The compound interest for a fixed, period of time on Rs 1800 at the rate of 10% per annum is Rs 378. Then, find the time (in years).

Options

A. `2.8`
B. `3.0`
C. `2.5`
D. `2.0`

Answer

To solve the problem of finding the time (in years) for a compound interest of Rs 378 on a principal amount of Rs 1800 at a rate of 10% per annum, we can follow these steps:

### Step 1: Understand the formula for Compound Interest
The formula for compound interest (CI) is given by:
\[ CI = P \left(1 + \frac{r}{100}\right)^t - P \]
where:
- \( CI \) = Compound Interest
- \( P \) = Principal amount
- \( r \) = Rate of interest per annum
- \( t \) = Time in years

### Step 2: Substitute the known values into the formula
Given:
- \( CI = 378 \)
- \( P = 1800 \)
- \( r = 10 \)

We can substitute these values into the formula:
\[ 378 = 1800 \left(1 + \frac{10}{100}\right)^t - 1800 \]

### Step 3: Simplify the equation
Rearranging the equation gives:
\[ 378 + 1800 = 1800 \left(1 + \frac{10}{100}\right)^t \]
\[ 2178 = 1800 \left(1 + 0.1\right)^t \]
\[ 2178 = 1800 \left(\frac{11}{10}\right)^t \]

### Step 4: Isolate the exponential term
Dividing both sides by 1800:
\[ \frac{2178}{1800} = \left(\frac{11}{10}\right)^t \]

### Step 5: Simplify the left side
Calculating \( \frac{2178}{1800} \):
\[ \frac{2178}{1800} = 1.21 \]

### Step 6: Rewrite the equation
Now we have:
\[ 1.21 = \left(\frac{11}{10}\right)^t \]

### Step 7: Recognize that \( 1.21 \) can be expressed as \( \left(\frac{11}{10}\right)^2 \)
Since \( 1.21 = \left(\frac{11}{10}\right)^2 \), we can equate the exponents:
\[ \left(\frac{11}{10}\right)^t = \left(\frac{11}{10}\right)^2 \]
Thus, we have:
\[ t = 2 \]

### Conclusion
The time \( t \) is 2 years.