Question

1. sandy made several investments. she bought 1000 shares of a company’s stock for $8.60/share, she bought a bond with a face value of $2500 and a coupon rate of 7%, and she invested $5000 into a fund that is expected to grow by 3.5% per year. (a) sandy pays a broker a commission of $14 to buy and sell stock. after one year, sandy sold all her shares, when they were worth $9.15/share. what was her net gain or loss? show your work. (b) the bond sandy purchased will mature in 10 years. how much interest will she receive semiannually? (c) how long will it take the fund she invested in to be worth $10,000? (d) sandy’s gross annual income is $51,350. she is paid biweekly and has 5% deducted from her paycheck for her 403(b). her employer matches her deduction, up to 4%. how much is deposited into sandy’s 403(b) each payday? answer:

Explanation:

Part (a)

Step1: Calculate cost to buy stocks

Cost to buy = (Number of shares × Price per share) + Commission to buy
Number of shares = 1000, Price per share = $8.60, Commission to buy = $14
Cost to buy = \( 1000 \times 8.60 + 14 = 8600 + 14 = 8614 \) dollars

Step2: Calculate revenue from selling stocks

Revenue from sell = (Number of shares × Selling price per share) - Commission to sell
Selling price per share = $9.15, Commission to sell = $14
Revenue from sell = \( 1000 \times 9.15 - 14 = 9150 - 14 = 9136 \) dollars

Step3: Calculate net gain/loss

Net gain = Revenue - Cost = \( 9136 - 8614 = 522 \) dollars

Part (b)

Step1: Recall bond interest formula

Semiannual interest = \( \frac{\text{Face Value} \times \text{Coupon Rate}}{2} \)
Face Value = $2500, Coupon Rate = 7% = 0.07

Step2: Calculate semiannual interest

Semiannual interest = \( \frac{2500 \times 0.07}{2} = \frac{175}{2} = 87.5 \) dollars

Part (c)

Step1: Use compound growth formula

The fund grows with compound interest: \( A = P(1 + r)^t \), where \( A = 10000 \), \( P = 5000 \), \( r = 3.5\% = 0.035 \), solve for \( t \).
\( 10000 = 5000(1 + 0.035)^t \)

Step2: Simplify and solve for \( t \)

Divide both sides by 5000: \( 2 = (1.035)^t \)
Take natural log: \( \ln(2) = t \ln(1.035) \)
\( t = \frac{\ln(2)}{\ln(1.035)} \approx \frac{0.6931}{0.0344} \approx 20.15 \) years

Part (d)

Answer:

s:
(a) Net gain of $\boldsymbol{522}$ dollars
(b) Semiannual interest of $\boldsymbol{87.5}$ dollars
(c) Approximately $\boldsymbol{20.15}$ years
(d) $\boldsymbol{177.75}$ dollars deposited per payday