Chapter 3 3.1: Simple Interest Formula
These problems involve calculating a simple interest on money. If you invest a principal amount of money, at an interest rate of r %, you will gain a simple interest on your investment. Hence, the simple interest, I, on principal, P, invested for t years at rate r, is given by I = P • r • t. Example 1: Find I given that P = $500, r = 15% or .15, and t = 2 years. I = P • r • t = (500)(.15)(2) = 150. Hence, the interest gained for the $500 at a rate of 15% for 2 years is $150.00.
Example 2: John borrowed $5,000.00 at a rate of 5% for 30 days to start his Internet business. What is the amount of interest due and the total amount to be paid after 30 days? So we have the following information: Borrowed money = $5,000 1.) Rate = 5% or .05 Time = 30 =
30 365
30 30 I = (5,000)(.05) = (250) = (250)(.08) = 20. So John will own $20 365 365
interest on the principal of $5,000 borrowed for 30 days. Total amount John will pay back is $5,000 + $20.00 = $5,020.00.
2.)
Geometric Formulas
A polygon is a many-sided figure. A quadrilateral is the term used to describe a figure that has four sides. The perimeter of a polygon is the distance around a twodimensional figure. To find the perimeter, sum up the lengths of all the sides of the figure.
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6 4 4 perimeter = 4 + 4 + 6 + 6 = 20
6 Example 3: Find the perimeter of the following rectangle, thus finding the general formula.
l
Here the length is l and width is w
w w so the perimeter = l + l + w + w
= 2l + 2 w
l
Hence, the general formula for the perimeter of a rectangle = 2l + 2w
The area of a two-dimensional figure is the number of square units (inches, feet, miles) that fit inside the figure. We calculate the areas of the following figures, using b = base, h = height, A = area, b1 = base one and b 2 = base two.
h b
A=
1 (b) (h) 2
h b
A = (b) (h)
2
h b
b1
A = (b) (h) parallelogram
h
A=
1 ( b1 + b2 ) h 2 trapezoid
b2 Example 4: Find the area of the rectangular region, excluding the part enclosed by the parallelogram.
10’
8'
5’
13’ The area we are asked to find will be =(Total area of rectangle) – (Area of parallelogram) Area of rectangle = (10)(13) = 130sq.ft. Area of parallelogram = (8)(5) = 40sq.ft. Hence the area we are asked to find is = 130 – 40 = 90sq.ft.
A circle is the set of points in the plane that are equidistant from a fixed point in the plane. The fixed point is the center and the distance is the radius. Any line segment that connects two points on the circle and passes through the center is called a diameter, which is twice the length of the radius.
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For a circle with diameter d and radius r, we have d = 2 • r ⇒ r =
d . The 2
circumference of a circle is the distance around it. Hence, the circumference of a circle
is, circumference = C = π • d where π ≈ 3.14 . Now, since d = 2 • r, we can also write the circumference as, C = 2π r. The area of a circle with radius of r is given by, A=
π • r 2 . The number “pi” is an irrational number, and π ≈ 3.141592653... .
Circumference of a circle with radius radius r is
C = 2π r
The area of a circle with radius r is circumference
A = π • r2
Example 5: A circular swimming pool has diameter 50 feet and is centered in a fenced-in square region measuring 80 feet by 80 feet. A concrete sidewalk 5 feet wide encircles the pool, and the rest of the region is grass: a.) b.) Find the area of the concrete sidewalk. Find the area of the grass.
sidewalk pool
grass
Solution:
a.) To find the area of the concrete sidewalk, we need to find the area of the circle and subtract the area of the pool from it. Hence, we know the radius of the large circle is 25 + 5 = 30, so the area is A = π ( 30 ) = 2827.35 . Hence the area of the sidewalk is
2
2 2827.35 – (area of the pool) = 2827.35 - π ( 25 ) = 2827.35 - 1963.5 = 863.85 square
feet.
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b.)The area of the grass is the total area of the square minus the area of the circle enclosing the pool and sidewalk. Area of the square is 80 times 80 which is 6400. Hence the total area of the grass is 6400 – 2827.4 = 3572.6 square feet.
The volume of a rectangular solid is the number of unit cubes needed to fill it.
The unit cube.
height
width length The unit cube has dimensions 1 x 1 x 1. Notice that it has three dimensions. The cube is
length x width x height, hence it is a 3-dimensional object. Therefore, the volume of a rectangular solid is also a three dimensional measurement.
Volume of a rectangular solid = V = length x width x height = l • w • h
Example 6: Find the volume of the rectangular solid with length 6”, width 6”, and height 8”.
