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MODULE 5 : MATHEMATICS OF FINANCE
Our target learning outcomes are a) Use mathematical concepts and tools in other
areas such as finance and business; b) Differentiate compound interest from simple interest;
c) Apply the interest and annuity formulas to cases of loans, credits, stocks bonds, property
purchases, and investment problems
UNIT 1: SIMPLE INTEREST AND COMPOUND INTEREST
Important Terms
Principal. It refers to the original sum of money borrowed in a loan or put into an
investment.
Interest is the charge for the privilege of borrowing money. From the investor’s
viewpoint, interest is the income from an invested amount at a given rate for a given time.
From the debtor’s viewpoint, interest is the money paid for the use of borrowed
money.
Interest Rate. The interest rate is the amount a lender charges for the use of money
expressed as a percentage of the principal. The interest rate is typically noted on an annual
basis known as the annual interest rate.
Time. This is the period from the beginning when the money was borrowed (or
invested) to the period when the money should be returned with the additional amount
(interest). This is also called the term of loan or term of investment.
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A. SIMPLE INTEREST
PowerPoint 14: Simple interest
It refers to the interest paid on the original principal. It is also characterized by a fixed
amount earned over time. Usually, simple interest is associated with loans and investments
which are short term in nature.
The formula in computing for the simple interest is given by
where: I – the Interest
P– the Principal
r – the rate of interest
t – the time period
We have to note that the time t should be expressed in years. Unless otherwise stated,
it will be assumed that the interest rate is an annual interest.
Example 61 : Find the simple interest earned in an account where ₱4,500 is on deposit for
4 years at 3 1/4% annual interest.
Solution:
Given: 𝑃 = ₱4, 500 , 𝑡 = 4 , 𝑟 = 3
1
4
Unknown: 𝐼
Solution:
Answer: 𝐼 = ₱
Example 62 : Find the simple interest for a loan of ₱12,400 due at the end of 8 1/4 years at
4 1/2% annual interest.
Solution:
Given: 𝑃 = ₱12, 400 , 𝑡 = 8
1
4
1
2
Unknown: 𝐼
Solution: 𝐼 = 𝑃𝑟𝑡 = 12400
Answer: 𝐼 = ₱4, 603. 5
Example 63 : Find the principal necessary to earn ₱814 in simple interest if the money is to
be left on deposit for 5 years and 3 months and earns 5 1/2% annual interest.
Solution:
Given: 𝐼 = ₱814, 𝑡 = 5
3
12
1
2
Unknown: 𝑃
Solution: Using 𝐼 = 𝑃𝑟𝑡, we obtain 𝑃 =
𝐼
𝑟𝑡
814
- 055 ( 5. 25 )
Answer: 𝑃 = ₱2, 819. 05
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Methods of computing for the simple interest when time is given between two dates
- Ordinary Interest for actual number of days
𝑂
This is referred to as the Banker’s Rule.
- Ordinary Interest for approximate number of days
𝑂
- Exact Interest for actual number of days
𝑒
- Exact interest for approximate number of days
𝑒
Note: Use the Banker’s rule unless otherwise specified. When we count the number of days
in between two dates we do not include the first day but we include the last day. Take note
that February may either have 29 or 28 days depending on if it is a leap year or not.
Example 67 : Calculate the simple interest due on a loan of ₱2000, at 6.5% simple interest,
which was availed on July 12, 2019, and to be repaid on December 12, 2019. Use the four
methods of computing for simple interest.
Solution:
Given: 𝑃 = ₱2, 000 , 𝑟 = 0. 065
𝑡: July 12, 2019 → December 12, 2019
Unknown: 𝐼 using the four different methods
Solution: First we count the number of days between July 12, 2019 and December
12, 2019 using actual time and approximate time.
Actual Time Approximate Time
July 31 - 12=19 July 30 - 12=
August 31 August 30
September 30 September 30
October 31 October 30
November 30 November 30
December 12 December 12
Total 153 Total 150
a) Ordinary Interest for actual number of days (Banker’s Rule)
𝑂
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b) Ordinary Interest for approximate number of days
𝑂
c) Exact Interest for actual number of days
𝑒
d) Exact Interest for approximate number of days
𝑒
Note: Among the four methods of computing simple interest, the Banker’s rule will usually
give the highest interest.
Final Amount. The sum of the principal and the interest which is accumulated at a
certain time. Final amount could be the Future Value or the Maturity Value
Future Value (of an investment). A term used to refer to the total amount on
deposit after the interest earned has been added to the principal.
Maturity Value (of the loan). A term used to refer to the total amount to be
repaid to the lender where the amount is the interest due on the loan plus the
principal.
The formula in computing for the Final Amount is given as:
where: F - the Final Amount.
P - the Principal.
I – the Interest.
Example 68 : Calculate the maturity value of simple interest, eight-month loan of ₱8,000 if
the interest rate is 9.75%.
