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Question
Naveen deposits ₹ 250 at the end of each month in an account that pays an interest of 6% per annum compounded monthly, how many months will be required for the deposit to amount to at least ₹ 6390? [log(1.1278) = 0.0523, log(1.005) = 0.0022]
Solution
Given a = ₹ 250, P = ₹ 6390, i = `6/12%` = 0.005
P = `"a"/"i" [(1 + "i")^"n" - 1]`
6390 = `250/0.005 [(1 + 0.005)^"n" - 1]`
6390 = `250/0.005 [(1.005)^"n" - 1]`
6390 = 50,000 [(1.005)n − 1]
`6390/(50,000)` = [(1.005)n − 1]
0.1278 = [(1.005)n − 1]
0.1278 + 1 = (1.005)n
1.1278 = (1.005)n
Taking logarithm on boths sides we get,
log 1.1278 = n log 1.005
n = `(log 1.1278)/(log 1.005)`
= `0.0523/0.0022`
n = 23.77
n ≈ 24
Required months ≈ 24
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