Question

1. q: a $438 amortized amount borrowed for a payment of quarterly at 6.35% for 14 yrs? a: $1947.16 15. q: a loan of $5000 at 5.5% interest paid monthly for 3 yrs. a: $150.98 16. principal = $250,000, interest rate = 7.35% (4), term = 7.5 yrs a: $5701.03

Explanation:

Step1: Identify formula for Q14

The amortization payment formula is $R = \frac{P \times i}{1-(1+i)^{-n}}$, where $P$ is principal, $i$ is periodic rate, $n$ is total periods.

Step2: Calculate values for Q14

$P=438$, $i=\frac{0.0625}{4}=0.015625$, $n=14\times4=56$
$R = \frac{438 \times 0.015625}{1-(1+0.015625)^{-56}}$

Step3: Compute Q14 result

$1+0.015625=1.015625$, $(1.015625)^{-56}\approx0.4149$, $1-0.4149=0.5851$, $438\times0.015625=6.84375$, $R=\frac{6.84375}{0.5851}\approx11.70$
(Note: Handwritten had a miscalculation; corrected here)

Step4: Identify formula for Q15

Use loan payment formula $R = \frac{P \times i}{1-(1+i)^{-n}}$

Step5: Calculate values for Q15

$P=5000$, $i=\frac{0.055}{12}\approx0.004583$, $n=3\times12=36$
$R = \frac{5000 \times 0.004583}{1-(1+0.004583)^{-36}}$

Step6: Compute Q15 result

$1+0.004583=1.004583$, $(1.004583)^{-36}\approx0.8468$, $1-0.8468=0.1532$, $5000\times0.004583=22.915$, $R=\frac{22.915}{0.1532}\approx150.09$

Step7: Identify formula for Q16

Use loan payment formula $R = \frac{P \times i}{1-(1+i)^{-n}}$

Step8: Calculate values for Q16

$P=250000$, $i=\frac{0.0735}{4}=0.018375$, $n=7.5\times4=30$
$R = \frac{250000 \times 0.018375}{1-(1+0.018375)^{-30}}$

Step9: Compute Q16 result

$1+0.018375=1.018375$, $(1.018375)^{-30}\approx0.5827$, $1-0.5827=0.4173$, $250000\times0.018375=4593.75$, $R=\frac{4593.75}{0.4173}\approx11008.27$
(Note: Handwritten had a miscalculation; corrected here)

Answer:

1. $\$11.70$ (corrected from handwritten error)
2. $\$150.09$ (matches handwritten approximate value)
3. $\$11008.27$ (corrected from handwritten error)