Cliff Branch bought a home with a 10.5% adjustable rate mortgage for 30 years. He paid $9.99 monthly per thousand on his original loan. At the end of 3 years he owes the bank $65,000. Since interest rates have risen to 12.5%, the bank will renew the mortgage at this rate, or Cliff can pay the bank $65,000. He decides to renew and will now pay $10.68 monthly per thousand on his loan. (You can ignore the small amount of principal paid during the 3 years.)
What was the old monthly payment?
What is the new monthly payment?
What is the percent increase in his monthly payment to the nearest10th

Question

Cliff Branch bought a home with a 10.5% adjustable rate mortgage for 30 years. He paid $9.99 monthly per thousand on his original loan. At the end of 3 years he owes the bank $65,000. Since interest rates have risen to 12.5%, the bank will renew the mortgage at this rate, or Cliff can pay the bank $65,000. He decides to renew and will now pay $10.68 monthly per thousand on his loan. (You can ignore the small amount of principal paid during the 3 years.)
What was the old monthly payment? 
What is the new monthly payment? 
What is the percent increase in his monthly payment to the nearest10th

amistre64 · Answer

hmm, seems workable to me.

we would prolly need to know the original loan amount.

amistre64 · Answer

i have my own equations that i use to play with ... the textbook ones i never liked

amistre64 · Answer

$$B_n=B_ok^n-P\left(\frac{k^n-1}{k-1}\right)$$
$$B_n+P\left(\frac{k^n-1}{k-1}\right)=B_ok^n $$
$$B_nk^{-n}+Pk^{-n}\left(\frac{k^n-1}{k-1}\right)=B_o $$

amistre64 · Answer

considering that P = Bo 9.99/1000 i might want to reconsider how i solve that