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If 1000be invested at interest rate 5% and the interest be added to the principal every 10 years, then the number of years in which it will amount to 2000?
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If 1000be invested at interest rate 5% and the interest be added to th...
Understanding the Problem:
To solve this problem, we need to calculate the number of years it will take for an initial investment of $1000 at an interest rate of 5%, where the interest is added to the principal every 10 years, to grow to $2000.
Calculating the Growth:
- After the first 10 years, the investment will grow to $1000 + 5% of $1000 = $1050.
- After the second 10 years, the investment will grow to $1050 + 5% of $1050 = $1102.50.
- This pattern continues, with the investment growing by 5% every 10 years.
Finding the Number of Years:
To find out how many years it will take for the investment to reach $2000, we can set up an equation:
$1000 * (1 + 0.05)^n = $2000
Solving this equation will give us the number of years (n) it takes for the investment to double.
Solving the Equation:
$1000 * (1.05)^n = $2000
(1.05)^n = 2
Taking the log of both sides:
n * log(1.05) = log(2)
n = log(2) / log(1.05)
n ≈ 14.21
Conclusion:
It will take approximately 14.21 years for the initial investment of $1000 at an interest rate of 5%, with interest added to the principal every 10 years, to grow to $2000.
To solve this problem, we need to calculate the number of years it will take for an initial investment of $1000 at an interest rate of 5%, where the interest is added to the principal every 10 years, to grow to $2000.
Calculating the Growth:
- After the first 10 years, the investment will grow to $1000 + 5% of $1000 = $1050.
- After the second 10 years, the investment will grow to $1050 + 5% of $1050 = $1102.50.
- This pattern continues, with the investment growing by 5% every 10 years.
Finding the Number of Years:
To find out how many years it will take for the investment to reach $2000, we can set up an equation:
$1000 * (1 + 0.05)^n = $2000
Solving this equation will give us the number of years (n) it takes for the investment to double.
Solving the Equation:
$1000 * (1.05)^n = $2000
(1.05)^n = 2
Taking the log of both sides:
n * log(1.05) = log(2)
n = log(2) / log(1.05)
n ≈ 14.21
Conclusion:
It will take approximately 14.21 years for the initial investment of $1000 at an interest rate of 5%, with interest added to the principal every 10 years, to grow to $2000.
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If 1000be invested at interest rate 5% and the interest be added to the principal every 10 years, then the number of years in which it will amount to 2000?
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If 1000be invested at interest rate 5% and the interest be added to the principal every 10 years, then the number of years in which it will amount to 2000? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about If 1000be invested at interest rate 5% and the interest be added to the principal every 10 years, then the number of years in which it will amount to 2000? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 1000be invested at interest rate 5% and the interest be added to the principal every 10 years, then the number of years in which it will amount to 2000?.
If 1000be invested at interest rate 5% and the interest be added to the principal every 10 years, then the number of years in which it will amount to 2000? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about If 1000be invested at interest rate 5% and the interest be added to the principal every 10 years, then the number of years in which it will amount to 2000? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 1000be invested at interest rate 5% and the interest be added to the principal every 10 years, then the number of years in which it will amount to 2000?.
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