Amount = Rs. 25000 x 1 + 12 3 100 = Rs. 25000 x 28 x 28 x 28 25 25 25 = Rs. 35123.20

C.I. = Rs. (35123.20 - 25000) = Rs. 10123.20

Answer : Option C

Explanation :

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Solution 1
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Let the sum be P and rate of interest be R% per annum.
Amount after 2 year = 8240
$MF#%\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 8240\\\\ \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^2 = 8240 \quad \color{#F00}{\text{--- ( 1)}}$MF#%
Amount after 3 year = 9888
$MF#%\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 9888\\\\ \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^3 = 9888 \quad \color{#F00}{\text{--- (2)}}$MF#%
$MF#%\color{#F00}{\text{(2)÷(1) =>}} \dfrac{\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^3 }{\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^2} = \dfrac{9888}{8240}\\\\ 1 + \dfrac{\text{R}}{100} = \dfrac{9888}{8240}\\\\ \dfrac{\text{R}}{100} = \left(\dfrac{9888}{8240} - 1\right) = \dfrac{1648}{8240} = \dfrac{1}{5}\\\\ \text{R} = \dfrac{100}{5} = 20\%$MF#%
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Solution 2
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If a certain sum of money at compound interest amounts to Rs.x in t₁ years and Rs.y in t₂ years, then the rate of interest per annum can be given by
$MF#%\text{R} = \left[\left(\dfrac{y}{x}\right)^{\text{1}/(\text{t}_2-\text{t}_1)}- 1\right] \times 100 \%$MF#%

Answer : Option D

Explanation :

Given that simple interest for 2 years is Rs.800
i.e., Simple interest for 1st year is Rs.400
and simple interest for 2nd year is also Rs.400
Compound interest for 1st year will be 400
and Compound interest for 2nd year will be 832 - 400 = 432
you can see that compound interest for 2nd year is more than simple interest for 2nd year by 432 - 400 = Rs.32
i.e, Rs. 32 is the interest obtained for Rs.400 for 1 year
$MF#%\text{Rate, R = }\dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times 32}{400 \times 1} = 8\%$MF#%
Difference between compound and simple interest for the 3rd year
= Simple Interest obtained for Rs.832
$MF#%= \dfrac{\text{PRT}}{100} = \dfrac{832 \times 8 \times 1}{100} = \text{Rs. 66.56}$MF#%
Total difference between the compound and simple interest for 3 years
= 32 + 66.56 = Rs.98.56

S.I on Rs.7350 for 1 year=Rs.(8575-7350)=Rs.1225
Rate=(100*1225/7350*1)%= 20/3%
Let the sum be Rs.x.
X(1+50/3*100)^2=7350
X*7/6*7/6=7350
X=(7350*36/49)=5400
Sum=Rs.5400

Answer : Option D

Explanation :

Let principal, P be Rs.100
Amount after 1 year on Rs.100 at 6% per annum when interest is compounded half-yearly
$MF#%=\text{P}\left(1 + \dfrac{\text{(R/2)}}{100}\right)^\text{2T} = 100\left(1 + \dfrac{(6/2)}{100}\right)^{2 \times 1}\\\\= 100\left(1 + \dfrac{3}{100}\right)^2= 100\left(\dfrac{103}{100}\right)^2= \dfrac{100 \times 103 \times 103}{100 \times 100}= \dfrac{103 \times 103}{100} = 106.09$MF#%
This means, if interest is compounded half-yearly at 6%, Rs.100 becomes Rs.106.09 after 1 year
Now, we need to find out the rate of interest on which Rs.100 becomes Rs.106.09 after 1 year
when the interest is compounded annually
$MF#%\begin{align}&\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 106.09\\\\ &100\left(1 + \dfrac{\text{R}}{100}\right)^1 = 106.09\\\\ &100\left(1 + \dfrac{\text{R}}{100}\right) = 106.09\\\\ &100 + \text{R} = 106.09\\\\ &\text{R} = 106.09 - 100 = 6.09\%\end{align}$MF#%
i.e, effective annual rate of interest is 6.09%

