An engineer who believes in “save now; play later” wants to retire in 30 years with $2.0 million. At 8% per year interest, the amount the engineer will have to invest each year (starting in year 1) to reach the $2 million goal is closest to:
(a) $17,660
(b) $28,190
(c) $49,350
(d) $89,680
A = 2,000,000(A/F,8%,30)
= 2,000,000(0.00883)
= $17,660
Answer is (a)
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