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ALGEBRA
- solved problems
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Interest
calculations |
Compound interest |
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38. |
If $10,000 is invested for
five years at 6% of the interest rate, find the accumulated or
final value and
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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 |
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Solution:
Given,
P
= $10,000, p%
= 6%
and
n =
5 years, A and
I
=?
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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. |
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39. |
After how many years will
deposit double at an interest rate of 6%.
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Solution: |
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40. |
At what annual interest
has to be deposited $5,000 for four years to grow to $8,000.
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Solution: |
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Periodic compounding |
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41. |
The principal amount of
$2,000 is invested for five years in a compound interest account
paying 6%
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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
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Solution: |
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Continuous compounding |
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42. |
Suppose $5,000 is deposited into an account that pays interest
compounded continuously at an
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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) |
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then
by substituting |
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and,
when m ®
oo
then x ®
oo, |
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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: |
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Exponential
growth and decay, application of the natural exponential
function |
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43. |
Suppose that
microorganisms in a culture dish grow exponentially. At the
start of an experiment
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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. |
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44. |
After 800
years a sample of radioisotope radium-226 has decayed to 70.71%
of its initial mass, find
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the half-life of radium-226. |
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·N0 into
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 |
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