P = Rs.68000,R = 50/3% p.a and T = 9/12 years = 3/4years.
S.I. = (P*R*T)/100 = Rs.(68,000*(50/3)*(3/4)*(1/100))= Rs.8500

Answer : Option A

Explanation :

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Solution 1
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The amount needs to be repaid in 4 years = Rs.6450
Suppose Rs.x is paid annually to repay this debt
Then, amount paid after 1st year = Rs.x
$MF#%\text{Interest for this Rs.x for the remaining 3 years = }\dfrac{x \times 5 \times 3}{100}= \dfrac{15x}{100}$MF#%
Then, amount paid after 2nd year = Rs.x
$MF#%\text{Interest for this Rs.x for the remaining 2 years = }\dfrac{x \times 5 \times 2}{100}= \dfrac{10x}{100}$MF#%
Amount paid after 3rd year = Rs.x
$MF#%\text{Interest for this Rs.x for the remaining 1 year = }\dfrac{x \times 5 \times 1}{100}= \dfrac{5x}{100}$MF#%
Amount paid after 4th year = Rs.x and this closes the entire debt
$MF#%\begin{align}&=> x + \dfrac{15x}{100} + x + \dfrac{10x}{100} + x + \dfrac{5x}{100} + x = 6450\\\\ &=> 4x + \dfrac{30x}{100} = 6450\\\\ &=> 4x + \dfrac{3x}{10} = 6450\\\\ &=> 40x + 3x = 64500\\\\ &=> 43x = 64500\\\\&=> x = 1500\end{align}$MF#%
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Solution 2 (using formula)
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The annual instalment which will discharge a debt of D due in T years at R% simple interest per annum
$MF#%= \dfrac{\text{100D}}{\text{100T} + \dfrac{\text{RT(T-1)}}{2}}$MF#%

Answer : Option D

Explanation :

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Solution 1
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Let Vikas, Vijay and Viraj gets Rs.x, Rs.y and Rs.z respectively.
Simple Interest on x at 5% for 1 year
= Simple Interest on y at 5% for 2 year
= Simple Interest on z at 5% for 3 year
$MF#%\begin{align}&\Rightarrow \dfrac{x \times 5 \times 1}{100} = \dfrac{\text{y} \times 5 \times 2}{100} = \dfrac{\text{z} \times 5 \times 3}{100}\\\\ &\Rightarrow5x = 10\text{y} = 15\text{z}\\\\ &\Rightarrow x = 2\text{y} = 3\text{z}\\\\ &\Rightarrow \text{y} = \dfrac{x}{2} \text{ and }\text{z} = \dfrac{x}{3} \quad \color{#F00}{\text{--- (1)}}\\\\ &\text{x + y + z = 7700 } \quad \color{#F00}{\text{(∵ the total amount is Rs. 7700)}} \\\\ &\Rightarrow x +\dfrac{x}{2}+\dfrac{x}{3} = 7700 \quad \color{#F00}{(∵ \text{substituted the values of y and z from from equation 1)}}\\\\ &\Rightarrow \dfrac{11x}{6} = 7700\\\\ &\Rightarrow \dfrac{x}{6} = 7000\\\\ &\Rightarrow x = 4200\\\\ &\text{z} = \dfrac{x}{3} = \dfrac{4200}{3} = 1400\end{align}$MF#%
i.e, Vikas gets Rs.4200 and Viraj gets Rs.1400
Share of Vikas is more than that of Viraj by (4200 - 1400) = 2800
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Solution 2
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If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R₁, R₂, ... , R_n respectively and time periods are T₁, T₂, ... , T_n respectively, then the ratio in which the sum will be divided in n parts can be given by
$MF#%\dfrac{1}{\text{R}_1\text{T}_1} : \dfrac{1}{\text{R}_2\text{T}_2} : \cdots \dfrac{1}{\text{R}_\text{n}\text{T}_\text{n}}$MF#%

Answer : Option B

Explanation :

Simple Interest, SI = 220 - 150 = Rs.70
$MF#%\text{R = }\dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times 70}{150 \times 10} = \dfrac{2 \times 7}{3} = \dfrac{14}{3} \%$MF#%

Answer : Option D

Explanation :

