Difficulty: Medium
Correct Answer: 4
Explanation:
Introduction / Context:
This question checks your understanding of growth under compound interest and how to estimate the time taken for an amount to more than double. The rate of interest is fixed at a relatively high value of 20% per annum, and you need to find the smallest integer number of years such that the amount exceeds twice the principal.
Given Data / Assumptions:
Concept / Approach:
For compound interest compounded annually, the amount after n years is:
A = P * (1 + r / 100)^n
We want:
P * (1.20)^n > 2 * P
Since P is positive, we can divide both sides by P and solve:
(1.20)^n > 2
We then check successive integer values of n until this inequality holds.
Step-by-Step Solution:
For n = 1: (1.20)^1 = 1.20, which is less than 2.
For n = 2: (1.20)^2 = 1.44, still less than 2.
For n = 3: (1.20)^3 = 1.728, still less than 2.
For n = 4: (1.20)^4 = 1.728 * 1.20 = 2.0736.
Now (1.20)^4 = 2.0736, which is greater than 2, so the amount is more than double after 4 years.
Therefore, the least number of complete years required is 4.
Verification / Alternative Check:
You can also use logarithms to solve (1.2)^n = 2 and then round up, but checking small integer powers is faster in this case.
Since for n = 3 the amount is still below double, and for n = 4 it exceeds double, the answer is confirmed.
Why Other Options Are Wrong:
3 years is too small because the amount becomes only 1.728 times the principal.
5 years and 6 years are larger than necessary; the amount has already more than doubled at 4 years.
Common Pitfalls:
A common mistake is to use simple interest and assume doubling happens when r * t = 100, which would give 5 years, but that is incorrect for compound interest.
Some learners assume doubling happens at around 3 years due to the high rate, without checking the exact powers of 1.2.
Final Answer:
The least number of complete years for the amount to become more than double is 4 years.
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