Question

Find the compound interest on 4000 for 2*1/2 years at 10% p.a.
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Answer 1

Given :

• Principal (P) = Rs. 4000
• Time (n) = 2 ½ = 5/2 = 2.5 years
• Rate (R) = 10% per annum compunded annually

To Find :

• Compound Interest (CI) = ?

Formula used :

• Amount (A) = P(1 + R/100)ⁿ
• Simple Interest = PRT ÷ 100
• CI = A - P

Solution :

Calculating the Amount for 2 years :

→ A = P(1 + R/100)ⁿ

→ A = 4000(1 + 10/100)²

→ A = 4000(1 + 1/10)²

→ A = 4000(11/10)²

→ A = 4000(1.1)²

→ A = 4000 × 1.21

→ A = Rs. 4840

Therefore,Amount for 2 years is Rs.4840.

Calculating the Simple interest on Rs.4840 for last 0.5 years :

➻ Simple Interest = PRT ÷ 100

➻ Simple Interest = 4840 × 10 × 0.5 ÷ 100

➻ Simple Interest = 48400 × 0.5 ÷ 100

➻ Simple Interest = 24200 ÷ 100

➻ Simple Interest = Rs. 242

Calculating Amount after 2.5 years :

➺ Amount after 2.5 years = Amount for 2 years + simple interest

➺ Amount after 2.5 years = 4840 + 242

➺ Amount after 2.5 years = Rs.5082

Now, let's calculate the Compound Interest :

⋙ CI = A - P

⋙ CI = Rs. 5082 - Rs. 4840

⋙ CI = Rs. 1082

• Therefore,the required Compound Interest is Ra. 1082.

Answer 2

To Find:-

• Find the compound interest.

Solution:-

P = 4000

T = 2.5 years

R = 10%

A = P( 1 +  r/100 )^n

4000 (11/10)^2

4000 * 121/100

40*121

4840

Now,

We have to find the simple interest for last 1/2 years:-

(4840*10)/100  * 1/2

48400/100 *1/2

484/2

242

Amount after 2.5 year = 4840 + 242

=5082

Compound interest = 5082 − 4000

CI = 1082