1. When in cell D1 the interest rate is change from 15% to 5% the future worth o
ID: 1160106 • Letter: 1
Question
1. When in cell D1 the interest rate is change from 15% to 5% the future worth of this alternative is $64,647.
a. True
b. False
2. When in cell D1 the interest rate is change from 15% to 25% the future worth of this alternative is $64,647.
a. True
b. False
3. When in cell D1 the interest rate is changed to 15% to 5% the present worth of this alternative is $48,241.
a. True
b. False
4. When in cell D1 the interest rate is changed to 15% to 25% the present worth of this alternative is -$19,360
a. True
b. False
5. In an engineering economic problem the higher the rate of return (i), the higher the future quantity and the lower the present worth quantity?
a. True
b. False
interest rate 15% 2 Year Cash flow future worth 4 110000 254436.68 60340.72 52470.19 45626.25 39675.00 34500.00 30000.00 8000.00 4 10 12 13 8000 Net present value$6,993.10 Net future value$16,175.47 15 16Explanation / Answer
1. When in cell D1 the interest rate is change from 15% to 5% the future worth of this alternative is $64,647.
Future Worth = -110,000 (F/P, 5%, 6) + 30,000 (F/A, 5%, 6) + 8,000
Future Worth = -110,000 (1.3401) + 30,000 (6.8019) + 8,000 = 64,647
a. True
2. When in cell D1 the interest rate is change from 15% to 25% the future worth of this alternative is $64,647.
Future Worth = -110,000 (F/P, 25%, 6) + 30,000 (F/A, 25%, 6) + 8,000
Future Worth = -110,000 (3.8147) + 30,000 (11.2588) + 8,000 = -73,853
b. False
3. When in cell D1 the interest rate is changed to 15% to 5% the present worth of this alternative is $48,241.
Future Worth = 64,647
Present Worth = 64,647 (P/F, 5%, 6)
Present Worth = 64,647 (0.7462) = 84,241 OR
PW = -110,000 + 30,000 (P/A, 5%, 6) + 8,000 (P/F, 5%, 6)
PW = -110,000 + 30,000 (5.0757) + 8,000 (0.7462) = 48,241
a. True
4. When in cell D1 the interest rate is changed to 15% to 25% the present worth of this alternative
is -$19,360
Future Worth = -73,853
Present Worth = -73,853 (P/F, 25%, 6)
Present Worth = -73,853 (1+.25) -6= -19,360 OR
PW = -110,000 + 30,000 (P/A, 25%, 6) + 8,000 (P/F, 25%, 6)
PW = -110,000 + 30,000 (2.9514) + 8,000 (0.2621) = -19,360
a. True
5. In an engineering economic problem the higher the rate of return (i), the higher the future quantity and the lower the present worth quantity?
a. False
In an engineering economic problem the higher the rate of return (i), the future worth and present worth both will be lower (decrease).
At 5%
At 15%
Present Worth
48,241
6993
Future Worth
64,647
16,175
At 5%
At 15%
Present Worth
48,241
6993
Future Worth
64,647
16,175
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