The least number of complete years in which a sum of money invested at...
P(1 + 20/100)n > 2P (6/5)n > 2
Now, (6/5 × 6/5 × 6/5 × 6/5) > 2
So, n = 4 years
View all questions of this test
The least number of complete years in which a sum of money invested at...
To find the least number of complete years in which a sum of money invested at 20% compound interest will be more than double, we need to use the compound interest formula:
A = P(1 + r/n)^(nt)
Where:
A = the final amount
P = the principal amount (initial investment)
r = the annual interest rate (as a decimal)
n = the number of times that interest is compounded per year
t = the number of years
In this case, we want to find the value of t when the final amount A is more than double the principal amount P. So, we can set up the following equation:
2P = P(1 + 0.20/n)^(nt)
Simplifying the equation, we get:
2 = (1 + 0.20/n)^(nt)
To find the least number of complete years, we can start by plugging in different values of n and solving for t. We will start with n = 1, which means the interest is compounded annually:
2 = (1 + 0.20/1)^(1t)
2 = (1 + 0.20)^t
2 = 1.20^t
Taking the logarithm of both sides, we get:
log 2 = log 1.20^t
log 2 = t log 1.20
t = log 2 / log 1.20
Using a calculator, we find that t ≈ 3.8 years. Since we are looking for the least number of complete years, we round up to the nearest whole number, which is 4.
Therefore, the correct answer is option B) 4. It will take at least 4 complete years for the sum of money invested at 20% compound interest to be more than double.
To make sure you are not studying endlessly, EduRev has designed Class 7 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 7.