Answer : Option A
Explanation :
Let the sum be Rs. P.
Amount after 2 years at 10% per annum when interest is compounded annually
$MF#%= \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = \text{P}\left(1 + \dfrac{10}{100}\right)^2
= \text{P}\left(\dfrac{110}{100}\right)^2 = \text{P}\left(\dfrac{11}{10}\right)^2\\\\ \text{Compound Interest = }\text{P}\left(\dfrac{11}{10}\right)^2 - \text{P} = \text{P}\left[\left(\dfrac{11}{10}\right)^2-1\right]$MF#%
Given that compound interest = 525
$MF#%\begin{align}&\Rightarrow \text{P}\left[\left(\dfrac{11}{10}\right)^2-1\right] = 525\\\\ &\Rightarrow \text{P}\left[\dfrac{121}{100}-1\right] = 525\\\\ &\Rightarrow \text{P}\times \dfrac{21}{100} = 525\\\\ &\Rightarrow \text{P} = 525\times \dfrac{100}{21} = 25 \times 100 = 2500\end{align}$MF#%
Simple interest on the same sum(Rs.2500) for 4 years at 5%
$MF#%= \dfrac{\text{PRT}}{100}=\dfrac{2500 \times 5 \times 4}{100} = 25 \times5 \times 4 = 25 \times 20 =\text{Rs. 500}$MF#%
Answer : Option C
Explanation :
Present worth of Rs. x due T years hence is given by
$MF#%\text{Present Worth (PW) = }\dfrac{x}{\left(1 + \dfrac{\text{R}}{100}\right)^\text{T}}$MF#%
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
Answer : Option C
Explanation :
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Solution 1
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Let the rate of interest per annum be R%
$MF#%\begin{align}&\text{Amount after 2 years on Rs.15000 when interest is compounded annually }\\\\ &= \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 15000\left(1 + \dfrac{\text{R}}{100}\right)^2\\\\&\text{Compound Interest =}15000\left(1 + \dfrac{\text{R}}{100}\right)^2 - 15000
= 15000\left[\left(1 + \dfrac{\text{R}}{100}\right)^2 - 1\right]\\\\ &\text{Simple Interest = }\dfrac{\text{PRT}}{100} = \dfrac{15000 \times \text{R} \times 2}{100} = \text{300R}\end{align}$MF#%
Difference between compound interest and simple interest = Rs.96
$MF#%\begin{align}&15000\left[\left(1 + \dfrac{\text{R}}{100}\right)^2 - 1\right] - \text{300R} = 96\\\\ &15000\left[1 + \dfrac{\text{2R}}{100} + \left(\dfrac{\text{R}}{100}\right)^2 - 1\right] - \text{300R} = 96\\\\ &15000\left[\dfrac{\text{2R}}{100} + \left(\dfrac{\text{R}}{100}\right)^2\right] - \text{300R} = 96\\\\ &300\text{R} + 15000\left(\dfrac{\text{R}}{100}\right)^2 - \text{300R} = 96\\\\ &15000\left(\dfrac{\text{R}}{100}\right)^2= 96\\\\ &15000\left(\dfrac{\text{R}^2}{10000}\right)= 96\\\\ &3\left(\dfrac{\text{R}^2}{2}\right)= 96\\\\ &\text{R}^2 = 64\\\\ &\text{R}= 8\end{align}$MF#%
Rate of interest per annum = 8%
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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
$MF#%= \text{P}\left(\dfrac{\text{R}}{100}\right)^2$MF#%
Answer : Option C
Explanation :
Let the sum be P
The sum P becomes 3P in 4 years on compound interest
$MF#%3\text{P} = \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^4\\\\ \Rightarrow 3 = \left(1 + \dfrac{\text{R}}{100}\right)^4$MF#%
Let the sum P becomes 81P in n years
$MF#%81\text{P} = \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^n\\\\ \Rightarrow 81 = \left(1 + \dfrac{\text{R}}{100}\right)^n \\\\ \Rightarrow (3)^4 = \left(1 + \dfrac{\text{R}}{100}\right)^n \\\\ \Rightarrow \left(\left(1 + \dfrac{\text{R}}{100}\right)^4\right)^4 = \left(1 + \dfrac{\text{R}}{100}\right)^\text{n} \\\\ \Rightarrow \left(1 + \dfrac{\text{R}}{100}\right)^{16}= \left(1 + \dfrac{\text{R}}{100}\right)^\text{n} \\\\ \text{n} = 16$MF#%
i.e, the sum will become 81 times in 16 years
Answer : Option C
Explanation :
Let sum be P
$MF#%\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 5458.32\\\\ \text{P}\left(1 + \dfrac{14}{100}\right)^2 = 5458.32\\\\ \text{P}\left(\dfrac{114}{100}\right)^2 = 5458.32\\\\ \text{P} = \dfrac{ 5458.32 \times 100 \times 100}{114 \times 114} = \dfrac{ 47.88 \times 100 \times 100}{114} = 0.42 \times 100 \times 100 = 4200$MF#%
Answer : Option D
Explanation :
Let the sum be P
Compound Interest on P at 10% for 2 years when interest is compounded half-yearly
$MF#%=\text{P}\left(1 + \dfrac{\text{(R/2)}}{100}\right)^\text{2T} - \text{P}
= \text{P}\left(1 + \dfrac{(10/2)}{100}\right)^{2 \times 2} - \text{P} = \text{P}\left(1 + \dfrac{1}{20}\right)^4- \text{P} = \text{P}\left(\dfrac{21}{20}\right)^4- \text{P}\\\\ \text{Simple Interest on P at 10% for 2 years = }\dfrac{\text{PRT}}{100} = \dfrac{\text{P} \times 10 \times 2}{100}=\dfrac{\text{P}}{5}$MF#%
Given that difference between compound interest and simple interest = 124.05
$MF#%=> \text{P}\left(\dfrac{21}{20}\right)^4- \text{P} - \dfrac{\text{P}}{5} = 124.05\\\\ => \text{P}\left[\left(\dfrac{21}{20}\right)^4 - 1 - \dfrac{1}{5}\right]= 124.05\\\\ => \text{P}\left[\dfrac{194481}{160000} - 1 - \dfrac{1}{5}\right]= 124.05\\\\ => \text{P}\left[\dfrac{194481 - 160000 - 32000}{160000}\right]= 124.05\\\\ => \text{P}\left[\dfrac{2481}{160000}\right]= 124.05\\\\ => \text{P} = \dfrac{124.05 \times 160000}{2481}=\dfrac{160000}{20}=8000$MF#%
Answer : Option A
Explanation :
Present worth of Rs. x due T years hence is given by
$MF#%\text{Present Worth (PW) = }\dfrac{x}{\left(1 + \dfrac{\text{R}}{100}\right)^\text{T}}$MF#%