Sum = Rs. 50 x 100 = Rs. 500. 2 x 5

Amount = Rs. 500 x 1 + 5 2 100 = Rs. 500 x 21 x 21 20 20 = Rs. 551.25

C.I. = Rs. (551.25 - 500) = Rs. 51.25

Answer : Option D

Explanation :

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Solution 1
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Let the sum be Rs.x
Amount after 3 years on Rs.x at 20% per annum when interest is compounded annually
$MF#%= \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = \text{x}\left(1 + \dfrac{20}{100}\right)^3 = \text{x}\left(\dfrac{120}{100}\right)^3 = \text{x}\left(\dfrac{6}{5}\right)^3\\\\ \text{Compound Interest = }\text{x}\left(\dfrac{6}{5}\right)^3 - x = x\left[\left(\dfrac{6}{5}\right)^3 - 1\right] = x\left[\dfrac{216}{125} - 1\right] = \dfrac{91x}{125}\\\\ \text{Simple Interest = }\dfrac{\text{PRT}}{100} = \dfrac{x \times 20 \times 3}{100} = \dfrac{3x}{5} $MF#%
Given that difference between compound interest and simple interest is Rs.48
$MF#%\dfrac{91x}{125} - \dfrac{3x}{5} = 48\\\\ \dfrac{91x - 75x}{125} = 48\\\\ \dfrac{16x}{125} = 48\\\\ x = \dfrac{48 \times 125}{16} = 3 \times 125 = \text{Rs. 375}$MF#%
i.e, the sum is Rs.375
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Solution 2
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The difference between compound interest and simple interest on Rs. P for 3 years at R% per annum
$MF#%= \text{P}\left(\dfrac{\text{R}}{100}\right)^2\left(\dfrac{\text{R}}{100}+3\right)$MF#%

Answer : Option A

Explanation :

Let the sum be P
$MF#%\text{Amount After 2 years }= \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = \text{P}\left(1 + \dfrac{5}{100}\right)^2=\text{P}\left(\dfrac{105}{100}\right)^2=\text{P}\left(\dfrac{21}{20}\right)^2$MF#%
Given that amount After 2 years = 882
$MF#%=> \text{P}\left(\dfrac{21}{20}\right)^2 = 882\\\\ => \text{P} = \dfrac{882 \times 20 \times 20}{21 \times 21} = 2\times 20 \times 20 = \text{Rs. 800}$MF#%

Answer : Option B

Explanation :

$MF#%\begin{align}&\text{Amount after 2 years = }\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T}\\\\ &= 20000\left(1 + \dfrac{4}{100}\right)^2 = 20000\left(\dfrac{104}{100}\right)^2 \\\\&= \dfrac{20000 \times 104 \times 104}{100 \times 100} = 2 \times 104 \times 104 = \text{Rs. }21632\end{align}$MF#%

Answer : Option A

Explanation :

$MF#%\text{Rs.20000 after 4 years } = 20000\left(1 + \dfrac{10}{100}\right)^4 = 20000\left(\dfrac{11}{10}\right)^4 = \text{Rs. 29282}\\\\ \text{Rs.20000 after 3 years} = 20000\left(1 + \dfrac{10}{100}\right)^3 = 20000\left(\dfrac{11}{10}\right)^3 = \text{Rs. 26620}\\\\ \text{Rs.20000 after 2 years } = 20000\left(1 + \dfrac{10}{100}\right)^2 = 20000\left(\dfrac{11}{10}\right)^2 = \text{Rs. 24200}\\\\ \text{Rs.20000 after 1 year } = 20000\left(1 + \dfrac{10}{100}\right)^1 = 20000\left(\dfrac{11}{10}\right) = \text{Rs. 22000}$MF#%
Total amount after 4 years = 29282 + 26620 + 24200 + 22000 = Rs. 102102