Introduction to interests - class-VIII

Principle$=Rs1770\quad\quad Time=2years$

S.I. for $1$ year $= Rs. (854-815) = Rs. 39$

Interest $= 200.....(700-500=200)$

T = 8 years; N = 2; R = ?
R $\times T$ = 100 $\times (N - 1)$

Principal = Rs 2000
Time = 4 years
Interest = Rs 400
Now, $Interest = \frac{Principal \times Rate \times Time}{100}$
$400 = \frac{2000\times R\times 4}{100}$
$R = \frac{400 \times 100}{2000 \times 4}$

$SI=\frac {2}{5}P, t=10, r=?$
$\frac {Ptr}{100}=\frac {2}{5}P$
or $\frac {10\times r}{100}=\frac {2}{5}$
or $r=\frac {20}{5}=4$%

increase in rate of interest

$Ratio\quad in\quad which\quad they\quad should\quad share\quad their\quad profits=Raio\quad of\quad investments\times \quad Time\quad period$

Simple interest will be(576-450)=Rs126

The rate of interest is the percentage of a sum of money charged for its use.

$\Rightarrow$   Let the principal be $Rs.x$.

Here, $S.I.=$ Rs. $100,,P=$ Rs. $600$ and $T=6$ months $=\dfrac{1}{2}$ year

Under simple interest,

Let,

Using the rule of weighted average,
The average rate of interest = $\displaystyle \dfrac{\dfrac{1}{2} , \times , 10 , + , \dfrac{1}{3} , \times , 9 , + , \dfrac{1}{6} , \times , 12}{\dfrac{1}{2} , + , \dfrac{1}{3} , + ,\dfrac{1}{6}}$
= 5 + 3 + 2 = 10%

$I = ₹300$ 

Total sum of money paid = Rs 120

We know that $I=\dfrac{PTR}{100}$

In case of Simple interest, total amount $A$ is,

According to question, we have:

Given that the output is doubled in last two years
Let the output before $2$ years be$= x$
Hence after two years it will be$= 2x$

We know that 

$A=3P$

Simple Interest $ I = \cfrac {PNR}{100} $
Given,
$ P = Rs  3,750 $
$ I = \cfrac {1}{5}  \times P =  Rs  750 $
$ N = 4  years $
$ R = ? $
So, $ => I = \cfrac {PNR}{100} $
$ => 750  = \cfrac {3,750 \times 4 \times R }{100} $
$ =>R = 5 $ %
And for $ 15 $ years, interest
$ I =  \cfrac {PNR}{100} $
$ => I  = \cfrac {3,750 \times 15 \times 5 }{100} = Rs. 2,812.5 $
And Amount $ = I + P = Rs.  3,750  + Rs.  2,812.5  = Rs.  6,562.5  $

Let the initial sum of money be $P$.

Amount in $2$ years $=$ Rs. $756$

Let the interest rate at which $Rs.,1000$ is invested is $r%$ 
Then Rs. $1500$ is invested at $(r+2)%$

Let the whole sum be Rs. $100$. 

Let the sum be Rs. $P$ and rate of interest per annum be $R%$
Then $\displaystyle P\left [ \left ( 1+\frac{R}{100} \right )^{2}-1 \right ]-\frac{2PR}{100}=$ Rs. $832-$ Rs. $800=$ Rs. $32$
$\displaystyle \Rightarrow P\left [ 1+\frac{2R}{100}+\frac{R^{2}}{10000}-1 \right ]-\frac{2PR}{100}=32$
$\displaystyle \Rightarrow \frac{PR^{2}}{10000}=32$

We know that

 A = Rs.3993, P = Rs.3000, n = 3, r = ?
$\displaystyle \therefore A=P\left ( 1+\frac{r}{100} \right )\Rightarrow 3993=3000\left ( 1+\frac{r}{100} \right )^{3}\Rightarrow \frac{3993}{3000}=\left ( 1+\frac{r}{100} \right )^{3}\Rightarrow \frac{1331}{1000}=\left ( 1+\frac{r}{100} \right )^{3}$
$\displaystyle \Rightarrow \left ( \frac{11}{10} \right )^{3}=\left ( 1+\frac{r}{100} \right )^{3}\Rightarrow 1+\frac{r}{100}=\frac{11}{10}\Rightarrow \frac{r}{100}=\frac{11}{10}-1=\frac{1}{10}$
$\displaystyle \therefore r=\frac{100}{10}=10%: : p.a.$

P = Rs.8000 C.I. = Rs. 1261

 Let the principal be Rs.P and rate of interest p.a. = r% Then
$\displaystyle P\left ( 1+\frac{r}{100} \right )^{2}=4840...........(i)$ and $\displaystyle P\left ( 1+\frac{r}{100} \right )^{3}=5324...........(ii)$
$\displaystyle \Rightarrow \frac{5324}{4840}=\frac{(1+r/100)^{3}}{(1+r/100)^{2}}\Rightarrow 1+\frac{r}{100}=\frac{1331}{1210}$
$\displaystyle \Rightarrow \frac{r}{100}=\frac{1331}{1210}-1=\frac{121}{1210}=\frac{1}{10}\Rightarrow r=\frac{1}{10}\times 100=10%: p.a.$

