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Present Worth of Annual Savings

Easypresent worthuniform series
A pump upgrade saves $12,000 at the end of each year for 10 years. At an interest rate of 6% per year, the present worth of the savings is most nearly:
  1. A$73,735
  2. B$120,000
  3. C$88,321
  4. D$158,170
Show solution

Answer: C. $88,321

Equation: present worth of a uniform series, P = A(P/A, i, n), with (P/A) = [1 − (1 + i)^−n]/i. Values: A = $12,000 is the yearly saving, i = 6% and n = 10 years. Step 1: (P/A, 6%, 10) = [1 − 1.06−10]/0.06 = 7.3601. Step 2: P = 12,000 × 7.3601 = $88,321. $120,000 adds the payments without discounting. $158,170 is the future worth at year 10. $73,735 uses 10% instead of 6%.
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