Financial Management Answers to Test by aft19038

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									        Test #1-C
       Spring 2005
      Prof: David Eagle
FINC335, Financial Management
   Answers and Explanations
                  Problem 1(5 points)
              
                            .982 15
           .982 
                                      .7615
                                           42 .31
                     t

          t 15            1  .982    .018

                  Problem 2 (5 points)
                   2X
         t
      2X                               2X 
             2X  Y 
                             2X
 2X  Y 
     
                    2X
                                   1 
                           2X  Y  2X  Y 
                                              
t 1
               1
                   2X  Y
    2X    2X  Y    2X       2X        Y      2X   2X  Y 2X
                                            *       
2X  Y  2X  Y 2X  Y  2X  Y 2X  Y 2X  Y          Y     Y
         Problem 3
0                          12

                       24,000

         1.03812



    24,000
         12
             $15,340.62
    1.038
                         Problem 11
0       1       2    3      4         11     12    13         

      180 210 210 210 …               210   305   305    …   305
    1.054
     (1.054)

                         (1.054)11


                         210        1       305
                              1        
                 180    .054  1.05410 
                                           .05412
                1.054          1.054        1.054
                         1590.52 5648.15
                170.78           
                          1.054     1.05411
                160.98  1589.45  3167.08  $4,846.89
                           Problem 5
r = 6.6% 12 = .55%
n=?
C = $800

             800        1    
140000           1         
           .0055     1.0055n 
                                  n ln 1.0055  ln 26.67
                1    
.9625  1                          ln 26.67   3.28
             1.0055n            n                    598.62
                                     ln 1.0055 .00548
   1
       n
           .0375
1.0055
              1
1.0055 
       n
                    26.67        Rounding up = 599 payments
            .0375
                           Problem 6
    -15                                               -3

   35,000                                            67,000
                          *(1+r)12

                                                       67 ,000
35,000 (1  r )   12
                        67 ,000     (1  r )   12
                                                     
                                                       35,000
                                                               1

          (1  r )12  1.914               1  r  1.914      12




             1  r  1.0556                      r  5.56%
                 Problem 7
 0                                           72

3000
                    *(1+r)72


              r = 6.8% 4 = 1.7%
              n = 18 * 4 = 72


       3000 (1.017 )   72
                             $10 ,097 .87
                       Problem 8
 0...

$4000
         *1.0410
                           *1.01510
     r = 8% 2 = 4%                             *1.00610
     n = 5 * 2 = 10    r = 6% 4 = 1.5%
                       n = 7 * 4 = 28        r = 7.2% 12 = .6%
                                             n = 13 * 12 = 156



        $4,000 (1.04)10 (1.015)28 (1.006)156 = $22,841.65
                      Problem 9
   0                    17   18    19         

                             $30   $30       $30



           (1+r)17

                      C    30
                               6000
r = 6% 12 = .005     r   .005
                               17
                                       $5,512.24
                          1.005
                       Problem 10

r = 6.6%  12 = .55%
n = 8 * 12 = 96
                        20000 
                                   C
                                 .0055
                                                  
                                        1.005596  1

                        110  C (.6931)
                             110
                        C          $158.71
                            .6931
                         Problem 11
 April 25,2005 1     2     3                  April 25,2020

      0       $39   $39   $39…                  ...$39


       1.03430
                                                 $1,000
                            1.03430

                                          7.8%(1000)
 r = 7.8% 2 = 3.4% n = 15*2 = 30      C             39
                                              2
 $39             $1,000
                             $1,147.05.6332  $366.76
            1
     1      30 
                   
.034  1.034  1.034     30


 $726.36  $366.76  $1,093.12
                       Problem 12

              C      1             
   80,000       1                (1.024)
            .024  1.024 24         
   1920  C (.4444)
      1920                             1st payment due immediately
   C        $4,320.43
      .4444                                    Annuity due


For alternative approach see explanation for Test 1-D, Spring 2005.

								
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