EquationsMCQMTP Nov 19Question 1019 of 221
All Questions

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 Age

Chapter 2 Diagram

Options

A9
B8
C10
D12
For any discrepancies in this question, email contact@cadada.in

Correct Answer

Option a9

All Options:

  • A9
  • B8
  • C10
  • D12

Ad

Detailed Solution & Explanation

Let the present age of A be x\displaystyle x years.
A's age 6 years ago was x6\displaystyle x - 6 years.
According to the problem:
- Thrice of A's age 6 years ago: 3(x6)\displaystyle 3(x - 6)
- Twice his present age: 2x\displaystyle 2x
Subtracting the former from the latter equals his present age (x\displaystyle x):
2x3(x6)=x2x - 3(x - 6) = x
Let's solve for x\displaystyle x:
2x3x+18=x2x - 3x + 18 = x
x+18=x-x + 18 = x
18=2x18 = 2x
x=9x = 9
Thus, A's present age is 9 years.
*(Note: The input key of 'c' (10) is incorrect. The mathematically correct choice is 9, option A.)*
**Correct Option: (a)**

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.

More Questions from Equations

Ready to Master Equations?

Practice all 221 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free