John invested money i

Question

John invested money in two schemes A and B offering compound interest @ 5 p.c.p.a. and 10 p.c.p.a. respectively. If the total amount of interest accrued through two schemes together in two years was Rs. 2075 and the total amount invested was Rs. 15,000, find out the amount invested in Scheme A?

Options:

A . Rs. 10000

B . Rs. 8000

C . Rs. 12000

D . Rs. 14000

Answer: Option A

Answer : Option A

Explanation :

Let the investment in scheme A be Rs.x
Then the investment in scheme = Rs.(15000-x)
$MF#%\text{Compound interest for Rs.x = }\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} - \text{P} = x\left(1 + \dfrac{5}{100}\right)^2 - x\\\\ \text{Compound interest for Rs. (15000-x) = }\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} - \text{P} = (15000-x)\left(1 + \dfrac{10}{100}\right)^2 - (15000-x)\\\\ \text{Total compound interest = }x\left(1 + \dfrac{5}{100}\right)^2 - x + (15000-x)\left(1 + \dfrac{10}{100}\right)^2 - (15000-x)\\\\ = x\left(1 + \dfrac{10}{100} + \dfrac{25}{10000}\right) - x + (15000-x)\left(1 + \dfrac{20}{100} + \dfrac{100}{10000}\right) - (15000-x)\\\\ = x\left(\dfrac{10}{100} + \dfrac{25}{10000}\right) + (15000-x)\left(\dfrac{20}{100} + \dfrac{100}{10000}\right) \\\\ = \dfrac{1025x}{10000} + (15000-x)\left(\dfrac{2100}{10000}\right) \\\\ = \dfrac{1025x}{10000} + 15000 \times \dfrac{2100}{10000} - \dfrac{2100x}{10000} \\\\ = \dfrac{-1075x}{10000} + 3150$MF#%
Given that total compound interest = Rs. 2075
$MF#%\dfrac{-1075x}{10000} + 3150 = 2075\\\\ => \dfrac{1075x}{10000} = 1075\\\\ => \dfrac{x}{10000} = 1\\\\ => x =10000$MF#%
i.e, Amount invested in Scheme A = Rs. 10000

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