finanical math
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Mawarid Induction Program
Mathematics & Algebra
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Topics to be Covered
• Unit 1 Operations with Whole Numbers
• Unit 2 Fractions• Unit 3 Decimals & Percentages
• Unit 4 Exponents or Powers
• Unit 5 Solving Algebraic Equations
• Unit 6 Solving Word Problems
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Unit 1
Operations with Whole Numbers
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The Order of Operations
• B - Brackets (innermost first)
• E - Exponents
• D - Division (from left to right)
• M - Multiplication (from left to right)
• A - Addition (from left to right)• S - Subtraction (from left to right)
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Examples
Ex 1: 3 - 4 + 2 = ?
-1 + 2 = 1
Ex 2: 2 + 6 x 4 2 = ?
2 + 6 x 2 = ?
2 + 12 = 14
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Examples (cont’d)
Example 3: 2 x (3 + 5) 4 + 2
2 x 8 4 + 2
2 x 2 + 2
4 + 2 = 6
Example 4: 2 + (3 - (2 + 5) x 2) = ?
2 + (3 - 7 x 2) =2 + (3 - 14) =
2 +(- 11) = -9
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Examples (cont’d)
Example 5: 42 - (3 - 23 x 4) = ?
42 - (3 - 8 x 4) =
42 - (3 - 32) =
42 - (-29) =
16 + 29 = 45
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Unit 2
Fractions
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Improper Fractions and
Mixed Numbers• An improper fraction is where the
numerator is larger than the denominator
(Ex: 9/6)• A mixed number is where there is a whole
number and a fraction put together
– (Ex: 2 3/4)
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Changing Improper Fractions into
Mixed Numbers• Divide the denominator into the numerator
to determine the whole number part of the
mixed number• If there is a remainder, it should be used as
the numerator which goes over the existing
denominator – Example: 9/4 becomes 2 1/4
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Changing Mixed Numbers to
Improper Fractions• Multiply the whole number part of the
mixed number by the denominator, then add
the numerator• Place that number over the denominator
– Example: 3 4/5 becomes 19/5
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Adding & Subtracting Fractions
• We must make a common denominator in
order to add & subtract fractions
• Once the common denominator isdetermined, the fractions must be changed
into equivalent fractions
• Then the numerators are added orsubtracted and placed over the common
denominator
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Adding & Subtracting Fractions
• Example: 2/3 + 3/4 = ?
• Solution:
– The lowest common denominator is 12
– The fractions are converted to 8/12 + 9/12
– The answer is 17/12 or 1 5/12
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Multiplication of Fractions
• Multiply the numerator of the first fraction by
the numerator of the second fraction
• Multiply the denominator of the first fraction by the denominator of the second fraction
– Example: 3/5 x 1/4
– Solution: Numerator 3 x 1 = 3
– Denominator 5 x 4 = 20
– therefore 3/20 is the solution
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Division with Fractions
• Turn the second fraction upside down, then
follow the rules for multiplication of fractions
– Example: 3/4 2/5
– Solution:
1. Turn the second fraction upside down (3/4 x 5/2)
2. Multiply numerators and denominators3. The answer is therefore 15/8 or 1 7/8
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Unit 3
Decimal Numbers & Percentages
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Decimal Numbers
• They are a different way of representing an
amount other than in fraction form
• They are a series of digits separated by a dot
• The digits to the left of the dot represent a
whole number
• The digits to the right of the dot represent
an amount that is less than one
– Example: 25.55
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Percentages (%)
• Is a method of expressing an amount in
terms of a part of one hundred
• It is like a fraction that has 100 as thedenominator (25/100 = 25%)
• 100% of an amount means the entire
amount
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Converting Fractions to Decimals
• Simply divide the numerator by the
denominator
– Example: 3/5
– Solution: 3 5 = .6
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Converting a Fraction
to a Percent• Convert the fraction to an equivalent
fraction that has a denominator of 100
• Place the % sign to the right of thenumerator amount
– Example: 3/5 = 60/100 = 60%
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Converting a Decimal
to a Percent• Multiply the decimal number by 100, then
place a % sign after it
• Or, move the decimal point two places tothe right, then place a % sign after it
– Example: 0.57
– Solution: 0.57 x 100 = 57%
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Converting a Percent
to a Decimal• Divide the percentage by 100 and take away
the % sign
• Or, move the decimal point two places tothe left, then remove the % sign
– Example: 28.0 %
– Solution: 28.0/100 = .28
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Unit 4
Exponents or Powers
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What is an Exponent?
