Variable Rate of Compound Interest

We will discuss here how to use the formula for variable rate of compound interest.


When the rate of compound interests for successive/consecutive years are different (r \(_{1}\)%, r \(_{2}\)%, r \(_{3}\)%, r \(_{4}\)%, .................. ) then:

A = P( 1 + \(\frac{r_{1}}{100}\))(1 + \(\frac{r_{2}}{100}\))(1 + \(\frac{r_{3}}{100}\)) .............

Where,

A = amount;

P = principal;

r \(_{1}\), r \(_{2}\), r \(_{3}\), r \(_{4}\).......... = rates for successive years.

Word problems on variable rate of compound interest:

1. If the rate of compound interest for the first, second and third year be 8%, 10% and 15% respectively, find the amount and the compound interest on $ 12,000 in 3 years.

Solution:

The man will receive an interest of 8% in the first year, 10% in the second year and 15% in the third year.

Therefore,

Amount = P( 1 + \(\frac{r_{1}}{100}\))(1 + \(\frac{r_{2}}{100}\))(1 + \(\frac{r_{3}}{100}\))

⟹ A = $ 12,000(1 + \(\frac{8}{100}\))(1 + \(\frac{10}{100}\))(1 + \(\frac{15}{100}\))

⟹ A = $ 12,000 (1 + 8/100)(1 + 10/100)(1 + 15/100)

⟹ A = $ 12,000 × 267/25 × 11/10 × 23/20

⟹ A = $ 12,000 × \(\frac{6831}{5000}\)

⟹ A = $ 16,394.40

Therefore, the required amount = $ 16,394.40

Therefore, the compound interest = Final amount - Initial principal

                                              = $ 16,394.40 - $ 12,000

                                              = $ 4,394.40

 

2. Find the compound interest accrued by Aaron from a bank on $ 16000 in 3 years, when the rates of interest for successive years are 10%, 12% and 15% respectively.

Solution:

For the first year:

Principal = $ 16,000;

Rate of interest = 10% and

Time = 1 years.

Therefore, interest for the first year = \(\frac{P × R × T}{100}\)

                                                 = $ \(\frac{16000 × 10 × 1}{100}\)

                                                 = $ \(\frac{160000}{100}\)

                                                 = $ 1,600

Therefore, the amount after 1 year = Principal + Interest

                                                = $16,000 + $ 1,600

                                                = $ 17,600

For the second year, the new principal is $ 17,600

Rate of interest = 12% and

Time = 1 years.

 

Therefore, the interest for the second year = \(\frac{P × R × T}{100}\)

                                                           = $ \(\frac{17600 × 12 × 1}{100}\)

                                                           = $ \(\frac{211200}{100}\)

                                                           = $ 2,112

Therefore, the amount after 2 year = Principal + Interest

                                                = $ 17,600 + $ 2,112

                                                = $ 19,712

For the third year, the new principal is $ 19,712

Rate of interest = 15% and

Time = 1 years.

Therefore, the interest for the third year = \(\frac{P × R × T}{100}\)

                                                       = $ \(\frac{19712 × 15 × 1}{100}\)

                                                       = $ \(\frac{295680}{100}\)

                                                       = $ 2,956.80

Therefore, the amount after 3 year = Principal + Interest

                                                = $ 19,712 + $ 2,956.80

                                                = $ 22,668.80

Therefore, the compound interest accrued = Final amount - Initial principal

                                                         = $ 22,668.80 - $ 16,000

                                                         = $ 6,668.80

 

 

3. A company offers the following growing rates of compound interest annually to the investors on successive years of investment.

4%, 5% and 6%

(i) A man invests $ 31,250 for 2 years. What amount will he receive after 2 years?

(ii) A man invests $ 25,000 for 3 years. What will be his gain?

Solution:

The man will get 4% for the first year, which will be compounded at the end of the first year. Again for the second year, he will get 5%. So,

A = P( 1 + \(\frac{r_{1}}{100}\))(1 + \(\frac{r_{2}}{100}\))

⟹ A = $ 31250(1 + \(\frac{4}{100}\))(1 + \(\frac{5}{100}\))

⟹ A = $ 31250 × 26/25 × 21/20

⟹ A = $ 34,125

Therefore, at the end of 2 years he will receive $ 34125.

(ii) The man will receive an interest of 4% in the first year, 5% in the second year and 6% in the third year.

Therefore,

Amount = P( 1 + \(\frac{r_{1}}{100}\))(1 + \(\frac{r_{2}}{100}\))(1 + \(\frac{r_{3}}{100}\))

⟹ A = $ 25000(1 + \(\frac{4}{100}\))(1 + \(\frac{5}{100}\))(1 + \(\frac{6}{100}\))

⟹ A = $ 25000 × 26/25 × 21/20 × 53/50

⟹ A = $ 28,938

Therefore, he gain = Final amount - Initial principal

                         = $ 28,938 - $ 25000

                         = $ 3,938

Compound Interest

Compound Interest

Compound Interest with Growing Principal

Compound Interest with Periodic Deductions

Compound Interest by Using Formula

Problems on Compound Interest

Practice Test on Compound Interest


Compound Interest - Worksheet

Worksheet on Compound Interest

Worksheet on Compound Interest with Growing Principal

Worksheet on Compound Interest with Periodic Deductions




8th Grade Math Practice 

From Variable Rate of Compound Interest to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Types of Fractions |Proper Fraction |Improper Fraction |Mixed Fraction

    Mar 02, 24 05:31 PM

    Fractions
    The three types of fractions are : Proper fraction, Improper fraction, Mixed fraction, Proper fraction: Fractions whose numerators are less than the denominators are called proper fractions. (Numerato…

    Read More

  2. Subtraction of Fractions having the Same Denominator | Like Fractions

    Mar 02, 24 04:36 PM

    Subtraction of Fractions having the Same Denominator
    To find the difference between like fractions we subtract the smaller numerator from the greater numerator. In subtraction of fractions having the same denominator, we just need to subtract the numera…

    Read More

  3. Addition of Like Fractions | Examples | Worksheet | Answer | Fractions

    Mar 02, 24 03:32 PM

    Adding Like Fractions
    To add two or more like fractions we simplify add their numerators. The denominator remains same. Thus, to add the fractions with the same denominator, we simply add their numerators and write the com…

    Read More

  4. Comparison of Unlike Fractions | Compare Unlike Fractions | Examples

    Mar 01, 24 01:42 PM

    Comparison of Unlike Fractions
    In comparison of unlike fractions, we change the unlike fractions to like fractions and then compare. To compare two fractions with different numerators and different denominators, we multiply by a nu…

    Read More

  5. Equivalent Fractions | Fractions |Reduced to the Lowest Term |Examples

    Feb 29, 24 05:12 PM

    Equivalent Fractions
    The fractions having the same value are called equivalent fractions. Their numerator and denominator can be different but, they represent the same part of a whole. We can see the shade portion with re…

    Read More