Question

find compound anually
P=8000
R=12
N(years)=2​

Answer 1

Given :

$\diamond \rm \: Principal =Rs \: 8000$

$\diamond \rm \: Rate = 12\%$

$\diamond \rm \: Time = 2 \: years$

To Find :

$\diamond \rm \: Amount \: from \: the \: given \: data$

Solution :

$\star \tt \: C.I = P[ (1 + \dfrac{r}{100} ) ^{t} - 1]$

Here,

P = Principal

P = PrincipalR = Rate

P = PrincipalR = RateT = Time

Putting Values :

$\hookrightarrow \sf \: 8000\Big[\Big(1 + \cancel \dfrac{12}{100} \Big)^{2} - 1\Big]$

$\hookrightarrow \sf \: 8000\Big[\Big(1 \dfrac{3}{25} \Big) ^{2} - 1\Big]$

$\hookrightarrow \sf \: 8000\Big[\Big( \dfrac{25 + 3}{25}\Big) ^{2}\: -\:1\Big]$

$\hookrightarrow \sf\:8000 \Big[\Big(\dfrac{28}{25}\Big)^{2} - 1\Big]$

$\hookrightarrow \sf\: 8000\Big[\Big(\dfrac{28}{25} \times\dfrac{28}{25} \Big)- 1\Big]$

$\hookrightarrow \sf\: 8000\Big( \dfrac{784}{625} - 1\Big)$

$\hookrightarrow \sf \: 8000\Big( \dfrac{784 - 625}{625}\Big)$

$\hookrightarrow \sf \: 8000\Big(\dfrac{159}{625}\Big)$

$\hookrightarrow \sf \: 8000 \times\dfrac{159}{ 625}$

$\hookrightarrow \sf \: \cancel\dfrac{1272000}{625}$

$\hookrightarrow \sf \large \boxed{ \sf \: Rs \: 2035.2}$

Answer 2

$\huge{\boxed{\underline{\underline{\mathbb{\red{SolUtion}}}}}}$

Given →

• Principal amount = Rs 8,000
• Rate = 12%
• Time = 2 years

_______________________

$\mathbb \blue{total \: amount \: = p(1 + { \frac{r}{100} )}^{time} }$

❏ Calculation

$> > \: 8000(1 + { \frac{12}{100} )}^{2}$

$> > 8000( { \frac{112}{100} )}^{2}$

$> > 8000 \times \frac{112}{100} \times \frac{112}{100}$

$> > 8 \times 112 \times \frac{112}{10}$

$> > \frac{100352}{10}$

$> > 10035.2$

So, Total amount = Rs 10,035.2

But we have to find out the value of interest then :-

$\mathcal \pink{total \: amount - principal \: amount}$

$\mathbb \green{10035.2 - 8000}$

$\mathbb \green{2035.2}$