Question

A loan of ₹15000 was taken on compound interest. If the rate of compound interest is 12 p. c. p.a. find the amount of settle the loan after 3years​

Answer 1

AnswEr :

$\bf{ Given}\begin{cases}\sf{Principal = Rs. \:15000}\\\sf{Rate = 12 \: \footnotesize\text{percent compounded per annum}}\\ \sf{Time = 3 \:Years}\\\sf{Amount = ?} \end{cases}$

• Let's Head to the Question Now :

$\longrightarrow \sf{Amount = P \times \bigg(1 +\dfrac{r}{100} \bigg)^{t}} \\ \\\longrightarrow \sf{Amount = 15000 \times \bigg(1 + \cancel\dfrac{12}{100} \bigg)^{3}} \\ \\\longrightarrow \sf{Amount = 15000 \times \bigg(1 + \dfrac{3}{25} \bigg)^{3}} \\ \\\longrightarrow \sf{Amount = 15000 \times \bigg(\dfrac{25 + 3}{25} \bigg)^{3}} \\ \\\longrightarrow \sf{Amount = 15000 \times \bigg(\dfrac{28}{25} \bigg)^{3}} \\ \\\longrightarrow \sf{Amount = \cancel{15000} \times \dfrac{784}{\cancel{625}} \times \dfrac{28}{25}} \\ \\\longrightarrow \sf{Amount = 24 \times784 \times \dfrac{28}{25}} \\ \\\longrightarrow \sf{Amount = \dfrac{526848}{25}}$

$\begin{array}{r|l} &\sf 21073.92 \\\cline{1-2} 25& 5\: 2\: 6\:8\:4\:8\\ &5\:0\: \\ \cline{2-2} & \quad2\:6\\&\quad2\:5\\\cline{2-2} &\quad\:\:\:1\:8\:4\\&\quad\:\:\:1\:7\:5\\\cline{2-2}&\quad \quad\quad\:9\:8\\&\qquad\quad\:7\:5\\\cline{2-2}&\qquad\quad\:2\:3\:0\\&\qquad\quad\:2\:2\:5\\\cline{2-2}&\qquad\qquad\:\:\:5\:0\\&\qquad\qquad\:\:\:5\:0\\\cline{2-2}&\qquad\qquad\:\:\boxed{0\:0}\end{array}$

$\longrightarrow\large\boxed{\sf{Amount =Rs. \: 21073.92}}$

⠀

∴ Rs. 21073.92 is amount to settle loan.

#answerwithquality #BAL

Answer 2

$\begin{lgathered} \red{\bf{Given}}\begin{cases} \sf \: principal = 15000 \\ \sf \: rate = 12\% \\ \sf \: time = 3 \: years\\ \green{\bf amount =?} \: \end{cases}\end{lgathered}$

$\Large\bold\star\underline{\underline\textbf{Solution(1)}}$

$\blue{ \boxed{\sf \: A = p(1 + \dfrac{rate}{100} )^{time}}}$

Putting values we get,,

$\red\leadsto \sf A \: = 15000(1 + \dfrac{12}{100} )^{3} \\ \\ \red\leadsto \sf A \: = 15000 \: \times \frac{28}{25} \times \frac{28}{25} \times \frac{28}{25} \\ \\ \red\leadsto \red{\boxed{\sf \green{A} \: \blue{=} \pink{rs.} \orange{21073.92}}}$

Hence, The amount that settle the loan is Rs.21073.92.

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$\Large\underline\mathfrak{Solution(2)}$

Lets Try it with Ratio method Now, ,

$\bf \: Rate = 12\% = \dfrac{12}{100} = \blue{\dfrac{3}{25}}$

Now, Let us assume that, Principal is Rs.25, Interest is Rs.3 .

→ Than our Amount will be = P + interest = Rs.28 ( in one yr)

$\green{\mathfrak Hence,}$

$25 - - - - - - - 28 \\ \\ {25}^{3} - - - - - - {28}^{3} \\ \\ \red{15625(p) - - - - - 21952(a)}$

So, we can say that,

when Principal is Rs.15625, at 12% our amount will be Rs.21952 .

so,

when Principal is Rs.15000 , at 12% , our amount will be

$\red{\boxed\implies} \: \bf A = \dfrac{21952 \times 15000}{15625} \\ \\ \red{\boxed\implies} \: \boxed{ \bf \red{A} \blue{=} \green{Rs}. \orange{21073.92}}$

Hence, amount settle the loan will be Rs.21073.92 .

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$\rule{200}{4}$

$\Large\bold\star\underline{\underline\textbf{Extra\:Brainly\:Knowledge}}$

For a principal amount of P at the interest rate of r per annum in t years if the interest is I then ...

1) For Simple Interest

$\bf\rightarrow \boxed{I=\frac{P\times r\times t}{100}}$

2)For Compound Interest

$\bf\rightarrow \boxed{I=P(1+\frac{r}{100}){}^{nt}}$

where, n=interest period

Ex. If the principal is compounded

annually(per year) then ,

n=1

If the principal is compounded

half yearly(in every 6 months) then ,

n=2

If the principal is compounded

quarterly (in every 3 months) then ,

n=4....

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#BAL