Question

you purchase a home for $125,000. the value of the property increases by 1.9% every year. after ten years, you decide to sell the home. in the meantime, you have made renovations and improvements to the house which increase its sale value by $48,300. how much will you sell the house for, to the nearest hundred dollars?
a. $257,500
b. $209,200
c. $150,900
d. $199,200

please select the best answer from the choices provided

a
b
c
d

Explanation:

Step1: Recall compound growth formula

The formula for compound growth is $A = P(1 + r)^t$, where $P$ is the principal amount, $r$ is the annual growth rate (in decimal), and $t$ is the time in years. Here, $P = 125000$, $r = 0.019$, and $t = 10$.

Step2: Calculate the value after 10 years due to growth

Substitute the values into the formula: $A = 125000(1 + 0.019)^{10}$. First, calculate $(1 + 0.019)^{10}\approx1.019^{10}\approx1.2079$. Then, $125000\times1.2079 = 150987.5$.

Step3: Add the renovation value

Add the renovation increase of $48300$: $150987.5 + 48300 = 199287.5$.

Step4: Round to nearest hundred

Rounding $199287.5$ to the nearest hundred gives $199300$? Wait, no, wait, maybe miscalculation. Wait, let's recalculate $(1.019)^{10}$. Let's compute step by step:

Year 1: $125000\times1.019 = 125000 + 125000\times0.019 = 125000 + 2375 = 127375$

Year 2: $127375\times1.019 = 127375 + 127375\times0.019 = 127375 + 2419.125 = 129794.125$

Year 3: $129794.125\times1.019 \approx 129794.125 + 129794.125\times0.019 \approx 129794.125 + 2466.088 = 132260.213$

Year 4: $132260.213\times1.019 \approx 132260.213 + 132260.213\times0.019 \approx 132260.213 + 2512.944 = 134773.157$

Year 5: $134773.157\times1.019 \approx 134773.157 + 134773.157\times0.019 \approx 134773.157 + 2560.690 = 137333.847$

Year 6: $137333.847\times1.019 \approx 137333.847 + 137333.847\times0.019 \approx 137333.847 + 2609.343 = 139943.19$

Year 7: $139943.19\times1.019 \approx 139943.19 + 139943.19\times0.019 \approx 139943.19 + 2658.919 = 142602.109$

Year 8: $142602.109\times1.019 \approx 142602.109 + 142602.109\times0.019 \approx 142602.109 + 2709.440 = 145311.549$

Year 9: $145311.549\times1.019 \approx 145311.549 + 145311.549\times0.019 \approx 145311.549 + 2760.919 = 148072.468$

Year 10: $148072.468\times1.019 \approx 148072.468 + 148072.468\times0.019 \approx 148072.468 + 2813.377 = 150885.845$

Ah, so more accurately, after 10 years, the value from growth is approximately $150885.845$. Then add the renovation: $150885.845 + 48300 = 199185.845$. Rounding to the nearest hundred: look at the tens digit, which is 8, so we round up the hundreds digit. The number is $199185.845$, so to the nearest hundred, it's $199200$? Wait, no, $199185.845$: the hundreds digit is 1 (wait, 199185: digits are 1 (100000), 9 (10000), 9 (1000), 1 (100), 8 (10), 5 (1)). Wait, no, 199185 is 1 hundred thousand, 9 ten thousands, 9 thousands, 1 hundred, 8 tens, 5 ones. So to round to nearest hundred, we look at the tens digit (8) which is ≥5, so we add 1 to the hundreds digit (1) to get 2, and set tens and ones to 0. So $199200$. Wait, but the option d is $199,200$. So that's the answer.

Wait, maybe my initial formula calculation was slightly off due to approximation, but the step-by-step year calculation gives us to around $199185$, which rounds to $199200$, which is option d.

Answer:

d. $199,200$