Originally Posted by

**jonah** From January 1, 1979 to January 1, 1980 is 1 year.

The price of the car is = a down payment/deposit of 3000 + remaining balance of $\displaystyle

500 \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}

{4}} \right)^{ - \left( {5*4} \right)} }}

{{\left( {\tfrac{{.10}}

{4}} \right)}}

$

Since the man did not bother to make any of the $500 quarterly payments until April 1, 1980, the accumulated amount (future value) of his remaining balance on January 1,1980 is

$\displaystyle

\left[ {500 \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}

{4}} \right)^{ - \left( {5*4} \right)} }}

{{\left( {\tfrac{{.10}}

{4}} \right)}}} \right]\left( {1 + \tfrac{{.10}}

{4}} \right)^{\left( {1*4} \right)}

$

Accordingly, “the rest of the 5 year period” is 5-1.

The new quarterly payments can then be determined by solving for R in the following equation:

$\displaystyle

\left[ {500 \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}

{4}} \right)^{ - \left( {5*4} \right)} }}

{{\left( {\tfrac{{.10}}

{4}} \right)}}} \right]\left( {1 + \tfrac{{.10}}

{4}} \right)^{\left( {1*4} \right)} = R \cdot \frac{{1 - \left( {1 + \tfrac{{.10}}

{4}} \right)^{ - \left[ {\left( {5 - 1} \right)*4} \right]} }}

{{\left( {\tfrac{{.10}}

{4}} \right)}}

$

By my reckoning, the answer on you tutorial sheet is correct.