Answer : Option A

Explanation :

$MF#%\text{Amount after 3 years = }\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 40000\left(1 + \dfrac{11}{100}\right)^3 \\\\= 40000\left(\dfrac{111}{100}\right)^3 = \dfrac{40000 \times 111 \times 111 \times 111}{100 \times 100 \times 100} = \dfrac{4\times 111 \times 111 \times 111}{ 100} = 54705.24$MF#%
Compound Interest = 54705.24 - 40000 = Rs. 14705.24

Let each installment be Rs.x
(P.W. of Rs.x due 1yr hence)+(P.W of Rs.x due 2yr hence)+(P.W of Rs. X due 3yr hence)=7620.
x/(1+(50/3*100))+ x/(1+(50/3*100))^2 + x/(1+(50/3*100))^3=7620
(6x/7)+(936x/49)+(216x/343)=7620
294x+252x+216x=7620*343
x=(7620*343/762)=3430
Amount of each installment=Rs.3430

Answer : Option A

Explanation :

Amount after 1¹â„₂ years when interest is compounded yearly
$MF#%= 5000 \times \left(1 + \dfrac{4}{100}\right)^1\times \left(1 + \dfrac{\dfrac{1}{2} \times 4}{100}\right)
= 5000 \times \dfrac{104}{100} \times \left(1 + \dfrac{2}{100}\right) \\\\ = 5000 \times \dfrac{104}{100} \times \dfrac{102}{100} = 50 \times 104 \times \dfrac{51}{50} \\\\ = 104 \times 51 = \text{Rs. }5304$MF#%
Compound Interest for 1 ¹â„₂ years when interest is compounded yearly
= Rs.(5304 - 5000)
Amount after 1¹â„₂ years when interest is compounded half-yearly
$MF#%= \text{P}\left(1 + \dfrac{\text{(R/2)}}{100}\right)^\text{2T} = 5000\left(1 + \dfrac{(4/2)}{100}\right)^{2 \times \frac{3}{2}} = 5000\left(1 + \dfrac{2}{100}\right)^3\\\\ = 5000\left(\dfrac{102}{100}\right)^3 = 5000\left(\dfrac{102}{100}\right)\left(\dfrac{102}{100}\right)\left(\dfrac{102}{100}\right) = 50 \times 102 \times \dfrac{51}{50}\times \dfrac{51}{50} \\\\ = 102 \times 51 \times \dfrac{51}{50} = 51 \times 51 \times \dfrac{51}{25} = \text{Rs. } 5306.04$MF#%
Compound Interest for 1 ¹â„₂ years when interest is compounded half-yearly
= Rs.(5306.04 - 5000)
Difference in the compound interests = (5306.04 - 5000) - (5304 - 5000) = 5306.04 - 5304 = Rs. 2.04