Engineering Board Exam Reviewer · Rate of Return Analysis · medium
A machine costing P200,000 saves P60,000 per year for 6 years and has a salvage value of P20,000 at the end of year 6. Find the rate of return.
- 30%
- 15%
- 21.21%
- 20%
Answer
C. 21.21%
Solution
Solve 0=-200,000+60,000(P/A,i,6)+20,000(P/F,i,6). At i=21% the net value is +P1,049.53 and at i=22% it is -P3,919.37, so the root is i 21.21%.
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