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 with Growing Principal**

**Compound Interest with Periodic Deductions**

**Compound Interest by Using Formula**

**Practice Test on Compound Interest**

**● Compound Interest - Worksheet**

**Worksheet on Compound Interest**

**Worksheet on Compound Interest with Growing Principal**

**8th Grade Math Practice****From Variable Rate of Compound Interest to HOME PAGE**

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