Volume = l • w • h = 6 • 6 • 8 = 288in.3
The volume of a circular cylinder is the product of the base b and the height h. However, the base b is actually the area of the circle that makes up the base. Hence the volume of a circular cylinder is: V = b• h = π • r 2 • h height
base
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Example 7: Find the volume of the circular cylinder with radius 9ft. and height 10ft. V = π • r 2 • h = π • 92 • 10 = ( 3.14 )( 81)(10 ) = 254.34ft.3
The volume of a circular cone, with base radius of r, is
1 the base area times the height: 3
1 1 V = • b • h = •π • r2 • h 3 3
height
base
Example 8: Find the volume of the cone that has base radius of 4ft. and height of 8ft. V=
1 1 3.14 •128 401.92 401.92 100 40192 • π • 42 • 8 = • π •16 • 8 = = = • = = 133.973ft.3 3 3 3 3 3 100 300
A sphere is a three dimensional circle. The volume of a sphere with radius of r is .
V=
4 •π • r3 3
Sphere
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Example 9: Find the volume of the globe, given that the radius is 6 inches.
4 4 4 • 3.14 • 216 V= • π • 63 = • 3.14 • 216 = = 4 • 3.14 • 72 = 904.32in.3 3 3 3
Solving for a Variable in a Formula
Sometimes you will need to solve for a certain variable in a formula. Solve for the variable as you have done before. Isolate the variable on one side of the equation by performing the same operation on each side of the equals sign. Example 10: The slope of a line is given by the letter m and is related to other variables by the formula y = mx + b. Solve for m.
Solution: I am going to solve for m by isolating it on one side of the equal sign. I
assume that all of the other variables are nonzero. What I do to one side of the equation I must also do to the other. Hence I get the following:
y = mx + b y − b = mx + b − b y − b = mx y−b =m x y b − =m x x
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3.2: Changing Application Problems into Equations
Algebra is used to help solve problems. Hence, we must learn how to set up problems then solve the problems. Translating a problem is perhaps the hardest part. Below are some common phrases and how they can be translated into mathematical expressions.
Word or Phrase Operation Statement Algebraic Form
Added to More than Increased by The sum of Subtracted from Less than Decreased by The difference between Multiplied by The product of Twice a number,etc. Of, when used in a percent or fraction Divided by The quotient of One-half of a number, etc.
7 added to a number 5 more than a number A number increased by 3 The sum of a number and 4 6 subtracted from a number Subtraction 7 less than a number A number decreased by 5 The difference between a number and 5 A number multiplied by 6 The product of 4 and a number Multiplication Twice a number 20% of a number Addition A number divided by 8 Division The quotient of a number and 6 One-seventh of a number
x+7 x+5 x+3 x+4 x–6 x–7 x–5 x–9 6x 4x 2x .20x
x 8 x 6 x 7
Keep in mind that the above phrases maybe combined in different ways with each other. Example 11: Convert the following phrases to their algebraic form. a.) b.) c.) Twice a number, decreased by 8 is 12, is written as, 2x – 8 = 12. One more that four times a number, is written as, 1 + 4x. Three times the sum of a number and 2, is written as, 3(x + 2). 8
Write Expressions Involving Percent
Percents are often used in word problems. When we are given that something has increased/decreased by n%, we change the percent into decimal form by dividing n by 100. For example, 6% would be
6 = .06 . 100
Example 12: Sue Ann bought a car. The cost of the car and a 8% sales tax was $25,000. Express this in algebraic form.
Solution: If the cost of the car is x and there was a 7% sales tax on x, which give .07x.
We have to add both quantities together to get the sum of $25,000. Hence, we have
x + .07x = $25,000.
Notice how we correctly write the percentage of a cost, c, as the percentage in decimal form times the cost c.
Phrase Correct Incorrect
.075c 1 A 7 % sales tax on c 2 dollars The cost, c, increased by a c + .075c 1 7 % sales tax 2 The cost, c, reduced by 25% c - .25c
.075
c + .075
c - .25
Express the Relationship between Two Related Quantities
When we are dealing with numbers that are related to each other in some way, we express the simplest number as a variable, and the other number contains that variable with whatever conditions stated applied to it.
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Example 13: a.) Two numbers differ by 4. One number will be x and the other number will be
x – 4.
b.) The sum of two numbers is 15. Let this expression be written as x + y = 15. Then let x be one number. If we solve for y in terms of x, we have y = 15 – x. Hence, the second number is expressed as 15 – x. Notice, if one number is 7, then the other number is 15 – 7 = 8.
