If nominal rate of return is 10% per annum and annual effective rate of interest is 10.25% per annum, determine the frequency of compounding:
Answers
Given : nominal rate of return is 10% per annum and annual effective rate of interest is 10.25% per annum,
To Find : determine the frequency of compounding:
Solution:
Let sat Amount = P
Rate of interest = 10 %
Let say frequency compounding be n
Then rate of interest = 10/n
Time = n
A = P(1 + 10/100n)ⁿ = P (1 + 10.25/100)
=> (1 + 10/100n)ⁿ = 1.1025
Taking log both sides
=> n log ( 1 + 1/10n) = log 1.1025
=> log ( 1 + 1/10n) = (1/n) log 1.1025
=> log ( 1 + 1/10n) = log ( 1.1025)^(1/n)
=> 1 + 1/10n = ( 1.1025)^(1/n)
on solving we get n = 2
Verification
(1 + 5/100)² = 1.1025
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Answer:
Step-by-step explanation:
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