Question

Antonio has just graduated from four years of college. For the last two years, he took out a Stafford loan to pay for his tuition. Each loan had a duration of ten years and interest compounded monthly. Antonio will pay each of them back by making monthly payments, starting as he graduates. Antonio’s loans are detailed in the table below. Year Loan Amount ($) Interest Rate (%) Subsidized? Junior 5,894 6.9 Y Senior 5,258 7.5 N Once all of his loans are paid off, what will Antonio’s total lifetime cost be? Round all dollar values to the nearest cent. a. $16,246.80 b. $17,804.40 c. $7,593.16 d. $9,874.76

Answer 1

Answer: Option a - $16,246.80.

The exact lifetime costs of both the loans put together works out to $ 15,665.32. However, none of the options march this answer. In this case, that option is a  - $16,246.80. Hence this is the answer.

We arrive at the answer as follows:

The total lifetime cost of a loan refers to the total amount a borrower repays in over the loan period in order to close the loan.

We can calculate that as:

$\mathbf{Lifetime Cost = EMI* Loan tenure (in months)}$ ------ (1)

We can calculate the EMI of a loan as :

$\mathbf{EMI = \left [\frac{Loan Value}{\frac{1-(1+r)^{-n}}{r}}\right]}$ ---- (2)

where

r = interest rate per month

n = number of months

For Junior Loan

Substituting the values for the junior loan in equation 2 we get an EMI of,

$\mathbf{EMI = \left[\frac{5894} {\frac{1-(1+0.00575)^{-120}}{0.00575}\right]}$

$\mathbf{EMI = \left[\frac{5894} {\frac{1-0.502568253}{0.00575}\right]}$

$\mathbf{EMI = \left [\frac{5894}{\frac{0.497431747}0.00575}}\right]}$

$\mathbf{EMI = \left [\frac{5894}{86.50986907}\right] = 68.13095504}$

Next, we susbsitute the EMI in equation 1 to get

$\mathbf{Lifetime Cost = (68.13* 120) = 8,175.71}$

For Senior Loan

We calculate the EMI from equation 2 as follows:

$\mathbf{EMI = \left[\frac{5258} {\frac{1-(1+0.00625)^{-120}}{0.00625}\right]}$

$\mathbf{EMI = \left[\frac{5258} {\frac{1-0.473470358}{0.00625}\right]}$

$\mathbf{EMI = \left [\frac{5258}{\frac{0.526529642}{0.00625}}\right]}$

$\mathbf{EMI = \left [\frac{5258}{84.24474271}\right] = 62.41339021}$

Next, we susbsitute the EMI in equation 1 to get

$\mathbf{Lifetime Cost = (62.41* 120) = 7,489.61}$

Total Lifetime costs

Finally we calculate the total lifetime costs as:

$\mathbf{Total Lifetime Costs = 8,175.71 + 7,489.61 = 15665.32}$

Answer 2

Answer:  

        OPTION  (A)  $ 16.246.83

Detailed Solution:  

Given Loan details:

Year    Amount  Interest  Subsidized?  Period (months)  Compounding  

Junior  $5,894   6.9% pa      Y  ........      120 months          monthly

Senior  $5,258   7.5% pa     N   ........     120 months          monthly

Subsidized loan means the interest on the total initial loan taken is paid by the education department of the government during the remaining period of education.

Formulas:

P = principal loan amount.       n = period or installments

i = interest rate pm

Factor D = [ (1+i)ⁿ - 1 ] / [ i * (1+i)ⁿ ]

EMI = monthly installment = P / D

Total Lifetime Cost estimate = EMI * n

Junior year Loan:

   D = [ (1+6.9/1200)¹²⁰ - 1 ] / [ 6.9/1200 * (1+6.9/1200)¹²⁰ ]

   D = 86.50987

   EMI = 5,894 /86.50987 = $ 68.130954

   Lifetime cost = $ 68.120954 * 120 = $ 8,175.71

Senior year Loan:

  The principal amount is revised with compound interest during 1 year as it is not subsidized.

  Amount accrued after compounding = P = $ 5,258 [ 1 + 7.5/1200 ]¹²

  P = $ 5,666.1922

  Factor D = [ (1 + 7.5/1200)¹²⁰ - 1 ] / [ 7.5/1200 * (1 + 7.5/1200)¹²⁰ ]

                 = 84.244

  EMI = P/D = $ 67.2593

  Lifetime cost = 67.2593 * 120 = $ 8,071.1156

Total lifetime cost of two loans = $16, 246.8256

So Option (A).

Derivation for the Earnest monthly Installment (EMI) :

Amount Accrued = P * (1 + i)ⁿ

If each of the n EMIs is deposited in a bank at the same interest i then the compounded total of all those will become:

   EMI * [ (1+i)ⁿ⁻¹ + (1+i)ⁿ⁻² + (1+i)ⁿ⁻³ + ..... + (1+i) + (1+i)⁰ ]

      = EMI * [ (1+i)ⁿ - 1 ] / { 1+ i) - 1 ]

Equating both amounts we get the  EMI.

  EMI = P * i * (1+i)ⁿ / [ (1+i)ⁿ - 1 ]