In what time will Rs. 8000 amount change to Rs. 8820 if it is compounded at a rate of 5 % annually.
Answer
641.1k+ views
Hint: The formula for compound interest is \[A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}\] where A is the final amount, P is the principal amount, r is the interest rate per year, n is the number of years. Apply this to the given data of the question to get the answer.
Complete step by step answer:-
Compound interest is the interest that is calculated on the principal amount along with the interest accumulated over the previous period or year.
The formula to calculate compound interest with a principal amount P, at an annual rate r for n years is given as follows:
\[A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}............(1)\]
In this problem, the principal amount P is given as Rs. 8000.
\[P = 8000...........(2)\]
The amount is compounded at a rate of 5 % annually.
\[r = 100...........(3)\]
We need to find the number of years n to accumulate an amount of Rs. 8820.
\[A = 8820..........(4)\]
Using equations (2), (3), and (4) in equation (1), we have:
\[8820 = 8000{\left( {1 + \dfrac{5}{{100}}} \right)^n}\]
Simplifying the terms in the bracket we get:
\[8820 = 8000{\left( {1.05} \right)^n}\]
Taking 8000 to the other side and dividing with 8820, we get:
\[\dfrac{{8820}}{{8000}} = {\left( {1.05} \right)^n}\]
\[1.1025 = {\left( {1.05} \right)^n}\]
Apply log to both the sides of the equation, to get as follows:
\[\log (1.1025) = \log ({1.05^n})\]
We know that the value of \[\log {x^n}\] is equal to \[n\log x\].
\[\log (1.1025) = n\log (1.05)\]
We know that the square of 1.05 is 1.1025, then, we have:
\[\log {\left( {1.05} \right)^2} = n\log (1.05)\]
\[2\log \left( {1.05} \right) = n\log (1.05)\]
Solving for n, we have:
\[n = 2\dfrac{{\log (1.05)}}{{\log (1.05)}}\]
\[n = 2\]
Hence, it takes 2 years for the amount Rs. 8000 to change to Rs. 8820 when it is compounded at the rate of 5 % annually.
Complete step by step answer:-
Compound interest is the interest that is calculated on the principal amount along with the interest accumulated over the previous period or year.
The formula to calculate compound interest with a principal amount P, at an annual rate r for n years is given as follows:
\[A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}............(1)\]
In this problem, the principal amount P is given as Rs. 8000.
\[P = 8000...........(2)\]
The amount is compounded at a rate of 5 % annually.
\[r = 100...........(3)\]
We need to find the number of years n to accumulate an amount of Rs. 8820.
\[A = 8820..........(4)\]
Using equations (2), (3), and (4) in equation (1), we have:
\[8820 = 8000{\left( {1 + \dfrac{5}{{100}}} \right)^n}\]
Simplifying the terms in the bracket we get:
\[8820 = 8000{\left( {1.05} \right)^n}\]
Taking 8000 to the other side and dividing with 8820, we get:
\[\dfrac{{8820}}{{8000}} = {\left( {1.05} \right)^n}\]
\[1.1025 = {\left( {1.05} \right)^n}\]
Apply log to both the sides of the equation, to get as follows:
\[\log (1.1025) = \log ({1.05^n})\]
We know that the value of \[\log {x^n}\] is equal to \[n\log x\].
\[\log (1.1025) = n\log (1.05)\]
We know that the square of 1.05 is 1.1025, then, we have:
\[\log {\left( {1.05} \right)^2} = n\log (1.05)\]
\[2\log \left( {1.05} \right) = n\log (1.05)\]
Solving for n, we have:
\[n = 2\dfrac{{\log (1.05)}}{{\log (1.05)}}\]
\[n = 2\]
Hence, it takes 2 years for the amount Rs. 8000 to change to Rs. 8820 when it is compounded at the rate of 5 % annually.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

How many of the following diseases can be studied with class 11 biology CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

Which of the following enzymes is used for carboxylation class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

