# 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

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