At what rate percent per annum will Rs. 3000 amount to Rs. 3993 in 3 years, if the interest is compounded annually?
A) 9 % p.a.
B) 10 % p.a.
C) 12 % p.a.
D) 15 % p.a.
Answer
621k+ views
Hint: To calculate the final amount with compound interest we require principal balance, interest rate and time period. We are given the final amount, principal amount and time period. We have to calculate the rate of interest using the amount after compound interest formula.
Formula used: Compound interest formula $A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$ where A is the final amount, P is the principal amount, r is the rate of interest and n is the time period.
Complete step-by-step Solution:
We are given that the interest is compounded annually for a principal amount of Rs. 3000 in 3 years.
We have to calculate the rate of interest (interest rate) so that the amount Rs. 3000 will become Rs. 3993 in 3 years (compound interest).
$A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$ Where A=Rs. 3993, P=Rs. 3000, n=3 years, r=?
$
A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n} \\
3993 = 3000{\left( {1 + \dfrac{r}{{100}}} \right)^3} \\
$
Send 3000 to the LHS
$
\dfrac{{3993}}{{3000}} = {\left( {1 + \dfrac{r}{{100}}} \right)^3} \\
\dfrac{{1331}}{{1000}} = {\left( {1 + \dfrac{r}{{100}}} \right)^3} \\
\dfrac{{{{\left( {11} \right)}^3}}}{{{{\left( {10} \right)}^3}}} = {\left( {1 + \dfrac{r}{{100}}} \right)^3} \\
{\left( {\dfrac{{11}}{{10}}} \right)^3} = {\left( {1 + \dfrac{r}{{100}}} \right)^3} \\
$
Apply cube root on both sides
$
\dfrac{{11}}{{10}} = 1 + \dfrac{r}{{100}} \\
\dfrac{{11}}{{10}} - 1 = \dfrac{r}{{100}} \\
\dfrac{{11 - 1}}{{10}} = \dfrac{r}{{100}} \\
\dfrac{1}{{10}} = \dfrac{r}{{100}} \\
r = \dfrac{{100}}{{10}} = 10 \\
$
The value of r is 10.
Therefore, from among the options given in the question Option B is correct, which is At 10% p.a. the amount Rs. 3000 will be Rs. 3993 in 3 years compoundly.
Note: Compound interest is adding interest to the principal sum of a loan or a deposit. It is the result of reinvesting interest, rather than paying it out, so the interest in the next period is then earned on the principal sum plus previously accumulated interest.
Formula used: Compound interest formula $A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$ where A is the final amount, P is the principal amount, r is the rate of interest and n is the time period.
Complete step-by-step Solution:
We are given that the interest is compounded annually for a principal amount of Rs. 3000 in 3 years.
We have to calculate the rate of interest (interest rate) so that the amount Rs. 3000 will become Rs. 3993 in 3 years (compound interest).
$A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$ Where A=Rs. 3993, P=Rs. 3000, n=3 years, r=?
$
A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n} \\
3993 = 3000{\left( {1 + \dfrac{r}{{100}}} \right)^3} \\
$
Send 3000 to the LHS
$
\dfrac{{3993}}{{3000}} = {\left( {1 + \dfrac{r}{{100}}} \right)^3} \\
\dfrac{{1331}}{{1000}} = {\left( {1 + \dfrac{r}{{100}}} \right)^3} \\
\dfrac{{{{\left( {11} \right)}^3}}}{{{{\left( {10} \right)}^3}}} = {\left( {1 + \dfrac{r}{{100}}} \right)^3} \\
{\left( {\dfrac{{11}}{{10}}} \right)^3} = {\left( {1 + \dfrac{r}{{100}}} \right)^3} \\
$
Apply cube root on both sides
$
\dfrac{{11}}{{10}} = 1 + \dfrac{r}{{100}} \\
\dfrac{{11}}{{10}} - 1 = \dfrac{r}{{100}} \\
\dfrac{{11 - 1}}{{10}} = \dfrac{r}{{100}} \\
\dfrac{1}{{10}} = \dfrac{r}{{100}} \\
r = \dfrac{{100}}{{10}} = 10 \\
$
The value of r is 10.
Therefore, from among the options given in the question Option B is correct, which is At 10% p.a. the amount Rs. 3000 will be Rs. 3993 in 3 years compoundly.
Note: Compound interest is adding interest to the principal sum of a loan or a deposit. It is the result of reinvesting interest, rather than paying it out, so the interest in the next period is then earned on the principal sum plus previously accumulated interest.
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