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Find the least number of integral years in which a sum of money invested at 20% compound interest per annum will become more than double itself.
- a)1
- b)4
- c)2
- d)3
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Find the least number of integral years in which a sum of money invest...
Given Data:
Compound Interest Rate (r): 20% or 0.20
The sum of money is more than doubled.
The sum of money is more than doubled.
Concept:
The formula for compound interest is A = P(1+r/n)(nt), where P is the principal amount, r is the annual interest rate, n is the number of times interest is compounded per year, t is the time the money is invested for in years. When compounded annually (n=1), this simplifies to A = P(1+r)t.
Solution:
We know that the formula for compound interest, when compounded annually, is A = P(1 + r)^t. Here, we want to find the smallest whole number for time t so that A > 2P.
Substituting A as 2P and r as 0.20 into the formula, we get, ⇒ 2 = (1 + 0.20)t ⇒ 2 = 1.20t
We can now try different values for t until we reach a number where 1.20t > 2.
Using t = 1, we get 1.201 = 1.20 which is not greater than 2.
Using t = 2, 1.202 = 1.44 which is not greater than 2.
Using t = 3, 1.203 = 1.728 which is not greater than 2.
Using t = 4, 1.204 = 2.0736 which is greater than 2.
Therefore, the least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is 4 years.
Most Upvoted Answer
Find the least number of integral years in which a sum of money invest...
Given:
Rate of interest = 20%
We have to find the least number of integral years in which a sum of money invested at 20% compound interest per annum will become more than double itself.
Solution:
Let us assume that the principal amount invested is P.
After n years, the amount becomes P(1+20/100)^n = P(6/5)^n
Now, we need to find the least value of n such that P(6/5)^n > 2P
Simplifying this inequality, we get (6/5)^n > 2
Taking logarithm on both sides, we get n log(6/5) > log2
Therefore, n > log2/log(6/5)
n > 3.8
The least value of n is 4.
Hence, the correct answer is option (B) 4.
Rate of interest = 20%
We have to find the least number of integral years in which a sum of money invested at 20% compound interest per annum will become more than double itself.
Solution:
Let us assume that the principal amount invested is P.
After n years, the amount becomes P(1+20/100)^n = P(6/5)^n
Now, we need to find the least value of n such that P(6/5)^n > 2P
Simplifying this inequality, we get (6/5)^n > 2
Taking logarithm on both sides, we get n log(6/5) > log2
Therefore, n > log2/log(6/5)
n > 3.8
The least value of n is 4.
Hence, the correct answer is option (B) 4.
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Find the least number of integral years in which a sum of money invested at 20% compound interest per annum will become more than double itself.a)1b)4c)2d)3Correct answer is option 'B'. Can you explain this answer?
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