| $ \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 |
| $\Rightarrow$ R2 =$ \left(\dfrac{96 \times 2}{3} \right) $= 64 |
$\Rightarrow$ R = 8.
$\therefore$ Rate = 8%.
Let the rate be R% per annum
$ P\left(1 + \dfrac{R}{100}\right)^T $= 1573.04
$ 1400\left(1 + \dfrac{\text{R}}{100}\right)^2$ = 1573.04
$\left(1 + \dfrac{R}{100}\right)^2$ =$ \dfrac{1573.04}{1400}$ = $\dfrac{157304}{140000}$ = $\dfrac{11236}{10000}$
$\left(1 + \dfrac{R}{100}\right) $= $\sqrt{\dfrac{11236}{10000}}$ = $\dfrac{\sqrt{11236}}{\sqrt{10000}}$ =$\dfrac{106}{100} $
$\dfrac{R}{100} $= $ \dfrac{106}{100} - 1 $=$ \dfrac{6}{100}$
R = 6%
Let the sum be Rs. $ x $. Then,
C.I. =$ \left[x\left(1 +\dfrac{4}{100} \right)^2- x\right] $=$ \left(\dfrac{676}{625} x- x\right) $=$ \dfrac{51}{625} x $.
S.I. =$ \left(\dfrac{x \times 4 \times 2}{100} \right) $=$ \dfrac{2x}{25} $.
$\therefore \dfrac{51x}{625} $-$ \dfrac{2x}{25} $= 1
$\Rightarrow x $ = 625.
| C.I. | = Rs.$ \left(4000 \times\left(1 +\dfrac{10}{100} \right)^2- 4000\right) $ |
| = Rs.$ \left(4000 \times\dfrac{11}{10} \times\dfrac{11}{10} - 4000\right) $ | |
| = Rs. 840. |
| $\therefore$ Sum = Rs.$ \left(\dfrac{420 \times 100}{3 \times 8} \right) $= Rs. 1750. |
| Sum = Rs.$ \left(\dfrac{50 \times 100}{2 \times 5} \right) $= Rs. 500. |
| Amount | = Rs.$ \left[500 \times\left(1 +\dfrac{5}{100} \right)^2\right] $ |
| = Rs.$ \left(500 \times\dfrac{21}{20} \times\dfrac{21}{20} \right) $ | |
| = Rs. 551.25 |
$\therefore$ C.I. = Rs. (551.25 - 500) = Rs. 51.25
| S.I. = Rs$ \left(\dfrac{1200 \times 10 \times 1}{100} \right) $= Rs. 120. |
| C.I. = Rs.$ \left[1200 \times\left(1 +\dfrac{5}{100} \right)^2- 1200\right] $= Rs. 123. |
$\therefore$ Difference = Rs. (123 - 120) = Rs. 3.
Let the sum be Rs. P.
| Then,$ \left[P\left(1 +\dfrac{10}{100} \right)^2- P\right] $= 525 |
| $\Rightarrow$P$ \left[\left(\dfrac{11}{10} \right)^2- 1\right] $= 525 |
| $\Rightarrow$ P =$ \left(\dfrac{525 \times 100}{21} \right) $= 2500. |
$\therefore$ Sum = Rs . 2500.
| So, S.I. = Rs.$ \left(\dfrac{2500 \times 5 \times 4}{100} \right) $= Rs. 500 |
Let the sum be P
Amount After 2 years = P$\left(1 + \dfrac{R}{100}\right)^T$ = P$\left(1 + \dfrac{5}{100}\right)^2$=P$ \left(\dfrac{105}{100}\right)^2$=P$\left(\dfrac{21}{20}\right)^2$
Given that amount After 2 years = 882
=> P$\left(\dfrac{21}{20}\right)^2$ = 882
=> P = $\dfrac{882 \times 20 \times 20}{21 \times 21} = 2\times 20 \times 20 = Rs.800$
substitute FV=2000 ,PV=1,000 and n=4
therefore r=$[\dfrac{(1000)^{\dfrac{1}{4}}}{FV}]-1$
=$(2)^1/4 -1$
=1.1892-1
=0.1892
=18.92%
substitute FV=4000 ,PV=2,500 and n=10
therefore r=$[\dfrac{(2,500)^{\dfrac{1}{n}}}{4,000}]-1$
=$(1.6)^0.1 -1$
=1.0481-1
=0.0481
=4.81%