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Solution 1
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Let the principal = Rs.x
and time = y years
Principal,x amounts to Rs.400 at 10% per annum in y years
Simple Interest = (400-x)
$MF#%\begin{align}&\text{Simple Interest = }\dfrac{\text{PRT}}{100}\\\\ &\Rightarrow (400-x) = \dfrac{x \times 10 \times y}{100}\\\\ &\Rightarrow (400-x) = \dfrac{xy}{10} \quad \color{#F00}{\text{--- (equation 1)}}\end{align}$MF#%
Principal,x amounts to Rs.200 at 4% per annum in y years
Simple Interest = (200-x)
$MF#%\begin{align}&\text{Simple Interest = }\dfrac{\text{PRT}}{100}\\\\ &\Rightarrow (200-x) = \dfrac{x \times 4 \times y}{100}\\\\ &\Rightarrow (200-x) = \dfrac{xy}{25} \quad \color{#F00}{\text{--- (equation 2)}}\end{align}$MF#%
$MF#%\begin{align}&\color{#F00}{\dfrac{\text{(equation 1)}}{\text{(equation 2)}} \Rightarrow} \dfrac{400-x}{200-x} = \dfrac{\left(\dfrac{xy}{10}\right)}{\left(\dfrac{xy}{25}\right)}\\\\ &\Rightarrow \dfrac{400-x}{200-x} = \dfrac{25}{10}\\\\ &\Rightarrow \dfrac{400-x}{200-x} = \dfrac{5}{2}\\\\ &\Rightarrow 800 - 2x = 1000 - 5x\\\\ &\Rightarrow 200 = 3x\\\\ &\Rightarrow x = \dfrac{200}{3}\\\\ &\color{#F00}{\text{Substituting this value of x in Equation 1, we get,}}\left(400 - \dfrac{200}{3}\right) = \dfrac{\left(\dfrac{200}{3}\right)y}{10}\\\\ &\Rightarrow \left(400 - \dfrac{200}{3}\right) = \dfrac{20y}{3}\\\\ &\Rightarrow 1200 - 200 = 20y\\\\ &\Rightarrow 1000 = 20y\\\\ &y = \dfrac{1000}{20} = 50\text{ years}\end{align}$MF#%
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Solution 2
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If a certain sum of money P lent out for a certain time T amounts to P₁ at R₁% per annum and to P₂ at R₂% per annum, then
$MF#%\text{P = }\dfrac{\text{P}_2\text{R}_1 - \text{P}_1\text{R}_2}{\text{R}_1-\text{R}_2}$MF#%
$MF#%\text{T = }\dfrac{\text{P}_1 - \text{P}_2}{\text{P}_2\text{R}_1 - \text{P}_1\text{R}_2} \times 100 \text { years} $MF#%

Answer : Option D

Explanation :

This means, simple interest at 4% for that principal is Rs.120
$MF#%\text{P} = \dfrac{100 \times \text{SI}}{\text{RT}} = \dfrac{100 \times 120}{4 \times 6} = \dfrac{100 \times 30}{6} = 100 \times 5 = 500$MF#%

Answer : Option D

Explanation :

P = Rs. 500
SI = Rs.500
T = 4
R = ?
$MF#%\text{ R } = \dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times 500}{500 \times 4} = \dfrac{100}{4} = 25\%$MF#%

Answer : Option A

Explanation :

Let the sum of Rs.725 is lent out at rate R% for 1 year
Then, at the end of 8 months, ad additional sum of 362.50 more is lent out at rate 2R% for remaining 4 months(1/3 year)
Total Simple Interest = 33.50
$MF#%\begin{align}&=> \dfrac{725 \times \text{R} \times1 }{100} + \dfrac{362.50 \times \text{2R} \times \dfrac{1}{3}}{100} = 33.50\\\\ &=> \dfrac{725 \times \text{R} \times1 }{100} + \dfrac{362.50 \times \text{2R} }{300} = 33.50\\\\ &=> \dfrac{725\text{R}}{100} + \dfrac{725\text{R} }{300} = 33.50\\\\ &=> 725\text{R}\left(\dfrac{1}{100} + \dfrac{1}{300}\right) = 33.50\\\\ &=> 725\text{R}\left(\dfrac{4}{300}\right) = 33.50\\\\ &=>725\text{R} \times 4 = 10050\\\\ &=>725\text{R} = 2512.5\\\\ &=>145\text{R} = 502.5\\\\ &=>\text{R} = \dfrac{502.5}{145} = 3.46\%\end{align}$MF#%

Answer : Option D

Explanation :

$MF#%\text{P} = \dfrac{100 \times \text{SI}}{\text{RT}} = \dfrac{100 \times 360}{12 \times 4} = \dfrac{100 \times 30}{4}= 25 \times 30 = 750$MF#%

Answer : Option A

Explanation :

$MF#%\text{Time, T = (22 + 30 + 21)days = 73 days = }\dfrac{73}{365}\text{ year} = \dfrac{1}{5}\text{ year}\\\\ \text{Rate, R = }7\dfrac{1}{2}\% = \dfrac{15}{2}\%\\\\ \text{SI = }\dfrac{\text{PRT}}{100} = \dfrac{1820 \times \dfrac{15}{2} \times \dfrac{1}{5}}{100} = \dfrac{1820 \times \dfrac{3}{2} }{100} = \dfrac{910 \times 3}{100} = \dfrac{2730}{100}=27.30$MF#%