Question

Shekhar deposited ₹ 20000 in a bank for 1 year,compounded half yearly at 14%per annum .Find the compound interest he gets​

Answer 1

Answer:

Rs 2898

Step-by-step explanation:

Pincipal = Rs 20000

Time = 1 years

Rate = 14%

It is given Interest is Compounded Half Yearly :

$\textbf{\underline{\underline{According\:To\:Question :}}}$

Rate = $\dfrac{14}{2}$ % = 7%

Time = 1 × 2 years = 2 years

COMPOUND INTEREST :

$\sf \: Amount \: = P \: ( 1 + \dfrac{r}{100} ) ^{n}$

$\sf \: Amount \: = 20000(1 + \dfrac{7}{100} ) ^{2}$

$\sf \: Amount = 20000( \dfrac{107}{100} ) ^{2}$

$\sf \: Amount = 20000 \times \dfrac{107}{100} \times \dfrac{107}{100}$

$\sf \: Amount = 22898$

C.I = Amount - Principal

C.I = Rs 22898 - Rs 20000

C.I = Rs 2898

Answer 2

Question :

⠀

Shekhar deposited ₹ 20000 in a bank for 1 year,compounded half yearly at 14%per annum .Find the compound interest he gets.

⠀

AnswEr :

⠀

As Amount is Compounded Half Yearly. So We Have to Reduced Rate by Half and Increase Time by Twice.

⠀

• $\bf{Principal = Rs. 20,000}$

• $\bf{Rate= \cancel \dfrac{14}{2}=7\% }$

• $\bf{Time = 1 \times 2 = 2 Years}$

⠀

⇒ $\mathsf{CI = P \bigg[ \bigg(1 + \dfrac{r}{100} \bigg) ^{t} - 1 \bigg]}$

⇒ $\mathsf{CI = 20000 \bigg[ \bigg(1 + \dfrac{7}{100} \bigg) ^{2} - 1 \bigg]}$

⇒ $\mathsf{CI = 20000 \bigg[ \bigg(\dfrac{100 + 7}{100} \bigg) ^{2} - 1 \bigg]}$

⇒ $\mathsf{CI = 20000 \bigg[ \bigg(\dfrac{107}{100} \bigg) ^{2} - 1 \bigg]}$

⇒ $\mathsf{CI = 20000 \bigg(\dfrac{11449}{10000} - 1 \bigg)}$

⇒ $\mathsf{CI = 20000 \bigg(\dfrac{11449 - 10000}{10000} \bigg)}$

⇒ $\mathsf{CI = 2\cancel{0000} \times \dfrac{1449}{1 \cancel{0000}} }$

⇒ $\mathsf{CI = 2 \times 1449}$

⇒ $\mathsf{CI = Rs.2898}$

⠀

$\therefore$ Rs.2898 is the Compound Interest.