• It is when a number is multiplied by itself
“n” times, if “n” is the exponent
• Given the expression 23:
– The 2 is called the “base”
– The 3 is called the “exponent”
– The expression 23 is called a “Power”
– The 2 is multiplied by itself 3 times 2 x 2 x 2
– The answer is 8
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Negative Exponents
• Exponents (n) can be negative numbers
• It means that the number raised to the
negative power (-n) is a denominator raisedto the positive power (n) where one (1) is
the numerator
– Example: 2-4 = 1/24 = 1/(2x2x2x2) = 1/16
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Powers of Ten (10)
• Given the expression 103 , we simply need
to write a 1 with 3 - 0’s beside it in order to
get the solution 1000101 = 10
102 = 100
104 = 10000100 = 1 (any number with a 0 exponent = 1)
10-6 = 1/1000000
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Operations with Exponents
Having the Same Base• Addition/Subtraction
– if we have the expression 23 + 24 , we must
calculate each expression before adding themtogether (solution 8 + 16 = 24)
– if we have the expression 23
- 22
, we mustcalculate each expression before subtracting
them (solution 8 - 4 = 4)
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Operations with Exponents
Having the Same Base (Cont’d) • Multiplication
– If we have the expression 32 x 34 , we can
simplify the expression by adding theexponents (n) together and using the sum as the
exponent for the base of 3 as follows:
32 x 34 = 32 +4 = 36 = 729
It is simply (3 x 3) x (3 x 3 x 3 x 3) = 36
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Operations with Exponents
Having the Same Base (Cont’d) • Division
– If we have the expression, 34 32 , we can
simplify the expression by subtracting thesecond exponent from the first one and using
the result as the exponent (n) for the base of 3
as follows:
34 32 = 34-2 = 32 = 9
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Unit 5
Solving Algebraic Equations
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Equations with one
Unknown Variable• An equation is an equality between two
mathematical terms
• These terms can be numbers, variables(expressed with letters), or a combination of
both
• For Example: 3n + 5 = 14 is an algebraicequation with “n” being the unknown
variable
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Solving Equations with one
Unknown Variable• First we need to collect alike terms on each
side of the equation
• Second, we need to perform operations toisolate numbers on one side of the equation
and variables on the other side
• In order for the equation to remainunaffected, whatever we do to one side of the
equation, we must also do to the other side
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Examples
#1. 2n + 3n + 4n = 3 + 5 + 10
Solution:
Step 1. Collect alike terms 9n = 18
Step 2. Divide both sides by 9 n = 2
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Examples
#2. 3n + 5 = 20
Solution:
Step 1. Subtract 5 from both sides 3n = 15
Step 2. Divide both sides by 3 n = 5
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Examples
#3. 2n + 4 = 3n - 2
Solution:
Step 1. Subtract 2n from both sides
4 = n - 2
Step 2. Add 2 to each side
6 = n or n = 6
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Examples
# 4. 3(2n + 5) = n + 20
Solution:
Step 1. Multiply the terms in the brackets by 36n + 15 = n + 20
Step 2. Subtract n from both sides
5n + 15 = 20Step 3. Subtract 15 from both sides 5n = 5
Step 4. Divide both sides by 5 n = 1
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Unit 6
Solving Word Problems
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Percentage Increases &
Decreases• When prices or interest rates go up or down
we must be able to compute the new price
or the rate of change etc.• The formula for solving problems like these
is as follows:
New Amount - Original Amount = Rate of ChangeOriginal Amount
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Example #1
If the original price of a clock was $80 and then
it went up by 20%, what was the new price of
the clock?Solution: Let X = New Amount
Original Amount = $80
Rate of Change = 20% or .20
X - 80 = .20 > X - 80 = 16 > X = 96
80
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Example #2
A stock price went up by 25% to $100.
What was the price before it went up?
Solution: Let X = Original Amount
New Amount = 100 Rate of Change = 25%
100 - X = .25 > 100 - X = .25X
X
100 = 1.25X > X = 80