In: Accounting
Joseph is considering a used car currently valued at $12,000. He can get an interest rate of 2.5% annually for a 5 year car loan. Joseph currently has $14,000 in a savings account and wants to use some of his savings for a down payment. If Joseph decided to put $3,000 down as a down payment, what are his monthly payments? If Joseph decided to put $5,000 down as a down payment, what are his monthly payments? How much does Joseph save in interest if he puts $5,000 down compared to $3,000? If Joseph decides to pay for the whole thing what could he potentially save in interest payments? If Joseph can invest $12,000 in an 8% annual investment. Alternatively he is considering just paying cash for the car leaving him no payment. Which would you recommend and why? Include the numbers!!
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Value of Car | 12,000.00 | ||||||
Rate of Interest | 2.50% | ||||||
Period (Years) | 5.00 | ||||||
Down payment | 3,000.00 | Down payment | 5,000.00 | ||||
Loan Value | 9,000.00 | Loan Value | 7,000.00 | ||||
We Know, | We Know, | ||||||
Monthly Payment= | PV*i | Monthly Payment= | PV*i | ||||
1-(1+i)^-n | 1-(1+i)^-n | ||||||
PV | Loan Value | 9,000.00 | PV | Loan Value | 7,000.00 | ||
Instalments in a year | 12.00 | Instalments in a year | 12.00 | ||||
n | 12*5 | 60.00 | n | 12*5 | 60.00 | ||
i | 2.5%/12 | 0.21% | 0.00208333333333333 | i | 2.5%/12 | 0.21% | 0.00208333333333333 |
Monthly Payment= | 9000*0.00208333333333333 | Monthly Payment= | 7000*0.00208333333333333 | ||||
1-(1+0.00208333333333333)^-60 | 1-(1+0.00208333333333333)^-60 | ||||||
Monthly Payment= | 18.75 | Monthly Payment= | 14.58333333 | ||||
1-(1.00208333333333333)^-60 | 1-(1.00208333333333333)^-60 | ||||||
Monthly Payment= | 18.75 | Monthly Payment= | 14.58333333 | ||||
1-0.882611659150112 | 1-0.882611659150112 | ||||||
Monthly Payment= | 18.75 | Monthly Payment= | 14.58333333 | ||||
0.11738834 | 0.117388341 | ||||||
Monthly Payment ($) | 159.73 | Monthly Payment ($) | 124.23 | ||||
Total payments | 9,583.58 | 159.73*60 | Total payments | 7,453.89 | 124.73*60 | ||
Loan Value | 9,000.00 | Loan Value | 7,000.00 | ||||
Interest paid | 583.58 | Interest paid | 453.89 | ||||
Difference in Interest paid | 129.68 | ||||||
Joseph will save $ 129.68 in interest if he puts $5,000 down compared to $3,000. | |||||||
If Joseph decides to pay for the whole thing he could have saved $ 453.89 in interest payments. | |||||||
If Joseph can invest $12,000 in an 8% annual investment. | |||||||
P | 12,000.00 | ||||||
i | 8% | 0.08 | |||||
Years | 5.00 | ||||||
Interest amount | P*((1+i)^n-1) | ||||||
Interest amount | 12000*((1+.08)^5-1) | ||||||
Interest amount | 5,631.94 | ||||||
Its better to take loan against car because Joseph can invest $ 12,000 which will earn him interest of $ 5,631.94. | |||||||
It is much more than the total interest he has to pay on loan. | |||||||