Simple Interest.docx

GENERAL INSTUCTION: THE FOLLOWING ARE YOUR LECTURE FOR SIMPLE INTEREST, SOLUTIONS ARE PROVIDED FOR YOUR SELF STUDY. AT THE END OF THIS LECTURE, PLEASE SOLVE THE ATTACHED ASSIGNMENTS WHICH IS TO BE SUBMITTED ONMARCH 9, 2019. SIMPLE INTEREST INTEREST It may be defined as the share or percentage of money or goods being invested or deposited which serves as incentive to clients. Examples: Pawned jewelries which are earning interest, royalty from sales of certain goods, withholding tax, and interest on money deposited in a bank, etc., Kinds of Interest 1. Simple Interest – refers to share or percentage that does not earn another interest. It is computed by finding the product of the original capital and the rate of interest and the length of time expressed in terms of years, in symbol I = P x r x t = Prt Where P – original principal, r – rate of interest and t – time in terms of years 2. Compound Interest – refers to the sum of all interest due at the end of the period be it monthly, quarterly, biannually or annually with the principal amount always combined in the computation with the interest. Thus, interest earns another interest. Examples: SIMPLE INTEREST 1. Find the interest given the following: a. Php 10,000 at 4% for 2 years Solution: 𝑰 = 𝑷 𝒓 𝒕 = (𝟏𝟎, 𝟎𝟎𝟎)(𝟎. 𝟎𝟒)(𝟐) = ₱𝟖𝟎𝟎. 𝟎𝟎 b. Php 275,000 at 12% in 5 years Solution: 𝑰 = 𝑷 𝒓 𝒕 = (𝟐𝟕𝟓, 𝟎𝟎𝟎)(𝟎. 𝟏𝟐)(𝟓) = ₱𝟏𝟔𝟓, 𝟎𝟎𝟎. 𝟎𝟎 SIMPLE INTEREST FOR PERIODS OF TIME EXPRESSED IN DAYS OR MONTHS Interest is usually computed on the basis of 360-day/year which is equivalent to 12 months with each month having 30 days. The interest for 30 days is computed as 30/360 or 1/12 of the interest for one year, for 60 days is 60/360 or 1/6 of the interest for one year, etc. Examples: 1. Find the interest given the following conditions: a. Php 750 at 10% for 60 days Solution: 1 6

𝐼 = 𝑃 𝑟 𝑡 = (750)(0.1) ( ) = ₱ 12.50 b. Php 6,500 at 15% interest for 2 years and 3 months Solution: Solve first for t: 𝑡 =2+

3 12

= 2 + 0.25 = 2.25 𝑦𝑟𝑠

𝐼 = 𝑃 𝑟 𝑡 = (6,500)(0.15)(2.25) = ₱ 2,193.00

2. Which will give more interest? An investment of Php 2,500 at 8.5% for 120 days or Php 2,500 at 9% for 60 days? Solution: Solving for the interest in each of the given Given : P = ₱ 2,500

P = ₱ 2,500

r = 8.5%

r = 9%

t = 120 days

t = 60 days

120

60

𝐼 = 𝑃 𝑟 𝑡 = (2,500)(0.085) (360) = ₱ 70.83

𝐼 = 𝑃 𝑟 𝑡 = (2,500)(0.09) (360) = ₱ 37.50

FINDING THE TOTAL AMOUNT The total amount due after t years can be computed as follows: S=P+I Where: S – total amount, P – principal and I – total interest earned for the given number of years. Examples: 1. Find the total amount earned after investing Php 15,000 at 8% interest for 3 years. Solution: 𝐼 = 𝑃 𝑟 𝑡 = (15,000)(0.08)(3) = ₱ 3,600.00 𝑆 = 𝑃 + 𝐼 = 15,000 + 3,600 = ₱ 𝟏𝟖, 𝟔𝟎𝟎. 𝟎𝟎

2. At what rate must Php 4,500 be invested in order to accumulate to Php 4,900 in 6 months? Solution: We first solve for the interest as follows, then solve for the rate;

Then

From 𝑆 = 𝑃 + 𝐼

𝐼 = 𝑆 − 𝑃 = 4,900 − 4,500 = ₱ 400

From 𝐼 = 𝑃 𝑟 𝑡

𝑟 = 𝑃𝑡 =

𝐼

400 6 12

4,500( )

