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MATHEMATICS (Section 1 - General Mathematics) Pgs 1 - 15
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Key Concepts:
Terms in this set (21)
Simple Interest
> What a situation is described as if the amount of interest charged does not change from time period to time period
An Example of Simple Interest
You just received a cell phone bill that costs $50
BUT
If you do not pay on its due time, there is a $5 fee for every month you postpone your payment!
Let's say you wait 6 months after it's due. The cost will now be $80
Notice that the amount of interest continually stays at $5 until you make your payment!
Simple Interest Formula
Interest = P
r
t
P = money loaned
r = interest rate
t = time periods
How do you find the total amount of money owed at the end of the loan?
Use this formula!
P+P⋅r⋅t
or
P⋅(1+r⋅t)
Using the Simple Interest Formula with the Same Cell Phone Bill Scenario
You're being charged 10% interest (since 10% of $50 is $5) per
month.
If it takes six months for you to pay back the company, you would owe them a total of
50 + 50 · (.1) · 6 = 50 · (1.6) = $80 dollars
P+P⋅r⋅t
Simple Interest Practice Problem:
You borrow $250 at 5% simple interest per week. If you ended up owing $350,
how long did it take for you to pay back the loan?
Since $250 was borrowed and $350 was paid back, the interest charged was $100. 5% of $250 is $12.50, which is the interest charged each week.
This means it took 8 weeks to pay back
the loan
350 = 250 + 250 ⋅ (.05) ⋅ t
OR
350 = 250 + 2.5 ⋅ t
Compound Interest
If the interest earned during each time period is then added to the principal amount when determining the amount of interest earned during the next
time period
Short Definition:
Interest is earned on interest
Example of Compound Interest
How much more money would you have to pay back at the end of the six months if you borrow $50 at 10% interest per month?
Remember! You didn't pay anything back until the end of the 6 months
first month -> $50 + (.1) ⋅ 50 = $55
second month -> $55 + (.1) ⋅ 55 = $60.50
third month -> $60.50 + (.1) ⋅ 60.50 = $66.55
fourth month -> $66.55 + (.1) ⋅ 66.55 = $73.205
fifth month -> $73.205 + (.1) ⋅ 73.205 = $80.5255
sixth month -> $80.5255 + (.1) ⋅ 80.5255 = $88.57805 or $88.58
Earning interest on the interest turned into an additional $8.58 on the loan
Annual Percentage Rate (APR)
The interest rate earned over the entire year, but the majority of banks
compound interest either monthly or daily
Example of APR
APR compounded monthly will earn
.05/12 = 0.416667% at each compounding
To be charged with 10% interest per month, the company would charge 120% APR compounded monthly
Practice Problem (Involving APR)
If $200 is invested at 2.5% APR compounded quarterly (four times per year),
How much is the investment worth at the end of seven years?
First Quarter Investment:
$200 + $200
(.025/4) = $200
(1 + (.025/4)) = $201.25
~~~~~~~~~~~~~~~~~~~~~~~~~~
Since interest compounded four times a year for 7 years, there will be 28 compoundings. At the end of the 7 years, investment is:
$200 * (1 + (.025/4)) ^ 28 = $238.119
OR
$238.12
Compound Interest Formula
P * (1 + r/n) ^ nt
P = Amount of Money Invested
r = Interest Rate
n = compounded times per year
t = total of years
At the end of 5 years, how much will an initial investment of $4000 be worth
if it earns 3.5% APR compounded monthly?
Since the money is being invested for 5 years and is compounded monthly, there are a total of 60 compoundings.
3.5% APR results in 0.2916666% per month, so at the end of the five
years, the investment will be worth:
4000 * (1 + .002916666)^60 = $4763.77
How much money did I invest initially if at the end of 8 years I have $5500
in an account that earned 2.8% APR compounded quarterly?
Quarterly compounding for 8 years gives a total of 32 compoundings at an interest rate of 0.7%
5500 = P * (1 + .007)^32
5500 = P ⋅ 1.250095
P = 4399.66522.
So, my initial investment was
$4399.67
Multiplication Rule (1st Form)
When selecting two objects, the total number of possible ways to select both of them at the same time is the product of the number of ways to select each object
Multiplication Rule (2nd Form)
When selecting two objects from different groups or two objects from the same group without repetition, the total number of possibilities is the product of the number of ways to select each object
Examples of Multiplication Rule (1st Form)
If I own 16 different shirts and 9 different pairs of pants, then this allows me to wear 144 different outfits
16 * 9 = 144
If there are options of 7 different types of pizza crusts and 11 different toppings. then we can have up to 77 different pizza orders
7 * 11 = 77
Each employee at a local business is given a unique personal identification code consisting
of two letters of the alphabet.
How many employees can be hired before the company runs out of codes?
We have 26 choices for the first letter and 26 choices for the second letter,
(26 letters in the alphabet)
The answer is 26^2 = 676 different employee ID codes
Example of Multiplication Rule (2nd Form)
If the pizza restaurant extends its menu and now offers four SIZES of pizza along with the seven crust options and eleven possible toppings
How many possible pizzas can be created?
Since each of the 77 possibilities can be made in any of four sizes, there are now
77 * 4 = 308 possible pizzas
Multiplication Rule (Final Form)
When selecting k objects from different groups or when the order of selection matters, the total
number of possibilities is
n1
n2
n3 .... nk-1 * nk
nk is the number of ways to select the kth object
Final Form Multiplication Rule Example
I have 16 different shirts, 9 different pairs of pants, 2 pairs of shoes, 3 belts, and
12 ties. Assuming I ignore such factors as "color coordination" and "matching,"
How many different outfits can I create?
16 ⋅ 9 ⋅ 2 ⋅ 3 ⋅ 12 to get 10,368 possible outfits
Look at it in another way:
With just shirts and pants -> 144 shirt-pants pairings (16 * 9)
Each of the 144 shirt-pants pairings can be paired with the shoes -> 288 outfits (144 *2)
Each of the 288 outfits can have one of the three belts added -> 846 outfits
(288 * 3)
Finally, a tie can make the outfit complete -> 10,368 outfits
(846 *12)
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