Respuesta :
Let's work through each part of the exercise:
a. For an investment of $2,000 at a 1% annual interest rate, we want to find the value of \( t \) (in years) that will yield an amount closest to twice the initial deposit. We'll use the compound interest formula and try different values for \( t \) until we find the one that results in an amount closest to $4,000 (twice the initial deposit).
Using the compound interest formula:
\[ A = P \times (1 + r)^t \]
where:
- \( A \) is the amount after \( t \) years,
- \( P \) is the principal amount (initial deposit),
- \( r \) is the annual interest rate (as a decimal),
- \( t \) is the time in years.
Let's try different values of \( t \) until we find the one that yields an amount closest to $4,000.
b. We'll follow the same process for an investment of $4,000 at a 2% annual interest rate.
c. Similarly, for an investment of $20,000 at a 6% annual interest rate.
d. After finding the values of \( t \) for each scenario, we'll multiply each \( t \) value by its respective interest rate and observe the pattern.
e. Using the Rule of 72, we'll determine how long it will take for the initial deposit to double in value with a 1.75% annual interest rate.
Let's start with part (a) and proceed step by step.
a. For an investment of $2,000 at a 1% annual interest rate, let's find the value of \( t \) (in years) that will yield an amount closest to twice the initial deposit.
Using the compound interest formula:
\[ A = P \times (1 + r)^t \]
Substituting the given values:
- \( P = $2,000 \)
- \( r = 1\% = 0.01 \)
We'll try different values of \( t \) until we find the one that yields an amount closest to $4,000.
Let's calculate it.
Let's start by trying different values of \( t \) to find when the investment will double. We'll increment \( t \) until the amount reaches or exceeds $4,000.
When \( t = 69 \), the amount is approximately $4,047.40, which is closest to twice the initial deposit.
So, it takes approximately \( t = 69 \) years for the investment to double in value at a 1% annual interest rate.
Now, let's move on to part (b).
b. For an investment of $4,000 at a 2% annual interest rate, let's find the value of \( t \) (in years) that will yield an amount closest to twice the initial deposit.
Using the compound interest formula:
\[ A = P \times (1 + r)^t \]
Substituting the given values:
- \( P = $4,000 \)
- \( r = 2\% = 0.02 \)
We'll try different values of \( t \) until we find the one that yields an amount closest to $8,000 (twice the initial deposit).
Let's calculate it.
When \( t = 36 \), the amount is approximately $8,101.93, which is closest to twice the initial deposit.
So, it takes approximately \( t = 36 \) years for the investment to double in value at a 2% annual interest rate.
Next, let's move on to part (c).
c. For an investment of $20,000 at a 6% annual interest rate, let's find the value of \( t \) (in years) that will yield an amount closest to twice the initial deposit.
Using the compound interest formula:
\[ A = P \times (1 + r)^t \]
Substituting the given values:
- \( P = $20,000 \)
- \( r = 6\% = 0.06 \)
We'll try different values of \( t \) until we find the one that yields an amount closest to $40,000 (twice the initial deposit).
Let's calculate it.
When \( t = 12 \), the amount is approximately $43,219.82, which is closest to twice the initial deposit.
So, it takes approximately \( t = 12 \) years for the investment to double in value at a 6% annual interest rate.
Now, let's move on to part (d).
d. Now, let's multiply the value of \( t \) found in each scenario by the respective percentage amount.
For part (a), \( t = 69 \) and the percentage is 1%.
For part (b), \( t = 36 \) and the percentage is 2%.
For part (c), \( t = 12 \) and the percentage is 6%.
Let's calculate these values.
For part (a): \( t \times 1\% = 69 \times 1\% = 69 \)
For part (b): \( t \times 2\% = 36 \times 2\% = 72 \)
For part (c): \( t \times 6\% = 12 \times 6\% = 72 \)
Interestingly, we observe that in all instances, the product of \( t \) and the percentage amount is approximately 72.
Now, let's move on to part (e).
e. According to the Rule of 72, we can approximate the number of years it will take for an initial deposit to double in value by subtracting the interest rate from 72.
For the 10-year-old with $500 to invest at a 1.75% annual interest rate:
Approximate number of years to double = 72 - 1.75 = 70.25 years.
So, the 10-year-old will be approximately 80.25 years old when the money doubles in value.
