Answered: Tara’s student loan of $22,000 at 3.82%…

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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Question

Tara’s student loan of $22,000 at 3.82% compounded quarterly was amortized over 6 years with payments made at the end of every month. What was the principal balance on the loan after 4 years? $0.00 Round to the nearest cent.

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Expert Solution

Step 1

A bank loan is when, over a certain amount of time, a provider gives to give loans to customers.

Step 2

The equations are;

$\begin{array}{rcl} p a y m e n t p e r i o d & < & c o m p o u n d i n g p e r i o d \\ i_{e f f} & = & {(1 + r)}^{\frac{1}{m}} - 1 \\ r & = & i n t e r e s t r a t e p e r c o m p u n d i n g p e r i o d . \\ m & = & p a y m e n t p e r i o d s i n a c o m p o u n d i n g p e r i o d \\ L o a n p a y m e n t s (A) & = & L o a n v a l u e (L V) [\frac{i \times {(1 + i)}^{n}}{{(1 + i)}^{n} - 1}] \\ T h e u n p a i d l o a n a f t e r t p e r i o d s & = & t \times A - (A - i \times L V) [\frac{{(1 + i)}^{t} - 1}{i}] \end{array}$

$\begin{array}{rcl} L V & = & 22000 \\ n & = & 6 y e a r s, 6 \times 12 m o n t h s = 72 \\ I n t e r e s t r a t e & = & 3.83 % \\ c o m p o u n d q u a r t e l r y & = & \frac{0.0382}{4} = 0.00955 p e r q u a r t e r \\ t & = & 4 y e a r s, 48 m o n t h s \\ i_{e f f} & = & {(1 + 0.00955)}^{\frac{1}{3}} - 1 \\ = & 0.003173253 \end{array}$

$\begin{array}{rcl} L o a n p a y m e n t & = & 22000 [\frac{0.003173253 {(1 + 0.003173253)}^{72}}{{(1 + 0.003173253)}^{72} - 1}] \\ = & 22000 \times 0.015558 \\ = & 342.27 \end{array}$

$\begin{array}{rcl} U n p a i d l o a n a f t e r 4 y e a r s; (t = 48) \\ = & 48 \times 342.27 - (342.27 - 0.003173257 \times 22000) [\frac{{(1 + 0.003173257)}^{48} - 1}{0.003173257}] \\ = & 7897.46 \end{array}$

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