Interest factor tables - guide to interest rate PDF

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Interest factor tables - a guide to interest rate and tables...

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The use of formulas and interest factor tables 1. I have $1,000 and place it in a savings account that pays 12%p.a. interest compounding annually. How much will be in the account at the end of three years? Formula approach ฀฀฀฀ = ฀฀฀฀฀฀(1 + ฀฀)฀฀

฀฀฀฀ = $1,000฀฀(1 + 0.12)3 ฀฀฀฀ = $1,000฀฀1.4049 ฀฀฀฀ = $1,404.90

Interest Factor table approach ฀฀฀฀ = ฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀%,฀฀ ฀฀฀฀ = $1,000฀฀฀฀฀฀฀฀฀฀12%,3 ฀฀฀฀ = $1,000฀฀1.4049 ฀฀฀฀ = $1,404.90

2. I have $1,000 and place it in a savings account that pays 12%p.a. interest compounding monthly. How much will be in the account at the end of three years? Formula approach ฀฀฀฀฀฀฀฀ ฀฀฀฀ = ฀฀฀฀฀฀ �1 + � ฀฀ ฀฀฀฀ = ฀฀฀฀฀฀ �1 +

12%3฀฀12 � 12

฀฀฀฀ = $1,000฀฀(1 + 0.01)36 ฀฀฀฀ = $1,000฀฀1.4308 ฀฀฀฀ = $1,430.80

Interest Factor table approach ฀฀฀฀ = ฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀%,฀฀ Note: to select the appropriate coordinate in the table you first must determine the period interest rate (12% ÷ 12 = 1%) and the total number of periods (3 x 12 = 36). ฀฀฀฀ = $1,000฀฀฀฀฀฀฀฀฀฀1%,36 ฀฀฀฀ = $1,000฀฀1.4308 ฀฀฀฀ = $1,430.80

3. I need $2,000 to pay for a new television in exactly two years time. How much will I need to place in a deposit account that pays 8%p.a. compounding annually? Formula approach ฀฀฀฀ =

฀฀฀฀ (1 + ฀฀)฀฀

฀฀฀฀ =

$2000 (1 + 0.08)2

฀฀฀฀ = $2,000 ÷ 1.1664 ฀฀฀฀ = $1,714.67 Interest Factor table approach ฀฀฀฀ = ฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀%,฀฀ ฀฀฀฀ = ฀฀฀฀฀฀฀฀฀฀฀฀฀฀8%,2 ฀฀฀฀ = $2,000 ฀฀ 0.8573 ฀฀฀฀ = $1,714.60 Note: sometimes there will be a small rounding difference between the formula calculation and the tables calculation.

4. I need $2,000 to pay for a new television in exactly two years time. How much will I need to place in a deposit account that pays 8%p.a. compounding quarterly? Formula approach ฀฀฀฀ = ฀฀฀฀ = ฀฀฀฀ =

฀฀฀฀ ฀฀ ฀฀฀฀฀฀ �1 + ฀฀� $2,000 �1 +

. 08 2฀฀4 � 4

$2,000 1.17166

฀฀฀฀ = $1,706.98

Interest Factor table approach ฀฀฀฀ = ฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀%,฀฀ Note: to select the appropriate coordinate in the table you first must determine the period interest rate (8% ÷ 4 = 2%) and the total number of periods (2 x 4 = 8). ฀฀฀฀ = ฀฀฀฀฀฀฀฀฀฀฀฀฀฀2%,8 ฀฀฀฀ = $2,000฀฀0.8535

฀฀฀฀ = $1,707.00

Note: sometimes there will be a small rounding difference between the formula calculation and the tables calculation.

5. I save $400 at the end of every month into an account that pays 6%p.a. compounding monthly. How much will be in the account at the end of four years? Formula approach ฀฀฀฀ = ฀฀฀฀ ฀฀

[(1 + ฀฀)฀ ฀ − 1] ฀

฀

Remember that ฀฀ (6% ÷ 12 = ½%) is the interest rate per period and ฀฀ (4 x 12 = 48) is the total number of periods. ฀฀฀฀ = $400 ฀฀

