The present age of A is six years less than twice the age of B. The present age of C is 40% the present age of A. If difference between the age of B and C is 6, then find the age of A three years ago.

Correct option is C

​$$\text{Let the present age of B be } b \text{ years}\\[10pt]\text{And present age of A } = (2b - 6) \text{ years}\\[10pt]\text{Present age of C } = \frac{40}{100} \times (2b - 6) \text{ years}\\[10pt]\text{ATQ,}\\[5pt]b - \left( \frac{40}{100} \times (2b - 6) \right) = 6\\[10pt]b - \left( \frac{4b}{5} - \frac{12}{5} \right) = 6\\[5pt]\frac{b}{5} + \frac{12}{5} = 6\\[5pt]\frac{b + 12}{5} = 6\\[5pt]b + 12 = 30\\[5pt]b = 18\\[10pt]\text{Age of A three years ago} = (2 \times 18 - 6) - 3 = 27 \text{ years}$$​​