Question #9e9ef

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Question #9e9ef

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Answer 1 · 2016-05-11T12:49:50

8.997%

Explanation:

I am assuming the amount is compounded annually.

Note:- In this explanation, A is amount, P in Principal amount, R is rate of interest, and t is the time

Suppose I have $P$ $P$ $. Now if the amount is compounded annually, then

After first year, $A = P + \frac{P \times R}{100}$ $A = P + (PxxR)/100$ , where $R$ $R$ is the rate of interest.

After second year, the rate of interest in on the CURRENT amount, not the original amount, so the previous A becomes the new P

$A_{new} = A_{old} + \frac{A_{old} \times R}{100}$ $A_"new" = A_"old" + (A_"old" xx R)/100$
$A_{new} = A_{old} (1 + \frac{R}{100})$ $A_"new" = A_"old" (1 + R/100)$

Putting value of $A_{old}$ $A_"old"$ , we get,
$A_{new} = P (1 + \frac{R}{100}) \times (1 + \frac{R}{100})$ $A_"new" = P(1 + R/100)xx(1 + R/100)$
$A_{new} = P {(1 + \frac{R}{100})}^{2}$ $A_"new" = P (1 + R/100)^2$

Extending this for $t$ $t$ time,

$A = P {(1 + \frac{R}{100})}^{t}$ $A = P (1 + R/100)^t$

Here, we have to find $R$ $R$ , so
P = 13000$
A = 20000$
t = 5 years

Putting this in $A = P {(1 + \frac{R}{100})}^{t}$ $A = P (1 + R/100)^t$ , we get

$20000 = 13000 {(1 + \frac{R}{100})}^{5}$ $20000 = 13000 (1 + R/100)^5$

$\frac{20}{13} = {(1 + \frac{R}{100})}^{5}$ $20/13 = (1 + R/100)^5$

Now we have to use a calculator. Calculate the $5^{th}$ $5^"th"$ root of $\frac{20}{13}$ $20/13$

So it came out to be
$1.08997$ $1.08997$

$1.08997 = 1 + \frac{R}{100}$ $1.08997 = 1 + R/100$

So, R comes out to be 8.997%

P.S. Round it off as per your requirements.

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Question #9e9ef

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