Question

Xu owns two investments, A and B, that have a combined total value of $40,000. Investment A is expected to pay $28,000 in 3 years from today and has an expected return of 7.1 percent per year. Investment B is expected to pay $36,000 in T years from today and has an expected return of 5.5 percent per year. What is T, the number of years from today that investment B is expected to pay $36,000?

Answer 1

Answer:

The number of years is $T =13 \ years$

Explanation:

From the question we are told that

       The total value of the investment A and B is  $k =$$40, 000

       The future value of A is $F_A =$$28,000

       The time period is  t = 3

       The expected return of A is  $e_A =$ 7.1 % =  0.071

       The future value of  B is  $F_B =$$36,000

        The time period for  B  is T

       The expected return of B is $e_B =$5.5 % = 0.055

     

The present value of investment A is mathematically represented as

        $A = \frac{F_A }{(1 + e_A) ^t}$

substituting values

      $A = \frac{ 28000 }{(1 + 0.071) ^3}$

       $A =$$ 22792.38

The present value of B is mathematically evaluated as

      $B = k - A$

  substituting values

     B  =  40, 000 - 22792.38

      B  =  $17,208

The future value of B is

      $F_B = B * (1 + e_B)^T$

 substituting values

     $36,000 =17,208 * (1 + 0.055)^T$

     $2.0921 = (1.055)^T$

take log of both sides

    $log(2.0921) =log (1.055)^T$

   $0.32057 = T log (1.055)$

=>   $T = \frac{0.3206}{0.0232}$

     $T =13 \ years$