S.I. for 1 year = Rs. $\left(854 - 815\right)$ = Rs. 39.
S.I. for 3 years = Rs.$\left(39 \times 3\right)$ = Rs. 117.
$\therefore$ Principal = Rs. $\left(815 - 117\right)$ = Rs. 698.
Principal = Rs.$ \left(\dfrac{100 \times 4016.25}{9 \times 5} \right) $
= Rs.$ \left(\dfrac{401625}{45} \right) $
= Rs. 8925.
Time =$ \left(\dfrac{100 \times 81}{450 \times 4.5} \right) $years= 4 years.
S.I. = Rs. $\left(15500 - 12500\right)$ = Rs. 3000.
Rate =$ \left(\dfrac{100 \times 3000}{12500 \times 4} \right) $%= 6%
Principal = Rs.$ \left(\dfrac{100 \times 5400}{12 \times 3} \right) $= Rs. 15000.
We need to know the S.I., principal and time to find the rate.
Since the principal is not given, so data is inadequate.
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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 = $\left(3x - x\right)$ = 2x
$\text{Simple Interest = }\dfrac{\text{PRT}}{100}$
$2x =\dfrac{x \times \text{R} \times 40}{100}$
$2 =\dfrac{\text{R} \times 40}{100}$
$200 =40\text{R}$
$\text{R} = 5\%$
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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
$\text{R = }\dfrac{100(\text{n}-1)}{\text{T}}\%$
n = 3
T = 40
$\text{R = }\dfrac{100(\text{n}-1)}{\text{T}} = \dfrac{100(3-1)}{40} = \dfrac{100 \times 2}{40} = \dfrac{100}{20} = 5\%$
P = Rs. 500
SI = Rs.500
T = 4
R = ?
$\text{ R } = \dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times 500}{500 \times 4} = \dfrac{100}{4} = 25\%$
This means, simple interest at 4% for that principal is Rs.120
$\text{P} = \dfrac{100 \times \text{SI}}{\text{RT}} = \dfrac{100 \times 120}{4 \times 6} = \dfrac{100 \times 30}{6} = 100 \times 5 = 500$
Let the Principal [P] be x
Then, Simple Interest [SI] = x/5
Time [T] = 4 years
$\text{Rate of interest per annum(R) = }\dfrac{100 \times \text{SI}}{\text{PT}} = \dfrac{100 \times \dfrac{x}{5}}{x \times 4} = \dfrac{20}{4}= 5\%$