Question

Kindly Answer the questions in the attachment without spamming :)​

Answer 1

$\Large \rm \tealB{\mp\widetilde{Answer \ 1}:-}$

Given:

• P = ₹10500
• R = 5%
• Time = 1 year

Use the relation,

$\ddag \red{\boxed{\boxed{A = P\left( 1 +\dfrac{R}{100}\right)^n}}}$

$\rm :\longmapsto A = 10600 \times \left(1 + \dfrac{\cancel{5}}{\cancel{100}_{20}}\right)^1$

$\rm :\longmapsto A = 10500 \left(1 +\dfrac{1}{20}\right)$

$\rm :\longmapsto A = 10500 \left( \dfrac{20 + 1}{20}\right)$

$\rm :\longmapsto A = 10500 \times \dfrac{21}{20}$

$\star\red{\boxed{\rm :\longmapsto A = 11025}}$

$\Large \rm \tealB{\mp\widetilde{Answer\ 2}:-}$

As the interested is compounded in 3 months, for a year )12 months) it will be compounded $\red{\rm 4 \ times.}$

Answer 2

Solution 1:-

Given:-

• Principal = ₹10,500
• Rate = 5%
• Time = 1 year

Formula:-

${ \star{ \underline{\boxed{\tt \pmb {A= P( 1 +{ \frac{R}{100}) }^{n}}}}}}$

Here,

• P stands for Principal
• R stands for Rate
• n stands for time

Putting the values:-

$\tt \hookrightarrow{A = 10500(1 + {\frac{5}{100})}^{1} }$

$\tt \hookrightarrow{A = 10500(1 + {\frac{\cancel5}{\cancel{100}_{20}})}^{1} }$

$\tt \hookrightarrow{A = 10500( \frac {20 + 1}{20})}$

$\tt \hookrightarrow{A = 10500( \frac {21}{20})}$

${ \star{ \underline{\boxed{ \sf \pmb {A= ₹ \: 11025}}}}}$

Solution 2:-

Given that the interest is compounded after every 3 months in a year.

[∵ 1 year = 12 months]

The interest will be compounded 4 times.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