Sum=Rs.(50 x 100/2x5)
=Rs.500.
Amount=[Rs.(500 x(1+5/100)²]
=Rs(500x21/20x21/20).
=Rs.551.25
C.I=Rs.(551.25 - 500)
=Rs.51.25

Answer : Option B

Explanation :

Amount after 1 year on Rs. 1600 (deposited on 1st Jan) at 5% when interest calculated half-yearly
$MF#%= \text{P}\left(1 + \dfrac{\text{(R/2)}}{100}\right)^\text{2T}
= 1600\left(1 + \dfrac{\text{(5/2)}}{100}\right)^{2 \times 1}
= 1600\left(1 + \dfrac{1}{40}\right)^2$MF#%
Amount after 1/2 year on Rs. 1600 (deposited on 1st Jul) at 5% when interest calculated half-yearly
$MF#%= \text{P}\left(1 + \dfrac{\text{(R/2)}}{100}\right)^\text{2T} = 1600\left(1 + \dfrac{\text{(5/2)}}{100}\right)^{2 \times \frac{1}{2}} = 1600\left(1 + \dfrac{1}{40}\right)$MF#%
Total Amount after 1 year
$MF#%=1600\left(1 + \dfrac{1}{40}\right)^2 + 1600\left(1 + \dfrac{1}{40}\right)
= 1600\left( \dfrac{41}{40}\right)^2 + 1600\left(\dfrac{41}{40}\right)
= 1600\left( \dfrac{41}{40}\right)\left[1 + \dfrac{41}{40}\right]\\\\ = 1600\left( \dfrac{41}{40}\right)\left( \dfrac{81}{40}\right) = 41 \times 81 = \text{Rs. }3321$MF#%
Compound Interest = Rs.3321 - Rs.3200 = Rs.121

Let the sum be Rs. x. Then,

C.I. = x 1 + 4 2 - x = 676 x - x = 51 x. 100 625 625

S.I. = x x 4 x 2 = 2x . 100 25

51x - 2x = 1 625 25

x = 625.

Answer : Option C

Explanation :

This problem is similar to the problems we saw in compound interest. We can use the formulas of compound interest here as well.
In compound interest, interest (a certain percentage of the principal) will be added to the principal after every year. Similarly, in this problem, a certain count(a certain percentage of the population) will be decreased from the total population after every year
$MF#%\text{i.e., the formula becomes, A = }\text{P}\left(1 - \dfrac{\text{R}}{100}\right)^\text{T}$MF#%

where Initial population = P, Rate = R% per annum, Time = T years and A = the population after T years
Please note that we have to use the -ve sign here instead of the + sign as the population gets decreased

Amount=Rs.[8000x(1+5/100)²]
=Rs.
[8000 x 21/20x21/20]
=Rs.8820.

Answer : Option A

Explanation :

Let the sum be P and Rate of Interest be R% per annum
Simple Interest on Rs.P for 2 years = 80
$MF#%\dfrac{\text{PRT}}{100} = 80\\\\ \dfrac{\text{PR}\times 2}{100} = 80\\\\ \dfrac{\text{PR}}{50} = 80\\\\ \text{PR} = 4000 \quad \color{#F00}{\text{--- (equation 1)}}$MF#%
$MF#%\text{Compound Interest = }\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} - \text{P} \\\\= \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^2 - \text{P} \\\\ = \text{P}\left[\left(1 + \dfrac{\text{R}}{100}\right)^2 - 1\right]
= \text{P}\left[\left(1 + \dfrac{\text{2R}}{100} + \dfrac{\text{R}^2}{10000}\right) - 1\right]
= \text{P}\left( \dfrac{\text{2R}}{100} + \dfrac{\text{R}^2}{10000}\right) \\\\ = \dfrac{\text{2PR}}{100} + \dfrac{\text{PR}^2}{10000}\\\\ = \dfrac{\text{2PR}}{100} + \dfrac{\text{PR} \times \text{R}}{10000}\\\\ = \dfrac{2 \times 4000}{100} + \dfrac{4000 \times \text{R}}{10000} \quad \color{#F00}{\text{ (∵ substituted the value of PR from equation 1)}}\\\\ = 80 + 0.4\text{R}$MF#%
Given that compound interest = Rs.80.80
=> 80 + 0.4R = 80.80
=> 0.4R = 0.80
$MF#%=> \text{R} = \dfrac{0.80}{0.4} = 2\%$MF#%
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Solution 2
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The difference between compound interest and simple interest on Rs. P for 2 years at R% per annum
$MF#%= \text{P}\left(\dfrac{\text{R}}{100}\right)^2$MF#%

Answer : Option D

Explanation :

Let the period be n years.
Amount after n years = Rs.20000 + Rs.3328 = Rs. 23328
$MF#%\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 23328\\\\ 20000\left(1 + \dfrac{8}{100}\right)^\text{n} = 23328\\\\ 20000\left(\dfrac{108}{100}\right)^\text{n} = 23328\\\\ \left(\dfrac{108}{100}\right)^\text{n} = \dfrac{23328}{20000} = \dfrac{11664}{10000} = \left(\dfrac{108}{100}\right)^2\\\\ \text{n = 2 years}$MF#%

Difference in C.I and S.I for 2 years= Rs(696.30-660)
=Rs.36.30.
S.I for one years= Rs330.
S.I on Rs.330 for 1 year=Rs.36.30
Rate=(100x36.30/330x1)%
=11%.

Answer : Option A

Explanation :

This means that, simple Interest on Rs.400 for 1 year = 420 - 400 = 20
$MF#%\text{Rate = }\dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times 20}{400 \times 1}= 5\%$MF#%
Rs.400 is the interest on the sum for 1st year
$MF#%\text{Hence, sum = }\dfrac{100 \times \text{SI}}{\text{RT}}= \dfrac{100 \times 400}{5 \times 1} = \text{Rs. 8000}$MF#%

Answer : Option D

Explanation :

$MF#%\text{Principal, P} = \dfrac{100 \times \text{SI}}{\text{RT}} = \dfrac{100 \times 80}{20 \times 2} = \text{Rs. 200}$MF#%
Amount after 2 year on Rs.200 at 20% per annum when interest is compounded annually
$MF#%= \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 200\left(1 + \dfrac{20}{100}\right)^2 = 200\left(\dfrac{120}{100}\right)^2 = \dfrac{200 \times 120 \times 120}{100 \times 100}\\\\ = \dfrac{2 \times 120 \times 120}{100}= 2 \times 12 \times 12 = \text{Rs. 288}$MF#%
Compound Interest = 288 - 200 = Rs.88