A sum of money invested at simple interest-doubles itself in 3 yrs and 4 months. Find in how many years, it will become 7 times of itself at the same rate? (in years)

Options

A. A)20
B. B)18
C. C)15
D. D)21

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the problem
We know that a sum of money doubles itself in 3 years and 4 months. We need to find out how long it will take for the same sum to become 7 times itself at the same rate of simple interest.

### Step 2: Convert time into years
First, we need to convert 3 years and 4 months into years.

- 3 years = 3 years
- 4 months = 4/12 years = 1/3 years

So, total time in years = 3 + 1/3 = 10/3 years.

### Step 3: Determine the interest earned
When the principal amount (P) doubles, the amount (A1) becomes 2P. The interest (I1) earned in this time is:

\[ I_1 = A_1 - P = 2P - P = P \]

This means that in 10/3 years, the interest earned is equal to the principal amount (P).

### Step 4: Calculate the time for 6 units of interest
Now, we need to find out how long it will take to earn 6 times the principal amount in interest (i.e., from P to 7P):

- We already know that it takes 10/3 years to earn 1 unit of interest (which is equal to P).
- To earn 6 units of interest, we multiply the time taken for 1 unit of interest by 6:

\[ \text{Time for 6 units of interest} = \frac{10}{3} \times 6 = \frac{60}{3} = 20 \text{ years} \]

### Step 5: Conclusion
Thus, it will take **20 years** for the sum of money to become 7 times itself at the same rate of simple interest.

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