To determine the exact time between two dates, there are two methods that we may use. On method is simply determining the number of days from month to moth from the first date to the second date of the term. Another method is by using Table 1 (found at the end of the module), the number of each day of the year.
Example 1.
Determine the exact number of days from July 2, 1986 to December 20, 1986
Solution:
a. Determine the number of days from month to month
July = 29
August = 31
September = 30
October = 31
November = 30
December = 20 (last day of the transaction) Total No. = 171 days.
b. Using Table 1
December 20 = 345th day of the year Less: July 2 = 290th day of the year
Exact No. of days = 171 days.
Example 2:
Find the different types of interest on P5,600 at 5% fromOctober 17, 1987 to June 26, 1988.
Solution
a. Find the exact time using the Table:
Number of day from October 17, 1987 to December 31, 1987
December 31 - 365th day of the year October 17 - 290th day of the year
Exam time = 75 days
There are 177 days from January 1, 1988 to June 26, 1988. Hence, the exact number of days from October 17, 1987 to June 26, 1988 is 252. That is 75 days plus 177 days.
b. Find the approximate time:
From October 17, 1987 to June 17, 1988 is 8 months. This is equivalent to 240 days. From June 17 to June 26 is 9 days. So the approximate number of days from October 17,1987 to June 26, 1988 is 240 days plus 9 days equals 249 days
Another method is done by subtracting the first date from the second date.
Year Month Date
Second date 1987 18 26
First date 1987 10 17
Approximate time = 8 mos. 9 days
Approximate number of days = (8 x 30) + 9 = 249 days. Note that 10 cannot be subtracted from 6, so borrow 1 year from 1988 convert to months and add to the column of month, hence 6 plus 12 equals 18.
To find the simple interest between two dates, gives us four types of interest as follows:
1. Ordinary interest on exact number of days I = PRT
= P5,600 (.05) (252/360) = P196
2. Ordinary interest on approximate number of days I = PRT
= P5,600 (.05) (249/360) = P193.67
3. Exact interest on exact number of days I = PRT
= P5,600 (.05) (252/365) = P193.32
4. Exact interest on approximate number of days I = PRT
= P5,600 (.05) (249/365) = P191.01
Among the four types of interest, the ordinary interest on exact number of days yields the highest interest. This type of interest is otherwise known as the Banker’s Rule which is widely used in business
A. Using the table, find the exact number of days between the following dates:
1) May 17, 1986 to December 3, 1986 2) November 15,1986 to August 9, 1987 3) January 22, 1984 to March 3, 1985 4) June 12, 1983 to July 15, 1986
B. Find the approximate number of days 1) April 16, 1982 to October 4, 1982
2) February 11, 1986 to September 30, 1986 3) November 28, 1985 to May 25, 1986 4) October 15, 1983 to July 24, 1987
C. Using the exact time, find the ordinary interest on
1) P 3,265 at 7% from August 14, 1986 to November 28, 1987 2) P 25,500 at 91/2% from April 20,1983 to September 26, 1986
D. Find the exact interest using approximate time:
1) P7,200 at 6% from February 10,1980 to April 12, 1982 2) P10,500 at 6 ½% from August 24,1983 to May 14, 1985
E. Find the ordinary interest using approximate number of days.
1) P825 at 8 1/2% from March 30, 1984 to October 3, 1984 2) P3,450 at 10% from September 14, 1985 to April 24, 1986.
SELF-CHECK 1.7
F. Find the exact interest using the exact number of days:
1) P5,300 at 12 ½% from May 5, 1984 to November 18, 1984 2) P16,800 at 9 ½% from December 3, 1985 to August 18, 1987
G. Solve the following problems:
1) On November 18, 1986, Mrs. Acero borrowed P4,500 at 8% simple interest. If the loan was paid on May 27, 1987, how much interest should she pay?
2) Mrs. Paras, who needed an additional amount of P8,260 for her business, borrowed from Mrs. Santos the said amount at 9% for 264 days. How much interest did Mrs. Paras pay Mrs. Santos?
3) On December 15, 1986, Mrs Lee loaned P5,000 to pay the hospital bills.
She promised to pay the amount plus the interest on April 9, 1987. how much was the interest, if the amount is worth 9 ½% simple interest?
4) At the start of business, Mr. Liptona loaned P50,500 at ABC Bank which charges 12% simple interest. He promised to pay the amount plus the interest at the end of one year and 275 days. How much was the interest paid?
1.8 6%, 6-Day Mehod for Computing Simple Interest
The use of the 6%, 6-day method of computing interest makes computation easier. This method has varied application as illustrated by the following examples:
Example 1: Find the interest earned from P1,200 at 6% for a. 6 days
b. 60 days
c. 600 days d. 6,000 days
Solution:
a) I = PRT
= (P1,200) (.06) (6/360) = (P1,200) (.001)
= P 1.20
b) I = PRT
= (P1,200) (.06) (60/360) = (P1,200) (.01)
= P 12
c) I = PRT
= (P1,200) (.06) (600/360) = (P1,200) (.1)
= P 120
d) I = PRT
= (P1,200) (.06) (6000/360) = (P1,200) (1)
= P 1,200
Take note that the interest is simply found by multiplying the given principal by .001 for 6 days, .01 for 60 days, .1 for 600 days and by 1 for 6,000 days.
Example 2.
Find the interest earned by P5,250 at 12% for a. 6 days b. 6000 days
Solution:
a. I = (P5,250) (.001)
= P5.25 (interest at 6% for 6 days)
Since, 12% = 2 (6%), then the interest at 12% for 6 days is equals to the interest at 6% for 6 days multiplied by 2. Hence,
I = (P5.25) (2)
= P10.50 (interest at 12% for 6 days
b. I = (P5,250) (1)
= P5250 (interest at 6% for 6 days)
Since, 12% = 2 (6%), then the interest at 12% for 6 days is equals to the interest at 6% for 6 days multiplied by 2. Hence,
I = (P5,250) (2)
= P10,500 (interest at 12% for 6 days
A. Using the 6%-6 day method, discount each of the following amounts 1. P620 at 6% for 15 days
2. P3,847 at 6% for 130 days 3. P192.15 at 3% for 240 days 4. P427.85 at 12% for 380 days 5. P33,712.20 at 15% for 72 days
SELF-CHECK 1.8
2.1 Simple Discount
Normally, when a person applies for loans from banks or credits institutions, the interest is collected in advanced which we call discount. To discount an amount is to find its value at a period earlier than its maturity date. Let S be the sum of money to be discounted, d the rate of discount per annum and t the term of discount in years. Then , the amount of discount represented by D is.
D = SDT Example:
Discount the amount, P2,400 at 10% discount rate for 150 days.
Solution:
D = SDT
= P2,400 (.10) (150/360) = P100
After the amount is discounted by the bank, the amount which the borrower recives is the present value (P) of the amount (S). To find the present value P.
P = S – D So,
P = P2,400- 100 = P2,300
INFORMATION SHEET 2.1 SIMPLE DISCOUNT
2.2 Solve for the Discount rate (d)
To solve for the discount rate (r), divide the discount D by the product of the total amount (S) and the term of discount. (t). Hence,
d = D S t
Example:
Find the discount rate if P820 yields a discount of P147.60 for 3 years.
Solution:
d = D S t = P147.60 P820
2.3 Solve for the Term of Discount (t)
To solve for the term of discount (t), divide the discount (D) by the product of the total amount (S) and the discount rate d. Thus,
t = D S d