Writing Expressions Involving Multiplication
If we want to write an algebraic expression with multiplication, we need to be aware of all the ways multiplication can be worded. Example 14: a.) The cost of purchasing x notebooks at $3.00 each. Is written algebraically as 3x. b.)An average chicken egg contains about 275 milligrams (mg) of cholesterol and an ounce of chicken contains about 25mg of cholesterol. Write an expression that represents the amount of cholesterol in x chicken eggs and y ounces of chicken. Since each chicken egg contains 275mg of cholesterol, the total amount in x eggs is 275x. Now, since an ounce of chicken contains 25mg of cholesterol, the total amount in y ounces is 25y. Hence, the total amount of cholesterol in x many eggs and y ounces of chicken is 275x + 25y.
We sometimes use consecutive integers. Consecutive integers are integers that differ by 1. Hence, numbers like 4 and 5 are consecutive. Consecutive odd or even integers are
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numbers that differ by 2. Two consecutive even integers are represented by x and x + 2. The word “is” usually denotes equality. Hence, in applications, we put an equals sign for the word “is”. Example 15: For two consecutive integers, the sum of the smaller and twice the larger is 29.
Solution: Since we are dealing with consecutive integers, name one x and the other will
be x + 1. Since the problem says the sum, the operation is addition. However, it also states “twice the larger” so we have 2(x + 1). So the total expression is x + 2(x + 1) = 29.
Example 16: In 1998, there were 122 less than twice the number of roller coasters in North America than there were in 1994. The difference in the number of roller coasters in 1998 and 1994 was 112.
Solution: Let n represent the number of roller coasters in 1994. In 1998, there were 122
less than twice the number in 1994, or 2n, so this phrase is 122 – 2n. The last statement has as its operation “difference” or subtraction, in which the difference “is” 112. The total expression is (122 – 2n) – n = 112.
3.3: Solving Application Problems
In the last section we learned how to translate expressions into algebraic expressions. Now we are going to look at how to combine expressions to solve a multistep problem. The following guidelines can help guide you through solving problems: I.) Understand the problem, by identifying the quantity you are asked to find.
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II.)
Translate as before the problem into an algebraic problem. Choose a variable to represent one quantity. If other quantities are part of the problem, write them as an expression of this variable.
III.) IV.) V.)
Solve the equation by carrying out the computations. Check the answer in the original equation. State the answer to the question.
Example 17: The sum of two odd consecutive integers is 104. Find the numbers.
Solution: Let x represent one of the integers, then since they are odd consecutive
integers, the second integer is x + 2. Since their sum is 104 we have the equation
x + (x + 2) = 104. Solving this equation for x we get:
2 x + 2 = 104 2 x = 104 − 2 2 x = 102 102 x= = 51 2 Hence, one integer is 51 and the other is 53. Checking we have 51 + 53 = 104. Therefore, the two odd consecutive integers that sum up to 104 are 51 and 53.
Example 18: Irene Doo worked a 55-hour week last week. She is not sure of her hourly
1 rate, but knows that she is paid 1 times her regular hourly rate for all hours worked over 2
a 40-hour week. Her pay was $600. What is her hourly rate?
Solution: Let x represent her hourly rate. Irene worked 15 hours overtime last week, so
she had 15(1.5)x that went into overtime pay, plus the 40 hours of regular pay. So all together she had 40x + 22.5x = 600. Solving for x we have:
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40 x + 22.5 x = 600 62.5 x = 600 x= 600 = 9.60 62.5
Checking we see that 40(9.60) + 22.5(9.60) = 384 + 216 = 600. Therefore, Irene’s hourly rate is $9.60
Selecting a Mortgage
When you buy a house, many banks use a table to figure out what the mortgage on the house will be. The mortgage payment on a house includes the principal and interest paid on the house in monthly payments, over so many years. There are different mortgage rates for different terms of the loan. The monthly payment does not usually include the taxes and insurance that is also paid on a house. Sometimes banks charge “points” when they give a loan. One point is 1% of the mortgage. Example 19: Leeanne Fisher is considering two banks for a 30-year $75,000 mortgage. NationsBank is charging 9.5% interest with 1 point and Bank America is charging 9.25% interest with 2 points. How long would it take for the total cost of the two mortgages to be the same?
Solution: With the 9.5% interest, Leeanne’s monthly payment will be
(75)(8.41)=$630.75, in addition the one point will add (.01)(75000) = $750. With the 9.25% interest the payment will be (75)(8.23)=$617.25, in addition she will have to pay the two points charged on the 9.25% rate, which is (.02)(75,000) = $1500. The quantities
8.41 and 8.23 are found on a mortgage rate table like the one in the text on page 195.
We want to know when these two quantities will be the same. So let x be the number of
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months when total payments from both mortgages are equal. We have 630.75x for the 9.5% loan and 617.25x for the 9.25% loan. Hence we can set up the equation: 630.75 x + 750 = 617.25 x + 1500 + 150 630.75 x − 617.25 x = 1650 − 750 13.5 x = 900 900 x= ≈ 66.66 13.5 So it will take approximately 66.7 months or 5.6 years.
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