1. When a new baby is born to the Johnsons, the family decides to invest $5000 in an account that earns 7% interest as a way to start the baby's college fund. If they do not touch that investment for 18 years, how much will there be in the college fund?

Question

2. The local animal rescue league is trying to reduce the number of stray dogs in the county. They estimate that there are currently 400 stray dogs and that through their efforts they can place about 8% of the animals each month. How many stray dogs will remain in the county 12 months after the animal control effort has started?

3. A basket of groceries costs $96.50. Assuming an inflation rate of 1.8% per year, how much will that same basket of groceries cost in 20 years?

Answer 1

Let's solve each of the problems step by step.

Problem 1: Future Value of an Investment

To calculate the future value of the investment, we can use the formula for compound interest:

\[ FV = P(1 + r)^n \] Where:

• \( FV \) is the future value of the investment
• \( P \) is the principal amount ($5000)
• \( r \) is the annual interest rate (7% or 0.07)
• \( n \) is the number of years the money is invested (18 years)

Substituting the values into the formula:

\[ FV = 5000(1 + 0.07)^{18} \]

Calculating the expression inside the parentheses:

\[ FV = 5000(1.07)^{18} \]

Now calculating \( (1.07)^{18} \):

\[ (1.07)^{18} \approx 3.386 \]

Now, we multiply this by the principal:

\[ FV \approx 5000 * 3.386 \approx 16930 \]

Therefore, the amount in the college fund after 18 years will be approximately $16,930.00.

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Problem 2: Remaining Stray Dogs

The number of stray dogs decreases by 8% each month. We can use the formula for exponential decay:

\[ N = N_0(1 - r)^t \] Where:

• \( N \) is the remaining quantity (number of dogs after 12 months)
• \( N_0 \) is the initial quantity (400 dogs)
• \( r \) is the fraction that is placed (8%, or 0.08)
• \( t \) is the time in months (12 months)

Substituting the values:

\[ N = 400(1 - 0.08)^{12} \]

Calculating \( (1 - 0.08) = 0.92 \):

\[ N = 400(0.92)^{12} \]

Calculating \( (0.92)^{12} \):

\[ (0.92)^{12} \approx 0.368 \]

Now, we multiply this by the initial amount:

\[ N \approx 400 * 0.368 \approx 147.2 \]

Since we can't have a fraction of a dog, we round down to 147. Therefore, the estimated number of stray dogs remaining after 12 months will be 147.

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Problem 3: Future Cost of Groceries

To calculate the future cost of groceries considering the inflation rate, we use a similar formula to that of compound interest:

\[ FV = P(1 + r)^n \]

Where:

• \( FV \) is the future cost
• \( P \) is the current cost ($96.50)
• \( r \) is the annual inflation rate (1.8% or 0.018)
• \( n \) is the number of years (20 years)

Substituting the values:

\[ FV = 96.50(1 + 0.018)^{20} \]

Calculating the expression inside the parentheses:

\[ FV = 96.50(1.018)^{20} \]

Now calculating \( (1.018)^{20} \):

\[ (1.018)^{20} \approx 1.48595 \]

Now we multiply this by the current cost:

\[ FV \approx 96.50 * 1.48595 \approx 143.74 \]

Thus, the estimated cost of that same basket of groceries in 20 years will be approximately $143.74.

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Final Answers:

1. Approximately $16,930.00
2. Approximately 147 stray dogs
3. Approximately $143.74