Question

Rachel deposited $5,960.32 into a savings account with an interest rate of 4.2% compounded twice a year. About how long will it take for the account to be worth $9,000? (3 points)
Select one:
a. 21 years, 1 month
b. 18 years, 0 months
c. 19 years, 10 months
d. 9 years, 11 months

Answer 1

Answer:

9 years , 11 months

Step-by-step explanation:

Rachel deposited $5,960.32 into a savings account with an interest rate of 4.2% compounded twice a year

Apply compound interest formula

$A= P(1+\frac{r}{n})^{nt}$

Where  P is the initial amount=$5,960.32

'r' is the rate of interest=4.2%=0.042

'n' is the compounding period=2

t is number of years

A is the amount after t years= 9000

Plug in all the values in the formula

$9000=5960.32(1+\frac{0.042}{2})^{2t}$

Divide both sides by 5960.32

$\frac{9000}{5960.32} =(1.021)^{2t}$

Take ln on both sides

$ln(\frac{9000}{5960.32}) =ln((1.021)^{2t})$

$ln(\frac{9000}{5960.32}) =2tln((1.021))$

Now divide both sides by ln(1.021)

19.82916538=2t

Divide both sides by 2

t=9.91458

1 year = 12 months

0.91458 times 12 = 11

So 9 years , 11 months

Answer 2

You need the compound interest formula solved for time:
Years = log (total / principal) / n * log (1 + rate / n)
Years = log (9,000 / 5,960.32) / 2 * log (1 + .042/2)
Years = log ( 1.509986041) / 2 * log (1.021)
Years =  0.1789729325 / 2 * 0.0090257420869
Years =  0.1789729325 / 0.0180514842
Years =  9.9145826746
the answer is "d"

Source:
http://www.1728.org/compint.htm