EquationsMCQMTP Jun 23 - Series 1Question 1028 of 221
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If thrice of A's age 6 years ago be subtracted from twice his present age, the result would be equal to his present age. Find A's present age.

Chapter 2 Diagram

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option c — 9

All Options:

  • A7
  • B8
  • C9
  • D6

Detailed Solution & Explanation

Let the present age of A be x\displaystyle x years.
A's age 6 years ago was x−6\displaystyle x - 6 years.
According to the problem, thrice of A's age 6 years ago, $3(x-6)$, is subtracted from twice his present age, $2x$, resulting in his present age, x\displaystyle x:
2x−3(x−6)=x2x - 3(x - 6) = x
Let's solve for x\displaystyle x:
2x−3x+18=x2x - 3x + 18 = x
−x+18=x-x + 18 = x
18=2x18 = 2x
x=9 yearsx = 9 years\displaystyle \text{ years}
Therefore, A's present age is 9 years.
**Correct Option: (c)**

About This Chapter: Equations

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Linear, Quadratic and Cubic Equations

This chapter covers Linear, Quadratic and Cubic Equations and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

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