An engineer deposits $25,000 into an account when the market interest rate is 8% per year and the inflation rate is 3% per year. The account is left undisturbed for 15 years.
The purchasing power of the accumulated dollars in terms of today’s dollars is closest to:
Solution calculated with interest factor with 4 decimal places.
Question 19 options:
$79,305 |
|
$16,046 |
|
$40,161 |
|
$50,902 |
Ans: $50,902
Explanation:
FV = P(1 + i)n
= 25000(1 + 0.08)15
= 79,304.2278
Purchasing power = 79,304.2278 / (1 + 0.03)15
= 79,304.2278 / 1.55797
= $50,902
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