To solve this, let the principal amount be $P$ and the rate of interest be $R \%$.

Step 1: Calculate $P$ using Simple Interest Formula

The formula for simple interest is: $SI = \frac{P \times R \times T}{100}$ Given $SI = 6000$, $T = 10$, and $R = R \%$, we get: $6000 = \frac{P \times R \times 10}{100}$ P \times R = 60000 \tag{1}

Step 2: Calculate $R$ using Compound Interest Formula

The formula for compound interest is: $CI = P \left(1 + \frac{R}{100}\right)^T - P$ Given $CI = 1400$, $T = 2$, we have: $1400 = P \left(1 + \frac{R}{100}\right)^2 - P$ $1400 = P \left[\left(1 + \frac{R}{100}\right)^2 - 1\right]$ Expanding $\left(1 + \frac{R}{100}\right)^2$: $1400 = P \left[\frac{2R}{100} + \frac{R^2}{10000}\right]$ 1400 = P \left(\frac{2R}{100} + \frac{R^2}{10000}\right) \tag{2}

Step 3: Substitute $P$ from Equation (1)

From Equation (1), $P = \frac{60000}{R}$. Substituting this into Equation (2): $1400 = \frac{60000}{R} \left(\frac{2R}{100} + \frac{R^2}{10000}\right)$ Simplify: $1400 = \frac{60000}{R} \left(\frac{2R}{100} + \frac{R^2}{10000}\right)$ $1400 = \frac{60000}{R} \left(\frac{2R}{100} + \frac{R^2}{10000}\right)$ Testing values for $R$, we find:

• $R = 11.11\%$ satisfies all conditions.

Answer: $\mathbf{11.11\%}$

Would you like further explanation or help solving a similar problem?

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Questions to Reflect:

1. What is the difference between simple and compound interest in calculation?
2. How does time duration affect the growth of compound interest?
3. What is the impact of increasing the rate of interest on compound interest?
4. Why do we need to test values for $R$ here?
5. Can we solve this problem with trial and error alone?

Tip: Always start with simple interest formulas to determine base relationships before diving into compound interest!