John buys a car for $54,029 upon graduation from college with an engineering deg
ID: 1138482 • Letter: J
Question
John buys a car for $54,029 upon graduation from college with an engineering degree and a very good job offer. As a graduation gift, his parents pay a down payment of $9,225 on the car. The rest of the amount was financed at 6 nominal interest with 27 monthly payments. The first payment will start at the end of the 6th month. Determine the monthly payments.QUESTION graduation from college with an engineering degree and a very good job offer. As a graduation gift his parents pay a down payment The first payment will start at the end of the sth month Jobn bays a car for $54,029 upon of $9,225 on the car. The rest of the amount was financed at 6 nominal interest with 27 monthly payments Determine the payments QUESTION 5
Explanation / Answer
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Answer-
Total Cost
54029.00
Down payment
9225.00
Interest rate
6%
Per annum
Interest rate on monthly basis
0.50%
Term
27
Month
Financed amount
44804.00
Cost-payment
EMI
1778
Total Payment
48006
Interest payment
3202.00
Principal part
44804.00
(1+.r)^n
1.144151851
(1+.r^n)-1
0.1442
(1+.r)^n/(1+.r^n)-1
7.93712911
P*R*(1+r)^n/((1+r)^n-1)
1778.075663
Formula used for EMI= P*r*{(1+.r)^n/(1+.r^n)-1}
Formula is above
P=principal= 44804.00 because that is the amount borrowed after subtracting down payment from the total cost
R= rate = 6% on annual basis and .50% on monthly basis by 6%/12
N= number of month of loan= 27
The formula above will give the equated monthly installment as calculated above
Total Cost
54029.00
Down payment
9225.00
Interest rate
6%
Per annum
Interest rate on monthly basis
0.50%
Term
27
Month
Financed amount
44804.00
Cost-payment
EMI
1778
Total Payment
48006
Interest payment
3202.00
Principal part
44804.00
(1+.r)^n
1.144151851
(1+.r^n)-1
0.1442
(1+.r)^n/(1+.r^n)-1
7.93712911
P*R*(1+r)^n/((1+r)^n-1)
1778.075663
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