Question

A car loan is taken for $13,000 to be paid back in 5 years, with monthly payments of $495. What nominal annual interest rate is being charged in this loan?

1) 27.42%

2) 42.26%

3) 39.00%

4) 2.29%

Answer 1

7t 7n

here we have loan amount is 13000

PV=13000

PMT=495

n=12 for monthly compound

t=5 years

T )-12.5 12 13000 495 12

| 1-0+ r-60 ) 13000 495 12

$13000=495\cdot 12\left[\frac{1-\left(1+\frac{r}{12}\right)^{-60}}{r}\right]$

$13000=5940\left[\frac{1-\left(1+\frac{r}{12}\right)^{-60}}{r}\right]$...............(1)

as we can see here equation is complex so we cannot solve for variable

to find the value of "r" we have to use either trial and error method or we can check all 4 option

.

so check first value of r=27.42%

r=27.42% = 0.2742

put it back in equation 1

1- (1+0.2742)-60 5940 | |-at- 1 3000 0.2742

$13000= 5940\left[\frac{1-\left(1+0.02285\right)^{-60}}{0.2742}\right]$

$13000= 5940\left[\frac{1-\left(1.02285\right)^{-60}}{0.2742}\right]$

$13000= 5940\left[\frac{1-0.2578}{0.2742}\right]$

$13000= 5940\left[\frac{0.7421}{0.2742}\right]$

$13000= 5940\left[2.7067\right]$

${\color{Blue} 13000 \neq 16078.29}$

here both sides are not equal so first option is incorrect

.

.

.

.

check second value of r=27.42%

r=27.42% = 0.2742

put it back in equation 1

$13000=5940\left(\frac{1-\left(1+\frac{0.4226}{12}\right)^{-60}}{0.4226}\right)$

$13000= 5940\left(\frac{1-\left(1+0.03521\right)^{-60}}{0.4226}\right)$

$13000= 5940\left(\frac{1-\left(103521\right)^{-60}}{0.4226}\right)$

$13000= 5940\left(\frac{1-0.1253}{0.4226}\right)$

( 746) 1300-5940 0.8740 13000 5940

$13000= 5940\left(2.069\right)$

${\color{Blue} 13000 \neq 12293.94}$

here both sides are not equal so second option is incorrect

.

.

.

.

check third value of r=27.42%

r=27.42% = 0.2742

put it back in equation 1

$13000= 5940\left(\frac{1-\left(1+\frac{0.39}{12}\right)^{-60}}{0.39}\right)$

$13000= 5940\left(\frac{1-\left(1+0.0325\right)^{-60}}{0.39}\right)$

$13000= 5940\left(\frac{1-\left(1.0325\right)^{-60}}{0.39}\right)$

$13000= 5940\left(\frac{1-0.1467}{0.39}\right)$

$13000= 5940\left(\frac{0.8532}{0.39}\right)$

$13000= 5940\left(2.18856\right)$

13000 13000

now both sides are equal

so answer is option C .........r=39%