Solution:
Given: 𝑡 =
8
12
2
3
Unknown: 𝐹
Solution: 𝐹 = 𝑃( 1 + 𝑟𝑡) = 8 , 000 [ 1 + 0. 0975 (
2
3
)] = ₱8, 520
Answer: 𝐹 = ₱8, 520
Example 69 : What principal will accumulate to ₱135,000 in 2 years at 15% simple interest?
Solution:
Given: 𝐹 = ₱135, 000 , 𝑡 = 2 , 𝑟 = 0. 15
Unknown: 𝑃
Solution: Using 𝐹 = 𝑃( 1 + 𝑟𝑡) we obtain 𝑃 =
𝐹
1 +𝑟𝑡
𝐹
1 +𝑟𝑡
135 , 000
1 +( 2 )( 0. 15 )
Answer: 𝑃 = ₱103, 846. 15
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*Conversion or Compounding periods
Conversion/Compounding Period Value of m
Annually 1
Semi-annually 2
Quarterly 4
Monthly 12
Simple Interest and Compound Interest Compared
To better understand the difference between simple and compound interest, let us
use the following information.
Given: P=₱1,000,
Simple Interest
Year Principal (amount
earning an interest)
Interest for the period
covered
Total interest for the
period covered
Final Amount at the end of 3 years where F = P (1 +rt)
F = 1000 [1 +(0.1)(3)] = ₱1,
*The same ₱1,000 earns an interest per period.
Compound Interest
Year Principal (amount earning
an interest)
Interest for the period
covered
Total interest for the
period covered
₱ 1000 ₱ 100 (a) ₱ 100
1000 100 (b)
100 (a) 10 (c) ₱ 110
₱ 1000 ₱ 100 (d)
100 (a) 10 (e)
100 (b) 10 (f)
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100 (c) 1 (g) ₱ 121
Final Amount at the end of 3 years where 𝐹 = 𝑃( 1 + 𝑖)
𝑛
3
F 1000 1 0. 1
1 , 331
*The same ₱1,000 earns an interest per period but the interests also earn interest.
Comparing simple and compound interest there is a ₱31 discrepancy. Said amount
is the accumulated interest of the interests.
Example 71 : Find the compound amount and interest if ₱1,000 is invested at 2%
compounded quarterly for 1 year and 6 months.
Solution:
Given: 𝑃 = ₱1, 000 , 𝑟 = 0. 02 , 𝑚 = 4 , 𝑡 = 1. 5
Unknown: 𝐹, 𝐼
Solution: 𝐹 = 𝑃 ( 1 +
𝑟
𝑚
𝑚𝑡
- 02
4
4 ( 1. 5 )
Answer: 𝐹 = ₱1, 030. 38 and 𝐼 = ₱30. 38
Cash Value. In some business transactions involving money, especially during the
onset of the purchase of an expensive good or service, a down payment is required and in
most cases, the purchaser makes financing arrangements to cover the remaining amount
owed to the seller. ( Investopedia). In the case that we want to compute for the cash value
then we use
CASH VALUE = Down payment(DP) + Present value of all future payments
If no down payment is required, then
CASH VALUE = Present value of all future payments
Example 72 : An equipment is purchased on installment by your company. Your company
paid an amount of ₱205,000 and owes a balance of ₱500,000 to be paid at the end of
seven years. Find the cash value of the equipment if money is worth 5% compounded
quarterly.
Solution:
Given: 𝐷𝑃 = ₱205, 000 , 𝐹 = ₱500, 000 , 𝑟 = 0. 05 , 𝑚 = 4 , 𝑡 = 7
Unknown: 𝐶𝑎𝑠ℎ 𝑉𝑎𝑙𝑢𝑒
To solve for the cash value, we have to determine the present value 𝑃 of the future
payment of ₱500, 000.
−𝑚𝑡
− 4 ( 7 )
Then , 𝐶𝑎𝑠ℎ 𝑣𝑎𝑙𝑢𝑒 = ₱205, 000 + ₱353, 109. 26 = ₱558, 109. 26
Answer: 𝐶𝑎𝑠ℎ 𝑣𝑎𝑙𝑢𝑒 = ₱558, 109. 26
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- Present Value of Ordinary Annuity
𝑜𝑟𝑑
= 𝑅 [
1 −( 1 +𝑖)
−𝑛
𝑖
] = 𝑅 [
1 −( 1 +
𝑟
𝑚
)
−𝑚𝑡
𝑟
𝑚
]
where: 𝑆 – the Sum or Amount of annuity
𝐴 – the Present value of an annuity
𝑅 – Periodic payment
𝑡 – the term of the annuity
𝑖 – the periodic rate
𝑟 – the rate of interest of an annuity
𝑚 – number of conversion periods.
Example 73 : A man deposits ₱12,200 every end of 6 months in an account paying 2%
compounded semi-annually. What amount is in the account at the end of 9 years and 6
months?
