CHAPTER 2
Interest and Money-Time Relationship
Solved Supplementary Problems
Problem 2.1
What is the annual rate of interest if P265 is earned in four months on an
investment of P15, 000?
Solution:
Let ‘n’ be the number of interest periods. Thus, on the basis of 1 year (12 mo.), the
interest period will be,
Hence, the rate of interest given by the formula, i= , is computed as
i= = 0.053 or 5.3%
( )( )
Thus, the annual rate of interest is 5.3%
2-2. A loan of P2, 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 is the future amount is due at the end of the loan period?
Solution:
( )
( )( )
Answer:
2-3. If you borrow money from your friend with simple interest of 12%, find the present worth of P20,
000, which is due at the end of nine months.
Given:
Future worth: F = P20, 000
Number of interest period: n=
Simple interest i = 12%
Solution:
( )
, ( )( )
Answer:
2-4. Determine the exact simple interest on P5, 000 for the period from Jan.15 to Nov.28, 1992, if the
rate of interest is 22%.
Given:
P= P5, 000
i= 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 days
Exact simple interest = Pin
= 5000×318∕366×0.22
= P955.74
Answer: P955.74
2-5. A man wishes his son to receive P200, 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?
Given:
F= P200, 000;
For compound interest:
i= 10%; n=5
For compound interest
i= 12%∕4= 3%; n= 5×4=20
, Solution:
P2= F (1+ i)-n
= 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
Answer: P68, 757.82
200,000 110 735.15
0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5
P1 P2 P1
2-6. By the condition of a will the sum of P25, 000 is left to be held in trust by her guardian until it
amounts to P45, 000. When will the girl receive the money if the fund is invested at 8%
compounded quarterly?
Given:
P = P25, 000 i = 8%∕4= 2%
F = P45, 000
Solution:
F = P (1+i) n
45000= 25000 (1+0.02)4n
45000∕25000= (1.02)4n
1.8= (1.02)4n
In (1.8) = 4nln (1.02)
29.682 = 4n
n = 7.42 years
Answer: 7.42 years
2-7. At a certain interest rate compounded semiannually P5, 000 will amount to P20, 000 after 10
years. What is the amount at the end of 15 years?
, Given:
P = P5, 000
n1 = 10
F1 = P20, 000
n2 = 15
F2=?
Solution:
At n1 = 10, F1= P20, 000
( )
F1 = P( )
( )
P20, 000 = P5, 000( )
= 14.35%
At n2 = 15
( )
F2 = P( )
( )
F2= P5, 000( )
F2= P39, 973.74
Answer: P39, 973.74
2-8. Jones Corporation borrowed P9, 000 from Brown Corporation on Jan. 1, 1978 and P12, 000 on
Jan. 1, 1980. Jones Corporation made a partial payment of P7, 000 on Jan. 1, 1981. It was agreed
that the balance of the loan would be amortizes by two payments one of Jan. 1, 1982 and the
other on Jan. 1, 1983, the second being 50%larger than the first. If the interest rate is 12%. What
is the amount of each payment?
Given: =12%
₱12,000
₱9,000
1978 1979 1980 1981 1982 1983
₱7,000
X
50%X+X
Solution:
Interest and Money-Time Relationship
Solved Supplementary Problems
Problem 2.1
What is the annual rate of interest if P265 is earned in four months on an
investment of P15, 000?
Solution:
Let ‘n’ be the number of interest periods. Thus, on the basis of 1 year (12 mo.), the
interest period will be,
Hence, the rate of interest given by the formula, i= , is computed as
i= = 0.053 or 5.3%
( )( )
Thus, the annual rate of interest is 5.3%
2-2. A loan of P2, 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 is the future amount is due at the end of the loan period?
Solution:
( )
( )( )
Answer:
2-3. If you borrow money from your friend with simple interest of 12%, find the present worth of P20,
000, which is due at the end of nine months.
Given:
Future worth: F = P20, 000
Number of interest period: n=
Simple interest i = 12%
Solution:
( )
, ( )( )
Answer:
2-4. Determine the exact simple interest on P5, 000 for the period from Jan.15 to Nov.28, 1992, if the
rate of interest is 22%.
Given:
P= P5, 000
i= 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 days
Exact simple interest = Pin
= 5000×318∕366×0.22
= P955.74
Answer: P955.74
2-5. A man wishes his son to receive P200, 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?
Given:
F= P200, 000;
For compound interest:
i= 10%; n=5
For compound interest
i= 12%∕4= 3%; n= 5×4=20
, Solution:
P2= F (1+ i)-n
= 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
Answer: P68, 757.82
200,000 110 735.15
0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5
P1 P2 P1
2-6. By the condition of a will the sum of P25, 000 is left to be held in trust by her guardian until it
amounts to P45, 000. When will the girl receive the money if the fund is invested at 8%
compounded quarterly?
Given:
P = P25, 000 i = 8%∕4= 2%
F = P45, 000
Solution:
F = P (1+i) n
45000= 25000 (1+0.02)4n
45000∕25000= (1.02)4n
1.8= (1.02)4n
In (1.8) = 4nln (1.02)
29.682 = 4n
n = 7.42 years
Answer: 7.42 years
2-7. At a certain interest rate compounded semiannually P5, 000 will amount to P20, 000 after 10
years. What is the amount at the end of 15 years?
, Given:
P = P5, 000
n1 = 10
F1 = P20, 000
n2 = 15
F2=?
Solution:
At n1 = 10, F1= P20, 000
( )
F1 = P( )
( )
P20, 000 = P5, 000( )
= 14.35%
At n2 = 15
( )
F2 = P( )
( )
F2= P5, 000( )
F2= P39, 973.74
Answer: P39, 973.74
2-8. Jones Corporation borrowed P9, 000 from Brown Corporation on Jan. 1, 1978 and P12, 000 on
Jan. 1, 1980. Jones Corporation made a partial payment of P7, 000 on Jan. 1, 1981. It was agreed
that the balance of the loan would be amortizes by two payments one of Jan. 1, 1982 and the
other on Jan. 1, 1983, the second being 50%larger than the first. If the interest rate is 12%. What
is the amount of each payment?
Given: =12%
₱12,000
₱9,000
1978 1979 1980 1981 1982 1983
₱7,000
X
50%X+X
Solution:
