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A sum put out at 4% compound interest payable half-yearly amounts to Rs. 13265.10 in 11ΓΆΒβ€2 years. The sum is
Options:
A .  Rs. 12500
B .  Rs. 11200
C .  Rs. 8840
D .  Rs. 12600
Answer: Option A

Answer : Option A

Explanation :

Let the sum be P
$MF#%\text{Time, T} = 1\dfrac{1}{2}\text{ year} = \dfrac{3}{2}\text{ year}\\\\ \text{Amount after }1\dfrac{1}{2}\text{ years} = \text{P}\left(1 + \dfrac{\text{(R/2)}}{100}\right)^\text{2T} = \text{P}\left(1 + \dfrac{(4/2)}{100}\right)^{2 \times \frac{3}{2}} = \text{P}\left(1 + \dfrac{2}{100}\right)^3\\\\= \text{P}\left(\dfrac{102}{100}\right)^3=\text{P}\left(\dfrac{51}{50}\right)^3$MF#%
Given that amount after 11ΓΆΒβ€2 years = 13265.10
$MF#%=> \text{P}\left(\dfrac{51}{50}\right)^3 = 13265.10\\\\ => \text{P} = 13265.10\left(\dfrac{50}{51}\right)^3 = \dfrac{13265.10 \times 50 \times 50 \times 50}{51 \times 51 \times 51} = \dfrac{260.1 \times 50 \times 50 \times 50}{51 \times 51}= \dfrac{5.1 \times 50 \times 50 \times 50}{51}\\\\ = 0.1 \times 50 \times 50 \times 50 = \text{Rs. 12500}$MF#%



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