Solution:
Given: 𝑅 = ₱12, 200 , 𝑚 = 2 , 𝑟 = 0. 02 , 𝑡 = 9. 5 𝑦𝑟𝑠
Unknown: 𝑆
𝑜𝑟𝑑
Solution:
𝑜𝑟𝑑
= 𝑅 [
𝑚𝑡
] = 12 , 200 [
2 ( 9. 5 )
] = ₱253, 892. 92
Answer: 𝑆
𝑜𝑟𝑑
Example 74 : A home video entertainment set is offered for sale for ₱18,000 down payment
and ₱1,800 every three months for the balance, for 18 months. If interest is to be computed
at 10% converted quarterly, what is the cash price equivalent of the set?
Solution:
Given: 𝐷 = ₱18, 000 , 𝑅 = ₱1, 800 , 𝑡 =
18
12
Unknown: 𝐶𝑎𝑠ℎ 𝑉𝑎𝑙𝑢𝑒
Solution: To find the cash value, we need to solve first for the present value of future
payments. As the future payments are in the form of an ordinary annuity, we simply
solve for 𝐴
𝑜𝑟𝑑
𝑜𝑟𝑑
= 𝑅 [
−𝑚𝑡
] = 1 , 800 [
− 4 ( 1. 5 )
] = ₱9, 914. 63
Now for the cash value,
𝑜𝑟𝑑
Answer: 𝐶𝑎𝑠ℎ 𝑉𝑎𝑙𝑢𝑒 = ₱27, 914. 63
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Example 75 : How much monthly deposit must be made for 5 years and 5 months in order
to accumulate ₱120,000 at 3% compounded monthly?
Solution:
Given: 𝑡 = 5
5
12
𝑜𝑟𝑑
Unknown: 𝑅
Solution: From 𝑆
𝑜𝑟𝑑
= 𝑅 [
( 1 +
𝑟
𝑚
)
𝑚𝑡
− 1
𝑟
𝑚
] we can obtain
𝑜𝑟𝑑
[
( 1 +
𝑟
𝑚
)
𝑚𝑡
− 1
𝑟
𝑚
]
− 1
𝑜𝑟𝑑
[
𝑟
𝑚
( 1 +
𝑟
𝑚
)
𝑚𝑡
− 1
]
Then , 𝑅 = 𝑆
𝑜𝑟𝑑
[
𝑟
𝑚
( 1 +
𝑟
𝑚
)
𝑚𝑡
− 1
]
[
- 03
12
( 1 +
- 03
12
)
12 ( 5 +
5
12
)
− 1
]
Answer: 𝑅 = ₱1, 702. 52
Example 76 : Bebong wants to buy a car worth ₱740,000. He can pay 40% of the price as
down payment and the balance payable every end of the month for 60 months, how
much must he pay monthly at 9% compounded monthly?
Solution:
Given: 𝐶𝑎𝑠ℎ 𝑝𝑟𝑖𝑐𝑒 = ₱740, 000 , 𝐷𝑃 = ₱740, 000 ( 0. 4 ) = ₱296, 000
60
12
Unknown: 𝑅
Solution: From 𝐴
𝑜𝑟𝑑
= 𝑅 [
1 −( 1 +
𝑟
𝑚
)
−𝑚𝑡
𝑟
𝑚
] we can obtain 𝑅 = 𝐴
𝑜𝑟𝑑
[
𝑟
𝑚
1 −( 1 +
𝑟
𝑚
)
−𝑚𝑡
].
Since 𝐶𝑎𝑠ℎ 𝑝𝑟𝑖𝑐𝑒 = 𝐷 + 𝐴
𝑜𝑟𝑑
, this implies that
𝑜𝑟𝑑
Finally , 𝑅 = 𝐴
𝑜𝑟𝑑
[
𝑟
𝑚
1 −( 1 +
𝑟
𝑚
)
−𝑚𝑡
] = 444 , 000 [
- 09
12
1 −( 1 +
- 09
12
)
− 12 ( 5 )
] = ₱9, 216. 71
Answer: 𝑅 = ₱9, 216. 71
B. Annuity Due
An annuity where the periodic payments are made at the beginning of each
payment interval.
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- How much money should be invested in an account that earns 9.5% interest,
compounded monthly, in order to have ₱30,000 in 5 years?
- Accumulate ₱180,000 for 5 years at 13% compounded semi-annually. (Note:
accumulating an amount means finding its maturity value.)
- At what interest rate will a principal of ₱9,500 accumulate to ₱15,000 in 2 years
compounded semi-annually?
- Polly purchased a car. She paid ₱150,000 as down payment and pays ₱5,500 at the
end of each month for 48 months. If the interest is 7.8% compounded monthly, how
much was the car worth?
- A house and lot are worth ₱4.3 million in cash. A buyer pays a 40% down payment
and agrees to pay the balance by equal payments at the end of each month for 10
years at the rate of 9% compounded monthly. How much will be the monthly
payment?
- In order to have ₱1 million in a fund at the end of 15 years, how much must be
deposited in the fund every quarter if money can be invested at 10.5% compounded
quarterly?
Well done! It’s time to answer Quiz 2 (40 points) and prepare for the
MIDTERM EXAMINATION!
