Question

yearly.
7. Maria invested 8,000 in a business. She would be paid interest at 5% per annum
compounded annually. Find
(1) The amount credited against her name at the end of the second year.
(1) The interest for the 3rd year.​

Answer 1

${\pmb{\underline{\sf{ Required \ Solution ... }}}}$

• Principal = ₹8000
• Rate (r) = 5%

~ The amount credited against her name at the end of the second year.

• Time (t) = 2 years

$\\ \circ \ {\pmb{\underline{\boxed{\sf{ Amount = P \left( 1 + \dfrac{r}{100} \right)^n }}}}} \\ \\ \\ \colon\implies{\sf{ 8000 \left( 1 + \dfrac{5}{100} \right)^2 }} \\ \\ \\ \colon\implies{\sf{ 8000 \left( \cancel{ \dfrac{105}{100} } \right)^2 }} \\ \\ \\ \colon\implies{\sf{ 8000 \left( \dfrac{21}{20} \right)^2 }} \\ \\ \\ \colon\implies{\sf{ \cancel{8000} \times \dfrac{21}{ \cancel{20} } \times \dfrac{21}{ \cancel{20} } }} \\ \\ \\ \colon\implies{\sf{ 20 \times 21 \times 21 }} \\ \\ \\ \colon\implies{\sf{Rs. \ 8820 _{(Amount)} }}$

So, At the end of 2nd year, She has to pay ₹ 8820 as Amount.

~ Now, This time we've to find the Interest for the ${\pmb{\sf{3^{rd} }}}$ year as We know that :-

$\colon\implies{\boxed{\sf{ P \left( 1 + \dfrac{r}{100} \right)^n }}}$

We can use ${\pmb{\sf{2^{nd} }}}$ year Amount as the Principal for ${\pmb{\sf{3^{rd} }}}$ year.

$\colon\implies{\sf{ 8820 \left( 1 + \dfrac{5}{100} \right)^1 }} \\ \\ \\ \colon\implies{\sf{ 8820 \left( \cancel{ \dfrac{105}{100} } \right) }} \\ \\ \\ \colon\implies{\sf{ \cancel{8820} \times \dfrac{21}{ \cancel{20} } }} \\ \\ \\ \colon\implies{\sf{ 441 \times 21 }} \\ \\ \\ \colon\implies{\underline{\boxed{\sf{ Rs. \ 9261}}}}$

So, Amount For the ${\pmb{\sf{3^{rd} }}}$ year is ₹9261 .

We also know that:-

$\colon\implies{\sf{ Interest = Amount - Principal }} \\ \\ \colon\implies{\sf{ 9261 - 8820 }} \\ \\ \colon\implies{\underline{\boxed{\sf{ 441 }}}}$

Hence, Interest for the ${\pmb{\sf{3^{rd} }}}$ year is ₹441 .

Answer 2

$\boxed {\mathrm { \: Required \: answer : \: }}$

In the question ,

P = Rs.8000

R = 5% per annum

N = 2 years

$\therefore \: \sf \: Amount \: after \: 2 \: years \: = P(1 + \frac{r}{100n})$

Amount after 2 years =

$\sf \: Rs \: [8000 \times (1 + \frac{5}{100} ) {}^{2} ]$

Amount after 2 years =

$\sf \: Rs \: (8000 \times \frac{105}{100} \times \frac{105}{100} ) = Rs8820$

Now, the principal for the third year will be the amount at the end of the second year.

$\sf Intrest = \frac{P \: × R \: × T \: }{100}$

$\sf Intrest = \frac{8820 \times 5 \times 1}{100} = Rs \: 441$

Hence , the answer is 441