Answer : Option C

Explanation :

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Solution 1
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Simple interest for 2 years is Rs. 320
=> Simple interest for first year = 320/2 = 160
=> Similarly, simple interest for second year is also 160
Compound Interest for first year = 160
Compound Interest for second year = 340-160 = 180
we can see that compound Interest for second year is more than
simple interest for second year by 180-160 = 20
i.e., Rs.20 is the simple interest on Rs.160 for 1 year
$MF#%\text{R} = \dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times 20}{160 \times 1} = 12.5\%$MF#%
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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#% = \dfrac{\text{R} \times \text{SI}}{2 \times 100} $MF#%

Answer : Option B

Explanation :

Let principal be P
$MF#%\begin{align}&\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} > 2P\\\\ &\text{P}\left(1 + \dfrac{20}{100}\right)^\text{T} > 2P\\\\ &\left(1 + \dfrac{20}{100}\right)^\text{T} > 2\\\\ &\left(\dfrac{120}{100}\right)^\text{T} > 2\\\\ &1.2^\text{T} > 2\end{align}$MF#%
Now let's find out the minimum value of T for which the above equation becomes true
If T = 1, 1.2^T = 1.2¹ = 1.2
If T = 2, 1.2^T = 1.2² ≈ 1.4
If T = 3, 1.2^T = 1.2³ ≈ 1.7
If T = 4, 1.2^T = 1.2⁴ ≈ 2.07 which is greater than 2
Hence T = 4
i.e., required number of years = 4

Let the rate be R% p.a.
[ 18000 ( 1 + ( R / 100 )^2 ) - 18000 ] - ((18000 * R * 2) / 100 ) = 405
18000 [ ( 100 + (R / 100 )^2 / 10000) - 1 - (2R / 100 ) ] = 405
18000[( (100 + R )^ 2 - 10000 - 200R) / 10000 ] = 405
9R^2 / 5 = 405
R^2 =((405 * 5 ) / 9) = 225
R = 15.
Rate = 15%.

Answer : Option B

Explanation :

This problem is similar to the problems we saw in compound interest.
We can use the formulas of compound interest here as well.
$MF#%\text{Rate of increase = }\dfrac{1}{5} \times 100 = 20\%\\\\ \text{Height after 2 years = }\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 50\left(1 + \dfrac{20}{100}\right)^2 \\\\ = 50\left(1 + \dfrac{1}{5}\right)^2 = 50\left(\dfrac{6}{5}\right)^2 = \dfrac{50 \times 6\times 6}{5\times 5} = 2 \times 6 \times 6 = 72\text{ cm}$MF#%

Amount = Rs. 1600 x 1 + 5 2 + 1600 x 1 + 5 2 x 100 2 x 100 = Rs. 1600 x 41 x 41 + 1600 x 41 40 40 40 = Rs. 1600 x 41 41 + 1 40 40 = Rs. 1600 x 41 x 81 40 x 40 = Rs. 3321.

C.I. = Rs. (3321 - 3200) = Rs. 121

Answer : Option B

Explanation :

$MF#%\text{Simple Interest = }\dfrac{\text{PRT}}{100} = \dfrac{900 \times 10 \times 1}{100} = \text{ Rs. }90$MF#%
Amount after 1 year on Rs.900 at 10% per annum when interest is reckoned half-yearly
$MF#%= \text{P}\left(1 + \dfrac{\text{(R/2)}}{100}\right)^\text{2T} = 900\left(1 + \dfrac{(10/2)}{100}\right)^{2 \times 1} = 900\left(1 + \dfrac{5}{100}\right)^{2} = 900\left(\dfrac{105}{100}\right)^{2}\\\\ = \dfrac{900\times 105 \times 105}{100 \times 100} = \text{Rs. 992.25}$MF#%
Compound Interest = 992.25 - 900 = 92.25
Required difference between simple interest and compound interest = 92.25 - 90 = Rs.2.25

Answer : Option B

Explanation :

Let the sum be P and rate of interest be R% per annum.
Amount after 3 years = 2200
$MF#%\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 2200\\\\ \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^3 = 2200 \quad \color{#F00}{\text{--- ( 1)}}$MF#%
Amount after 6 years = 4400
$MF#%\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 4400\\\\ \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^6 = 4400 \quad \color{#F00}{\text{--- (2)}}$MF#%
$MF#%\color{#F00}{\text{(2)÷(1) =>}} \dfrac{\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^6 }{\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^3} = \dfrac{4400}{2200} = 2\\\\ => \left(1 + \dfrac{\text{R}}{100}\right)^3 = 2 \\\\ \color{#F00}{\text{(Substituting this value in equation 1) => }} \quad \text{P} \times 2 = 2200\\\\ \text{P} = \dfrac{2200}{2} = 1100$MF#%

Answer : Option C

Explanation :

Let the rate be R% per annum
$MF#%
\begin{align}&\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 1573.04\\\\ &1400\left(1 + \dfrac{\text{R}}{100}\right)^2 = 1573.04\\\\ &\left(1 + \dfrac{\text{R}}{100}\right)^2 = \dfrac{1573.04}{1400} = \dfrac{157304}{140000} = \dfrac{11236
}{10000}\\\\&\left(1 + \dfrac{\text{R}}{100}\right) = \sqrt{\dfrac{11236}{10000}} = \dfrac{\sqrt{11236}}{\sqrt{10000}} =\dfrac{106}{100} \\\\\ &\dfrac{\text{R}}{100} = \dfrac{106}{100} - 1 = \dfrac{6}{100}\\\\&\text{R} = 6\%\end{align}
$MF#%

Clearly, Rate = 5% p.a., Time = 3 years, S.I.= Rs. 1200
So principal=Rs. [100*1200]/3*5=Rs. 8000
Amount = Rs. 8000 x [1 +5/100]^3 - = Rs. 9261
C.I. = Rs. (9261 - 8000) = Rs. 1261

Let the sum be Rs.P
P(1+R/100)^3=6690…(i)
P(1+R/100)^6=10035…(ii)
On dividing,we get (1+R/100)^3=10025/6690=3/2.
Substituting this value in (i),we get:
P*3/2=6690
P=(6690*2/3)=4460
Hence,the sum is Rs.4460