Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
521
Solve the system using elimination.
Alice is currently 20 years old. To determine Mel’s age, substitute y = 20 into
either equation of the system and solve for x.
Mel is currently 45 years old, 25 years older than Alice.
Calculating Interest
Simple, compound, and continuously compounding
23.9 How much simple interest accrues on a loan of $1,200 at an annual rate of
4.5% over a three-year period?
The formula for simple interest is i = prt, where i is interest, p is the principal,
r is the annual interest rate expressed as a decimal, and t is the length of time
(in years) the principal accrues interest. Substitute p = 1,200, r = 0.045, and t = 3
into the formula to calculate i.
The total accrued interest is $162.
23.10 If you wish to borrow $3,000 from a friend who will charge you simple interest
at an annual rate of 7%, and you cannot exceed $3,300 in total debt. What
is the maximum length of time you have to repay the debt and the accrued
interest? Round the answer to the nearest whole day.
You intend to borrow $3,000 but your total debt cannot exceed $3,300.
Therefore, the maximum total interest you can afford to pay is
$3,300 – $3,000 = $300. Substitute i = 300, p = 3,000, and r = 0.07 into the
simple interest formula to calculate the number of years t it would take to
accrue $300 in interest.
Multiply
the second
equation by –1, add
the equations, and
solve for y.
Make sure
you answer the
question posed by
the word problem.
In this case, you’re
supposed to gure
out how much older
Mel is than Alice, so
subtract their ages:
x – y = 45 – 20 = 25.
The principal
is the dollar
amount that earns
interest. In this
case, the principal
is $1,200.
To change
4.5% into a
decimal, move the
decimal point two
places to the left:
0.045.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
522
To convert t into days, multiply by 365.
365(1.42857142857) ≈ 521.428571429 ≈ 521
The maximum length of time you can afford to borrow the money is 521 days.
23.11 Describe the difference between simple and compound interest.
Simple interest accrues only on the principal, whereas compound interest
transfers interest accrued into principal each time it is compounded. Consider
a savings account that pays only simple interest. No matter what length of time
the money remains in the account, interest is earned only on the principal
deposited initially. However, a compound interest account would earn interest
on all the money in the account.
23.12 If $500 is deposited into a savings account with a 3.75% annual interest rate
compounded monthly, what is the balance of the account ten years later?
Round the answer to the hundredths place.
The formula for compound interest is , where b is the balance,
p is the principal, r is the annual interest rate expressed as a decimal, n is the
number of times the interest is compounded in one year, and t is the length of
time (in years) interest is earned. Substitute p = 500, r = 0.0375, n = 12, and
t = 10 into the formula to calculate b.
The balance of the account is $727.07.
Because
r is an
annual
interest rate,
t is measured in
years. There are
365 days in a (non
leap) year, so
multiply t by
365.
Including
the interest
you’ve already
earned on your
initial investment.
Each time it
compounds, all the
interest you’ve earned
is added to your
initial investment,
and you start
earning interest
on that.
The balance
is the original
deposit (principal)
plus all the interest
it earned. In other
words, the balance
is the total amount
of money in your
account.
Interest
is compounded
monthly, a total of
n = 12 times per year.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
523
23.13 How long will it take an initial investment of $4,000 to grow to $5,000 at an
annual interest rate of 8% compounded quarterly? Express the answer in
years, rounded to the thousandths place.
Substitute b = 5,000, p = 4,000, r = 0.08, and n = 4 into the compound interest
formula.
Solve for t.
Take the natural logarithm of both sides of the equation.
According to a property of logarithms, the power of a logarithmic argument
can be removed from the argument and multiplied by the logarithm itself.
It will take approximately 2.817 years for the initial investment of $4,000 to grow
by $1,000.
The interest
is compounded
quarterly, n = 4
times per year.
This is the
property ln x
a
= a
ln
x
from Problem 19.18.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
524
23.14 If $900 is deposited into a savings account with an annual interest rate
of 5.5%, how much interest is earned over a two-year period if interest is
compounded continuously? Round the answer to the hundredths place.
The formula for continuously compounding interest is b = pe
rt
, where b is the
balance, p is the principal, e is Euler’s number (described in Problem 18.21), r is
the annual interest rate expressed as a decimal, and t is the length of time (in
years) interest accrues. Substitute p = 900, r = 0.055, and t = 2 into the formula to
calculate b.
The total balance of the account after two years is $1,004.65. To determine the
total amount of interest earned, subtract the balance from the principal.
i = $1,004.65 – $900 = $104.65
A total of $104.65 in interest is earned.
23.15 You wish to deposit $3,500 in a savings account with an annual interest rate of
6.25%. How much more interest will you earn over a 20-year period if interest
is compounded continuously rather than annually? Round each balance and
the final answer to the hundredths place.
Substitute p = 3,500, r = 0.0625, and t = 20 into the continuously compounding
interest formula and calculate b.
The continuously compounding interest account earns $12,216.20 – $3,500 =
$8,716.20 interest. Substitute p = 3,500, r = 0.0625, and t = 20 into the
compound interest formula to calculate the balance of the account if interest is
compounded annually (n = 1 time per year).
The more
times in a year
that interest is
compounded, the
more interest you
earn. You can’t
compound interest
more often than
“continuously.”
You need a
calculator to
evaluate e
0.11
.
The problem
doesn’t ask for the
total balance. It asks
how much money was
earned on the $900
investment, so subtract
900 from the balance
to calculate the
interest.
Recognize
the formula
b = pe
rt
? It’s
the exponential
growth formula rst
dened in Problem
19.36. Continuously
compounding interest
makes the
investment grow
exponentially.
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