Keilantra invested $60,000 in an account paying an interest rate of 2, one half221​% compounded quarterly. Cai invested $60,000 in an account paying an interest rate of 1, start fraction, 7, divided by, 8, end fraction187​% compounded continuously. To the nearest *hundredth of a year*, how much longer would it take for Cai's money to double than for Keilantra's money to double?

To determine how much longer it will take for Cai's money to double compared to Keilantra's money, given their respective interest rates and compounding methods, we will calculate the time required for each investment to double. 

### Step-by-Step Solution

1. **Determine the interest rate and compounding methods:**
   - Keilantra's interest rate: $ 2 \frac{1}{2} \% = 2.5\% $ compounded quarterly.
   - Cai's interest rate: $ 1 \frac{7}{8} \% = 1.875\% $ compounded continuously.

2. **Convert the annual interest rates to decimal form:**
   - Keilantra: $ r = 0.025 $
   - Cai: $ r = 0.01875 $

3. **Use the compound interest formula for Keilantra:**
   \[
   A = P \left(1 + \frac{r}{n}\right)^{nt}
   \]
   Where:
   - $ P = 60000 $
   - $ A = 2P = 120000 $
   - $ r = 0.025 $
   - $ n = 4 $ (quarterly compounding)

4. **Set up the equation for doubling the investment ( $ A = 2P $ ):**
   \[
   60000 \left(1 + \frac{0.025}{4}\right)^{4t} = 120000
   \]
   \[
   \left(1 + \frac{0.025}{4}\right)^{4t} = 2
   \]
   \[
   \left(1 + 0.00625\right)^{4t} = 2
   \]
   \[
   (1.00625)^{4t} = 2
   \]

5. **Solve for $ t $:**
   Taking the natural logarithm on both sides:
   \[
   4t \ln(1.00625) = \ln 2
   \]
   \[
   4t \cdot 0.0062341 = 0.693147
   \]
   \[
   t = \frac{0.693147}{4 \cdot 0.0062341} \approx \frac{0.693147}{0.0249364} \approx 27.80 \text{ years}
   \]

6. **Use the continuous compounding formula for Cai:**
   \[
   A = Pe^{rt}
   \]
   Where:
   - $ P = 60000 $
   - $ A = 2P = 120000 $
   - $ r = 0.01875 $

7. **Set up the equation for doubling the investment ( $ A = 2P $ ):**
   \[
   60000e^{0.01875t} = 120000
   \]
   \[
   e^{0.01875t} = 2
   \]

8. **Solve for $ t $:**
   Taking the natural logarithm on both sides:
   \[
   0.01875t = \ln 2
   \]
   \[
   t = \frac{\ln 2}{0.01875} \approx \frac{0.693147}{0.01875} \approx 36.97 \text{ years}
   \]

9. **Calculate the difference in time:**
   \[
   36.97 - 27.80 = 9.17 \text{ years}
   \]

Therefore, it will take approximately **9.17 years longer** for Cai's money to double than for Keilantra's money.

Would you like more details or have any questions about this solution?

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### Related Questions:
1. How would the calculations change if both interest rates were compounded annually?
2. What is the impact of a higher interest rate on the time required for an investment to double?
3. How can the continuous compounding formula be derived from the basic compound interest formula?
4. What would be the effect of different initial investments on the time to double at the same interest rates?
5. How does the frequency of compounding affect the total amount accrued over a fixed period?
6. What are the key differences between continuous compounding and discrete compounding?
7. How do you derive the natural logarithm values used in these calculations?
8. How does inflation affect the real value of money in the context of compound interest?

### Tip:
When comparing investment options, always consider both the interest rate and the compounding frequency to understand their true growth potential over time.

Explore how compound interest and continuous compounding affect investment growth rates. Learn through a detailed comparison of investment doubling times for different interest rates and compounding frequencies. Discover which method offers faster investment growth over time.

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