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Answers
Answer:
Given:
A sum of rupees 3500 becomes rupees 4390.40 at the compound interest of 12%
To find:
The number of years
Solution:
Let "n" years represent the no. of years.
To solve the above-given problem we will use the following formula:
\boxed{\bold{A = P[1 + \frac{R}{100} ]^n}}
A=P[1+
100
R
]
n
Here we have
P = Principal = Rs. 3500
R = rate of interest = 12%
A = amount = Rs. 4390.40
Now, by substituting the given values in the formula, we get
4390.40 = 3500 [1 + \frac{12}{100} ]^n4390.40=3500[1+
100
12
]
n
\implies \frac{4390.40}{3500} = [ \frac{112}{100} ]^n⟹
3500
4390.40
=[
100
112
]
n
\implies 1.2544 = [ 1.12 ]^n⟹1.2544=[1.12]
n
taking log on both sides
\implies log (1.2544) = log [ 1.12 ]^n⟹log(1.2544)=log[1.12]
n
using the formula → log aᵇ = b log a
\implies log (1.2544) = n \:log [ 1.12 ]⟹log(1.2544)=nlog[1.12]
\implies 0.098 = n \:[ 0.049 ]⟹0.098=n[0.049]
\implies n = \frac{0.098}{0.049}⟹n=
0.049
0.098
\implies \bold{n = 2\:years}⟹n=2years
Thus, the number of years after which a sum of rupees 3500 will become rupees 4390.40 at the compound interest of 12% is 2 years.
Step-by-step explanation:
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