Question

1. brad wants to save $35000 so he can purchase a new vehicle with cash. he can put $150 per week into an investment that earns an annual rate of 3.25% compounded quarterly. how many months until he will have enough money saved? complete the table and circle the value that was calculated.

pv $0
fv $35000
periods
rate
payment
pmt/yr
cmp/yr
how many years?____

Explanation:

Step1: Identify the values

• \( PV = 0\) (no initial amount)
• \( FV=\$35000\) (future value)
• \( PMT = 150\) (quarterly payment)
• \( r=3.25\%=0.0325\) (annual interest rate)
• \( CMP/Yr = 4\) (compounded quarterly)
• \( Rate=\frac{0.0325}{4}=0.008125\) (quarterly interest rate)

Step2: Use the future - value of an ordinary annuity formula

The formula for the future value of an ordinary annuity is \( FV = PMT\times\frac{(1 + i)^{n}-1}{i}\), where \( i\) is the interest rate per period and \( n\) is the number of periods.
Substitute the values: \( 35000=150\times\frac{(1 + 0.008125)^{n}-1}{0.008125}\)

Step3: Solve for \( n\)

First, simplify the equation:
\(\frac{35000\times0.008125}{150}=(1 + 0.008125)^{n}-1\)
\(\frac{284.375}{150}+1=(1.008125)^{n}\)
\(1.895833 + 1=(1.008125)^{n}\)
\(2.895833=(1.008125)^{n}\)

Take the natural logarithm of both sides: \( \ln(2.895833)=n\ln(1.008125)\)
\(n=\frac{\ln(2.895833)}{\ln(1.008125)}\)
\(\ln(2.895833)\approx1.063\), \(\ln(1.008125)\approx0.0081\)
\(n=\frac{1.063}{0.0081}\approx131.23\) (quarter - periods)

Step4: Convert \( n\) to years

Since there are 4 quarters in a year, the number of years \( t=\frac{n}{4}\)
\(t=\frac{131.23}{4}\approx32.81\) months.
Since \( 1\) year has \( 12\) months, \( t=\frac{32.81}{12}\approx2.73\) years

Answer:

Approximately \( 2.73\) years.