ALGEBRA - solved problems
Interest calculations
Compound interest
 38 If \$10,000 is invested for five years at 6% of the interest rate, find the accumulated or final value and
total interest earned at the end of the period under both, simple and compound interest.
In compound interest calculations, the interest earned in each period is added at the end of a period to the principal of the previous period, to become the principal for the next period.
The compounding periods can be yearly, semiannually, quarterly, or the interest can be compounded more frequently even continuously.
If P is the principal or initial value of investment,  A is the accumulated amount or final value of investment and the compound interest rate is i %
then,               A = P · r n,   where  r = 1 + i,    and where
 or , and Solution:   Given,  P = \$10,000,  p% = 6%  and  n = 5 years,  A and  I =?
a)  Under simple interest, the total interest earned in five years period is
I = i% · P · n     =>      I = 6/100 · 10000 · 5 = \$3000,
the accumulated value after five years period is
A = P + I = P (1 + i n)      =>     A = 10000 · (1 + 6/100 · 5) = 10000 · 1.3 = \$13000,
so that         I = A - P = 13000 -  10000 = \$3000.
b)  Under compound interest, the accumulated value after five years period is
A = P(1 + i%)n     =>    A = 10000 · (1 + 0.06)5 = 10000 · 1.338225 = \$13382.25
therefore, the total interest earned in five years period is
I = A - P = 13382.25 -  10000 = \$3382.25.
So, the interest compounding (or interest earned on interest) brings the extra \$382.25 in comparison with the simple interest.
 39 After how many years will deposit double at an interest rate of  6%.
 Solution:  40 At what annual interest has to be deposited \$5,000 for four years to grow to \$8,000.
 Solution:  Periodic compounding
 41 The principal amount of \$2,000 is invested for five years in a compound interest account paying 6%
compounded quarterly, find the final or accumulated amount in the account.
Note that for any given interest rate the investment grows more if the compounding period is shorter.
So, if P is an amount of money invested for n years at an interest rate i, compounded m times per year then, the total number of compounded periods is mn and the interest rate per period is i/m
 and the accumulated or future value is Solution: Continuous compounding
 42 Suppose \$5,000 is deposited into an account that pays interest compounded continuously at an
annual rate of 8%. How much will the account be worth in 20 years?
As for any given interest rate the investment grows more if the compounding period is shorter, we let the number of periods in a year approach infinity to compound the interest continuously, meaning that the balance grows by a small amount every instant.
Thus, to derive a formula for continuous compounding we need to evaluate the above formula when m tends to infinity (i.e., the year is divided into infinite number of periods) then by substituting and,  when m ® oo  then x ® oo, that is, the limit in the square brackets converges to the number e = 2.71828 . . . ,  thus obtained is
the continuous compounding formula     A = P · e i n,  where e is the base of natural logarithms.
 Solution: Exponential growth and decay, application of the natural exponential function
 43 Suppose that microorganisms in a culture dish grow exponentially. At the start of an experiment
there are 8,000 of bacteria, and two hours later the population has increased to 8,600.
How long will it take for the population to reach 20,000?
Solution: Given the initial population N0 = 8,000 and for t = 2 hours the population increased to N = 8,600.
We first find the growth rate k and then the time needed the population increases to 20,000. 44 After 800 years a sample of radioisotope radium-226 has decayed to 70.71% of its initial mass, find
Radioactive decay is a typical example to which the exponential decay model can be applied.
Solution: The half-life of a radioactive material is the amount of time required for half of a given sample to
decay.  First we plug the given information into the formula for exponential decay
N(t) = N0 ekt    to find the decay constant k.
Since N0 is the initial quantity, N(t) is the quantity after time t and k is the decay constant then,
by substituting   N(t = 800) = 0.7071·Ninto the formula
0.7071N0 = N0 e k · 800
and solving for k,                                         0.7071 e k · 800 | ln
ln 0.7071 = 800 · k
k = ln (0.7071) / 800
k = - 0.0004332217.
Thus, the formula for the amount of radium-226 present at a time t is
N(t) = N0 e - 0.0004332217 · t
As we want determine the half-life or the time for half of a substance to decay, we substitute N(t) = (1/2) N0
into the formula    Solved problems contents