SOLVING PROBLEMS ON
SIMPLE INTEREST
Let us define first the following terms:
A deposit is a financial term that means money is held at
a bank. It is a transaction involving a transfer of money to
another party for safekeeping.
A loan is a money (or property) given with the promise
that it will be paid back in the future, usually with interest.
Services are professional support to aid customers. This is
sometimes described as the intangible product.
Utilities refer to the basic amenities like electricity
and water.
Mortgage is a loan, secured by a collateral, that the
borrower is obliged to pay at specified terms.
Amortization is the process of reducing a cost or
total in regular small amounts.
Interest is described as the money paid regularly at a
particular rate for the use of money lent, or for delaying
the repayment of a debt.
Principal is the original amount invested or borrowed.
Rate is the amount of a charge or payment (usually in
percent) with reference to some basis of calculation.
Time can be defined as the duration or term used in
solving simple interest.
Let us have the following examples:
A. Matoy started his small business by borrowing an
amount of ₱20,000.00 payable for 3 years as a simple
interest rate of 8% annually. How much is the interest
that he has to pay?
Let us have the following examples:
A. Matoy started his small business by borrowing an
amount of ₱20,000.00 payable for 3 years as a simple
interest rate of 8% annually. How much is the interest
that he has to pay?
Simple Interest (I):
𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡 = 𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 × 𝑟𝑎𝑡𝑒 × 𝑡𝑖𝑚𝑒
Given:
𝑰=?
𝑷= ₱20,000.00
𝒕 = 3 𝑦𝑒𝑎𝑟𝑠
𝒓= 8%
Solution:
𝑰=𝑷𝒓𝒕 𝑰= (₱20,000) (0.08) (3)
𝑰 = ₱𝟒,𝟖𝟎𝟎.𝟎𝟎
B. Meena was given ₱200,000.00 by her parents and
she deposited it in a bank. Find the maturity value of
Meena’s money in five years if the bank offers a simple
interest of 1% per annum.
Maturity Value (F):
𝐹= 𝑃 (1 + 𝑟𝑡)
𝐹= 𝑃 + 𝐼
Given:
𝑭 = ? 𝑷 =
₱200,000.00
𝒓 = 1% 𝒕 = 5 𝑦𝑒
𝑎𝑟𝑠
Solution: 𝑭 = 𝑷 (𝟏 + 𝒓𝒕)
𝑭 = ₱200,000 [1 + (0.01) (5)]
𝑭 = ₱𝟐𝟏𝟎,𝟎𝟎𝟎.𝟎𝟎
C. Dereck borrowed ₱2,200,000.00 payable for 5
years to purchase a pick-up truck. A bank offered him
a car loan with simple interest rate of 6.75% per year.
What is the maturity value of the loan?
Given:
𝑭 = ? 𝑷 = ₱2,200,000.00
𝒓 = 6.5% 𝒕 = 5 𝑦𝑒𝑎𝑟𝑠
Solution: 𝑭= 𝑷 (𝟏 + 𝒓𝒕) 𝑭= ₱2,200,000
[1+(0.065)(5)] 𝑭= ₱𝟐,𝟗𝟏𝟓,𝟎𝟎𝟎.𝟎𝟎
D. If a house and lot is sold for ₱4,000,000.00 and the
bank requires 25% down payment, find the amount
of mortgage.
Mortgage:
Down payment = down payment rate × cash price
Amount of the loan (mortgage) = cash price − down payment
Given:
down payment = ? cash price = ₱4,000,000.00
mortgage = ? down payment rate = 25%
Solution:
𝐷𝑜𝑤𝑛 𝑝𝑎𝑦𝑚𝑒𝑛𝑡 = 0.25 × ₱4,000,000
 = ₱𝟏,𝟎𝟎𝟎,𝟎𝟎𝟎.𝟎𝟎
𝑀𝑜𝑟𝑡𝑔𝑎𝑔𝑒 = ₱4,000,000 – ₱1,000,000
 = ₱𝟑,𝟎𝟎𝟎,𝟎𝟎𝟎.𝟎𝟎
E. Using ordinary interest, what is the amount due
(maturity value) on ₱40,000.00 invested at 10%
simple interest for 6 months. Assume that 1 month is
equal to 30 days.
For Ordinary Interest: I = P × (r/360) × t
Given:
I = ? P = ₱40,000
r = 10% t = 6months or 180 days
Solution:
𝑰 = ₱40,000 (0.1/360) (180)
= ₱𝟐,𝟎𝟎𝟎.𝟎𝟎
𝑭 = 𝑃 + 𝐼
= ₱40,000 + ₱2,000 = ₱𝟒𝟐,𝟎𝟎𝟎.𝟎𝟎
 Activity : What’s the Problem?
 1. Keith borrowed an amount of ₱50,000.00 from the bank. How much interest will he
have to be pay on the loan for 3 years at a simple interest rate of 7.5% per year?
2. Pedro has been saving his lunch money and it reached an amount of ₱30,000.00 To
secure his money and gain a small interest, he deposited it in the bank. How much is
Pedro’s money after ten years if the bank offers a simple interest of 1.5% per annum.
 3. Popsie started a school bus business using a van worth ₱3.2M. He got a car from a
bank with a simple interest at a rate of 5% per year. What is the maturity value of the
loan after 5 years?
 4. As newlyweds, Mr. And Mrs. Sanchez wanted to purchase their dream house worth
₱5,000,000.00 and the bank requires 15% down payment, find the amount of
mortgage.
 5. Using ordinary interest, what is the future value on a ₱125,000.00 invested at 9%
simple interest for 5 months. Assume that 1 month is equal to 30 days.

simple interest.pptx

  • 1.
