Let the original rate be R%. Then, new rate = (2R)%.

Note:
Here, original rate is for 1 year(s); the new rate is for only 4 months i.e. year(s).

725 x R x 1 + 362.50 x 2R x 1 = 33.50 100 100 x 3

(2175 + 725) R = 33.50 x 100 x 3

(2175 + 725) R = 10050

(2900)R = 10050

R = 10050 = 3.46 2900

Original rate = 3.46%

Let principal = P. Then, S.l. = P and T = 16 yrs.
Rate = (100 x P)/(P*16)% = 6 1/4% p.a.

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).

Then, x x 14 x 2 + (13900 - x) x 11 x 2 = 3508 100 100

28x - 22x = 350800 - (13900 x 22)

6x = 45000

x = 7500.

So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

Principal = Rs. 100 x 5400 = Rs. 15000. 12 x 3

Answer : Option B

Explanation :

Let sum = x
Time = 20 years
Simple Interest = x
$MF#%\text{R = }\dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times x}{x \times 20} = \dfrac{100}{20} = 5\%$MF#%

Answer : Option D

Explanation :

Amount borrowed = Rs.10
Let rate of interest = R%
$MF#% \text{Simple Interest for Rs.10 for 6 months at R% = }\dfrac{10 \times \text{R} \times \dfrac{1}{2}}{100} = \dfrac{\text{R}}{20}\\
\text{i.e, 10 + }\dfrac{\text{R}}{20}\text{ is due in 6 months}$MF#%
Payment after 1st month = Rs.3
$MF#%\text{Interest for this Rs.3 for the remaining 5 months = }\dfrac{3 \times R \times \dfrac{5}{12}}{100}$MF#%
Payment after 2nd month = Rs.3
$MF#%\text{Interest for this Rs.3 for the remaining 4 months = }\dfrac{3 \times R \times \dfrac{4}{12}}{100}$MF#%
...
Payment after 5th month = Rs.3
$MF#%\text{Interest for this Rs.3 for the remaining 1 month = }\dfrac{3 \times R \times \dfrac{1}{12}}{100}$MF#%
Payment after 6th month = Rs.3 and this closes the loan
$MF#%=> (3 + 3 + 3 + 3 + 3 + 3) + \left(\dfrac{3 \times R \times \dfrac{5}{12}}{100} + \dfrac{3 \times R \times \dfrac{4}{12}}{100} + ... + \dfrac{3 \times R \times \dfrac{1}{12}}{100} \right) = \text{ 10 + }\dfrac{\text{R}}{20} \\\\ 18 + \dfrac{\dfrac{3\text{R}}{12}\left(5+4+...+1\right)}{100}=\text{ 10 + }\dfrac{\text{R}}{20}\\\\ 18 + \dfrac{15\text{R}}{400} = \text{ 10 + }\dfrac{\text{R}}{20} \\\\ 8 = \dfrac{\text{R}}{20} - \dfrac{15\text{R}}{400} = \dfrac{5\text{R}}{400}=\dfrac{\text{R}}{80}\\\\ \text{R} = 640\%$MF#%

Answer : Option D

Explanation :

Let the Principal(P) be x
Then, Simple Interest(SI) = x/5
Time(T) = 4 years
$MF#%\text{Rate of interest per annum(R) = }\dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times \dfrac{x}{5}}{x \times 4} = \dfrac{20}{4}= 5\%$MF#%

Answer : Option A

Explanation :

Let x , y and x be his investments in A, B and C respectively. Then
Then, Interest on x at 10% for 1 year
+ Interest on y at 12% for 1 year
+ Interest on z at 15% for 1 year
= 3200
$MF#%\dfrac{x \times 10 \times 1}{100} + \dfrac{y \times 12 \times 1}{100} + \dfrac{z \times 15 \times 1}{100} = 3200\\\\ \Rightarrow 10x + 12y + 15z = 320000 \quad \color{RED}{---(1)}$MF#%
Amount invested in Scheme C was 240% of the amount invested in Scheme B
$MF#%=> z = \dfrac{240y}{100} = \dfrac{60y}{25} = \dfrac{12y}{5} \color{RED}{---(2)}$MF#%
Amount invested in Scheme C was 150% of the amount invested in Scheme A
$MF#%=> z = \dfrac{150x}{100} = \dfrac{3x}{2}\\\\ => x = \dfrac{2z}{3} = \dfrac{2}{3} \times \dfrac{12y}{5} = \dfrac{8y}{5} \color{RED}{---(3)}$MF#%
From(1),(2) and (3),
10x + 12y + 15z = 320000
$MF#%10\left(\dfrac{8y}{5}\right) + 12y + 15\left(\dfrac{12y}{5}\right) = 320000\\\\ 16y + 12y + 36y = 320000\\\\ 64y = 320000\\\\ y=\dfrac{320000}{64}= \dfrac{10000}{2}=5000$MF#%
i.e.,Amount invested in Scheme B = Rs.5000

Answer : Option D

Explanation :

Amount borrowed = Rs.9
Let rate of interest = R%
$MF#%\text{Simple Interest for Rs.9 for 10 months at R% = }\dfrac{9 \times \text{R} \times \dfrac{10}{12}}{100} = \dfrac{90\text{R}}{1200}\\\\ \text{i.e., }9 + \dfrac{90\text{R}}{1200}\text{ is due in 10 months}$MF#%
Payment after 1st month = Rs.1
$MF#%\text{Interest for this Rs.1 for the remaining 9 months = }\dfrac{1 \times \text{R} \times \dfrac{9}{12}}{100} $MF#%
Payment after 2nd month = Rs.1
$MF#%\text{Interest for this Rs.1 for the remaining 8 months = }\dfrac{1 \times \text{R} \times \dfrac{8}{12}}{100}\\\\ \cdots$MF#%
Payment after 9th month = Rs.1
$MF#%\text{Interest for this Rs.1 for the remaining 1 month = }\dfrac{1 \times \text{R} \times \dfrac{1}{12}}{100} $MF#%
Payment after 10th month = Rs.1 and this closes the loan
$MF#%\begin{align}&9 + \dfrac{90\text{R}}{1200} = 10 \times 1 + \left(\dfrac{1 \times \text{R} \times \dfrac{9}{12}}{100} + \dfrac{1 \times \text{R} \times \dfrac{8}{12}}{100} + \cdots + \dfrac{1 \times \text{R} \times \dfrac{1}{12}}{100}\right)\\\\ &9 + \dfrac{90\text{R}}{1200} = 10 + \dfrac{ \text{R}}{1200} \left(9 + 8 + \cdots + 1 \right)\\\\ &9 + \dfrac{90\text{R}}{1200} = 10 + \dfrac{ \text{R}}{1200} \left(\dfrac{9 \times 10}{2}\right)\\\\ &9 + \dfrac{90\text{R}}{1200} = 10 + \dfrac{ \text{45R}}{1200}\\\\ &\dfrac{45\text{R}}{1200} = 1\\\\ &\text{R} = \dfrac{1200}{45} = 26.67\%\end{align}$MF#%

Answer : Option A

Explanation :

P = Rs.900
SI = Rs.81
T = ?
R = 4.5%
$MF#%\text{T= }\dfrac{100 ×\text{SI}}{\text{PR}} = \dfrac{100 × 81}{900 × 4.5} = 2 \text{ years}$MF#%