The least number of years in which a sum of money on 19% p.a. compound interest will be more than double is

Options

A. 3 years
B. 4 years
C. 5 years
D. 2 years

Answer

To solve the problem of finding the least number of years in which a sum of money on 19% p.a. compound interest will be more than double, we can follow these steps:

### Step 1: Understand the formula for compound interest
The formula for the amount (A) after time (t) years with principal (P) and rate of interest (r) is given by:
\[ A = P \left(1 + \frac{r}{100}\right)^t \]

### Step 2: Set up the equation
We want to find the time (t) when the amount is more than double the principal. Therefore, we set up the equation:
\[ A > 2P \]

Substituting the formula for A:
\[ P \left(1 + \frac{19}{100}\right)^t > 2P \]

### Step 3: Simplify the equation
We can cancel P from both sides (assuming P > 0):
\[ \left(1 + \frac{19}{100}\right)^t > 2 \]

### Step 4: Calculate the value of (1 + r/100)
Calculate \( 1 + \frac{19}{100} \):
\[ 1 + \frac{19}{100} = 1.19 \]

### Step 5: Rewrite the inequality
Now we have:
\[ (1.19)^t > 2 \]

### Step 6: Use logarithms to solve for t
To solve for t, we can take the logarithm of both sides:
\[ \log((1.19)^t) > \log(2) \]

Using the property of logarithms:
\[ t \cdot \log(1.19) > \log(2) \]

### Step 7: Isolate t
Now, isolate t:
\[ t > \frac{\log(2)}{\log(1.19)} \]

### Step 8: Calculate the values
Using a calculator, find the logarithmic values:
- \( \log(2) \approx 0.3010 \)
- \( \log(1.19) \approx 0.0792 \)

Now substitute these values into the inequality:
\[ t > \frac{0.3010}{0.0792} \approx 3.8 \]

### Step 9: Determine the least integer value for t
Since t must be a whole number and greater than 3.8, the least integer value for t is 4.

### Final Answer
The least number of years in which a sum of money on 19% p.a. compound interest will be more than double is **4 years**.

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