yearly.7. Maria invested 8,000 in a business. She would be paid interest at 5% per annumcompounded annually. Find(1) The amount credited against her name at the end of the second year.(1) The interest for the 3rd year. About the author Cora
[tex] {\pmb{\underline{\sf{ Required \ Solution … }}}} \\ [/tex] Principal = ₹8000 Rate (r) = 5% ~ The amount credited against her name at the end of the second year. Time (t) = 2 years [tex] \\ \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)} }} [/tex] 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 [tex] {\pmb{\sf{3^{rd} }}}[/tex] year as We know that :- [tex] \colon\implies{\boxed{\sf{ P \left( 1 + \dfrac{r}{100} \right)^n }}} \\ [/tex] We can use [tex] {\pmb{\sf{2^{nd} }}}[/tex] year Amount as the Principal for [tex] {\pmb{\sf{3^{rd} }}}[/tex] year. [tex] \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}}}} \\ [/tex] So, Amount For the [tex] {\pmb{\sf{3^{rd} }}}[/tex] year is ₹9261 . We also know that:- [tex] \colon\implies{\sf{ Interest = Amount – Principal }} \\ \\ \colon\implies{\sf{ 9261 – 8820 }} \\ \\ \colon\implies{\underline{\boxed{\sf{ 441 }}}} [/tex] Hence, Interest for the [tex] {\pmb{\sf{3^{rd} }}}[/tex] year is ₹441 . Reply
[tex]\boxed {\mathrm { \: Required \: answer : \: }} [/tex] In the question , P = Rs.8000 R = 5% per annum N = 2 years [tex]\therefore \: \sf \: Amount \: after \: 2 \: years \: = P(1 + \frac{r}{100n})[/tex] Amount after 2 years = [tex]\sf \: Rs \: [8000 \times (1 + \frac{5}{100} ) {}^{2} ] [/tex] Amount after 2 years = [tex]\sf \: Rs \: (8000 \times \frac{105}{100} \times \frac{105}{100} ) = Rs8820[/tex] Now, the principal for the third year will be the amount at the end of the second year. [tex]\sf Intrest = \frac{P \: × R \: × T \: }{100} [/tex] [tex]\sf Intrest = \frac{8820 \times 5 \times 1}{100} = Rs \: 441[/tex] Hence , the answer is 441 Reply
[tex] {\pmb{\underline{\sf{ Required \ Solution … }}}} \\ [/tex]
~ The amount credited against her name at the end of the second year.
[tex] \\ \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)} }} [/tex]
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 [tex] {\pmb{\sf{3^{rd} }}}[/tex] year as We know that :-
[tex] \colon\implies{\boxed{\sf{ P \left( 1 + \dfrac{r}{100} \right)^n }}} \\ [/tex]
We can use [tex] {\pmb{\sf{2^{nd} }}}[/tex] year Amount as the Principal for [tex] {\pmb{\sf{3^{rd} }}}[/tex] year.
[tex] \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}}}} \\ [/tex]
So, Amount For the [tex] {\pmb{\sf{3^{rd} }}}[/tex] year is ₹9261 .
We also know that:-
[tex] \colon\implies{\sf{ Interest = Amount – Principal }} \\ \\ \colon\implies{\sf{ 9261 – 8820 }} \\ \\ \colon\implies{\underline{\boxed{\sf{ 441 }}}} [/tex]
Hence, Interest for the [tex] {\pmb{\sf{3^{rd} }}}[/tex] year is ₹441 .
[tex]\boxed {\mathrm { \: Required \: answer : \: }} [/tex]
In the question ,
P = Rs.8000
R = 5% per annum
N = 2 years
[tex]\therefore \: \sf \: Amount \: after \: 2 \: years \: = P(1 + \frac{r}{100n})[/tex]
Amount after 2 years =
[tex]\sf \: Rs \: [8000 \times (1 + \frac{5}{100} ) {}^{2} ] [/tex]
Amount after 2 years =
[tex]\sf \: Rs \: (8000 \times \frac{105}{100} \times \frac{105}{100} ) = Rs8820[/tex]
Now, the principal for the third year will be the amount at the end of the second year.
[tex]\sf Intrest = \frac{P \: × R \: × T \: }{100} [/tex]
[tex]\sf Intrest = \frac{8820 \times 5 \times 1}{100} = Rs \: 441[/tex]
Hence , the answer is 441