400

= 2,250 = 17.78%

3. For what length of time must Php 11,200 be invested at 9% to have a sum of Php 12, 300? Solution: We first solve for I as follows then solve for t; From 𝑆 = 𝑃 + 𝐼

𝐼 = 𝑆 − 𝑃 = 12,300 − 11,200 = ₱1,100

From 𝐼 = 𝑃 𝑟 𝑡

𝑡 = 𝑃𝑟 = 11,200(0.09) = 1,008 = 1.09 𝑦𝑒𝑎𝑟𝑠

𝐼

1,100

1,100

Note that 0.09 year may be converted into months : 0.09 x 12 month = 1.08 month Therefore, 1.09 years is equal to 1 year and 1 month 4. I have Php 150,000 and invested half of it at 5% interest and the other half at 7% , find the accumulated value in 5 years and 4 months? Solution: Solving for the accumulated value for each investment separately; 4

For investment ₱ 75,000 at 5% in 5 years and 4 months or 5 + 12 = 5.33 𝑦𝑒𝑎𝑟𝑠 𝐼 = 𝑃 𝑟 𝑡 = (75,000)(0.05)(5.33) = ₱ 19, 987.5

𝑆 = 𝑃 + 𝐼 = 75,000 + 19,987.50 = ₱ 94,987.50 For investment ₱ 75,000 at 7% in 5 years and 4 months or 5 +

4 12

= 5.33 𝑦𝑒𝑎𝑟𝑠

𝐼 = 𝑃 𝑟 𝑡 = (75,000)(0.07)(5.33) = ₱ 27,982.50 𝑆 = 𝑃 + 𝐼 = 75,000 + 27,982.50 = ₱102,982.50 Therefore, the Accumulated Value after 5 years and 4 months; = 94,982.50 + 102,982.50 = ₱ 197,965.00 EXACT AND ORDINARY INTEREST; EXACT AND APPROXIMATE TIME The exact interest is based on using t as a fraction whose numerator is the number of days in the term of the transaction and whose denominator is the exact number of days in a year. If 360 days is used it is called ordinary interest. Exact time is the actual number of days in the term of the transaction. Approximate time arbitrarily uses 30 days for each month. Summarizing in the table that follows: Exact Interest 𝑁𝑜. 𝑜𝑓 𝑑𝑎𝑦𝑠 𝑒𝑥𝑎𝑐𝑡 𝑛𝑜. 𝑜𝑓 𝑑𝑎𝑦𝑠/𝑦𝑒𝑎𝑟

Ordinary Interest 𝑁𝑜. 𝑜𝑓 𝑑𝑎𝑦𝑠 360

Exact Time Actual no. of days

Approximate Time Uses 30 days for each month

For convention, the exact number of days between two dates shall be used while a whole number of month shall be considered between two months of the same date. For instance, from Jun 8 to October 12 is 126 (22+31+31+30+12) while from June 8 to Oct 8 is 4 months (July, Aug, Sept and Oct) Examples: 1. Find the different types of simple interest earned by Php 45,000.00 at 5% from May 17 to December 13, 2014. Solution: a. Using exact time and ordinary interest: Actual No. of Days : May 17 – Dec 13; (14 +30+31+31+30+31+30+13) = 210 days 210

𝐼 = 𝑃 𝑟 𝑡 = (45,000)(0.05) (360) = ₱ 1,312.50 b. Using exact time and exact interest 210 𝐼 = 𝑃 𝑟 𝑡 = (45,000)(0.05) ( ) = ₱ 1,294.52 365 c. Using approximate time and ordinary interest. Approximate time from May 17 – Dec 13; (13 +30+30+30+30+30+30+13)=206 206 𝐼 = 𝑃 𝑟 𝑡 = (45,000)(0.05) ( ) = ₱ 1,287.50 360 d. Using approximate time and exact interest 206 𝐼 = 𝑃 𝑟 𝑡 = (45,000)(0.05) ( ) = ₱ 1,269.86 365

2. Find the exact number of days between the given dates a. May 24, 2015 to Nov, 5, 2015

b. Dec. 10, 2014 to Feb. 18, 2015

Solution: For Your Reference, consider the table below Jan Feb March Apr May Jun July Aug Sep Oct Nov Dec Total

