Explanation

For a zero-coupon bond, the price is calculated as:

$P = \frac{F}{(1 + \frac{r}{m})^{m \times n}}$

Where:

• $P$ = price = 90
• $F$ = face value = 100 (assumed for zero-coupon bonds)
• $r$ = annual yield-to-maturity (what we're solving for)
• $m$ = compounding frequency = 4 (quarterly compounding)
• $n$ = years to maturity = 3

Rearranging the formula:

$90 = \frac{100}{(1 + \frac{r}{4})^{4 \times 3}}$ $90 = \frac{100}{(1 + \frac{r}{4})^{12}}$ $(1 + \frac{r}{4})^{12} = \frac{100}{90} = 1.111111$ $1 + \frac{r}{4} = (1.111111)^{\frac{1}{12}}$ $1 + \frac{r}{4} = 1.008797$ $\frac{r}{4} = 0.008797$ $r = 0.035188$ $r = 3.52\%$

This is closest to 3.5% (Option C).

Verification: $(1 + \frac{0.035}{4})^{12} = (1.00875)^{12} = 1.1100$ $\frac{100}{1.1100} = 90.09 \approx 90$

Therefore, the correct answer is C. 3.5%.