Question

Juan is leasing a new car. The price of the car is 35,000. The terms of
the lease go as follows. There are 120 monthly payments with the first payment
being one month from now. The nominal rate of interest is 8.4% convertible
monthly. The payments in the first year are X. In the second year, the payments
are 1.01X, in the third 1.01 X^2 and so on. Just after his 60th payment, Juan
sells his car for Y dollars. Before Juan receives a check, he first must pay off the outstanding loan balance still owed just after his 60th payment to the lender.
What is the smallest value of Y such that Juanís check is at least 20000?

(a) 15,000-15,500
(b) 23,500-24,000
(c) 34,500-35,000
(d) 41,500-42,000
(e) 44,000-44,500

Answer 1

Solution:

Given:

The Price of Car i.e. Initial Investment = $ 35,000

Total Number of Monthly Payments = 120 Months

Annual Rate of Interest = 8.4 %

Payment in the 1st Year = X

Payment in the 2nd Year = 1.01 X

Payment in the 3rd Year = 1.01 X^2 and So on…till 10 Years as the number of months given are 120, so 12 Months each year which is equal to 10 Years.

Therefore, Total Number of Payments = 120 Months or 10 Years.

And Number of Payments per year = 12 per years

Juan sells his car just after his 60th Payment i.e. after 5 Years, as 60 Months is equal to 5 Years.

The amount at which he sells the car = Y dollars

To Calculate:

The value of Y.

Tabulation of Lease Amortization Schedule on per year basis i.e. 10 Years:

Payment Number	Opening Amount of Payment	Repayment per Year 35000/ 10= $ 3500 per year	Interest @ 8.4 % on Opening Amount of Payment	Closing Amount= Initial Investment $ 35,000 -Repayment Amount
1	$ 35,000	$ 3,500	2,940	$ 31,500
2	$ 31,500	$ 3,500	2,646	$ 28,000
3	$ 28,000	$ 3,500	2,352	$ 24,500
4	$ 24,500	$ 3,500	2,058	$ 21,000
5	$ 21,000	$ 3,500	1,764	$ 17,500
Total		$ 17,500	11,760
6	$ 17,500	$ 3,500	1,470	$ 14,000
7	$ 14,000	$ 3,500	1,176	$ 10,500
8	$ 10,500	$ 3,500	8,82	$ 7,000
9	$ 7,000	$ 3,500	5,88	$ 3,500
10	$ 3,500	$ 3,500	2,94	$ 0,000
Total		$ 17,500	$ 4,410

As given in the question that Juan sold his car after 60th payment i.e. after 5 years so the remaining amount of outstanding loan payment due is the Principal Amount for the next 5 years (from 6 to 10 years) + The interest due on the amount @ 8.4 % for the next 5 years (from 6 to 10 years), which is calculated in the table above and highlighted yellow.

Amount to be given by Juan to the lender = $ 17,500 + $ 4,410 = $ 21,910

The smallest value of Y such that Juan’s cheque is at least $ 20,000 is

$ 21,910 + $ 20,000 = $ 41,910

Y = $ 41,910

Since this value of $ 41,910 lies in option (d) so our answer option is option (d).

Ans: (d) 41,500-42,000