R $\times$ T = 100 $\times$ (N - 1)
$R \times 5 = 100 \times \left ( \displaystyle \frac {5}{4} - 1 \right )$
$R \times 5 = 100 \times \displaystyle \frac {1}{4}$
$R = 5%$

We know that Simple Interest $ = \cfrac {PNR}{100} $
Given,  $ \cfrac {7953 \times 2 \times R}{100}  + \cfrac {1800  \times 3 \times R}{100}  = Rs 2343.66 $
$=> 159.06R + 54R = Rs 2343.66 $
$ => 213.06R = 2343.66 $
$ => R = 11 %$ 

$S.I = \dfrac{P\times t\times r}{100}$

Amount $= Rs. 2662$
Principal $= Rs. 1600$
Time $= 1.5$ years $= 6$ quarters
$A = P\left(1 + \cfrac{R}{100}\right)^T$
$2662 = 1600 \left(1 + \cfrac{R}{100}\right)^6$
$1.088 = 1 + \cfrac{R}{100}$
$R = 8.8%$
Hence, annual rate of interest $=8.8\times 4 = 35%$

Given $P=1, 800, SI=432, T=3 years$
$R=\dfrac {SI\times 100}{P\times T}=\dfrac {432\times 100}{1800\times 3}=8$%

Rate $= \displaystyle \left ( \frac{100 \times 300}{5000 \times 3} \right )$% = 2%

Given, Sum (P) = Rs. 100,
Amount due (A) = Rs. 121
Time (n) = 2 years, Rate (r) = ?
We know $A = P \displaystyle \left ( 1 + \frac{r}{100} \right )$
$\therefore 121 = 100 \displaystyle \left ( 1 + \frac{r}{100} \right )^2$
or $\displaystyle \left ( 1 + \frac{r}{100} \right )^2 = \frac{121}{100}$
or $\displaystyle 1 + \frac{r}{100} = \frac{11}{10} = 1 + \frac{1}{10} = 1 + \frac{10}{100}$
$\therefore r = 10$%

$\displaystyle \frac{1200 \times R \times 3}{100} - \frac{1000 \times R \times 3}{100} = 50$
or $6 R = 50$
or $R = \displaystyle 8 \frac{1}{3}$%

Rs. 1000 + S.l. on Rs. 1000 for 11 months = Rs. 1000 + S.I. on Rs. 100 for (1 + 2 + 3 + 4 + ... + 10) months Rs. 1000 S.I. on Rs. 100 for 100 months 
= Rs.1000 + S.l. on Rs. 100 for 55 months
S.l. on Rs. 100 for 55 months = Rs. 100
$\therefore Rate = \displaystyle \left ( \frac{100 \times 100 \times 12}{100 \times 55} \right )$% $21 \displaystyle \frac{9}{11}$%

Interest on Rs. 10580 for 1 year = Rs. (12167 - 10580) = Rs. 1587
$\therefore Rate = \displaystyle \left ( \frac{100 \times 1587}{10580} \right )$% = 15%

Given,
Simple Interest for two years is 50
$\frac { P\times T\times R }{ 100 } =Simple\quad Interest$
$\frac { P\times 2\times R }{ 100 } =50$
 $PR=2500$
 $P=\frac { 2500 }{ R } $................EQ(1)
Compound Interest for two years will be 52
$P{ { { \left( 1+\frac { r }{ 100 }  \right)  }^{ 2 } } }-P=52$
$\Rightarrow P\left( 1+\frac { 2R }{ 100 } +\frac { { R }^{ 2 } }{ 10000 } -1 \right) =52$
 $\Rightarrow P\left( \frac { 2R }{ 100 } +\frac { { R }^{ 2 } }{ 10000 }  \right) =52$
$\Rightarrow P\left( \frac { 200R+{ R }^{ 2 } }{ 10000 }  \right) =52$
$\Rightarrow P\left( \frac { 200R+{ R }^{ 2 } }{ 10000 }  \right) =52$
$\Rightarrow P\left( 200R+{ R }^{ 2 } \right) =520000$
$\Rightarrow 200R+{ R }^{ 2 }=\frac { 520000 }{ P } $
 $\Rightarrow 200R+{ R }^{ 2 }=520000\times \frac { R }{ 2500 } $(TAKING EQUATION FROM SIMPLE INTEREST EQ(1))
 $\Rightarrow 200R+{ R }^{ 2 }=208R$
$\Rightarrow { R }^{ 2 }=(208R-200R)$
 $\Rightarrow { { R }^{ 2 } }=8R$
 $R=8$
Rate will be 8%

Simple Interest $ SI = \dfrac {PNR}{100} $

Difference between the C.I of two successive years=$7410-5700=Rs. 1710$

Amount in three year $=Rs. 20160$

Sum after 2 years=Rs.1089

Difference between the C.I of two successive half -years=$760.50-650=110.50$

Sum after 2 years $=Rs.5292$

$P=Rs.8000$

Let sum of money be $p$.

