Download Section 2.5: Percent Problems - Madison Area Technical College

College Mathematics FINAL EXAM Sample Equation Sheet Page 1 of 17

To convert one fraction to another equivalent fraction, you can either:
 Multiply both the numerator and denominator by the same number, or
 Divide both the numerator and denominator by the same number.
Reducing a Fraction involves writing the fraction as an equivalent fraction with smaller numbers in the
numerator and denominator.
To convert a mixed number to an improper fraction:
 Multiply the denominator and the whole number part.
 Add that answer to the numerator to get the improper numerator (the denominator stays the same).
To convert an improper fraction to a mixed number:
 Divide the denominator into the numerator.
 The quotient is the whole number part; the remainder is the numerator.
To add or subtract proper fractions:
 Convert both fractions to equivalent fractions with the same denominators.
 Add the numerators; the denominator is that common denominator.
 Reduce.
Section 1.1.5: Adding and Subtracting Mixed Numbers
To add or subtract mixed numbers
 First deal with the whole numbers, then deal with the fractions. Examples:
5
11
5
11
1
9
1
9
2 5  2 5
7 5 7 5
7
14
7
14
2
16
2
16
1
1
1
1
5 11
1 9
1  2  1  2 
7 
2 
2
3
2
3
7 14
2 16
1 1
10 11
8 9
 3 
7 
2 
2 3
14 14
16 16
16 8 9
3 2
21
1  
 3 
7
16 16 16
6 6
14
24 9
7
5
1

 7 1
 3
16 16
6
14
15
1
1
5
 8
3
16
2
6
15
1
1
8
16
2
College Mathematics FINAL EXAM Sample Equation Sheet Page 2 of 17
To multiply fractions:
 Take numerator times numerator and denominator times denominator. Example:
5 3 5  3 15
 

8 4 8  4 32
To divide fractions:
 Take the reciprocal of the second fraction, then multiply. Example:
5 3 5 4 5  4 5 1 5
   


8 4 8 3 83 23 6
To multiply or divide mixed numbers:
 Convert them to improper fractions, then use the regular rules for multiplying and dividing fractions just
covered.
To convert a decimal to a fraction:
5128
10,000
2  2  2  641

22225555
641

25555
641

1250
0.5128 
375
1000
3  125

8  125
3

8
0.375 
To convert a fraction to a decimal:
 To convert from fraction to a decimal, do long division. Stop when you get zero for a remainder or the
pattern of numbers repeats.
To add or subtract decimals:
 To add: Line up the decimal points and add the digits from right to left, carrying as necessary.
 To subtract: Line up the decimal points and subtract the digits from right to left, borrowing as necessary.
To multiply decimals:
 Multiply as if there were no decimals present
 The answer will have as many decimal places as all the decimal places of both factors added up.
To divide decimals:
 Move the decimal point the same number of places in the divisor and dividend until the divisor is a
whole number.
 Then do standard long division, putting a decimal point in the quotient right above the decimal point in
the dividend.
College Mathematics FINAL EXAM Sample Equation Sheet Page 3 of 17
Identifying and Writing Significant Digits:
 The significant digits (also called significant figures and abbreviated sig figs) of a number are those
digits that carry meaning contributing to its precision.

To identify the significant digits in a measurement:
o All digits are significant except leading zeros after a decimal point and trailing zeros before it.
Both types of zeros merely serve as placeholders.
o 100 m has 1 significant digit; 100. m has 3 significant digits; 100.0 has 4 significant digits; 0.001
m has 1 significant digit; 0.001 00 m has 3 significant digits; 100.001 00 m has 8 significant
digits
Section 1.3.2: Significant Digits after Adding or Subtracting
 When you add or subtract measurements, then you have to round your answer to the precision of the
least accurate measurement. That’s because the uncertainty in the least precise measurement
overwhelms the precision of the other measurements.
28.25 cm
+15.67 cm
43.92cm
114.37 cm
3.080 cm
+27.3
cm
144.75cm  144.8cm
Section 1.3.3: Significant Digits after Multiplying or Dividing
 Rule: Find the measurement with the fewest significant figures. Round your answer to that many
significant figures.
 Example: 10. cm  22.85 cm = 228.5 cm2  230 cm2 (2 sig figs.)
Section 1.5: Exponents
Section 1.5.1 Definition of Positive Integer Exponents
Exponents were originally developed as a shortcut notation for repeated multiplication:
2  2  2  2  2  25
Rules of Exponents
x n  x m  x n m
 x  y n  x n y n
 