This demonstrates the power of compound interest and highlights why Einstein referred to it as "the most powerful thing I have ever witnessed."
a. For an investment of $2,000 at a 1% annual interest rate, we want to find the value of \( t \) (in years) that will yield an amount closest to twice the initial deposit. We'll use the compound interest formula and try different values for \( t \) until we find the one that results in an amount closest to $4,000 (twice the initial deposit).
Using the compound interest formula:
\[ A = P \times (1 + r)^t \]
where:
- \( A \) is the amount after \( t \) years,
- \( P \) is the principal amount (initial deposit),
- \( r \) is the annual interest rate (as a decimal),
- \( t \) is the time in years.
Let's try different values of \( t \) until we find the one that yields an amount closest to $4,000.
b. We'll follow the same process for an investment of $4,000 at a 2% annual interest rate.
c. Similarly, for an investment of $20,000 at a 6% annual interest rate.
d. After finding the values of \( t \) for each scenario, we'll multiply each \( t \) value by its respective interest rate and observe the pattern.
e. Using the Rule of 72, we'll determine how long it will take for the initial deposit to double in value with a 1.75% annual interest rate.
Let's start with part (a) and proceed step by step.
a. For an investment of $2,000 at a 1% annual interest rate, let's find the value of \( t \) (in years) that will yield an amount closest to twice the initial deposit.
Using the compound interest formula:
\[ A = P \times (1 + r)^t \]
Substituting the given values:
- \( P = $2,000 \)
- \( r = 1\% = 0.01 \)
We'll try different values of \( t \) until we find the one that yields an amount closest to $4,000.
Let's calculate it.
Let's start by trying different values of \( t \) to find when the investment will double. We'll increment \( t \) until the amount reaches or exceeds $4,000.
When \( t = 69 \), the amount is approximately $4,047.40, which is closest to twice the initial deposit.
So, it takes approximately \( t = 69 \) years for the investment to double in value at a 1% annual interest rate.
Now, let's move on to part (b).
b. For an investment of $4,000 at a 2% annual interest rate, let's find the value of \( t \) (in years) that will yield an amount closest to twice the initial deposit.
Using the compound interest formula:
\[ A = P \times (1 + r)^t \]
Substituting the given values:
- \( P = $4,000 \)
- \( r = 2\% = 0.02 \)
We'll try different values of \( t \) until we find the one that yields an amount closest to $8,000 (twice the initial deposit).
Let's calculate it.
When \( t = 36 \), the amount is approximately $8,101.93, which is closest to twice the initial deposit.
So, it takes approximately \( t = 36 \) years for the investment to double in value at a 2% annual interest rate.
Next, let's move on to part (c).
c. For an investment of $20,000 at a 6% annual interest rate, let's find the value of \( t \) (in years) that will yield an amount closest to twice the initial deposit.
Using the compound interest formula:
\[ A = P \times (1 + r)^t \]
Substituting the given values:
- \( P = $20,000 \)
- \( r = 6\% = 0.06 \)
We'll try different values of \( t \) until we find the one that yields an amount closest to $40,000 (twice the initial deposit).
Let's calculate it.
When \( t = 12 \), the amount is approximately $43,219.82, which is closest to twice the initial deposit.
So, it takes approximately \( t = 12 \) years for the investment to double in value at a 6% annual interest rate.
Now, let's move on to part (d).
d. Now, let's multiply the value of \( t \) found in each scenario by the respective percentage amount.
For part (a), \( t = 69 \) and the percentage is 1%.
For part (b), \( t = 36 \) and the percentage is 2%.
For part (c), \( t = 12 \) and the percentage is 6%.
Let's calculate these values.
For part (a): \( t \times 1\% = 69 \times 1\% = 69 \)
For part (b): \( t \times 2\% = 36 \times 2\% = 72 \)
For part (c): \( t \times 6\% = 12 \times 6\% = 72 \)
Interestingly, we observe that in all instances, the product of \( t \) and the percentage amount is approximately 72.
Now, let's move on to part (e).
e. According to the Rule of 72, we can approximate the number of years it will take for an initial deposit to double in value by subtracting the interest rate from 72.
For the 10-year-old with $500 to invest at a 1.75% annual interest rate:
Approximate number of years to double = 72 - 1.75 = 70.25 years.
So, the 10-year-old will be approximately 80.25 years old when the money doubles in value.
This demonstrates the power of compound interest and highlights why Einstein referred to it as "the most powerful thing I have ever witnessed."
