We need to know the S.I., principal and time to find the rate.

Since the principal is not given, so data is inadequate.

Let the sum borrowed be x. Then,
(x*2*6)/100 + (x*9*3)/100 + (x*14*4)/100 = 11400
3x/25 + 27x/100 + 14x / 25) = 11400
95x/100 = 11400
x = (11400*100)/95 = 12000

Answer : Option B

Explanation :

Let the rate of interest per annum be R%
$MF#%\text{Simple Interest for Rs. 5000 for 2 years at rate R% per annum }\\\\ + \text{Simple Interest for Rs. 3000 for 4 years at rate R% per annum = Rs.2200}$MF#%
$MF#%\Rightarrow \dfrac{5000\times \text{R}\times 2}{100} + \dfrac{3000\times \text{R}\times 4}{100}=2200\\\\ \Rightarrow \text{100R + 120R} = 2200\\\\ \Rightarrow \text{220R} = 2200\\\\ \Rightarrow \text{R} = 10$MF#%
i.e, Rate = 10%.

S.I. for 3 years = Rs. (12005 - 9800) = Rs. 2205.

S.I. for 5 years = Rs. 2205 x 5 = Rs. 3675 3

Principal = Rs. (9800 - 3675) = Rs. 6125.

Hence, rate = 100 x 3675 % = 12% 6125 x 5

S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 - 117) = Rs. 698.

Answer : Option A

Explanation :

Let the parts be x, y and z
R = 5%
x + interest on x for 2 years = y + interest on y for 3 years = z + interest on z for 4 years
$MF#%\begin{align}&\left(x + \dfrac{x \times 5 \times 2}{100}\right) = \left(y + \dfrac{y \times 5 \times 3}{100}\right) = \left(z + \dfrac{z \times 5 \times 4}{100}\right)\\\\ &\left(x + \dfrac{x}{10}\right) = \left(y + \dfrac{3y}{20}\right) = \left(z + \dfrac{z}{5}\right)\\\\ &\dfrac{11x}{10} = \dfrac{23y}{20} = \dfrac{6z}{5}\\\\\\\\ &\text{ Let }\dfrac{11x}{10} = \dfrac{23y}{20} = \dfrac{6z}{5} = \text{ k} \quad \text{ (where k is a constant)}\\\\ &\text{Then, }x = \dfrac{10k}{11}, \quad y = \dfrac{20k}{23}, \quad z = \dfrac{5k}{6}\\\\ &\text{ we know that }x + y + z = 2379 \\\\ &\dfrac{10k}{11} + \dfrac{20k}{23} + \dfrac{5k}{6} = 2379\\\\ &10k \times 23 \times 6 + 20k \times 11 \times 6 + 5k \times 11 \times 23 = 2379 \times11 \times23 \times 6\\\\ &1380k + 1320k + 1265k = 2379 \times11 \times23 \times 6\\\\ &3965k = 2379 \times11 \times23 \times 6\\\\ &k = \dfrac{2379 \times11 \times 23 \times 6}{3965}\\\\ &\text{First part, x = }\dfrac{10k}{11} = \dfrac{10}{11} \times \dfrac{2379 \times11 \times 23 \times 6}{3965} = \dfrac{10 \times 2379\times 23 \times 6}{3965} \\\\&= \dfrac{2\times 2379\times 23 \times 6}{793} = 2 \times 3 \times 23 \times 6 = 828\end{align}$MF#%

Answer : Option D

Explanation :

Let rate = R%
Then, Time, T = R years
P = Rs.1400
SI = Rs.686
$MF#%\begin{align}&\text{SI= }\dfrac{\text{PRT}}{100} \\ \\
&\Rightarrow \text{686 = }\dfrac{\text{1400 × R × R}}{100} \\ \\
&\Rightarrow 686 = 14 \text{ R}^2 \\ \\
&\Rightarrow 49 = \text{R}^2 \\ \\
&\Rightarrow \text{R} = 7\end{align}$MF#%
i.e.,Rate of Interest was 7%

Answer : Option D

Explanation :

Let the sum of money be x
then
$MF#%\begin{align}&\dfrac{x \times 4 \times 8}{100 } = \dfrac{560 \times 12 \times 8}{100 }\\\\ &x \times 4 \times 8 = 560 \times 12 \times 8\\\\ &x \times 4 = 560 \times 12\\\\ &x = 560 \times 3 = 1680\end{align}$MF#%

Answer : Option A

Explanation :

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Solution 1
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Let the sum of money be Rs.x
After 40 years, this becomes 3x
Simple Interest = 3x - x = 2x
$MF#%\begin{align}&\text{Simple Interest = }\dfrac{\text{PRT}}{100}\\\\&2x =\dfrac{x \times \text{R} \times 40}{100}\\\\&2 =\dfrac{\text{R} \times 40}{100}\\\\&200 =40\text{R}\\\\&\text{R} = 5\%\end{align}$MF#%
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Solution 2
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If a sum of money becomes n times in T years at simple interest, then the rate of interest per annum can be given be
$MF#%\text{R = }\dfrac{100(\text{n}-1)}{\text{T}}\%$MF#%

Answer : Option D

Explanation :

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Solution 1
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The person borrows Rs. 5000 for 2 years at 4% p.a. simple interest
$MF#%\text{Simple interest that he needs to pay = }\dfrac{\text{PRT}}{100} = \dfrac{5000 \times 4 \times 2}{100} = 400$MF#%
He also lends it at 6¹â„₄% p.a for 2 years
$MF#%\text{Simple interest that he gets = }\dfrac{\text{PRT}}{100} = \dfrac{5000 \times \dfrac{25}{4} \times 2}{100} = 625$MF#%
His overall gain in 2 years = Rs.625 - Rs.400 = Rs.225
$MF#%\text{His overall gain in 1 year = }\dfrac{225}{2}\text{ = Rs.112.5}$MF#%
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Solution 2
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The person borrows Rs. 5000 for 2 years at 4% p.a. simple interest
He also lends it at 6¹â„₄% p.a for 2 years
$MF#%6 \dfrac{1}{4}\% - 4\% = 2 \dfrac{1}{4}\%$MF#%
So his gain in the transaction for 1 year
= The simple interest he gets for Rs.5000 for 1 year at 2 ¹â„₄% per annum
$MF#%= \dfrac{\text{PRT}}{100} = \dfrac{5000 \times \dfrac{9}{4} \times 1}{100} = 112.5$MF#%