xm
n
 x mn
College Mathematics FINAL EXAM Sample Equation Sheet Page 4 of 17
The Meaning of Zero and Negative Exponents:
16  24
8  23
4  22
2  21
1  20
1
 21
2
1
 22
4
1
 23
8
1
 24
16
x 3  1xxx (1 times 3 factors of x)
x 2  1xx (1 times 2 factors of x)
x1  1x (1 times 1 factors of x)
x 0  1 (1 times no factors of x)
Rules of Exponents
x n  x m  x n m
n
x
xn

 
yn
 y
xn
 x n m
m
x
 
xm
n
 x mn
 x  y n  x n y n
x n 
1
xn
College Mathematics FINAL EXAM Sample Equation Sheet Page 5 of 17
Order of Operations:
1.
Perform all operations that appear in parentheses (or other grouping symbols) first. If grouping symbols
are nested, do the innermost first.
2.
Raise all bases to powers in the order encountered moving from left to right.
3.
Perform all multiplications/divisions in the order encountered moving from left to right.
4.
Perform all additions/subtractions in the order encountered moving from left to right.
Note that expressions like 2(3 + 7) can be evaluated in two ways: using either the order of operations, or the
“Distributive Property:” a(b + c) = ab + ac.
Simplifying Expressions:
 Use the distributive property
o -5(2x – 8) = -10x + 40
 Add like terms
o 4x + 5x = 9x
o 3a + 4b + 7a + 3b = 10a + 7b
Procedure for solving any linear equation
1) First simplify each side of the equation as much as possible
a) Use the distributive property to remove parentheses groups
b) Combine like terms on each side.
2) Use the addition/subtraction properties of equality to move like terms to the same side of the
equation. Your equation will have all variables on one side and all numbers on the other side at this
point. Combine those like terms.
3) Use the multiplication property of equality to remove any remaining coefficient on the variable term
by dividing both sides by the coefficient.
4) Check all solutions in the original equation to verify solutions. Correct for possible mistakes if your
answers do not satisfy the equation.
Solve 2(3h  4) – 9  h – 2  4 5h  1
2(3h  4) – 9  h – 2   4  5h  1
6h  8  9h  18  20h  4
3h  26  20h  4
3h  26  26  20h  4  26
3h  0  20h  22
3h  20h  20h  20h  22
23h  22
23h 22

23 23
h
22
23
College Mathematics FINAL EXAM Sample Equation Sheet Page 6 of 17
Section 2.3: Linear Inequalities in One Variable
If you multiply by a negative number, you have to reverse the inequality.
25
x  7
(1)(2)  (1)(5)
(1)( x)  (1)(7)
2  5
x  7
Steps for Solving a Linear Inequality:
1. Simplify each side separately.
2. Isolate the variable terms on one side using the addition property of inequalities.
3. Isolate the variable using the multiplication property of inequalities.
Six Steps for Doing Word Problems (Applied Problems):
1. Read the problem carefully, and draw a picture if you can. Label the picture with all given information.
2. Assign a variable, and write any other unknown quantities in terms of the variable.
3. Write an equation, starting with a verbal equation if it helps.
4. Solve the equation.
5. State the answer, and verify that it makes sense.
6. Check the answer in the words of the original problem.
Mixture Problems: Draw a picture with each container, a label written above each container, the amount held
under each container, and the percentage held inside each container.
Interest Problems: Interest = Principal  Interest Rate  I = pr.
Money Problems: number  value of one item = total value.
Speed Problems: distance = rate  time  d = rt. Draw a picture with all distances, rates (speeds), and times
labeled.
Section 2.5: Percent Problems
Definition of Percent: A percent (%) indicates the number of hundredths of a whole.
1
 0.01  1%
1 part of 100 parts 
100
To convert a decimal fraction to a percent: move the decimal two places to the right, then add the % symbol.
To convert common fractions and mixed numbers to a percent: use a calculator to express the number as a
decimal fraction, then convert that decimal to a percent.
To convert percents to decimal fractions: drop the % symbol, then move the decimal two places to the left.
To convert percents to common fractions: express the percent as a decimal fraction, then find the common
fraction equivalent of the decimal (and reduce).
College Mathematics FINAL EXAM Sample Equation Sheet Page 7 of 17
Basic Calculations of Percentages, Percents, and Rates
3 out of 4 parts is 75%
3
4
part
whole
 75%

