Lucas invests R4million into an account earning interest of 6% per annum compounded monthly.He withdraws an allowance of R30000 per month.The first withdrawal is exactly one month after he has deposited the R4million.How many such withdrawals will Lucas be able to make?
To answer this question, we need to calculate the balance of Lucas's account after each withdrawal until it reaches zero.
The formula to calculate the future value of an investment with monthly compounding interest is:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment
P = the principal amount (the initial investment)
r = annual interest rate (in decimal form)
n = number of times that interest is compounded per year
t = number of years
In this case:
P = R4,000,000 (initial investment)
r = 6% or 0.06 (annual interest rate)
n = 12 (monthly compounding)
t = unknown
We can rearrange the formula to solve for t:
A/P = (1 + r/n)^(nt)
Since the future value should be zero after all the withdrawals, we have:
0/R4,000,000 = (1 + 0.06/12)^(12t)
Next, we substitute the value of the allowance withdrawal, R30,000, into the formula:
A = P(1 + r/n)^(nt) - w
Where:
w = withdrawal amount
R30,000 = R4,000,000(1 + 0.06/12)^(12t) - R30,000
We can simplify the equation to:
150 = (1 + 0.06/12)^(12t) - 1
Now, we solve for t by isolating the exponent:
(1 + 0.06/12)^(12t) = 150 + 1
(1 + 0.06/12)^(12t) = 151
Next, we take the logarithm of both sides to solve for t:
log((1 + 0.06/12)^(12t)) = log(151)
12t * log(1 + 0.06/12) = log(151)
t = log(151) / (12 * log(1 + 0.06/12))
After evaluating this expression, we find that t is approximately 16.98 years.
However, since we are interested in the number of withdrawals, we multiply t by the number of withdrawals made in a year:
Withdrawals per year = 12 (monthly compounding)
Number of withdrawals = t * Withdrawals per year
Number of withdrawals ≈ 16.98 * 12
Therefore, Lucas will be able to make approximately 203 withdrawals.