Principal = Rs. 10000; Rate = 2% per half-year; Time = 2 years = 4 half-years.
Amount = Rs [10000 * (1+(2/100))^4] = Rs(10000 * (51/50) * (51/50) * (51/50) * (51/50))
= Rs. 10824.32
C.I. = Rs. (10824.32 - 10000) = Rs. 824.32.

Answer : Option C

Explanation :

$MF#%\text{Sum, P = }\dfrac{100 \times \text{SI}}{\text{RT}} = \dfrac{100 \times 60}{5 \times 2} = \text{Rs. }600$MF#%
Amount after 2 years on Rs.600 at 5% per annum when interest is compounded annually
$MF#%= \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 600\left(1 + \dfrac{\text{5}}{100}\right)^2\\\\ = 600\left(\dfrac{105}{100}\right)^2 = \dfrac{600 \times 105 \times 105}{100 \times 100} = \dfrac{6 \times 105 \times 105}{100} = \text{Rs. 661.5}$MF#%
Compound Interest = 661.5 - 600 = 61.5

Let the two parts be Rs. x and Rs. (1301 - x).
x(1+4/100)^7 =(1301-x)(1+4/100)^9
x/(1301-x)=(1+4/100)^2=(26/25*26/25)
625x=676(1301-x)
1301x=676*1301
x=676
So,the parts are rs.676 and rs.(1301-676)i.e rs.676 and rs.625.

Answer : Option A

Explanation :

$MF#%\text{Time, T = 2 years 73 days = }2\dfrac{1}{5}\text{ year}\\\\\text{Rate, R}= 6\dfrac{1}{4}\% = \dfrac{25}{4}\%$MF#%
$MF#%\text{ Amount after 2 years 73 days} = 20480\left(1 + \dfrac{\left(\dfrac{25}{4}\right)}{100}\right)^2 \times \left(1 + \dfrac{\dfrac{1}{5}\left(\dfrac{25}{4}\right)}{100}\right) \\\\ = 20480\left(1 + \dfrac{25}{4 \times 100}\right)^2 \times \left(1 + \dfrac{\left(\dfrac{5}{4}\right)}{100}\right) = 20480\left(1 + \dfrac{1}{16}\right)^2 \times \left(1 + \dfrac{1}{80}\right)\\\\ = 20480\left(\dfrac{17}{16}\right)^2 \times \left(\dfrac{81}{80}\right) = \dfrac{20480 \times 17 \times 17 \times 81}{16 \times 16\times 80}= \dfrac{256 \times 17 \times 17 \times 81}{16 \times 16}
\\\\= 17 \times 17 \times 81 = \text{Rs. 23409}$MF#%
Compound Interest = 23409 - 20480 = Rs.2929

Answer : Option D

Explanation :

Let P = Rs.100
Simple Interest = Rs. 80 ( ∵ 80% increase is due to the simple interest)
$MF#%\text{Rate of interest} =\dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times 80}{100 \times 8} = 10\%\text{ per annum}$MF#%
Now let's find out the compound interest of Rs. 14,000 after 3 years at 10%
P = Rs.14000
T = 3 years
R = 10%
$MF#%\text{Amount after 3 years } = \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 14000\left(1 + \dfrac{10}{100}\right)^3 \\\\ = 14000\left(\dfrac{110}{100}\right)^3 = 14000\left(\dfrac{11}{10}\right)^3 = 14 \times 11^3 = 18634$MF#%
Compound Interest = Rs.18634 - Rs.14000 = Rs.4634

Answer : Option A

Explanation :

Amount after 2 year on Rs.3000 at 10% per annum when interest is compounded annually
$MF#%= \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 3000\left(1 + \dfrac{10}{100}\right)^2 = 3000\left(\dfrac{110}{100}\right)^2 \\\\= \dfrac{3000 \times 110 \times 110}{100 \times 100} = 3 \times 11 \times 110 = 3630$MF#%
Compound Interest = 3630 - 3000 = Rs.630
Given that simple interest on a certain sum of money for 4 years at 5% per annum is half of the compound interest.
$MF#%\text{i.e., simple interest = }\dfrac{630}{2}\text{ = Rs.315}\\\\ \text{P} = \dfrac{100 \times \text{SI}}{\text{RT}} = \dfrac{100 \times 315}{5 \times 4} = \dfrac{20 \times 315}{4} = 5 \times 315 = \text{Rs. 1575}$MF#%

Amount of Rs. 100 for 1 year
when compounded half-yearly = Rs. 100 x 1 + 3 2 = Rs. 106.09 100

Effective rate = (106.09 - 100)% = 6.09%

P 1 + 20 n > 2P         6 n > 2. 100 5

Now, 6 x 6 x 6 x 6 > 2. 5 5 5 5

So, n = 4 years.

15000 x 1 + R 2 - 15000 - 15000 x R x 2 = 96 100 100

15000 1 + R 2 - 1 - 2R = 96 100 100

15000 (100 + R)² - 10000 - (200 x R) = 96 10000

R² = 96 x 2 = 64 3

R = 8.

Rate = 8%.

Answer : Option A

Explanation :

$MF#%\text{Amount after 3 years = }10000\left(1 + \dfrac{2}{100}\right)\left(1 + \dfrac{5}{100}\right)\left(1 + \dfrac{10}{100}\right) \\\\ = 10000\left(\dfrac{102}{100}\right)\left(\dfrac{105}{100}\right)\left(\dfrac{110}{100}\right)
= \dfrac{102 \times 105 \times11}{10} = \text{Rs. 11781}$MF#%