Answer : Option C
Explanation :
$MF#%\text{Amount after 3 years on Rs.20000 at 10% compound interest }\\\\=20000\left(1 + \dfrac{10}{100}\right)^3
= 20000\left(\dfrac{11}{10}\right)^3 = 26620 \quad \color{#F00}{\text{--- (1)}}$MF#%
He paid Rs.2000 after 1st year.
Hence Rs.2000 and its compound interest for 2 years (i.e., amount on 2000 after 2 year) need to be reduced from (1)
Similarly, he paid Rs.2000 after 2nd year.
Hence Rs.2000 and and its compound interest for 1 year (i.e., amount on 2000 after 1 year) need to be reduced from (1)
Similarly, he paid Rs.2000 after 3rd year.
Hence this Rs.2000 also need to be reduced from (1)
Hence, remaining amount
$MF#%= 26620 - 2000\left(1 + \dfrac{10}{100}\right)^2 - 2000\left(1 + \dfrac{10}{100}\right)^1 - 2000\\\\ = 26620 - 2000\left(\dfrac{11}{10}\right)^2 - 2000\left(\dfrac{11}{10}\right) - 2000\\\\ = 26620 - 2420 - 2200 - 2000 \\\\ = 20000$MF#%
i.e, he still owes Rs.20000 even after three installments
Answer : Option D
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 sum be x
Simple interest on x for 2 years = Rs.2000
$MF#%\text{Simple interest = }\dfrac{\text{PRT}}{100}\\\\ 2000 = \dfrac{x \times \text{R} \times 2}{100}\\\\ \Rightarrow x\text{R} = 100000 \quad \color{#F00}{\text{--- (1)}}$MF#%
Compound Interest on x for 2 years = 2400
$MF#%\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} - \text{P} = 2400\\\\ x\left(1 + \dfrac{\text{R}}{100}\right)^2 - x = 2400\\\\ x\left(1 + \dfrac{\text{2R}}{100} + \dfrac{\text{R}^2}{10000}\right) - x = 2400\\\\ x\left(\dfrac{\text{2R}}{100} + \dfrac{\text{R}^2}{10000}\right) = 2400\\\\ \dfrac{2x\text{R}}{100} + \dfrac{x\text{R}^2}{10000} = 2400 \quad \color{#F00}{\text{--- (2)}}\\\\$MF#%
Substituting the value of xR from (1) in (2) ,we get
$MF#%\dfrac{2 \times 100000}{100} + \dfrac{100000 \times \text{R}}{10000} = 2400\\\\ 2000 + 10 \text{R} = 2400\\\\ 10 \text{R} = 400\\\\ \text{R} = 40\%$MF#%
Answer : Option B
Explanation :
Let the period be n years
Then, amount after n years = Rs.(30000 + 4347) = Rs. 34347
$MF#%\begin{align}&\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 34347\\\\ &30000\left(1 + \dfrac{7}{100}\right)^\text{n} = 34347\\\\ &30000\left(\dfrac{107}{100}\right)^\text{n} = 34347\\\\ &\left(\dfrac{107}{100}\right)^\text{n} = \dfrac{34347}{30000} = \dfrac{11449}{10000} = \left(\dfrac{107}{100}\right)^2\\\\ &n = 2\text{ years}\end{align}$MF#%
Answer : Option C
Explanation :
$MF#%\text{P}\left(1 + \dfrac{10}{100}\right)^2 \times \left(1 + \dfrac{\dfrac{1}{2} \times 10}{100}\right) = 31762.5\\\\ \text{P}\left(\dfrac{11}{10}\right)^2 \times \left(\dfrac{21}{20}\right) = 31762.5\\\\ \text{P} = \dfrac{31762.5 \times 20 \times 10 \times 10}{21 \times11 \times 11} = \dfrac{1512.5
\times 20 \times 10 \times 10}{11 \times 11} = \dfrac{137.5 \times 20 \times 10 \times 10}{11} \\\\ = 12.5 \times 20 \times 10 \times 10 =\text{Rs. 25000}$MF#%
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
P(1+R/100)^15=2P
(1+R/100)^15=2P/P=2
LET P(1+R/100)^n=8P
(1+R/100)^n=8=23={(1+R/100)^15}^3
(1+R/100)^n=(1+R/100)^45
n=45
Answer : Option B
Explanation :
Amount after 1 year on Rs.2400 at 10% per annum when interest is reckoned half-yearly
$MF#%= \text{P}\left(1 + \dfrac{\text{(R/2)}}{100}\right)^\text{2T} = 2400\left(1 + \dfrac{(10/2)}{100}\right)^{2 \times 1}= 2400\left(1 + \dfrac{5}{100}\right)^2= 2400\left(\dfrac{105}{100}\right)^2 \\\\= \dfrac{2400 \times 105 \times 105}{100 \times 100} =\dfrac{24 \times 105 \times 105}{100} = \dfrac{24 \times 21 \times 105}{20}= \dfrac{24 \times 21 \times 21}{4} = 6 \times 21 \times 21 = 2646$MF#%
Compound Interest = 2646 - 2400 = Rs. 246
Simple Interest on Rs.2400 at 10% per annum for 1 year
$MF#%= \dfrac{\text{PRT}}{100} = \dfrac{2400 \times 10 \times 1}{100} = \text{Rs. }240$MF#%
Required difference = 246 - 240 = Rs.6
