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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 9 years ?
SOLUTION.—The sum of the installmenis paid on 1 share for 95 years or
$200-$114=$86 114 months, is $114; and the difference
115x114 between $200, the final value, and $114,
24 the amount paid, equals $86, which is the gain, or interest on the investment.
=15.7+% $1, the first payment, is on interest for
24 114 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) } of (1+114) x 114, months It of (1+114)x114, years; hence the interest on $1 for 1 year, or the
115x114 rate, is $86; =$.157+, or 15.7%.
24 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 87 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 share
723. To find the rate of interest paid by al borrower.
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 ?
SOLUTION.-The loan was 10 x ($200 --- $60) = $1400; $10 + $7
10x ($200—$60)=$1400. int.=$17, the monthly payment,
100x ($10+$7)=$1700. which in 100 mo. equals $1700.
101 x 100 Now the interest on the monthly $17x
24 payments (Case III.) is equivalent
101 x 100
$1700+$429.25==$2129.25. to the interest on $17 for
$729.25:-8}=$87.51. years at 6%, or $429.25; hence
$87.51 - $1400=.0625+. 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 85 years; and the interest for 1 year is $729.25=89, or $87.51. hence the rate is $87.51-$1400=.0625+ or 61%.
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 83 years ? Ans. Inst., 7.54%; Gross, 9.47%.
NOTE.-A more complete discussion of this subject will be found in Brooks's Higher Arithmetic.
724. Mensuration treats of the measurement of ger 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 direction 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 wben 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; 88, ABC.
MENSURATION OF SURFACES. 734. A Surface is that which has length and breadth without thickness. Surfaces are plane or curved.
73.), A Plane Surface is a surface such that if any iwo
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 lides, either straight or 'urved
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 jining the vertices of two angles not consecutive.
739. The Perimeter of a polygon is the am of its sides.
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; thels 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 wbicn 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 ine Rum of the sides, subtract from each side separately, multiply the half runn 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 rode und altitude 57 rods
Ans. 13 A. 571 P.
3. Required the area of a triangular field whose base is 965 rods and altitude 576 rods.
Ans. 1737 A. 4. What is the area of a field whose sides are respectively 20, 30, and 40 chains ?
Ans. 29 A. 8 P.-.
THE QUADRILATERAL. 741. A Quadrilateral is & polygon having four siden and therefore four angles There are three classes, the par. allelogram, trapezoid, and trapezium.
745. A Parallelogram is a quadrilateral whose opposite sides are parallel. The altitude is the perpendicular distance between its opposite sides.
746. A parallelogram which is right-angled is called a Rectangle.
When the four sides are equal it is called a Square.
747. An oblique-angled parallelogram is called a Rhomboid. An equilateral rhomboid is called a Rhombus.
Rule.- To find the area of a parallelogram, multiply the base by the altitude.
1. What is the area of a parallelogram 20 feet long and 18 feet wide ?
Ans. 40 sq. yd. 2. A has a rectangular lot 192 chains long and 65 chains wide; what is its area ?
Ans. 1248 acres. 3. What is the difference in the area of two lots, one being 245 rd. long, 42 rd. wide, and the other 85 chains long and 18 chains wide ?
Ans. 88 A. 110 P. 748. A Trapezoid is a quadrilateral which has two of its sides parallel. Its altitude is the perpendicular distance between its parallel sides.
Rule.— To find the area of a trapezoid, multiply one. half the sum of the parallel sides by the altitude.
1. Required the area of a trapezoid, one side being 120 in., the otber 96 in., and the altitude 48 in. Ans. 36 sq. feet.
9. What is the area of a trapezoid, the sides being 365 and 124 in., and the altitude 86 in. ?. Ans. 146 sq. ft. 3 Aq. in.