Answer : Option B

Explanation :

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Solution 1
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Let Principal = P
Rate of Interest = R%
$MF#%\text{Required Ratio = }\dfrac{\left(\dfrac{\text{PR} \times 5}{100}\right)}{\left(\dfrac{\text{PR}\times 15}{100}\right)}= \dfrac{5}{15}=\dfrac{1}{3}=1 : 3$MF#%
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Solution 2
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$MF#%\text{Simple Interest = }\dfrac{\text{PRT}}{100}$MF#%
Here Principal(P) and Rate of Interest (R) are constants
Hence, Simple Interest ∝ T
$MF#%\text{Required Ratio = }\dfrac{\text{Simple Interest for 5 years}}{\text{Simple Interest for 15 years}} = \dfrac{\text{T}_1}{\text{T}_2}=\dfrac{5}{15}=\dfrac{1}{3}=1 : 3$MF#%

Answer : Option A

Explanation :

Let the partial amount at 5% be x and the partial amount at 8% be (1550-x)
Interest on x at 5% for 3 years + interest on (1550-x) at 8% for 3 years = 300
$MF#%\begin{align}&\dfrac{\text{x} \times 5 \times 3}{100} + \dfrac{\text{(1550-x)} \times 8 \times 3}{100} = 300\\\\ &\dfrac{\text{x} \times 5}{100} + \dfrac{\text{(1550-x)} \times 8}{100} = 100\\\\ &5x + 8(1550-x) = 10000\\\\ &5x + 12400 - 8x = 10000\\\\ &3x = 2400\\\\ &x = 800\end{align}$MF#%
Required Ratio = x : (1550-x) = 800 : (1550-800) = 800 : 750 = 16 : 15

Let sum = Rs. x. Then, S.l. = Rs. 4x/9
Let rate = R% and time = R years.
Then, (x*R*R)/100=4x/9 or R2 =400/9 or R = 20/3 = 6 2/3%= 20/3%.
Rate = 6 2/3 % and Time = 6 2/3 years = 6 years 8 months.

Gain in 2 years = Rs. 5000 x 25 x 2 - 5000 x 4 x 2 4 100 100 = Rs. (625 - 400) = Rs. 225.

Gain in 1 year = Rs. 225 = Rs. 112.50 2

Answer : Option B

Explanation :

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Solution 1
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$MF#%\begin{align}&\dfrac{3000 \times 5 \times \text{T}}{100} - \dfrac{2000 \times 5 \times \text{T}}{100} = 50\\\\ &(3000-2000)\dfrac{5 \times \text{T}}{100} = 50\\\\ &1000 \times \dfrac{5 \times \text{T}}{100} = 50\\\\ &50 \text{T} = 50\\\\ &\text{T} = 1\text{ year}\end{align}$MF#%
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Solution 2
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Difference in principal = 3000-2000 = 1000
i.e., Simple Interest on Rs.1000 will be Rs.50
$MF#%\dfrac{1000 \times 5 \times \text{T}}{100} = 50\\\\ 50 \text{T} = 50\\\\ \text{T} = 1\text{ year}$MF#%

Answer : Option A

Explanation :

Let the sub be Rs.x and the initial rate be R%.Then
$MF#%\begin{align}&\dfrac{\text{x}\times (\text{R+4}) \times 2}{100} - \dfrac{\text{x}\times \text{R} \times 2}{100} = 60\\\\ &\Rightarrow \dfrac{\text{x}\times 4 \times 2}{100} = 60\\\\ &\Rightarrow \dfrac{\text{x} \times 2}{100} = 15\\\\ &\Rightarrow 2x = 1500\\\\ &\Rightarrow x = 750\end{align}$MF#%

Answer : Option D

Explanation :

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Solution 1
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Let his investments be Rs.12000 and Rs.x
Rs. 12000 is invested at the simple interest rate of 10% per annum for 1 year
$MF#%\text{Simple Interest = }\dfrac{\text{PRT}}{100} = \dfrac{12000 \times 10 \times 1}{100} = \text{Rs. 1200}$MF#%
Rs. x is invested at the simple interest rate of 20% per annum for 1 year
$MF#%\text{Simple Interest = }\dfrac{\text{PRT}}{100} = \dfrac{x \times 20 \times 1}{100} = \text{Rs.}\dfrac{x}{5}$MF#%
$MF#%\begin{align}&\text{Total interest = Rs.}\left(1200 + \dfrac{x}{5}\right)\\\\ &\text{i.e., Rs.}\left(1200 + \dfrac{x}{5}\right)\text{ is the simple interest for Rs.(12000 + x) at 14% per annum for 1 year}\\\\ &\Rightarrow \left(1200 + \dfrac{x}{5}\right) = \dfrac{(12000 + x) \times 14 \times 1}{100}\\\\ &\Rightarrow 120000 + 20x = 14 \times 12000 + 14x\\\\ &\Rightarrow 6x = 14 \times 12000 - 120000 = 48000\\\\ &\Rightarrow x = 8000\end{align}$MF#%
Total amount invested = 12000 + x = 12000 + 8000 = Rs. 20000
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Solution 2
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If an amount P₁ is lent out at simple interest of R₁% per annum and another amount P₂ at simple interest rate of R₂% per annum, then the rate of interest for the whole sum can be given by
$MF#%\text{R} = \dfrac{\text{P}_1\text{R}_1 + \text{P}_2\text{R}_2}{\text{P}_1+\text{P}_2}$MF#%

Answer : Option D

Explanation :

Simple Interest (SI) for 1 year = 854-815 = 39
Simple Interest (SI) for 3 years = 39 × 3 = 117
Principal = 815 - 117 = Rs.698

Answer : Option C

Explanation :

Let Simple Interest for Rs.150 at 6% for n years = Simple Interest for Rs.800 at 4½ % for 2 years
$MF#%\begin{align}&\dfrac{150 \times 6 \times n}{100} = \dfrac{800 \times \dfrac{9}{2} \times 2}{100} \\\\ &150 \times 6 \times n = 800 \times \dfrac{9}{2} \times 2 \\\\ &150 \times 6 \times n = 800 \times 9\\\\ &3 \times 6 \times n = 16 \times 9 \\\\
&6 \times n = 16 \times 3\\\\
&2 \times n = 16\\\\ &n = 8\text{ years }\end{align}$MF#%

Let the principal be P and rate of interest be R%.

Required ratio = P x R x 6 100 = 6PR = 6 = 2 : 3. P x R x 9 100 9PR 9