SOLUTION.-The monthly payment on 1 share equals $1 dues and $1 interest, or $2, and on 15 shares the payment is $30. The first installment is on interest 100 months, the second installment 99 months, and so on; hence the interest

of a payment of $1 for the different periods equals the interest of $1 for a number of months represented by an arithmetical series whose first term is 1, last term 100, and number of terms 100, or (Art. 665) † of (100+1)x100. The interest of $1 for 1 month is 9, and for the aggre101 × 100 gate months, of 100×101×2¢= =$25.25; and on $30 it is 4 $25.25x30 $757.50. The sum of the payments equals $30 × 100, or $3000; and the cost of the loan equals $3000+$757.50, or $3757.50.

Rule.-I. Multiply the number of months by the number of months increased by 1, and divide by 4, to find the interest at 6% on the aggregate monthly payments of $1.

II. Multiply the interest on the aggregate payments of $1, by the monthly payment, to find the interest on the payments. Add this interest to the sum of the payments; the result will be the cost of the loan.

NOTE. We have assumed that the monthly payments are entitled to simple interest from the time of their payment until the close of the series, in determining the actual cost of a loan. It would be more correct to reckon annual interest, but this makes the calculation rather difficult. To be strictly accurate, we should reckon compound interest.

WRITTEN EXERCISES.

2. Mr. Thomas bought a loan on 12 shares, new series, of El Paso Building Association, at $85 premium, Net Plan; if the series runs out in 9 years, what is the actual cost of his loan? Ans. $2597.427.

3. Mr. Burton bought a loan of a Philadelphia building association on 10 shares of a new series at 65 a month premiam; what is the actual cost of the loan if the series runs out in 9 years? Ans. $3806.46.

4. A rents a house at $12 a month, and at the end of 10 years buys it for $1200; B buys a house for $1200, borrow. ing money of a building association on 8 shares of a new series at $50 premium, Gross Plan, which runs out in 10 yr.. and paying an annual tax of $24 at the beginning of each year; which house cost the most? Ans. A's, $255.60.

CASE IV.

722. To find the rate of interest received by a nonborrower.

1. What rate of interest do I receive on 5 shares, dues $1 per share, if the series runs out in 91 years?

OPERATION.

$200-$114-$86

115x114

=equated time.

24

115X114

$86

=15.7+%

24

SOLUTION.-The sum of the installments paid on 1 share for 9 years or 114 months, is $114; and the difference between $200, the final value, and $114, the amount paid, equals $86, which is the gain, or interest on the investment. $1, the first payment, is on interest for 114 months, the second payment is on interest for 113 months, etc.; hence the interest on the installments for the different periods is equivalent to the interest on $1 for a number of months represented by the sum of an arithmetical series whose first term is 1 and last term 114, or (Art. 665) 1⁄2 of (1+114) x 114, months= of (1+114) × 114, years; hence the interest on $1 for 1 year, or the rate, is $86÷

115x114

24

$.157+, or 15.7%.

Rule.-I. Subtract the sum of the installments paid on one share from the final value of the share, and the difference will be the interest on the investment.

II. Multiply the number of payments by the number of payments increased by 1, and divide by 24, to find the equated time, or the number of years in which $1 will produce the same interest as the installments.

III. Divide the interest on the investment by the equated time; the quotient will be the equated rate per cent.

WRITTEN EXERCISES.

2. By the annual report of the Investment Building and Loan Association made at the end of the eighth year, the present value of the first series is $186.90; what is the equated rate of legal interest at that time? Ans. 23.43%.

3. It was estimated that the first series, including dues, would be worth $180.75 when 8 years old, but at the end of 8 years the association canceled the series by paying the estimated value, less the unpaid dues on each share; what rate % was realized by the stockholders? Ans. 20.29%.

NOTE.-In Prob. 2, $186.90 is regarded as the final value of the snare

CASE V.

723. To find the rate of interest paid by a bor

rower.

