Finding yield rate for 2 bonds An investor purchases two bonds with the following properties: 
Bond 1: Has a face value $1000 and is redeemable at par. Pays coupons annually at a rate of 8.1% annual and was purchased for 1117.19. 
Bond 2: Has a face value $1000 and is redeemable at par. Pays coupons annually at a rate of 6.5% annual and was purchased for 981.32. 
If both bonds mature in the same number of years and the investor yields the same rate on both bonds, find the yield rate.
I have tried my best to work it out but there are too many variables and not sure how to go about it. Please help me find the yield rate
 A: I would just make a spreadsheet to compute the return on each bond for different terms.  Most spreadsheets will compute rate of return for a cash flow.  The first bond will have a yield of $8.1\%$ for very long terms because the effect of the premium will be reduced.  At $1$ year the yield will be negative.  The second bond will have a yield of $6.5\%$ per year for long terms for the same reason, but it will rise for short terms because of the discount.  At $1$ year it will be something like $8.4\%$ because the discount will give an additional $1.9\%$ of yield.  You can then try different terms to see where the yields match.  I would guess around $8-9$ years as you have $13.6\%$ premium differential and $1.6\%$ yield differential, but that is a guess.
A: It's like a CFA question.
Strictly speaking, the valuation of bonds depends on how you model interest rate term structure (ho-lee, HJM, or LIBOR etc.). But here I guess we assume the interest rate will be constant throughout the investment horizon. Plus, the convention is that bond pays semi-annual coupon at the rate of $(1 + C)^{0.5} - 1$ (or approx. $0.5C$), but here I just assume it pays once a year, at the rate of $C$ for simplicity.
So here is the answer (hopefully I got the numbers right, but even if not, you get how to do it): 
$$1117.19 = \frac {1000}{(1+r)^n} + \frac{81}{(1+r)} + \frac{81}{(1+r)^2} + ... + \frac{81}{(1+r)^n}=\frac{1000}{(1+r)^n} + 81 \times \frac{(1+r)^n - 1}{(1+r)^n \times r}$$ 
$$981.32 = \frac{1000}{(1+r)^n} + \frac{65}{(1+r)} + \frac{65}{(1+r)^2}+...+\frac{65}{(1+r)^n}=\frac{1000}{(1+r)^n} + 65 \times \frac{(1+r)^n - 1}{(1+r)^n \times r}$$
let $x = \frac {1000}{(1+r)^n}$, and $y=\frac{(1+r)^n - 1}{(1+r)^n\times r}$, we have $$1117.19=x + 81 \times y$$
$$981.32=x+65 \times y$$
Solving x and y, we have:
$$\frac{1000}{(1+r)^n} =x= 429.348125$$
$$\frac{(1+r)^n - 1}{(1+r)^n \times r}=y=8.491875$$
Thus
$$(1+r)^n = 2.329112303$$
$$r=0.06719975=6.72\%$$
$$n=13$$
