Consider a student loan of $17,500 at a fixed APR of 12% for 20 years. What is the monthly payment?

Consider a student loan of $25,000 at a fixed APR of 6% for 15 years. What is the monthly payment?

Consider a home mortgage of $225,000 at a fixed APR of 4.5% for 20 years. What is the monthly payment?

Consider a home mortgage of $225,000 at a fixed APR of 6% for 30 years. What is the monthly payment?

Suppose that on January 1 you have a balance of $3000 on a credit card whose APR is 21%, which you want to pay off in 1 year. Assume that you make no additional charges to the card after January 1. Calculate monthly payment

Suppose that on January 1 you have a balance of $4500 on a credit card whose APR is 17%, which you want to pay off in 3 years. Assume that you make no additional charges to the card after January 1. Calculate monthly payment

Ap = Pr/[1-(1+i)^-n]
Assuming the interest compounds monthly
p = 175000(.01)/[1 - 1/(1.01)^240] = $1926.90

b) 240*1926.90 = $462,456

c) 175000/462456 = 37.84% is paid to principal and 100-37.84 = 62.16% is paid to interest.

Answer Link

chisnau chisnau · Answer 2 · 2019-07-22T10:34:25-04:00

Answer:

The EMI formula is :

$EMI=\frac{p\times r\times(1+r)^{n}}{(1+r)^{n}-1}$

Part 1.

p = 17500

r = $12/12/100=0.01$

n = $12\times20=240$

Putting values in formula we get

$\frac{17500\times0.01\times(1+0.01)^{240}}{(1+0.01)^{240}-1}$

=> $\frac{17500\times0.01\times(1.01)^{240}}{(1.01)^{240}-1}$

= $192.69

Part 2.

p = 25000

r = @@MATH_TOKEN_5@@

n = @@MATH_TOKEN_6@@

Putting values in formula we get

@@MATH_TOKEN_7@@

=> @@MATH_TOKEN_8@@

= $210.96

Part 3.

p = 225000

r = $4.5/12/100=0.00375$

n = $12\times20=240$

Putting values in formula we get

$\frac{225000\times0.00375\times(1+0.00375)^{240}}{(1+0.00375)^{240}-1}$

=> $\frac{225000\times0.00375\times(1.00375)^{240}}{(1.00375)^{240}-1}$

= $1423.46

Part 4.

p = 225000

r = @@MATH_TOKEN_13@@

n = @@MATH_TOKEN_14@@

Putting values in formula we get

@@MATH_TOKEN_15@@

=> @@MATH_TOKEN_16@@

= $1348.98

Part 5.

p = 3000

r = $21/12/100=0.0175$

n = 12

Putting values in formula we get

$\frac{3000\times0.0175\times(1+0.0175)^{12}}{(1+0.0175)^{12}-1}$

=> $\frac{3000\times0.0175\times(1.0175)^{12}}{(1.0175)^{12}-1}$

= $279.35

Part 6.

p = 4500

r = @@MATH_TOKEN_20@@

n = @@MATH_TOKEN_21@@

Putting values in formula we get

@@MATH_TOKEN_22@@

=> @@MATH_TOKEN_23@@

= $160.43