1. A buys a loan on 10 shares, Net Plan, at the beginning of a series, at $60 premium per share, and pays $10 dues and $7 interest on net sum received, for 8 years; what is the average or equated rate of interest?

years at 6%, or $429.25; hence the actual cost of the loan is $1700 +$429.25, or $2129.25; therefore

$2129.25-$1400, or $729.25, is the interest on the loan for 8 years; and the interest for 1 year is $729.25÷8, or $87.51 hence the rate is $87.51-$1400.0625+ or 6%.

Rule.-I. Find the sum of the installments, and the interest on the installments for the equated time at 6%; their sum will be the entire cost of the loan.

II. Subtract the amount of the loan from its entire cost; the remainder will be the interest on the loan for the period, from which the rate is readily found by the method of simple interest.

WRITTEN EXERCISES.

2. Mr. Jay borrows $4600, at 56 cents premium a month, on the Installment Plan; what sum do his monthly payments aggregate, and what equated rate % will he pay if the series runs out in 9 years? Ans $58.88; 9.25%.

3. I buy a loan of 10 shares, new series, in an association on the Installment Plan, at 60 cents a month premium, and in another, a loan of 10 shares on the Gross Plan at $60 premium; what rate % do I pay for each loan if each series runs out in 8 years? Ans. Inst., 7.54%; Gross, 9.47%.

NOTE-A more complete discussion of this subject will be found in Brooks's Higher Arithmetic.

SECTION XII.

MENSURATION.

724. Mensuration treats of the measurement of geo metrical magnitudes.

725. Geometrical Magnitudes consist of the Line, Surface, Volume, and Angle.

726. A Line is that which has length without breadth or thickness. Lines are either straight or curved.

727. A Straight Line is one that has the same direc tion at every point.

728. A Curved Line is one that changes its direction at every point. The word line used alone means a straight line.

729. Parallel Lines are those which have the same direction. Parallel lines, it is thus seen, will never meet.

730. One line is said to be perpendicular to another when the adjacent angles formed by the two lines are equal. 731. An Angle is the opening between two lines which diverge from a common point.

732. A Right Angle is an angle formed by one line perpendicular to another; as, ABC.

733. An Acute Angle is an angle less than a right angle; as, DEF. An Obtuse Angle is one larger than right angle; as, DEG.

B

D

E

MENSURATION OF SURFACES.

734. A Surface is that which has length and breadth without thickness. Surfaces are plane or curved.

735. A Plane Surface is a surface such that if any two

of its points be joined by a straight line, every part of that line will lie in the surface.

736. A Plane Figure is a plane surface bounded by lives, either straight or curved

737. A Polygon is a figure bounded by straight lines; as, ABCDE.

A Polygon

of three sides is called a Triangle, of four sides, a Quadrilateral, etc.

738. A Diagonal of a polygon is a line ja ning the vertices of two angles not consecutive.

739. The Perimeter of a polygon is the sides.

am of its

740. The Area of a plane figure is the number of square units in its surface.

NOTE.-The principles of mensuration are derived from geometry; their application to practical purposes is usually given in arithmetic.

THE TRIANGLE.

741. A Triangle is a polygon of three

sides and three angles; as, ABC.

742. The Base is the side upon whicn

it seems to stand; as, AB. The Altitude is a line perpendicular to the base, drawn from the angle opposite; as, CD.

743. An Equilateral Triangle is a triangle which has its three sides equal; when two sides are equal it is called isosceles; when its sides are unequal it is called scalene.

Rule. To find the area of a triangle, multiply the base by one-half of the altitude.

NOTE. If the three sides are given and not the altitude, take haif the sum of the sides, subtract from it each side separately, multiply the half surn and these remainders together, and take the square root of the product.

1. What is the area of a triangle whose base is 25 rods and altitude 18 rods? Ans. 225 sq. rd., or 1 A. 65 P.

2. Required the area of a triangle whose base is 75 rods and altitude 57 rods Ans. 13 A. 57 P.