CHAPTER 2
Interest and Money-Time Relationship
Solved Supplementary Problems
Problem 2.1
Solution:
What is the annual rate of interest if 265 is
... [Show More] earned in four months on an
investment of 15,000?
Let ‘n’ be the number of interest periods. Thus, on the basis of 1 year (12
mos.), the
interest period will be,
4
=
=
3
Hence, the rate of interest given by the formula,
i=
(
, is computed as
= 0.053 or 5.3%
)
,
i=
Thus, the annual rate of interest is 5.3%
Problem 2.2
A loan of 2, 000 is made for a period of 13 months, from January 1 to January 31
the
following year, at a simple interest of 20%. What future amount is due at the
end of
the loan period?
Solution:
For the period of 13 months, the number of interest periods ‘n’ on the basis of
1 year
(12 mos.) is calculated as
=
3
Using the formula for future worth,
given, the future amount is computed as
=
,
.
= 2, 000[1 +
= (1 +
1
(0.2)]
12
) with interest and principal
is the amount due at the end of the loan period.
Problem 2.3
Solution:
If you borrow money from your friend with simple interest of 12%, find the
present
worth of 20, 000, which is due at the end of nine months.
The present worth of the borrowed money at the end of nine months is computed
using the formula,
= (1 +
)−
The number of interest periods on the basis of 1 year (12mos.) is,
=
9
=
3
4
Then, with the simple interest, number of periods, and the future amount given,
substituting these values to the present worth formula, the principal amount is
calculated as,
= 20, 000[1 +
Hence,
=
,
.
(0.12)]−
is the principal amount/borrowed money
Problem 2.4
Determine the exact simple interest on 5,000 for the period from Jan.15 to
Nov.28,
1992, if the rate of interest is 22%.
Solution:
January 15 = 16 (excluding Jan.15)
February
= 29
March
= 31
April
= 30
May
= 31
June
= 30
July
= 31
August
= 31
September = 30
October
= 31
November 28 = 28
(including Nov.28)
318
In exact simple interest, 1 interest period is equal to 366 days for 1 leap
year.
Thus,
Using the formula,
=
=
3 8
3
days
, the exact simple interest is computed as
I = ( 5,000)(
I = P955.74
3 8
3
)(0.22)
Problem 2.5
A man wishes his son to receive 200, 000 ten years from now. What amount should
he invest if it will earn interest of 10% compounded annually during the first 5
years
and 12% compounded quarterly during the next 5 years?
Solution:
=
(1 + )−
= 200000 (1+0.03)-20
P2= P110,735.15
P1= P2 ( 1+i )-n
= 110,735.15 (1+0.10)-5
P1= P68,757.82
Problem 2.6
By the condition of a will, the sum of 25, 000 is left to be held in trust by
her
5, 000. When will the girl receive the money if the
guardian until it amounts to
fund is invested at 8% compounded quarterly?
Solution:
For compound interest, the rate of interest per interest period is given by the
formula
=
,
=
8%
If the nominal rate of interest is 8% compounded quarterly,
then,
4
=
. 8
4
= 0.02
Hence, the formula to be used is
= (1 + )
Substituting the given values to the formula where
5000 =
25000 (1 + 0.02)4
=
(compounded quarterly),
5000 ∕ 25000 = (1.02)4
1. = (1.02)4
Using algebra, multiply ‘ln’ on both sides
(1. ) =
(1.02)
2 .6 2 =
Thus, the number of years the girl will receive the money is
=
.
Problem 2.7
At a certain interest rate compounded semiannually 5,000 will amount to
after 10 years. What is the amount at the end of 15 years?
20,000
Solution:
First, compute for the interest rate that is compounded semiannually (
the formula,
= (1 +
)
With the given values of = 5000,
1 =
thus,
1+
20,000 =
i = 14.35%
i
(n )
5,000 1 +
(
= 20000, and
= 2) using
= 10 (after 10 years),
)
at the end of 15 years ( = 15), the future worth can be computed as [Show Less]