Compounding interest annually

$P\left(1-\dfrac{10}{100}\right)^3=729$

P=Rs.12000,Time=2 Years,Rate=5% and 6%
$A=P\left(1+\frac{R}{100}\right)^T$
$\Rightarrow 12000\left(1+\frac{5}{100}\right)\left(1+\frac{6}{100}\right)$
$\Rightarrow 12000\times \frac{105}{100}\times \frac{106}{100}$
$\Rightarrow 13356  Rs$
$C.I=A-P$
$\Rightarrow 13356-12000=1356  Rs.$

$\Rightarrow$ Let the sum = $Rs.x$.

$\Rightarrow$  $P=$Rs.$60000,,T=2,$years and $R=8%$

Given, $P=$ Rs. $8,000,,R=10%$.

Interest for the first year 

Here,$P=$ Rs $7000$ , $R=10%p.a$

In three years.a sum of money deposited by anyone will become $=2^3=8$times

$A=P(1+\cfrac{r}{100})^n$

Interest for first year  = $ \cfrac{12000 \times 1 \times 18}{100} = 2160$

Interest for the first year =$ \cfrac{12000 \times 1 \times 18}{100} = 2160$

Interest for the first year =$ \cfrac{25000 \times 1 \times 20}{100} = 5000$

$Principal = 64000$
$Rate = 7.5% ,p.a$
$Time = 3 years$ 
$\text{Where r is the rate and t is the time.}$
$CI = 64000 (1 + 7.5/100)^3 - 64000$
$= 64000 (1 + 75/1000)^3 - 64000$
$= 64000 (1 + 3/40)^3 - 64000$
$= 64000 (43/40)^3 - 64000$
$= 64000 * 43/40 * 43/40 * 43/40 - 64000$
$= 43 * 43 * 43 - 64000$
$= 79507 - 64000$
$= 15507$
$\text{Hence CI will be 15507 rupees}$

Interest is compounded annually,

Here, Principal =Rs. 62,500
Rate =8% per annum =4% half year
Time =$1\frac{1}{2}$ years =3 half years
$Amount = P(1 +\frac{R}{100})^T$
$\therefore Amount =Rs.[62500\times {(1+\frac{4}{100})}^{3}]$
$Amount = \displaystyle(62500\times \frac{104}{100}\times \frac{104}{100}\times \frac{104}{100}) =Rs. 70304$
$\therefore$ Compound interest =Rs.(70304-62500)=Rs.7804

Principal = Rs 1000
time = 1 year
Rate = 5 %

Let the amount Rs 100 and rate of interest r
By hypothesis
$200=100(1+\frac{r}{100})^{5}$
$\Rightarrow 2=(1+\frac{r}{100})^{5}$
$\Rightarrow (1+\frac{r}{100})=2^{\frac{1}{5}}$
Lets amount become 8 time in n years then
$800=100(1+\frac{r}{100})^{n}$
$\Rightarrow 8=(1+\frac{r}{100})^{n}$=$(2^{\frac{1}{5}})^{^{n}}$
$\Rightarrow (2)^{3}$=$ (2)^{\frac{n}{5}}$
$\therefore \frac{n}{5}=3$ 
OR n=15 years

For the first year

$P=Rs.48000$

$Sum=Rs.13500$

Principal$=Rs.12800$

$P=13500$

$Sum=rs.48000$

P $=Rs.12800$

machine value after 2 year=Rs.10240

$Sum=Rs.13500$

Amount in three year $=Rs. 20160$

Principal $=Rs.12800$

Rate of interest=$\frac{Difference  in  interest  of  two  consecutive periods\times 100}{C.I  of  preceding year\times Time}\times 100$

Given, Principal $P=2,000, r=5%, n=5$ years

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Given, $A =$ Rs. $4913$, $ r = 6%$, $n = 4$ and $t =$ $4\dfrac{1}{2}$ years
Therefore, $4913 = P\left (1+\dfrac{0.06}{1}\right)^{1\times 4.5}$
$\Rightarrow 4913 = P\times 1.06^{4.5}$
$\Rightarrow 4913 = P\times 1.2998$
$\Rightarrow P = \dfrac{4913}{1.2998}$
$\Rightarrow P =$ Rs. $3779.81$

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Given, $A =$ Rs. $4620$, $r = 10%$, $n = 1$ and $t =$ $4\dfrac{1}{2}$ years
Therefore, $4620 = P\left (1+\dfrac{0.1}{1}\right)^{1\times 4.5}$
$\Rightarrow 4620 = P\times 1.1^{4.5}$
$\Rightarrow 4620 = P\times 1.535561$
$\Rightarrow P = \dfrac{4620}{1.535561}$
$\Rightarrow P =$ Rs. $3008.67$

