Activity  4  Practice  Solutions  Sheet  P.  57-­‐8  

1.    Write  and  solve  and  equation  to  determine  the  balance  after  25  years  in  an  account  that  had  an  initial  investment  of  $18,000  at  3%  interest,  compounded  annually.    Compounded  annually  implies  we  use  the  formula  

A(t) = P(1+ r)t   NOTE:    This  is  the  same  as   A(t) = P(1+r

n)n*t    for  n=1.  

Here,  P  =  18,000,    r  =  0.03,  and  t  =  25.    So,    A(t)=18000(1+0.03)25  =  18000(1.03)25    =  $37,688      2.    Determine  the  balance  after  10  years  in  an  account  that  had  an  initial  investment  of  $25,000  at  5%  interest,  compound  annually.    Again,  since  it  is  compound  annually  we  use  A(t) = P(1+ r)t  Here,    P  =  25,000,    r  =  0.05,  and  t  =  10.    So,    A(t)  =  25000(1.05)10  =  $40,722    8.    How  much  interest  is  earned  on  an  investment  of  $5200  earning  7.5%  interest,  compounded  annually,  over  a  period  of  3  years.    Again,  compounded  annually  means  we  use  A(t) = P(1+ r)t    Note  that  the  problem  is  asking  ONLY  for  the  amount  of  interest  gained  !!!  So,    A(t)  =  5200(1.075)3  =  $6459.94  This  is  the  total  amount  including  the  Principal.    We  must  remove  the  Principal  to  see  how  much  was  from  interest.    $6459.94  –  $5200  =  $1259.94    (Multiple  choice  answer  B  )      12.    Determine  the  interest  earned  after  5  years  in  an  account  that  had  an  initial  investment  of  $25000  at  3.5%  interest  compounded  daily.    Interest  is  compounded  DAILY  so  we  use  

A(t) = P(1+ r

n)n*t   Where  P  =  25000,    r  =  0.035,    t  =  5  years,    and  n  =  365  since  it’s  compounded  daily.  

So,    𝐴 𝑡 = 25000 1+ !.!"#!"#

(!"#⋅!)  =  $29,780.90  

Again  here,  for  this  problem,  they  are  asking  for  the  amount  of  interest  only.    We  must  subtract  out  the  original  Principal  to  find  the  interest  only.    So,    $29,780  -­‐  $25,000  =  $4780.    

14.    Determine  the  amount  of  money  that  needs  to  be  invested  at  6%  interest  compounded  monthly,  to  have  $20,000  in  15  years.    We  know  that  in  15  years,  A(t)  [actually  called  A(15)  ]  =  $20,000.    Also  t  =  15,  r  =  0.06,  and  n=12  since  it  is  compound  monthly.    

So,    20,000 = 𝑃   1+ !.!"!"

(!"⋅!").    It  is  easiest  to  calculate  the  value  of  the  expression  after  P.    This  give  us:  

 

20,000 = 𝑃  (2.45409).    Then  divide  both  sides  by  2.4509  and  we  see  that  P  =  $8,150    15.    Complete  the  table  by  finding  each  balance:    P  =  $10,000,  and  r  =  0.05  .      In  top  row,  t  =  5.    The  compounding    and  formulas  for  first  row  are  as  follows:  annually    n=1,  so  A(t)  =  10,000 1.05 ! =  $12,763  

quarterly    n=4,  so  𝐴 𝑡 = 10,000 1+ !.!"!

!⋅!  =  $12,820  

monthly    n=12,  so  𝐴 𝑡 = 10,000 1+ !.!"!"

!"⋅!  =  $12,833  

weekly    n=52,    so    𝐴 𝑡 =  10,000 1+ !.!"!"

!"⋅!  =  $12,838  

 For  second  row,  t  =  10.    The  compounding  and  formulas  are  as  follows:    annually    n=1,  so  A(t)  =  10,000 1.05 !" =  $16,289  

quarterly    n=4,  so  𝐴 𝑡 = 10,000 1+ !.!"!

!⋅!"  =  $16,436  

monthly    n=12,  so  𝐴 𝑡 = 10,000 1+ !.!"!"

!"⋅!"  =  $16,470  

weekly    n=52,    so    𝐴 𝑡 =  10,000 1+ !.!"!"

!"⋅!"=  $16,483  

 Table  looks  like  this  $10,000   5%  

Annually  5%  Quarterly  

5%  monthly  

5%  weekly  

In  5  years  

$12,763   $12,820   $12,833   $12,838  

In  10  years  

$16,289   $16,436   $16,470   $16,483  

     20.    Determine  the  balance  after  20  years  in  an  account  that  had  an  initial  investment  of  $25,000  at  5%  interest,  compounded  continuously.    Here,  we  are  given  that  interest  is  compounded  CONTINUOUSLY,  so  we  use  the  formula:  

A(t) = Pert     So    𝐴 𝑡 = 25,000 ⋅ 𝑒(!.!"⋅!")  =  $67,957.          

24.    Write  and  solve  an  equation  to  determine  the  balance  after  15  years  in  an  account  that  had  an  initial  investment  of  $25,000  at  3.5%  interest  –  compounded  continuously.    Again,  note  that  interest  is  compounded  CONTINUOUSLY.  S0,    𝐴 𝑡 = 25,000𝑒!.!"#⋅!"  =  $42,261.      26.    A  new  car  was  purchased  in  2005  for  $20,000.    It  depreciates  at  a  rate  of  9%.    

a. Write  a  continuous  exponential  function  that  represents  the  value  of  the  car  after  t  years  of  ownership.  

b. When  will  the  car  have  a  value  of  $10,000    

a. Depreciation  means  that  the  r  value    is  stated  as  -­‐0.09.    So,  𝐴 𝑡 =  20,000𝑒!!.!"⋅!  b. Here  we  know  the  value  of  A(t),  which  is  $10,000.    We  need  to  find  the  value  of  t.  

So,    10,000 = 20,000𝑒!!.!"⋅! .  At  this  point,  the  easiest  way  to  solve  for  t  is  to  graph  the  function  on  your  calculator  as  𝑌 = 20,000𝑒(!!.!"!)  .    Then  find  the  value  of  X  when  Y  =  10,000.    Doing  this  we  see  the  value  is  very  close  to  7.7.    Since  we  used  X  for  t  ,  we  have  that  t  =  7.7  years.    

We  will  learn  more  about  ways  to  solve  this  equation  algebraically  in  the  next  couple  of  sections.