
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 

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 N_{0}
= 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 radium226 has decayed to 70.71%
of its initial mass, find


the halflife of radium226. 
Radioactive decay is a
typical example to which the exponential decay model can be
applied. 
Solution:
The halflife 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)
= N_{0 }e^{kt }to
find the decay constant k. 
Since
N_{0}
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·N_{0 }into
the formula 
0.7071N_{0
}= N_{0 }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 radium226 present at a time t
is 
N(t)
= N_{0 }e ^{
0.0004332217 · t} 
As
we want determine the halflife or the time for half of a substance to decay,
we substitute N(t)
= (1/2) N_{0} 
into
the formula 











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