equivalent value.

7. What is the value of of 1 of 7 of 3 of $ 18?

8. What is the value of of of of $72?

RULE 2. When there are any two numbers, one in the numerators, and the other in the denominators, which may be divided by a number without a remainder, the quotients arising from such division may be used in the operation of the question, instead of the original numbers. The quotients also may be cancelled, as other numbers.

1. Reduce ofofofto its lowest terms.

OPERATION.

2 7 1 4×14×21× 5

56 7×27×25×11 495

1 9 5

=

In performing this question, we find that 14 among the numerators, and 7 among the Ans. denominators, may be divided by 7, and that their quotients will be 2 and 1. We write

the 2 above the 14, and 1 below the 7. We also find a 21 among the numerators, and a 27 among the denominators, which may be divided by 3, and that their quotients will be 7 and 9. We write the 7 above the 21, and 9 below the 27. We again find a 5 among the numerators, and a 25 among the denomi nators, which may be divided by 5, and that their quotients will be 1 and 5. We write the 1 over the 5, and the 5 below the 25. We then multiply the 4, 2, 7, and 1 together for a numerator = 56, and the 1, 9, 5, and 11 for a denominator = 495. The answer will therefore be 56

495.

2. Reduce of off of to a simple fraction.