Principal = Rs. 16000; Time = 9 months =3 quarters;
Rate = 20% per annum = 5% per quarter.
Amount = Rs. [16000 x (1+(5/100))3] = Rs. 18522
CJ. = Rs. (18522 - 16000) = Rs. 2522
Principal = Rs. 500; Amount = Rs. 583.20; Time = 2 years.
Let the rate be R% per annum.
[ 500 (1+(R/100)^2 ] = 583.20 or
[ 1+ (R/100)]^2 = 5832/5000 = 11664/10000
[ 1+ (R/100)]^2 = (108/100)^2
1 + (R/100) = 108/100
R = 8
Let the sum be Rs. x.
C.I. = [ x * (1 + (( 50/(3*100))^3 - x ] = ((343x / 216) - x) = 127x / 216
127x /216 = 1270
x = (1270 * 216) / 127 = 2160
Thus, the sum is Rs. 2160
S.I.= Rs( 2160 * (50/3) * 3 * (1 /100 ) ) = Rs. 1080
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 :
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 B
Explanation :
Let the rate of interest be R% per annum.
Assume that Rs. 10000 amount to Rs. 160000 in T years
$MF#%10000\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 160000\\\\ => \left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = \dfrac{160000}{10000} = 16\\\\ => \left(1 + \dfrac{\text{R}}{100}\right)^\text{T/2} = \sqrt{16}= 4 \quad \color{#F00}{\text{--- (1)}}$MF#%
$MF#%\text{In T/2 years, Rs.10000 amounts to }10000\left(1 + \dfrac{\text{R}}{100}\right)^\text{T/2} \\\\ = 10000 \times 4 \quad \color{#F00}{ \text{[∵ from (1)]}}\\\\ = 40000$MF#%
Answer : Option C
Explanation :
Interest earned in 3rd year = Rs. 1596
Interest earned in 2nd year = Rs. 1400
i.e, in 3rd year, Andrews gets additional interest of (Rs. 1596 - Rs. 1400) = Rs.196
This means, Rs.196 is the interest obtained for Rs.1400 for 1 year
$MF#%\text{R} = \dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times 196}{1400 \times 1} = \dfrac{196}{14} = 14\%$MF#%
Answer : Option D
Explanation :
$MF#%1386 = \text{P}\left(1 + \dfrac{5}{100}\right)\left(1 + \dfrac{10}{100}\right)\left(1 + \dfrac{20}{100}\right)\\\\ 1386 = \text{P}\left(\dfrac{21}{20}\right)\left(\dfrac{11}{10}\right)\left(\dfrac{6}{5}\right)\\\\ \text{P} = \dfrac{1386 \times 20 \times 10 \times 5}{21 \times 11 \times 6} = \dfrac{66\times 20 \times 10 \times 5}{11 \times 6} = 20 \times 10 \times 5 = \text{Rs. 1000}$MF#%
i.e., the sum is Rs.1000
C.I. when interest
compounded yearly = Rs. 
5000 x 
1 + 4 
x 1 + 
x 4 
100 100 = Rs. 
5000 x 26 x 51 
25 50 = Rs. 5304.
C.I. when interest is
compounded half-yearly = Rs. 5000 x 1 + 2 3 100 = Rs. 5000 x 51 x 51 x 51 50 50 50 = Rs. 5306.04
Difference = Rs. (5306.04 - 5304) = Rs. 2.04