31 28 31 30 31 30 31 31 30 31 30 31 365

a. May 24, 2015 – Nov. 5, 2015: 7+ 30+31+31+30+31+5 = 165 days b. Dec. 10, 2014 to Feb. 18, 2015: 21+ 31+18 = 70 days SIMPLE INTEREST ON INSTALLMENT PAYMENTS THE MERCHANT RULE – The interest on money that is paid by installment is computed from the balance of the principal before each payment. Example 1: The amount Php 10, 000 at 6% payable in one was paid as follows: Php 3,000 was paid after 2 months, Php 4,000 after 5 months and Php 2,000 after 9 months, how much must be paid at the end of the year? Solution 2

𝐼1= 10,000 x .06 x 12 = 100 ; 10,000 – 3,000 = 7,000 𝐼2 = 7,000 𝑥 .06 𝑥 𝐼3 = 3,000 𝑥 .06 𝑥 𝐼2 = 1,000 𝑥 .06 𝑥

3 12 4 12 3 12

= 105; 7,000 – 4,000 = 3,000 = 60 ; 3,000 – 2,000 = 1,000 = 15 ;

Total Interest = ₱ 280 Total Amount to be paid at the end of the year = 1,000 + 280 = ₱ 1,280

Example 2: A 2 HP air conditioner whose cash value is Php 37,000 can be bought by installment on the following terms; down payment of Php 10,000 and monthly payment of Php 1650 for 18 months. Find the rate of simple interest. Solution: Consider the unpaid cash value of the air conditioner as a loan. Then find the average loan as the loan decreases every time a payment is made. Original loan = 37,000 – 10,000 = ₱27,000

Monthly Repayments =

27,000 18

= 1500

Interest on loan = Installment value –cash value I = 10,000 + (1650 x 18 ) – 37,000 = 39,700 – 37,000 = 2,700 The average loan can be computed by adding the first and last loan and dividing by 2, as follows: Average loan =

27,000+1,500 2 𝐼

Finally, the rate of interest, 𝑟 = 𝑃𝑡 =

= 14,250

2,700 14,250𝑥

2,700

18 12

= 14,250𝑥1.5 = 0.1263 = 𝟏𝟐. 𝟔𝟑%

Long method of computing the average loan 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 total

27000 25500 24000 22500 21000 19500 18000 16500 15000 13500 12000 10500 9000 7500 6000 4500 3000 1500 256500

𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑛 =

256500 = 14,250 18

MATH IN BUSINESS NAME__________________________________________ Score______________ Date ____________ Course, yr & Section____________________ Student no. _______________ Prof. Emelita A. Isaac ASSIGNMENT NO. 2 Solve the following problems. Write legibly. Submit on March 9, 2019. 1. A woman borrowed Php 15,000 at 12 % simple interest payable in six months with the option to give partial payments at will. She paid Php 4000, Php 3,000 and Php 5,500 at the end of the first month, 3 nd month and 5th month, how much more must he pay at the end of 6th month.

2. A Samsung cellphone whose cash value is Php 6,500 was bought from a local dealer. A down payment of Php 2,000 was paid and 3 monthly installments of Php 1, 500. Find the rate of installment plan.

3. An OFW borrowed Php 30,000 and promise to pay in equal installments for 10 months. If the rate of simple interest is 12%, how much is the monthly interest?

4. Find the present value of Php 12,500 due in 540 days at 7% discount,

5. Using exact time, find the ordinary interest on a. Php 11,200 at 7% from June 1 to Sept 5, 2014

b. Php 25,000 from April 20, 2015 to Aug 27, 2015.

6. Using exact interest and using approximate time at 8%. a. Php 5,500 from Feb 5 to April 10, 2015

b. Php 375,000 from August 15 to November 15, 2014.

7. At what rate of interest must be borrowed in order to double the amount of Php 75,000 in 5 years.

8. How much was borrowed if the interest earned is Php 2,500 at 6% after 45 days.

9. Find the compound amount and compound interest on the principal, Php 20,000 borrowed at 6% compounded annually for 3 years.

10. Find the compound amount which would be obtained from an interest of Php 2,000 at 6% compounded quarterly for 5 years.