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Two contracts pay Amani $100,000 in year 1 and $100,000 in year 2 versus $132,000 in year 1 and $66,000 in year 2. At what lowest interest rate does the present value of the second contract exceed the first (10%, 7%, 9%, or 8%)?

Answer

Compare present values: $PV_1=\frac{100{,}000}{1+r}+\frac{100{,}000}{(1+r)^2}$ and $PV_2=\frac{132{,}000}{1+r}+\frac{66{,}000}{(1+r)^2}$. The difference is $PV_2-PV_1=\frac{32{,}000}{1+r}-\frac{34{,}000}{(1+r)^2}$, which is positive when $r>6.25\%$. Therefore, the lowest listed rate where contract 2 has the higher present value is 7% (choice b).

Explanation

What you are comparing

Both contracts pay the same total ($200,000$), but the timing differs. The better contract depends on the discount rate $r$ because earlier money is worth more.

Write each contract as a present value

Contract 1: $$ PV_1=\frac{100{,}000}{1+r}+\frac{100{,}000}{(1+r)^2} $$ Contract 2: $$ PV_2=\frac{132{,}000}{1+r}+\frac{66{,}000}{(1+r)^2} $$

Find the interest rate where the second becomes better

Look at the difference: $$ \begin{aligned} PV_2-PV_1 &= \left(\frac{132{,}000}{1+r}-\frac{100{,}000}{1+r}\right)+\left(\frac{66{,}000}{(1+r)^2}-\frac{100{,}000}{(1+r)^2}\right)\\ &=\frac{32{,}000}{1+r}-\frac{34{,}000}{(1+r)^2} \end{aligned} $$ Let $x=1+r$. Then: $$ PV_2-PV_1=\frac{32{,}000x-34{,}000}{x^2} $$ Since $x^2>0$, the sign depends on the numerator: $$ 32{,}000x-34{,}000>0 \;\Rightarrow\; x>\frac{34}{32}=1.0625 \;\Rightarrow\; r>0.0625=6.25\% $$ So contract 2 has the higher PV for any rate above $6.25\%$.

Match to the answer choices

The smallest rate listed that is greater than $6.25\%$ is 7%, so the correct choice is b.

(Quick check at $7\%$: $PV_1\approx 180{,}802$, $PV_2\approx 181{,}012$, so contract 2 is slightly higher.)

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Skills You Achive
time value of money present value discounting cash flows inequality solving

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