Question

$\sf question -:$

Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find. (i) The amount credited against her name at the end of the second year

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Answer 1

⚘Given :-

• ➸ principal = 8000
• ➸ rate = 5%
• ➸ time = 2 years

⚘To find :-

• ➸ the amount after 2 years

⚘Solution :-

$\mapsto \sf\boxed{\bold{\pink{a = p(1 + \frac{r}{100}^{n}}}}$

where,

• ➸ a = amount
• ➸ p = principal = 8000
• ➸ r = rate = 5%
• ➸ n = 2 years

now by putting values :-

$\sf :⟼ 8000 \bigg(1 + \frac{5}{100} \bigg)^{2}$

$\sf :⟼8000 \bigg(1 + \frac{1}{20} \bigg)^{2}$

$\sf :⟼8000 \bigg( \frac{20 + 1}{20} \bigg)^{2}$

$\sf :⟼8000 \bigg( \frac{21}{20} \bigg)^{2}$

$\sf :⟼8000 \times \bigg( \frac{21 \times 21}{20 \times 20} \bigg)$

$\sf :⟼20 \times 21 \times 21$

$\sf :⟼20 \times 441$

$\sf :⟼amount = 8820$

so therefore amount after 2 years is rs 8820.

⚘ more info :-

• the interest is calculated on the amount of the previous year this is known as interest compounded or compound interest .
• compound interest is the interest calculated on the previous year's amount = (a = p + t).

Answer 2

Answer:

Gɪᴠᴇɴ :

• ➛ Principle = Rs.8000
• ➛ Time = 2 years
• ➛ Rate of Interest = 5% per annum

$\begin{gathered}\end{gathered}$

Tᴏ Fɪɴᴅ :

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

$\begin{gathered}\end{gathered}$

Cᴏᴄᴇᴘᴛ :

↝ Here the concept of Amount has been used. We have given that the Principal is Rs.8000, Time is 2 years and rate is 5 p.c.p.a . We have need to find the Amount.

↝ So, we will find out the Amount by substituting the values in the formula.

$\begin{gathered}\end{gathered}$

Usɪɴɢ Fᴏʀᴍᴜʟᴀ :

$\longrightarrow{\footnotesize{\underline{\boxed{\pmb{\sf{Amount={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}}$

Where :-

• » A = Amount
• » P = Principle
• » R = Rate
• » T = Time

$\begin{gathered}\end{gathered}$

Sᴏʟᴜᴛɪᴏɴ :

Finding the Amount by substituting the values in the formula :-

${\dashrightarrow{\small{\sf{Amount={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}$

${\dashrightarrow{\small{\sf{Amount={8000{\bigg(1 + \dfrac{5}{100}{\bigg)}^{2}}}}}}}$

${\dashrightarrow{\small{\sf{Amount={8000{\bigg(\dfrac{(1 \times 100) + (5 \times 1)}{100}{\bigg)}^{2}}}}}}}$

${\dashrightarrow{\small{\sf{Amount={8000{\bigg(\dfrac{100 + 5}{100}{\bigg)}^{2}}}}}}}$

${\dashrightarrow{\small{\sf{Amount={8000{\bigg(\dfrac{105}{100}{\bigg)}^{2}}}}}}}$

${\dashrightarrow{\small{\sf{Amount={8000{\bigg( \cancel{\dfrac{105}{100}}{\bigg)}^{2}}}}}}}$

${\dashrightarrow{\small{\sf{Amount={8000{\bigg({\dfrac{21}{20}}{\bigg)}^{2}}}}}}}$

${\dashrightarrow{\small{\sf{Amount={8000{\bigg({\dfrac{21}{20} \times \dfrac{21}{20}}{\bigg)}}}}}}}$

${\dashrightarrow{\small{\sf{Amount={8000{\bigg(\dfrac{441}{400} \bigg)}}}}}}$

${\dashrightarrow{\small{\sf{Amount={8000 \times \dfrac{441}{400}}}}}}$

${\dashrightarrow{\small{\sf{Amount={ \cancel{8000} \times \dfrac{441}{\cancel{400}}}}}}}$

${\dashrightarrow{\small{\sf{Amount={ 20 \times 441}}}}}$

${\dashrightarrow{\small{\sf{Amount={Rs.8820}}}}}$

${\longrightarrow{\small{\underline{\boxed{\pmb{\sf{Amount={Rs.8820}}}}}}}}$

∴ The amount credited against Maria name at the end of the second year is Rs.8820.

$\begin{gathered}\end{gathered}$

Lᴇᴀʀɴ Mᴏʀᴇ :

$\dashrightarrow{\small{\underline{\boxed{\sf{\purple{ Simple \: Interest = \dfrac{P \times R \times T}{100}}}}}}}$

$\dashrightarrow\small{\underline{\boxed{\sf{\purple{Amount={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}$

$\dashrightarrow\small{\underline{\boxed{\sf{\purple{Amount = Principle + Interest}}}}}$

$\dashrightarrow\small{\underline{\boxed{\sf{\purple{ Principle=Amount - Interest }}}}}$

$\dashrightarrow\small{\underline{\boxed{\sf{\purple{Principle = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}}}}}}$

$\dashrightarrow\small{\underline{\boxed{\sf{\purple{Principle = \dfrac{Interest \times 100 }{Time \times Rate}}}}}}$

${\underline{\overline{\rule{200pt}{2pt}}}}$