Question

23. need proper working

Y ound to nearest $1), $570 D) $900 ZZ) If you put $10 in a savings account at the beginning of each month for 15 years, how much money will be in the account at the end of the 10th year? Assume that the account earns12% compounded monthly and round to the nearest $1. X2 12 A) $1,200 B) $2,323 C) $5,046 D) $3,485 23) You have contracted to buy a house for $250,000, paying $30,000 down and taking out a fully amortizing loan for the balance, at a 5.7% annual rate for 30 years. What will your monthly payment be if they make equal monthly installments over the next 30 years (to the nearest dollar)? A) $1,035 -763 220v th 1-1.057 B) $1,123 C) $1,189 ouiz D) $1,277 DurPMT 24) Jimmy just bought a new Ford SUV for his business. The price of the vehicle was $40,000. Jimmy made a $5,000 down payment and took out an amortized loan for the rest. The car dealership made the loan at 8% interest compounded monthly for five years. He is to pay back the principal and interest in equal monthly installments beginning one month from now. Determine the amount of Jimmy's monthly payment. A) $634.56 R$709.67

Answer 1

22.

Amount deposited at starting of each month = P = $10

Number of months = n = 10*12 = 120

Monthly Interest Rate = r = 0.12/12 = 0.01

Amount in account at 10th Year = FV = P(1+r)ⁿ +....+ P(1+r)² + P(1+r)

= P [((1 + r)ⁿ - 1) / r])(1 + r)

= 10 [((1 + 0.01)¹²⁰ - 1) / 0.01])(1 + 0.01)

= $2323.39

= $2,323.39

23.

Value of the home = $250000

Downpayment = $30000

Loan Amount P = $250000 - $30000 = $220000

Interest Rate = 5.7% or 0.057/12 monthly

Number of payment periods = n = 30*12 = 360 months

Let monthly payments made be X

Hence, the sum of present value of monthly payments must be equal to the value of the loan amount

=> X/(1+r) + X/(1+r)² +....+ X/(1+r)^N = P

=> X[1- (1+r)^-N]/r = P

=> X = rP(1+r)^N/[(1+r)^N-1]

Hence, Monthly Payments =  rP(1+r)^N/[(1+r)^N-1]

= 220000*( 0.057/12)*(1+ 0.057/12)³⁶⁰/((1+ 0.057/12)³⁶⁰-1)

= $1276.88

= $1,277