Ishita invested a sum of Rs. $12, 000$ at $5 %$ per annum compound interest.She received an amount of Rs. $13, 230$ after years.Find the value of $n$ .

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Solution

$A = P (1 + \frac{R}{100})^{n}$
$13230 = 12000 (1 + \frac{5}{100})^{n}$
$\frac{1323}{1200} = (\frac{10.5}{100})^{n}$
$\Rightarrow (1.05)^{n} = 1.1025 = (1.05)^{2}$
On compairing we get n=2
Thus, the value of n is 2 years.

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Ishita invested a sum of Rs.12,000 at 5% per annum compound interest.She received an amount of Rs.13,230 after years.Find the value of n.

A=P(1+R100)n13230=12000(1+5100)n13231200=(10.5100)n⇒(1.05)n=1.1025=(1.05)2On compairing we get n=2Thus, the value of n is 2 years.

Ishita invested a sum of Rs. $12, 000$ at $5 %$ per annum compound interest.She received an amount of Rs. $13, 230$ after years.Find the value of $n$ .

$A = P (1 + \frac{R}{100})^{n}$
$13230 = 12000 (1 + \frac{5}{100})^{n}$
$\frac{1323}{1200} = (\frac{10.5}{100})^{n}$
$\Rightarrow (1.05)^{n} = 1.1025 = (1.05)^{2}$
On compairing we get n=2
Thus, the value of n is 2 years.