150% of $20
is
$30
150%    $20 

$30
%  whole

part
%
20 is the number
such that 10 is 50% of that number...
10
50%
part
whole 
%
20 
Section 2.6: Percent Problems from Finance
No problems from this section…
To solve a direct variation problem:
1) Write down the direct variation equation: Quantity1 = constant  Quantity2
2) Put given numbers in for Quantity1 and Quantity2, then solve for the constant.
3) Use the constant in the direct variation equation to find the desired quantity.
To solve a joint variation problem:
1) Write down the direct variation equation: Quantity1 = constant  Quantity2  Quantity3  …
2) Put given numbers in for the quantities, then solve for the constant.
3) Use the constant in the direct variation equation to find the desired quantity.
Inverse variation, or inverse proportion is the case where one quantity is proportional to the reciprocal of
1
k
another quantity. y  k   y 
x
x
To solve an inverse variation problem:
1) Write down the inverse variation equation: Quantity1 = constant / Quantity2
2) Put given numbers in for the quantities, then solve for the constant.
3) Use the constant in the direct variation equation to find the desired quantity.
College Mathematics FINAL EXAM Sample Equation Sheet Page 8 of 17
Section 3.1: Graphing a Linear Equation Using a Table of Values
When we graph “A vs. B”, the vertical axis represents the amount of A, and the horizontal axis
represents the amount of B. B is called the independent variable, and A is called the dependent variable.
Technique #1 for graphing a line: USING A TABLE OF VALUES
1. Solve the linear equation for the dependent variable (y if y and x are the only two variables in the
equation.)
2. Pick some random values of the independent variable (x if y and x are the only two variables in the
equation).
3. Calculate the values for the dependent variable using the equation; make a table of the values.
4. Graph the points and connect them with a straight line.
Technique #2 for graphing a line: SLOPE INTERCEPT METHOD
 Solve the linear equation for the dependent variable (i.e., solve for y if x and y are the only two variables
in the equation)
 The constant term is the vertical intercept, or the y-intercept. Plot it.
 The factor of the x term is the slope. Use it to count up and over (or down and backward) to plot more
points on the line.
Technique #3 for graphing a line: INTERCEPT METHOD
 Set the independent variable to zero, then solve the linear equation for the dependent variable (i.e., set x
= 0 and solve for y if x and y are the only two variables in the equation).
 Plot the point (0, y). This is called the y-intercept.
 Set the dependent variable to zero, then solve the linear equation for the independent variable (i.e., set y
= 0 and solve for x if x and y are the only two variables in the equation).
 Plot the point (x, 0). This is called the x-intercept.
 Draw the line between the two intercepts.
To graph a linear inequality:
1. Graph the boundary line by solving for the dependent variable.
a. If the inequality uses  or , then draw a solid line to show that the line itself satisfies the
inequality.
b. If the inequality uses just < or >, then draw a dashed line to show that the line does not satisfies
the inequality.
2. Shade the appropriate side.
a. If the inequality is < or , shade the graph below the line.
b. If the inequality is > or , shade the graph above the line.
College Mathematics FINAL EXAM Sample Equation Sheet Page 9 of 17
Solving a System of Two Linear Equations by Graphing

Graph both the lines.