Amount invested $= 1500$

Given:
Principal $=$ Rs. $ 4,000$
Interest $= 7200 - 4000 = 3200$
We know the formula,
Rate of interest, $R =$ $\dfrac{I\times 100}{PT}$
$\Rightarrow R =$ $\dfrac{3200\times 100}{4000\times 4}$
$\Rightarrow R = 20%$

Given:
Principal $= 5,000$
Interest $= 15000 - 5000 = 10000$
We know the formula,
Rate of interest, $R =$ $\dfrac{I\times 100}{PT}$
$\Rightarrow R =$ $\dfrac{5000\times 100}{10000\times 2}$
$\Rightarrow R = 25%$

If the principal amount is $p$, annual interest is $r$ $%$ then for simple interest after $n$ years the amount will be $p\left (1+\dfrac {nr}{100}\right)$.
Here $p=5000$, $n=\dfrac {1}{6}$, $r=4$ $%$
So, the amount after $2$ months is $5000\left (1+\dfrac {2}{300}\right) = 5000\left (\dfrac {151}{150}\right) = 5033.33$

$\left [15000\times \left (1 + \dfrac {R}{100}\right )^{2} - 15000\right ] - \left (\dfrac {15000\times R\times 2}{100}\right ) = 96$
$\Rightarrow 15000 \left [\left (1 + \dfrac {R}{100}\right )^{2} - 1 - \dfrac {2R}{100}\right ] = 96$
$\Rightarrow 15000 \left [\dfrac {(100 + R)^{2} - 10000 - (200\times R)}{10000}\right ] = 96$
$\Rightarrow R^{2} = \left (\dfrac {96\times 2}{3}\right ) = 64$
$\Rightarrow R = 8$.
$\therefore Rate = 8$%.

We have,

We have given that interest and principal are equal.

We have given that interest and borrowed principal are same.

We have given that interest and principal are equal.

$\Rightarrow$  $P=Rs.5000,,T=5,years $ and $S.I.=Rs.1500$

$\Rightarrow$  $P=Rs.3000,,T=3,years$ and $S.I.=Rs.1000$.

We have,

$SI=P/100; r=4t \implies t=r/4$

For simple interest, the interest earned is determined by the formula: I = PRT

Required sum
= $24000 \left ( 1 + \frac{5}{100} \right )\left ( 1 + \frac{10}{100} \right )\left ( 1 + \frac{4}{100} \right )$
= $Rs.24000 \times 1.05 \times 1.10 \times $ $1.04$
= $Rs. 28828.80$
Hence, Amount after three years is $Rs. 28828.80$

$A=P(1+it)$
$50,000=46,875\left(1\times i\times 1 \frac{8}{12}\right)$
$i=0.04$; rate $=4%$

We know $A=$P+ PIT$
i.e., $1050=1000+1000\times I\times 6/12$
$50=500I$
$I=1/10=10%$

Required rate of interest $= \dfrac {75}{750}\times 100 = 10$

Rs. 1000 + S.I. on Rs. 1000 for 11 months
$= Rs. 1000 +$ S.I. on Rs. 100 for $(1+ 2 + 3 + 4 + ... + 10)$ months
Rs. 1000 S.I. on Rs. 100 for 100 months
$= Rs. 1000 +$ S.I. on Rs. $100$ for $55$ months
S.I. on Rs. 100 for 55 months
$= Rs. 100$
$\therefore  Rate = (\cfrac{100\times 100\times 12}{100\times 55})$% $=21\cfrac{9}{11}$ %

interest is compounded, Amount $ A = P(1+ \frac {R}{100})^n $
So, for the first situation
$ 1375 = P \times (1+ \frac {R}{100})^5 $   - (1)

And for the first situation
$ 1980 = P \times (1+ \frac {R}{100})^7 $   ---- (2)

Dividing eqn 2 by eqn 1, we get
$ \frac {1980}{1375} =   (1+ \frac {R}{100})^2 $ 
$ => \frac {396}{275} =   (1+ \frac {R}{100})^2 $ 
$ => \frac {36}{25} =   (1+ \frac {R}{100})^2 $ 
$ => (1+ \frac {R}{100}) = \frac {6}{5} $
$ => \frac {R}{100} = \frac {1}{5} $
$ => R =20 $ %

Simple Interest $ SI = \frac {PNR}{100} $
So, $ SI = \frac {P \times 2 \times 8}{100} = Rs 0.16P $

When interest is compounded, Amount $ A = P(1+ \frac {R}{100})^n $
So, A $ = P \times (1+ \frac {8}{100})^2 = Rs  1.1664P  $
And $ CI = A - P = 0.1664P $

Si, difference $ CI - SI = Rs 0.1664P - Rs 0.16P = Rs 25.60 $
$ => 0.0064P = 25.60 $
$ => P = Rs  4000 $

$I=Interest$

$Interest=\dfrac { P\times R\times T }{ 100 } \ =\dfrac { Rs.2500\times 6\times 4 }{ 100 } \ =Rs.600$