A man invested $$\frac{1}{3}$$ of his income at 7% simple interest, $$\frac{1}{4}$$ of his income at 8% simple interest, and the remainder of his income at 10% simple interest. If the total interest at the end of one year was ₹561, what was his initital income?

Correct option is B

Given:

A man invested of$$\frac 13$$​ of his capital at 7%
$$\frac 14$$​ of capital at 8%
The rest of the capital is at 10%
Thereby getting an annual income of Rs. 561.
Formula Used:
​$$S.I = \frac{P × R × T}{100}$$​​
where P = Sum of money, r = Rate, and t = Time
Solution:
Let the total income be 12x
$$\frac 13$$​ of 12x
=> 12x × $$\frac 13$$​​
=> 4x
$$\frac 14$$​ of 12x
=> 12x × $$\frac 14$$​​
=> 3x
Remaining = 12x – 4x – 3x
=> 5x

As we know,
​$$\frac{4x \times 7 \times 1}{100} + \frac{3x \times 8 \times 1}{100} + \frac{5x \times 10 \times 1}{100} = 561$$​​
=> 28x + 24x + 50x = 56100
=> 102x = 56100
=> x = $$\frac{56100}{102}$$​​
Total sum = $$\frac{56100}{102}$$​ × 12 = Rs. 6600