Question

You're trying to save to buy a new car valued at $48,690. You have $38,000 today that can be invested at your bank. The bank pays 3.7 percent annual interest on its accounts. How long will it be before you have enough to buy the car for cash? Assume the price of the car remains constant. 5.13 years 9.29 years

Answer 1

Answer:

6.82 years

Explanation:

Future Value, F.V.=  $48,690

Present Value, P.V. = $38,000

Rate=3.7% = 0.037

$F.V=P.V(1+r)^n$

$48690=38000(1+0.037)^n\\\frac{48690}{38000} =(1.037)^n$

Changing to logarithm form

$n=log_{1.037}\frac{48690}{38000}\\=log_{1.037}1.2813\\=\frac{log 1.2813}{log 1.037} \\n=6.82$

After 6.82 years, the future value of $38000 will be $48,690/

Answer 2

Answer:

Time = 6.82 Years

Explanation:

Amount of investment now                          38,000.00  

Interest rate per year                                  3.7%  

Time Opening Balance      Interest            Closing Balance

1.00         38,000.00              1,406.00            39,406.00

2.00         39,406.00              1,458.02            40,864.02

3.00         40,864.02              1,511.97            42,375.99

4.00         42,375.99              1,567.91            43,943.90

....            ...                                  ....                          ....

7.00         47,255.91              1,748.47             49,004.38

Amount at end of 6 years             47,255.91  

Amount required for car                     48,690.00  

Interest required in 7th year             1,434.09  

Actual Interest earned                     1,748.47  

Time = 6 + 1434.09/1748.47    

Time = 6.82 Years