  • 2.
    Let us definefirst the following terms: A deposit is a financial term that means money is held at a bank. It is a transaction involving a transfer of money to another party for safekeeping. A loan is a money (or property) given with the promise that it will be paid back in the future, usually with interest. Services are professional support to aid customers. This is sometimes described as the intangible product.
  • 3.
    Utilities refer tothe basic amenities like electricity and water. Mortgage is a loan, secured by a collateral, that the borrower is obliged to pay at specified terms. Amortization is the process of reducing a cost or total in regular small amounts.
  • 4.
    Interest is describedas the money paid regularly at a particular rate for the use of money lent, or for delaying the repayment of a debt. Principal is the original amount invested or borrowed. Rate is the amount of a charge or payment (usually in percent) with reference to some basis of calculation. Time can be defined as the duration or term used in solving simple interest.
  • 5.
    Let us havethe following examples: A. Matoy started his small business by borrowing an amount of ₱20,000.00 payable for 3 years as a simple interest rate of 8% annually. How much is the interest that he has to pay?
  • 6.
    Let us havethe following examples: A. Matoy started his small business by borrowing an amount of ₱20,000.00 payable for 3 years as a simple interest rate of 8% annually. How much is the interest that he has to pay? Simple Interest (I): 𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡 = 𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 × 𝑟𝑎𝑡𝑒 × 𝑡𝑖𝑚𝑒
  • 7.
    Given: 𝑰=? 𝑷= ₱20,000.00 𝒕 =3 𝑦𝑒𝑎𝑟𝑠 𝒓= 8% Solution: 𝑰=𝑷𝒓𝒕 𝑰= (₱20,000) (0.08) (3) 𝑰 = ₱𝟒,𝟖𝟎𝟎.𝟎𝟎
  • 8.
    B. Meena wasgiven ₱200,000.00 by her parents and she deposited it in a bank. Find the maturity value of Meena’s money in five years if the bank offers a simple interest of 1% per annum. Maturity Value (F): 𝐹= 𝑃 (1 + 𝑟𝑡) 𝐹= 𝑃 + 𝐼
  • 9.
    Given: 𝑭 = ?𝑷 = ₱200,000.00 𝒓 = 1% 𝒕 = 5 𝑦𝑒 𝑎𝑟𝑠 Solution: 𝑭 = 𝑷 (𝟏 + 𝒓𝒕) 𝑭 = ₱200,000 [1 + (0.01) (5)] 𝑭 = ₱𝟐𝟏𝟎,𝟎𝟎𝟎.𝟎𝟎
  • 10.
    C. Dereck borrowed₱2,200,000.00 payable for 5 years to purchase a pick-up truck. A bank offered him a car loan with simple interest rate of 6.75% per year. What is the maturity value of the loan? Given: 𝑭 = ? 𝑷 = ₱2,200,000.00 𝒓 = 6.5% 𝒕 = 5 𝑦𝑒𝑎𝑟𝑠 Solution: 𝑭= 𝑷 (𝟏 + 𝒓𝒕) 𝑭= ₱2,200,000 [1+(0.065)(5)] 𝑭= ₱𝟐,𝟗𝟏𝟓,𝟎𝟎𝟎.𝟎𝟎
  • 11.
    D. If ahouse and lot is sold for ₱4,000,000.00 and the bank requires 25% down payment, find the amount of mortgage. Mortgage: Down payment = down payment rate × cash price Amount of the loan (mortgage) = cash price − down payment
  • 12.
    Given: down payment =? cash price = ₱4,000,000.00 mortgage = ? down payment rate = 25% Solution: 𝐷𝑜𝑤𝑛 𝑝𝑎𝑦𝑚𝑒𝑛𝑡 = 0.25 × ₱4,000,000  = ₱𝟏,𝟎𝟎𝟎,𝟎𝟎𝟎.𝟎𝟎 𝑀𝑜𝑟𝑡𝑔𝑎𝑔𝑒 = ₱4,000,000 – ₱1,000,000  = ₱𝟑,𝟎𝟎𝟎,𝟎𝟎𝟎.𝟎𝟎
  • 13.
    E. Using ordinaryinterest, what is the amount due (maturity value) on ₱40,000.00 invested at 10% simple interest for 6 months. Assume that 1 month is equal to 30 days. For Ordinary Interest: I = P × (r/360) × t
  • 14.
    Given: I = ?P = ₱40,000 r = 10% t = 6months or 180 days Solution: 𝑰 = ₱40,000 (0.1/360) (180) = ₱𝟐,𝟎𝟎𝟎.𝟎𝟎 𝑭 = 𝑃 + 𝐼 = ₱40,000 + ₱2,000 = ₱𝟒𝟐,𝟎𝟎𝟎.𝟎𝟎
  • 15.
     Activity :What’s the Problem?  1. Keith borrowed an amount of ₱50,000.00 from the bank. How much interest will he have to be pay on the loan for 3 years at a simple interest rate of 7.5% per year? 2. Pedro has been saving his lunch money and it reached an amount of ₱30,000.00 To secure his money and gain a small interest, he deposited it in the bank. How much is Pedro’s money after ten years if the bank offers a simple interest of 1.5% per annum.  3. Popsie started a school bus business using a van worth ₱3.2M. He got a car from a bank with a simple interest at a rate of 5% per year. What is the maturity value of the loan after 5 years?  4. As newlyweds, Mr. And Mrs. Sanchez wanted to purchase their dream house worth ₱5,000,000.00 and the bank requires 15% down payment, find the amount of mortgage.  5. Using ordinary interest, what is the future value on a ₱125,000.00 invested at 9% simple interest for 5 months. Assume that 1 month is equal to 30 days.