Question

A house was valued at $95,000 in the year 1991. The value appreciated to $165,000 by the year

2008.
Use the compound interest formula S = P(1 + r) to answer the following questions.
A) What was the annual growth rate between 1991 and 2008?
T =
Round the growth rate to 4 decimal places.
B) What is the correct answer to part A written in percentage form?
T =
%.
C) Assume that the house value continues to grow by the same percentage. What will the value
equal in the year 2011?
value = $
Round to the nearest thousand dollars.

Answer 1

A) 0.0330

B) 3.30%

C) $182,000

Explanation:

Given the compound interest formula S = P(1 +r)^t, you want to find ...

• r for S=165000, P=95000, t=17
• S for t=20, other values the same

A) Growth rate

The growth rate r in the formula can be found by solving for r:

S = P(1 +r)^t

S/P = (1 +r)^t

(S/P)^(1/t) = 1 +r

r = (S/P)^(1/t) -1

r = (165000/95000)^(1/17) -1

r ≈ 0.0330

B) Percent

Multiplying by 100%, we get ...

r = 0.0330 × 100% = 3.30%

The growth rate is about 3.30%.

C) 2011

The value of the house (in thousands) in 2011 will be ...

S = 95(165/95)^(20/17) ≈ 181.88 ≈ 182

The value of the house in 2011 will be about $182,0000.

__

Additional comment

Using 95(1+0.0330)^20 for the 2011 house value gives a different result in the 10s place, so rounding to thousands will give the same result. In general, we don't like to round intermediate results in a computation like this. That is why we used (165/95)^(1/17) for (1 +r) instead of (1 +0.033).

Here, it doesn't matter, but it does matter in many cases.

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