More solved problems on compound interest using formula are shown below.

**Solution:**

Given, SI = $ 6750, R = \(\frac{20}{3}\)% p.a. and T = 3 years.

sum = 100 × SI / R × T

= $ (100 × 6750 × ³/₂₀ × 1/3 ) = $ 33750.

Now, P = $ 33750, R = \(\frac{20}{3}\)% p.a. and T = 3 years.

Therefore, amount after 3 years

= $ {33750 × (1 + (20/3 × 100)}³ **[using A = P (1 + R/100)ᵀ]**

= $ (33750 × 16/15 × 16/15 × 16/15) = $ 40960.

Thus, amount = $ 40960.

Hence, compound interest = $ (40960 - 33750) = $ 7210.

**Solution:**

Let the sum be $ 100. Then,

SI = $ (100 × 6 × 2/100) = $ 12

and compound interest = $ {100 × (1 + 6/100)² - 100}

= $ {(100 × 53/50 × 53/50) - 100} = $ (2809/25 - 100) = $ 309/25

Therefore, (CI) - (SI) = $ (309/25 – 100) = $ 9/25

If the difference between the CI and SI is $ 9/25, then the sum = $ 100.

If the difference between the CI and SI is $ 18, then the sum = $ (100 × 25/9 × 18 )

= $ 5000.

Hence, the required sum is $ 5000.

**Alternative method**

Let the sum be $ P.

Then, SI = $ (P × 6/100 × 2) = $ 3P/25

And, CI = $ {P × (1 + 6/100)² - P}

= $ {(P × 53/50 × 53/50) - P} = $ (\(\frac{2809}{2500}\)P - P) = $ (309P/2500)

(CI) - (SI) = $ (309P/2500 – 3P/25) = $ (9P/2500)

Therefore, 9P/2500 = 18

⇔ P = 2500 × 18/9

⇔ P = 5000.

Hence, the required sum is $ 5000.

**Solution:**

Let the sum be $ 100. Then,

amount = $ {100 × (1 + 8/100)²}

= $ (100 × 27/25 × 27/25) = $ (2916/25)

If the amount is $ 2916/25 then the sum = $ 100.

If the amount is $ 72900 then the sum = $ (100 × 25/2916 × 72900) = $ 62500.

Hence, the required sum is $ 62500.

**Alternative method**

Let the sum be $ P. Then,

amount = $ {P × (1 + 8/100)²}

= $ {P × 27/25 × 27/25} = $ (729P/625)

Therefore, 729P/625 = 72900

⇔ P = (72900 × 625)/729

⇔ P = 62500.

Hence, the required sum is $ 62500.

**Solution:**

Let the required rate be R% per annum.

Here, A = $ 2205, P = $ 2000 and n = 2 years.

**Using the formula A = P(1 + R/100)ⁿ,**

2205 = 2000 × ( 1 + R/100)²

⇒ (1 + R/100)² = 2205/2000 = 441/400 = (21/20)²

⇒ ( 1 + R/100) = 21/20

⇒ R/100 = (21/20 – 1) = 1/20

⇒ R = (100 × 1/20) = 5

Hence, the required rate of interest is 5% per annum.

**Solution:**

Let the required time be n years. Then,

amount = $ {1000 × (1 + 10/100)ⁿ}

= $ {1000 × (11/10)ⁿ}

Therefore, 1000 × (11/10)ⁿ = 1331 **[since, amount = $ 1331 (given)] **

⇒ (11/10)ⁿ = 1331/1000 = 11 × 11 × 11/ 10 × 10 × 10 = (11/10)³

⇒ (11/10)ⁿ = (11/10)³

⇒ n = 3.

Thus, n = 3.

Hence, the required time is 3 years.

**● Compound Interest**

**Compound Interest with Growing Principal**

**Compound Interest with Periodic Deductions**

**Compound Interest by Using Formula**

**Compound Interest when Interest is Compounded Yearly**

**Compound Interest when Interest is Compounded Half-Yearly**

**Compound Interest when Interest is Compounded Quarterly**

**Variable Rate of Compound Interest**

**Difference of Compound Interest and Simple Interest**

**Practice Test on Compound Interest**

**Uniform Rate of Growth and Depreciation**

**● Compound Interest - Worksheet**

**Worksheet on Compound Interest**

**Worksheet on Compound Interest when Interest is Compounded Half-Yearly**

**Worksheet on Compound Interest with Growing Principal**

**Worksheet on Compound Interest with Periodic Deductions**

**Worksheet on Variable Rate of Compound Interest**

**Worksheet on Difference of Compound Interest and Simple Interest**

**Worksheet on Uniform Rate of Growth**

**Worksheet on Uniform Rate of Depreciation**

**Worksheet on Uniform Rate of Growth and Depreciation**

**8th Grade Math Practice** **From Problems on Compound Interest to HOME PAGE**

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