QUESTIONS 23 TO 30 P $100,000; SV = $10,000; AR $100,000 AC $50,000; N 10 years;
ID: 2797563 • Letter: Q
Question
QUESTIONS 23 TO 30 P $100,000; SV = $10,000; AR $100,000 AC $50,000; N 10 years; MARR = 10% PRESENT WORTH VALUES -15% | -10% -5% 10% Base Case 5% 15% P (S) SV ($) AR ($) AC (S) N (yrs) MARR 23. The Present Worth of cell AA is a)-100,000+ (100,000-50,000)PA,10%, 10)(1.15)+1 0,000PF,10%, 10) b)-100,000( 1.15)+( 100,000-50,000)(F/A, 10%, 10)+1 0,000(PM 10%, 10) c)-100,000(0.85) + (100,000-50,000)(P/A, 10%,10)+ 10,000(P/F,10%, 10) d)-100,000(0.85)+(100,000-50,000)(P/A 10%, 10)+10,000(0.85 )(PF, 10%, 10) e) None of the above answers 24. The Present Worth of cell BB is a)-100,000+000,000-50,000)(P/A,10%, 10)(1.15)+10,000(P/F,10%,10) b)-100,000( 1 .15)+(100,000-50.000)(F/A,10%, 10)+1 0,000(PM 10%, 10) c)-100,000 + (100,000( 1.1 5)-50,000)(P/A, 10%),10)+10,000( PF, 10%, 10) d)-100,000+000,000-50,000)(P/A, 10%(1. 1 ), 10)+10,000)(P/F, 10%(0.9), 10) e) None of the above answers 25. The Present Worth of cell CC is a)-100,000+(100,000-50,000)(P/A,10%, 10)(1.15)+10,000(PF,10%, 10)Explanation / Answer
1.
As -15% means initial investment would be reduced by 15% or 0.85 times the intiial..Option C..Why not option D because D assumes change in salvage values as well
2.
Option C..becausethis is the only option where revenues are increased by 15% or 1.15 times the intiial
3.
Option B..because this is the only option where time is reduced by 10% or 0.9 times the intiial
4.
Option A..because this is the only option where MARR is increased by 10% or 1.1 times the intiial
5.
Annual revenues
6.
Salvage value
7.
311100.. -IC+50000/1.1+50000/1.1^2+.......50000/1.1^10+10000/1.1^10=0
=>IC=311083
8.
65600....As -100000+(x-50000)/1.1+(x-50000)/1.1^2+.......(x-50000)/1.1^10+10000/1.1^10=0
=>x=65647