Answer : Option A

Explanation :

A's share after 5 years = B's share after 7 years
$MF#%\text{(A's present share)}\left(1 + \dfrac{5}{100}\right)^5 = \text{(B's present share)}\left(1 + \dfrac{5}{100}\right)^7\\\\ => \dfrac{\text{(A's present share)}}{\text{(B's present share)}}=\dfrac{\left(1 + \dfrac{5}{100}\right)^7}{\left(1 + \dfrac{5}{100}\right)^5} = \left(1 + \dfrac{5}{100}\right)^{(7-5)} = \left(1 + \dfrac{5}{100}\right)^2 = \left(\dfrac{21}{20}\right)^2 = \dfrac{441}{400}$MF#%
i.e, A's present share : B's present share = 441 : 400
$MF#%\text{Since the total present amount is Rs.3364, A's share = }3364 \times \dfrac{441}{(441+400)} \\\\ = 3364 \times \dfrac{441}{841} = 4 \times 441 = \text{ Rs. 1764}$MF#%
B's present share = 3364 - 1764 = Rs.1600

Amount = Rs [7500*(1+(4/100)^2] = Rs (7500 * (26/25) * (26/25)) = Rs. 8112
therefore, C.I. = Rs. (8112 - 7500) = Rs. 612.

Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.
Amount = Rs. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]
=Rs.[8000 * (23/20) * (23/20) * (21/20)]=Rs. 11109
C.I.=Rs.(11109-8000)=Rs.3109.

Clearly, Rate=5% p.a .,
Time=3 years
S.I =Rs.1200.
So,Principal=Rs.(100 x 1200/3x5)
=Rs.8000.
Amount=Rs.[8000 x (1+5/100)³]
=Rs(8000x21/20x21/20x21/20)
=Rs.9261
C.I=Rs.(9261-8000)
=Rs.1261.

Answer : Option B

Explanation :

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Solution 1
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$MF#%\text{Principal = }\dfrac{100 \times \text{SI}}{\text{RT}}= \dfrac{100 \times 60}{5 \times 2} = \text{Rs. 600}\\\\ \text{Compound Interest = }\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} - \text{P}\\\\ = 600\left(1 + \dfrac{5}{100}\right)^2 - 600 = 600\left(\dfrac{21}{20}\right)^2 - 600 = 661.5 - 600 = \text{Rs. 61.5}$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#% = \dfrac{\text{R} \times \text{SI}}{2 \times 100} $MF#%

Answer : Option D

Explanation :

$MF#%1386 = \text{P}\left(1 + \dfrac{5}{100}\right)\left(1 + \dfrac{10}{100}\right)\left(1 + \dfrac{20}{100}\right)\\\\ 1386 = \text{P}\left(\dfrac{21}{20}\right)\left(\dfrac{11}{10}\right)\left(\dfrac{6}{5}\right)\\\\ \text{P} = \dfrac{1386 \times 20 \times 10 \times 5}{21 \times 11 \times 6} = \dfrac{66\times 20 \times 10 \times 5}{11 \times 6} = 20 \times 10 \times 5 = \text{Rs. 1000}$MF#%
i.e., the sum is Rs.1000

Principal = Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a.
Let the time be n years. Then,
[ 1000 (1+ (10/100))^n ] = 1331 or (11/10)^n = (1331/1000) = (11/10)^3
n = 3 years

Answer : Option C

Explanation :

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Solution 1
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Simple Interest for 2 years = Rs.200
$MF#%\text{Sum, P} = \dfrac{100 \times \text{SI}}{\text{RT}} = \dfrac{100 \times 200}{7 \times 2} = \dfrac{100 \times 100}{7}\\\\ \text{Compound Interest = }\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} - \text{P}\\\\ = \text{P}\left(1 + \dfrac{7}{100}\right)^2 - \text{P} = \text{P}\left[\left(1 + \dfrac{7}{100}\right)^2 - 1\right] = \text{P}\left(1 + \dfrac{14}{100} + \dfrac{49}{10000} - 1\right)=\text{P}\left(\dfrac{14}{100} + \dfrac{49}{10000}\right) \\\\= \dfrac{100 \times 100}{7}\left(\dfrac{14}{100} + \dfrac{49}{10000}\right)=200 + 7 = \text{Rs. }207$MF#%
Required Difference = 207 - 200 = Rs.7
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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#% = \dfrac{\text{R} \times \text{SI}}{2 \times 100} $MF#%

Amount=Rs(30000+4347)
=Rs.34347.
Let the time be n years.
Then,30000(1+7/100)^n
=34347.
=34347/3000
=11449/1000
=(107/100)^n
n=2years.

Answer : Option D

Explanation :

Let the sum be Rs.1 which becomes Rs.2 after 4 years
$MF#%\Rightarrow 2 = 1\left(1 + \dfrac{\text{R}}{100}\right)^4 \quad \color{#F00}{\text{--- (equation 1)}}$MF#%
Let the sum of Rs.1 becomes Rs.8 after n years
$MF#%\Rightarrow 8 = 1\left(1 + \dfrac{\text{R}}{100}\right)^n \quad \color{#F00}{\text{--- (equation 2)}}\\\\ \Rightarrow (2)^3 = 1\left(1 + \dfrac{\text{R}}{100}\right)^n \\\\ \Rightarrow \left[1\left(1 + \dfrac{\text{R}}{100}\right)^4
\right]^3 = 1\left(1 + \dfrac{\text{R}}{100}\right)^n \quad \color{#F00}{\text{(∵ replaced 2 with the value in equation 1)}}\\\\ \left(1 + \dfrac{\text{R}}{100}\right)^{12} = \left(1 + \dfrac{\text{R}}{100}\right)^n\\\\ \text{n} = 12$MF#%
i.e., the sum amounts to 8 times in 12 years