[(1 + 0.005)48 − 1] 0.005

฀฀฀฀ = $400 ฀฀ 54.0978 ฀฀฀฀ = $21,639.13

Interest Factor table approach ฀฀฀฀ = ฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀%,฀฀ Remember that ฀฀ (6% ÷ 12 = ½%) is the interest rate per period and ฀฀ (4 x 12 = 48) is the total number of periods. ฀฀฀฀ = $400 ฀฀ ฀฀฀฀฀฀฀฀0.5%,48 ฀฀฀฀ = $400 ฀฀ 54.0978 ฀฀฀฀ = $21,639.13

6. I need $2,000 to pay for a new television in exactly two years time. How much will I need to save each month in account that pays 6%p.a. compounding monthly? Formula approach ฀฀฀฀ = ฀฀฀฀ ฀฀

[(1 + ฀฀)฀ ฀ − 1] ฀

฀

so ฀฀฀฀ = ฀฀฀฀ ÷

[(1 + ฀฀)฀ ฀ − 1] ฀

฀

Remember that ฀฀ (6%÷12 = ½%) is the interest rate per period and ฀฀ (2 x 12 = 24) is the total number of periods. ฀฀฀฀ = $2,000 ÷

[(1 + 0.005)24 − 1] 0.005

฀฀฀฀ = $2,000 ÷ 25.4320 ฀฀฀฀ = $78.64 Interest Factor table approach ฀฀฀฀ = ฀฀฀฀ ฀฀ ฀฀฀฀฀฀฀฀฀฀฀฀%,฀฀ ฀฀฀฀ = ฀฀฀฀ ÷ ฀฀฀฀฀฀฀฀฀฀฀฀%,฀฀ Remember that ฀฀ (6%÷12 = ½%) is the interest rate per period and ฀฀ (2 x 12 = 24) is the total number of periods. ฀฀฀฀ = $2,000 ÷ ฀฀฀฀฀฀฀฀฀฀0.5%,24 ฀฀฀฀ = $2,000 ÷ 25.4320 ฀฀฀฀ = $78.64

7. I have inherited a substantial sum and I want to put money into an investment account that pays 12%p.a. compounding annually so that I can withdraw $50,000 every year for the next 30 years during my retirement. How much will I need to deposit into the account. Formula approach ฀฀฀฀ = ฀฀฀฀ ฀฀

1 � + ฀฀)฀฀ ฀ ฀

�1 −(1

฀฀฀฀ = $50,000 ฀฀

1 � + 0.12)30 0.12

�1 −(1

฀฀฀฀ = $50,000 ฀฀ 8.055184 ฀฀฀฀ = $402,759.20

Interest Factor table approach ฀฀฀฀ = ฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀฀%,฀฀ ฀฀฀฀ = $50,000 ฀฀฀฀฀฀฀฀฀฀฀฀12%,30 ฀฀฀฀ = $50,000 ฀฀ 8.0552 ฀฀฀฀ = $402,760.00 Note: sometimes there will be a small rounding difference between the formula calculation and the tables calculation.

8. I borrow $10,000 for five years at an interest rate of 24%p.a. What equal monthly repayments would be needed to repay the loan in full by the end of five years? Formula approach 1 �1 − � (1 + ฀฀)฀฀ ฀฀฀฀ = ฀฀฀฀ ฀฀ ฀ ฀ so 1 �1 − � (1 + ฀฀)฀฀ ฀฀฀฀ = ฀฀฀฀ ÷ ฀ ฀

฀฀ (24%÷12 = 2%) is the interest rate per monthly period and ฀฀ (5 x 12 = 60) is the total number of monthly periods. ฀฀฀฀ = $10,000 ÷

�1 −

1 � (1 + 0.02)60 0.02

฀฀฀฀ = $10,000 ÷ 34.760887 ฀฀฀฀ = $287.68 Interest Factor table approach ฀฀฀฀ = ฀฀฀฀ ฀฀ ฀฀฀฀฀฀฀฀฀฀฀฀%,฀฀ so ฀฀฀฀ = ฀฀฀฀ ÷ ฀฀฀฀฀฀฀฀฀฀฀฀%,฀฀ ฀฀ (24%÷12 = 2%) is the interest rate per monthly period and ฀฀ (5 x 12 = 60) is the total number of monthly periods. ฀฀฀฀ = $10,000 ÷ ฀฀฀฀฀฀฀฀฀฀2%,60 ฀฀฀฀ = $10,000 ÷ 34.7609 ฀฀฀฀ = $287.68...