We know the formula,
$A = P(1+\frac{r}{n})^{n.t}$
Given, $P =$ Rs. $3500$, $r = 10%$, $n = 12$ and $t =$ $5\dfrac{1}{2}$ years
Therefore, we have $A = 3500\left (1+\dfrac{0.1}{12}\right)^{12\times 5.5}$
$\Rightarrow A = 3500\times 1.008^{66}$
$\Rightarrow A = 3500\times 1.72927$
$\Rightarrow A = 6052.45$
To find interest we use formula $A = P + I$, since $A = 6052.45$ and $P = 3500$, we have:
$A = P + I$
$\Rightarrow 6052.45 = 3500 + I$
$\Rightarrow I = 6052.45 - 3500 = 2552.45$
Interest, $I =$ Rs. $2552.45$

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Given:
$P =$ Rs. $4000, r = 3%, n = 1$ and $t = 3.5$ years
$\Rightarrow A = 4000\left (1+\dfrac{0.03}{1}\right)^{1\times 3.5}$
$\Rightarrow A = 4000\times 1.03^{3.5}$
$\Rightarrow A = 4000\times 1.108997$
$\Rightarrow A =$ Rs. $4435.99$
To find interest, we use formula 

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Where,
$A =$ total amount
$P =$ principal or amount of money deposited,
$r =$ annual interest rate
$n =$ number of times compounded per year
$t =$ time in years
Given:
$P =$ Rs. $600, r = 3%, n = 1 $ and $t =$ $6\dfrac{1}{2}$ years
$A = 600\left (1+\dfrac{0.03}{1}\right)^{1\times 6.5}$
$A = 600\times 1.03^{6.5}$
$A = 600\times 1.211831$
$A =$ Rs. $727.1$

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Where,
$A =$ total amount
$P =$ principal or amount of money deposited,
$r =$ annual interest rate
$n =$ number of times compounded per year
$t =$ time in years
Given: $P =$ Rs. $1200, r = 10%, n = 1$ and $t =$ $2\dfrac{1}{2}$ years
$A = 1200\left (1+\dfrac{0.1}{1}\right)^{1\times 2.5}$
$A = 1200\times 1.1^{2.5}$
$A = 1200\times 1.269059$
$A =$ Rs. $1522.87$

Given: $P = 3000, r = 5%, n = 10$ years
We know $A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
Putting all the given values in the formula, we get

Given: $P = 1000, r = 10%, n = 2$ years
We know the formula $A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
Putting all the given values in the formula, we get

Given, $P = 20000, r = 10%, n = 3$ years
$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
$A = 20000\left [\left (1+\dfrac{10}{100}\right)^3\right]$
$A =$ Rs. $26620$
Compound interest $=$ Amount $-$ Principal
CI $= 26620 - 20000$
CI $=$ Rs. $6620$

Given, $P =$ Rs. $3000$

Given, $P = 2500, r = 10%, n = 2$ years
We know, $A = P[(1+\dfrac{r}{100})^n]$
$\Rightarrow A = 2500[(1+\dfrac{10}{100})^2]$
$\Rightarrow A =$ Rs. $3025$

Given: $P = 1728, A = 3375, n = 23$ years

We have, $P=$ Rs. $25,000,,R=20%,p.a.=\dfrac{20}{2}%\ \text{,half-year}$ and $T=1\dfrac{1}{2}$ years $=3$ half-years.

$\Rightarrow$  C.I. when interest compounded yearly = $[10000\times (1+\dfrac{4}{100})^1\times (1+\dfrac{\dfrac{1}{2}\times 4}{100})]-10000$

Here $P=$ Rs. $15,000$ and $R=10%$.

$\Rightarrow$  Here, $P=$Rs.$1,20,000,,T=2,$years and $R=8%$

$\Rightarrow$  $A=P(1+\dfrac{R}{100})^T$

$\Rightarrow$  $A=P(1+\dfrac{R}{100})^T$

Here, $P=$ Rs. $20,000,,T=2$ years $73$ days $=2\dfrac{1}{5}$ years and $R=6\dfrac{1}{4}%=\dfrac{25}{4}%$

We have,

We have,

$\Rightarrow$  $P=Rs.20,500,,R=7%$ and $T=2,years$

Interst for the first year 

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Where,
$A =$ total amount
$P =$ principal or amount of money deposited,
$r =$ annual interest rate
$n =$ number of times compounded per year
$t =$ time in years
Given:
$P =$ Rs. $20000, r = 10%, n = 1$ and $t =$ $4\dfrac{1}{2}$ years
$A = 20000\left (1+\dfrac{0.1}{1}\right)^{1\times 4.5}$
$A = 20000\times 1.1^{4.5}$
$A = 20000\times 1.535561$
$A =$ Rs. $30711.22$
To find interest we use formula $A = P + I$, since $A = 30711.22$ and $P = 20000$ we have:
$A = P + I$
$30711.22 = 20000 + I$
$I = 30711.22 - 20000 = 10711.22$
Interest, I $=$ Rs. $10711.22$

Given: $P = 12000, r = 15%, n = 3$ years
$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
$A = 12000\left [\left (1+\dfrac{15}{100}\right)^3\right]$
$A =$ Rs. $18250.50$