Read the coordinates of the intersection point off the graph.

Check to see if those coordinates are the solution.
Solving a System of Two Linear Equations Using Algebra

Solve one of the equations for one of the variables; pick the easiest variable to solve for.

Replace that variable in the other equation with the expression you just derived.

Solve your new equation; it should have only one variable in it.

Substitute that answer into the first equation and solve it to find the value of the original variable.

Check to see if those coordinates are the solution.
Section 4.1.1: Conversions within the English System
(This table will be given to you along with the final exam)
Linear Measure :
Area Measure :
1 ft = 12 in
1 acre = 43,560 sq feet
1 yd = 3 ft
1 acre = 160 sq rods
1 mi = 5,280 ft
1 sq mile = 640 acres
1 mi = 1,760 yd
1 rod = 16.5 ft
1 furlong = 220 yd
Weight :
Time :
1 lb = 16 oz
1 min = 60 s
1 ton = 2,000 lb
1 hr = 60 min
1 stone = 14 lb
1 hr = 3600 s
1 slug = 32.174 049 lb
1 day = 24 hours
1 day = 1,440 min
1 day = 86,400 sec
1 year = 365.25 day
1 year = 31,557,600 sec
Volume Measure :
1 pt = 16 oz
1 qt = 2 pt = 32 oz
1 gal = 4 qt = 128 oz
1 gal = 231 in3
1 ft3 = 7.480 519 gal
Speed :
1 mi / hr = 1.466 467 ft / sec
College Mathematics FINAL EXAM Sample Equation Sheet Page 10 of 17
Prefix
Symbol
Power of Ten
Number per Base Unit
nano
n
109  0.000000001
109  1, 000, 000, 000
micro

