price of two bonds given the period, par value and coupon being (i + 0.04) and (i - 0.04) and that one bond is x amount greater than the other

Question

Given the following determine the price of each bond and the yield rate i.

i- each bond is 10-year bond with semiannual coupons redeemable at its par value 10,000 and is bought to yield an annual nominal rate i, convertible semi-annually.

ii- bond A has annual coupon rate of (i+.04) paid semi-annually iii- bond B has an annual coupon rate of (i-0.04) paid semiannualy.

iv - Price of bond A is 5,341.12 greater than the price of Bond B.

I solved most of the questions in my textbook for this section but I went on to take a practice exam and was dropped a bomb with this question.. In fact every question on the practice exam appears to be layered with increasing difficulty and are the type of questions I was never exposed to in any of my textbook problems.. I dont even know where to begin and this is extremely discouraging as I've already spent a great deal of time and effort studying up to this point and feel as if I havent taken any significant step.

Answer 1

Let $A$ and $B$ denote the prices of bonds $A$ and $B$, respectively. Part (i) is just giving us information. Parts (ii) and (iii) are intended for you to set up two equations and two unknowns.

Starting with bond A, we know $r=i+.04$. Since the face amount is 10,000, the coupons will be of size $10,000\frac{r}{2}=10,000(\frac{i+.04}{2}),$ where we divide by $2$ because of compounding semi-annually. These coupons are paid at times $t=1,2,\dots 20$, where time is in half-years. The face amount is then paid at time $t=20$. Hence, denoting $v=\frac{1}{1+i/2}$ the present value of $A$ and $B$ are: \begin{align} A&=10,000v^{20}+10,000\left(\frac{i+.04}{2}\right)(v+v^2+\dots+v^{20})\\ B&=10,000v^{20}+10,000\left(\frac{i-.04}{2}\right)(v+v^2+\dots+v^{20}) \end{align} Subtracting, we find \begin{align} A-B&=10,000(.04)(v+v^2+\dots+v^{20}) \\ &=400(v+v^2+\dots+v^{20}) \end{align} From (iii), we know $A-5341.12=B$, or equivalently, $A-B=5341.12$. It should be clear how to solve for $i/2$. Once you know $i/2$, just plug in to find the prices of $A$ and $B$.