A person invested in all ₹2600 at 4%.6% and 8% per annum simple interest. At the end of the year. he got the same interest in all the three cases. The money invested at 4% is :

Options

A. ₹200
B. ₹600
C. ₹800
D. ₹1,200

Explanation

To solve the problem step by step, we will denote the amounts invested at different interest rates as follows:

- Let the amount invested at 4% be \( x \).
- Let the amount invested at 6% be \( y \).
- Let the amount invested at 8% be \( z \).

Given that the total investment is ₹2600, we can write:

\[ x + y + z = 2600 \]

Since the interest earned from all three investments is the same at the end of the year, we can express the simple interest for each investment using the formula:

\[ \text{Simple Interest} = \frac{\text{Principal} \times \text{Rate} \times \text{Time}}{100} \]

For each investment, the interest can be expressed as:

1. Interest from \( x \) at 4%:
\[ \text{Interest}_1 = \frac{x \times 4 \times 1}{100} = \frac{4x}{100} \]

2. Interest from \( y \) at 6%:
\[ \text{Interest}_2 = \frac{y \times 6 \times 1}{100} = \frac{6y}{100} \]

3. Interest from \( z \) at 8%:
\[ \text{Interest}_3 = \frac{z \times 8 \times 1}{100} = \frac{8z}{100} \]

Since all three interests are equal, we can set them equal to each other:

\[ \frac{4x}{100} = \frac{6y}{100} = \frac{8z}{100} \]

Removing the common denominator of 100, we have:

\[ 4x = 6y = 8z \]

From these equations, we can express \( y \) and \( z \) in terms of \( x \):

1. From \( 4x = 6y \):
\[ y = \frac{4x}{6} = \frac{2x}{3} \]

2. From \( 4x = 8z \):
\[ z = \frac{4x}{8} = \frac{x}{2} \]

Now we can substitute \( y \) and \( z \) back into the total investment equation:

\[ x + \frac{2x}{3} + \frac{x}{2} = 2600 \]

To solve this, we need a common denominator. The least common multiple of 3 and 2 is 6. Rewriting each term with a denominator of 6, we get:

\[ x + \frac{4x}{6} + \frac{3x}{6} = 2600 \]

Combining the fractions:

\[ x + \frac{4x + 3x}{6} = 2600 \]

This simplifies to:

\[ x + \frac{7x}{6} = 2600 \]

To eliminate the fraction, multiply the entire equation by 6:

\[ 6x + 7x = 15600 \]

Combining like terms:

\[ 13x = 15600 \]

Now, divide by 13:

\[ x = \frac{15600}{13} = 1200 \]

Thus, the amount invested at 4% is:

\[ \boxed{1200} \]