106  0.000001
106  1,000,000
milli
m
103  0.001
103  1000
centi
c
102  0.01
102  100
deci
d
101  0.1
101  10
deca
da
101  10
hecto
h
102  100
kilo
k
103  1000
mega
M
106  1,000,000
giga
G
109  1, 000, 000, 000
1
 0.1
10
1
102 
 0.01
100
1
103 
 0.001
1000
1
106 
1, 000, 000
1
109 
 0.000000001
1, 000, 000, 000
101 
Section 4.2: Measuring Area
 In general, when converting area units, you have to square the conversion fraction because the units of
measurement are a dimension squared.
Section 4.3: Measuring Volume
 In general, when converting area units, you have to cube the conversion fraction because the units of
measurement are a dimension cubed.
College Mathematics FINAL EXAM Sample Equation Sheet Page 11 of 17
Section 4.4: Conversion Between Metric and English Units
(This table will be given to you along with the final exam)
Linear Measure :
Area Measure :
1 inch = 2.54 centimeters
1 in2 = 6.451 6 cm2
1 meter = 3.280 840 feet
1 m2 = 1.195 990 yd2
1 meter = 1.093 613 yard
1 mi2 = 2.589 988 km2
1 mile = 1.609 344 kilometer
1 acre = 40.468 5642 are
1 are = 100 square meters
1 hectare = 100 are=2.471 054 acre
Weight :
1 kilogram = 2.20462262 pounds
1 oz = 28.349 523 g
Speed :
1 m / s = 2.236 936 mph
1 m / s = 3.280 840 ft / s
1 mph = 1.609344 kph
Volume Measure :
1 oz = 29.573 530 mL
1 L = 1.056 688 qt
1 gal = 3.785 412 L
1 ft3 = 7.480 519 48 gal
1 ft3 = 28.316 846 59 L
1 L = 61.023 744 in3
1 m3 = 1000 L
1mL = 1 cm3
Temperature :
9F
F 
 TempC  32F
5C
5C
°C 
  Temp°F  32F 
9 F
°K 
1°K
 Temp°C  273.15°K
1°C
Angle facts:
 A right angle is one-fourth of a full rotation and measures 90.
 A straight angle is one-half of a full rotation and measures 180.
 An angle between 0 and 90 is called an acute angle.
 An angle between 90 and 180 is called an obtuse angle.
 Complementary angles have measures that add up to 90.
 Supplementary angles have measures that add up to 180.
Converting from degrees, minutes, and seconds (DMS) to decimal degrees (DD)
 Example: convert 322417 to DD
 Divide the number of second by sixty to convert it to an equivalent number in minutes.
 1 
o 17 
  0.283 333
 60 
 Add that to the number of minutes, then divide by sixty again to convert to an equivalent number in
degrees.
 1 
o 24.283 333 
  0.404 722
 60 
 Add that number to the number of degrees, and round to an appropriate number of places:
o 322417  32.405
College Mathematics FINAL EXAM Sample Equation Sheet Page 12 of 17
Converting from decimal degrees (DD) to degrees, minutes, and seconds (DMS)
 Example: convert 185.659 to DMS
 Multiply the decimal portion by sixty to convert it to an equivalent number in minutes.
 60 
o 0.6592 
  39.552
 1 
 Multiply the resulting decimal portion by sixty again to convert to an equivalent number in seconds.
 60 
o 0.552 
  33.12
 1 
 Round to an appropriate number of places:
o 185.659  1853933
Vertex ( or vertical ) angles. Are equal:
Parallel lines lie in the same plane and do not intersect.
A transversal is a line that intersects two parallel lines; a transversal creates eight angles, with two different
measures per angle.
College Mathematics FINAL EXAM Sample Equation Sheet Page 13 of 17
c  a 2  b2 .
The Pythagorean Theorem:
To calculate a leg, say a, knowing the hypotenuse and the other leg, b, rearrange the formula to
a  c2  b2
leg  hypotenuse2  other leg2
or
Some geometric formulas for perimeters:
Rectangle
P  2l  2w
C  2 r
Circle
(perimeter is called
circumference)
Triangle
Pi is a numeric
constant, and is
approximately:
  3.14
P  a bc
.
College Mathematics FINAL EXAM Sample Equation Sheet Page 14 of 17
Some geometric formulas for areas:
Rectangle
Arect  lw
Circle
Acirc  r 2
Parallelogram
Aparallelogram  bh
Triangle
Triangle
Trapezoid
Atri 
1
bh
2
Atri  s  s  a  s  b  s  c  ;
s  12  a  b  c 
Atrap 
1
 B  b h
2
 (measured in radians) 
s S

r R
Arc length  radius  central angle in radians
s  r 

measured in radians 
C  2πr  r   full revolution angle in radians 
so a full revolution of 360  2 radians
or more simply  rads  180
To convert decimal degree measurements to radians we use the conversion factor 1 
 radians
180
.
College Mathematics FINAL EXAM Sample Equation Sheet Page 15 of 17
To convert radian measure to decimal degrees we use the conversion factor 1 
180
.
 radians
B  W
P  2  2W
V  B  h  W  h
L  P  h  2  h  2W  h
A  L  2 B  2  h  2W  h  2  W
B   r2 
 D2
4
C   D  2 r
V  B  h   r 2h 
 D2h
4
L  C  h   Dh  2 rh
A  L  2 B  2 rh  2 r 2   Dh 
V
h  B h  L W

3
3
V
h  B  r 2h

3
3
 D2
2
College Mathematics FINAL EXAM Sample Equation Sheet Page 16 of 17
V
4 r 3  D3

3
6
A  4r 2  D 2
Some surface areas of solids:
Sphere
Asphere  4 r 2
Box
Abox  2lw  2lh  2wh
Cylinder
Acyl  2 rh  2 r 2
Some geometric formulas for volumes:
Box
Vbox  lwh
Cylinder
Vcyl  r 2 h
College Mathematics FINAL EXAM Sample Equation Sheet Page 17 of 17
Sphere
4
Vsphere  r 3
3
Pyramid
1
Bh
3
B is the base area.
The base can be any
shape (not just a
triangle, square, or
rectangle)!
V pyr 
Cone
Vcone 
1
1
Bh  r 2 h
3
3