P = Rs. 68000,        R = `50/3` % ,   p.a and T = `9/12` years = `3/4` years.

`:.`      S.I. =  `((P xx R xx T)/(100))` 

=Rs.`(68000 xx 50/3 xx 3/4 xx 1/100)` = Rs.8500

Answer : Option A

Explanation :

Let the sum lent to C be Rs. x
Simple Interest for Rs.1500 at 8% per annum for 4 years
+ Simple Interest for Rs.x at 8% per annum for 4 years = Rs.1400
$MF#%\Rightarrow \dfrac{1500 \times 8 \times 4}{100} +\dfrac{x\times 8 \times 4}{100}= 1400\\\\ \Rightarrow 480 +\dfrac{32x}{100}= 1400\\\\ \Rightarrow \dfrac{32x}{100} = 920\\\\ \Rightarrow x = \dfrac{920 \times 100}{32} = 2875$MF#%

S.I. = Rs. (15500 - 12500) = Rs. 3000.

Rate = 100 x 3000 % = 6% 12500 x 4

Answer : Option A

Explanation :

SI = Rs.929.20
P = ?
T = 5 years
R = 8%
$MF#%\text{P = }\dfrac{100 \times \text{SI}}{\text{RT}}=\dfrac{100 \times 929.20}{8 \times 5}\text{ = Rs.2323}$MF#%

Answer : Option D

Explanation :

Simple Interest, SI = (3875 - 2500) = Rs.1375
Principal, P = Rs. 2500
Time, T = 4 years
R = ?
$MF#%\text{R} = \dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times 1375}{2500 \times 4} = \dfrac{100 \times 1375}{10000} = 13.75\%$MF#%

Answer : Option D

Explanation :

Let x , y and x be his investments at 4%, 6% and 8% respectively
Simple Interest on x at 4% for 1 year
= Simple Interest on y at 6% for 1 year
= Simple Interest on z at 8% for 1 year
$MF#%\begin{align}&\dfrac{x \times 4 \times 1}{100} = \dfrac{y \times 6 \times 1}{100} = \dfrac{z \times 8 \times 1}{100}\\\\ &\Rightarrow 4x = 6y = 8z\\\\ &\Rightarrow 2x = 3y = 4z\\\\ &\text{Hence, we have, }y = \dfrac{2x}{3}\text{ and }z = \dfrac{2x}{4}=\dfrac{x}{2}\\\\ &\text{we know that x + y + z = }2600\\\\ &\Rightarrow x + \dfrac{2x}{3} + \dfrac{x}{2} = 2600\\\\ &\Rightarrow 6x + 4x + 3x = 2600 \times 6\\\\ &\Rightarrow 13x = 2600 \times 6\\\\ &\Rightarrow x = \dfrac{2600 \times 6}{13}= 200 \times 6 = 1200\\\\ &\text{i.e., Money invested at 4% = Rs.1200}\end{align}$MF#%

Time = 100 x 81 years = 4 years. 450 x 4.5

Answer : Option B

Explanation :

Let the sum of money be Rs.x
$MF#%\text{Amount after 4 years = }\dfrac{7x}{5}$MF#%
T = 4 years
R = ?
$MF#%\text{Simple Interest, SI = }\left(\dfrac{7x}{5} - x\right) = \dfrac{2x}{5} \\\\ \text{R = }\dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times \dfrac{2x}{5} }{x \times 4}= \dfrac{40}{4} = 10\%$MF#%

Answer : Option B

Explanation :

Simple Interest for 3 years = (Rs.12005 - Rs.9800) = Rs.2205
$MF#%\text{Simple Interest for 5 years = }\dfrac{2205}{3} \times 5 = \text{Rs.}3675$MF#%
Principal(P) = (Rs.9800 - Rs.3675) = Rs.6125
$MF#%\text{R = }\dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times 3675}{6125 \times 5}= 12\%$MF#%

Answer : Option B

Explanation :

Let the man invests Rs.x at 6% and Rs.y at 7%
Simple Interest on Rs.x at 6% for 2 years + Simple Interest on Rs.y at 7% for 2 years = Rs.354
$MF#%\begin{align}&\dfrac{\text{x} \times 6 \times 2}{100} + \dfrac{\text{y} \times 7 \times 2}{100} = 354\\\\ &\text{x} \times 6 \times 2 + \text{y} \times 7 \times 2 = 354 \times 100\\\\ &\text{x} \times 6 + \text{y} \times 7 = 177\times 100\\\\ &\text{6x} + \text{7y} = 17700 \quad \color{#F00}{\cdots \text{(1)}}\end{align}$MF#%
One-forth of the first sum is equal to one-fifth of the second sum
$MF#% => \dfrac{x}{4}= \dfrac{y}{5}\\\\ => x = \dfrac{4y}{5}\quad \color{#F00}{\cdots \text{(2)}}$MF#%
Solving (1) and (2),
$MF#%\begin{align}&\text{6x} + \text{7y} = 17700 \\\\ &6\left(\dfrac{4y}{5}\right) + 7y = 17700\\\\ &24y + 35y = 17700 \times 5\\\\ &59y = 17700 \times 5\\\\ &y = 300 \times 5 = 1500\\\\ &x = \dfrac{4y}{5} = \dfrac{4\times1500}{5} = 4 \times 300 = 1200\end{align}$MF#%
total sum invested = x + y = 1500 + 1200 = 2700