Homework Answers
Value of the machine now = $120000
Inflation = i = 6%
Hence, Value of the machine if purchased after 10 years = 120000*1.0610 = $214901.72
Value received by scrapping the first machine = $20000
Hence, additional amount required after 10 years to purchase the machine = 214901.72 - 20000 = $194901.72
Let the amount invested each year for 10 years be P
Rate of return = r = 4%
Future Value of the investment = P(1+r)n-1 +....+ P(1+r)2 + P(1+r) + P = P[(1+r)n -1]/r = P[(1+0.04)10 -1]/0.04
=> P[(1+0.04)10 -1]/0.04 = 194901.72
Solving for P gives
=> P = $16233.33
From the given options option (D) is correct i.e $16235
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