Check out the Variable Rate of Compound Interest in this article. It is a bit difficult to find out the Compound Interest When the Rate of Successive Years is Different. If you follow the entire article, it becomes easy to find the compound interest when the Variable Rate is given. Solve every problem and learn the concept in an easy way.
We explained each and every step to solve all the Problems related to compound interest variable rate. The complete concepts of Compound Interest are given in 10th Grade Math Compound Interest articles.
Also, read:
- Compound Interest when Interest is Compounded Yearly
- Compound Interest with Growing Principal
- Compound Interest as Repeated Simple Interest
- SHREECEM Pivot Point Calculator
How to Calculate Compound Interest with Variable Rates?
The detailed information regarding How to find the Compound Interest When Successive Year’s Rate of Interest is Different is mentioned here.
Let us consider the amount to be A and the Principal to be P, and the Rate of Compound Interest for Successive Years to be different i.e. r1%, r2%, r3%, r4%, …… then the Formula to calculate the amount is given by
A = P (1 + \(\frac { r1 }{ 100 } \)) (1 + \(\frac { r2 }{ 100 } \)) (1 + \(\frac { r3 }{ 100 } \)) (1 + \(\frac { r4 }{ 100 } \))……
Here A is the amount, P is the Principal, and r1, r2, r3, r4, … are the for successive years.
Variable Rate of Compound Interest Examples
Various problems on Compound interest variable rates are given with formula and explanation below.
Example 1. If the rate of compound interest for the first, second, and third years is 6%, 8%, and 9% respectively, find the amount and the compound interest on $ 6000 in 3 years.
Solution: Given that the rate of compound interest for the first, second, and third years is 6%, 8%, and 9% respectively.
r1 = 6%, r2 = 8%, and r3 = 9%.
Principal Amount P = $6000.
Firstly, find out the amount.
The formula to find the amount is A = P (1 + \(\frac { r1 }{ 100 } \)) (1 + \(\frac { r2 }{ 100 } \)) (1 + \(\frac { r3 }{ 100 } \)) where p is the principal amount, r1, r2, r3 are the Rate of interest for the successive years.
Substitute all the information in the above formula.
A = $ 6000 (1 + \(\frac { 6 }{ 100 } \)) (1 + \(\frac { 8 }{ 100 } \)) (1 + \(\frac { 9 }{ 100 } \))
A = $ 6000 (\(\frac { 106 }{ 100 } \)) (\(\frac { 108 }{ 100 } \)) (\(\frac { 109 }{ 100 } \))
A = 7486.992
Therefore, the required amount = $ 7486.992
Now, find the compound interest.
The formula to find the compound interest is compound interest = Final amount – Initial principal
compound interest = $ 7486.992 – $ 6000 = $1486.992
Therefore, the compound interest is $1486.992.
Example 2. Find the compound interest accrued by John from a bank on $ 8000 in 3 years, when the rates of interest for successive years are 3%, 5%, and 7% respectively.
Solution: Given that know the compound interest accrued by John from a bank on $ 8000 in 3 years when the rates of interest for successive years are 3%, 5%, and 7% respectively.
Given that for the first year:
Principal = $ 8000;
Rate of interest = 3% and
Time = 1 years
Therefore, interest for the first year = (P × R × T)/100 = (8000 × 3 × 1)/100 = $240
Therefore, the amount after 1 year = Principal + Interest = $8000 + $240 = $8240.
For the second year, the new principal is $ 8240.
Rate of interest = 5% and
Time = 1 years.
Therefore, the interest for the second year = (P × R × T)/100 = (8240 × 5 × 1)/100 = $412
Therefore, the amount after 2 year = Principal + Interest = $8240 + $412 = $8652.
For the third year, the new principal is $8652.
Rate of interest = 7% and
Time = 1 years.
Therefore, the interest for the second year = (P × R × T)/100 = (8652 × 7 × 1)/100 = $605.64
Therefore, the amount after 3 year = Principal + Interest = $8652 + $605.64 = $9257.64.
Therefore, the compound interest accrued = Final amount – Initial principal = $9257.64 – $8000 = $1257.64
Example 3. A company offers the following growth rates of compound interest annually to the investors on successive years of investment.
6%, 8% and 10%
(i) A man invests $ 40,000 for 2 years. What amount will he receive after 2 years?
(ii) A man invests $ 20,000 for 3 years. What will be his gain?
Solution:Â
(i) Given that the man will get 6% for the first year, which will be compounded at the end of the first year. Again for the second year, he will get 8%. So,
A = P (1 + \(\frac { r1 }{ 100 } \)) (1 + \(\frac { r2 }{ 100 } \))
A = $40,000Â (1 + \(\frac { 6 }{ 100 } \)) (1 + \(\frac { 8 }{ 100 } \))
A = $40,000Â (\(\frac { 106 }{ 100 } \)) (\(\frac { 108 }{ 100 } \))
A = $45792
Therefore, at the end of 2 years, he will receive $ 45792.
(ii) The man will receive an interest of 6% in the first year, 8% in the second year, and 10% in the third year.
A = P (1 + \(\frac { r1 }{ 100 } \)) (1 + \(\frac { r2 }{ 100 } \)) (1 + \(\frac { r3 }{ 100 } \))
A = 20000 (1 + \(\frac { 6 }{ 100 } \)) (1 + \(\frac { 8 }{ 100 } \)) (1 + \(\frac { 10 }{ 100 } \))
A = $20000 (\(\frac { 106 }{ 100 } \)) (\(\frac { 108 }{ 100 } \)) (\(\frac { 110 }{ 100 } \))
A = 25185.6
Therefore, he gains = Final amount – Initial principal = 25185.6 – 20000 = $5185.6
FAQs on Compound Interest Variable Rate
1. Does compound interest have a variable rate?
Yes, we will have a variable rate for compound interest.
2. How do you calculate the compound interest variable interest rate?
The formula to calculate successive rate of compound interest is A = P (1 + \(\frac { r1 }{ 100 } \)) (1 + \(\frac { r2 }{ 100 } \)) (1 + \(\frac { r3 }{ 100 } \)) (1 + \(\frac { r4 }{ 100 } \))……
where A is the amount, P is the Principal, and r1, r2, r3, r4, … are the for successive years.
3. What is a variable interest rate?
A variable rate is an amount charged to a borrower for a variable-rate loan, such as a mortgage.
4. What is the formula to find the compound interest?
The formula to find the compound interest is compound interest = Final amount – Initial principal.
Summary
Compound Interest Successive Rate definition, examples, formula, and faqs will let you understand the variable rate of compound interest. So, read every part of this article to learn it within less time.