A man inherited a regular endowment of P 100,000.00 every end of 3 months for x years. However, he may choose to get a single lump sum of P 3,702,939.80 at the end of 4 years. If the rate of interest was 14o/o compounded quarterly, what is the value of x?
To find the value of x, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the final amount after compound interest
P = the principal amount (initial endowment)
r = the annual interest rate (14%)
n = the number of times that interest is compounded per year (4 times per year, since it is compounded quarterly)
t = the number of years
For the first scenario, the value of x is the number of times the regular endowment is given in x years. The total amount after x years is given by:
A1 = P * (1 + r/n)^(nt)
For the second scenario, we are given the lump sum amount after 4 years. So,
A2 = P2 = P * (1 + r/n)^(nt)
We know that A2 = P2 = P2 = 3,702,939.80
Substituting the values:
3,702,939.80 = 100,000 * (1 + 0.14/4)^(4*4)
Simplifying:
36.65 = (1 + 0.14/4)^(16)
Taking the logarithm of both sides:
log(36.65) = log[(1 + 0.14/4)^(16)]
Using a calculator:
log(36.65) ≈ 16 * log(1.035)
Dividing both sides by 16:
log(36.65) / 16 ≈ log(1.035)
Using a calculator to find the value of the logarithms:
0.03024 ≈ 0.0042
Therefore, the value of x is approximately:
x = 4 years / 0.0042 ≈ 952.38 years
Since x represents the number of times the regular endowment is given in x years, it cannot have a fractional value. Therefore, the value of x is rounded down to the nearest whole number:
x = 952 years