EquationsMCQMTP Nov 18Question 1016 of 221
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The age of a person is twice the sum of the ages of his two sons and 5 years ago his age was thrice the sum of their ages. Find present age.

Chapter 2 Diagram

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

✅ Option d — 50 years

All Options:

  • A60 years
  • B52 years
  • C51 years
  • D50 years

Detailed Solution & Explanation

Let the present age of the person be P\displaystyle P years and the sum of the present ages of his two sons be S\displaystyle S years. According to the first condition: P=2S— (Equation 1)P = 2S \quad — (Equation 1)\displaystyle \text{--- (Equation 1)} 5 years ago: - The person's age was P−5\displaystyle P - 5 years. - The sum of the ages of his two sons was S−(2×5)=S−10\displaystyle S - (2 ×\displaystyle \times 5) = S - 10 years (since each son's age decreases by 5 years). According to the second condition: P−5=3(S−10)— (Equation 2)P - 5 = 3(S - 10) \quad — (Equation 2)\displaystyle \text{--- (Equation 2)} Substitute Equation 1 (P=2S\displaystyle P = 2S) into Equation 2: 2S−5=3(S−10)2S - 5 = 3(S - 10) 2S−5=3S−302S - 5 = 3S - 30 30−5=3S−2S30 - 5 = 3S - 2S S=25S = 25 Now find the present age of the person P\displaystyle P: P=2S=2(25)=50 yearsP = 2S = 2(25) = 50 years\displaystyle \text{ years} Thus, the person's present age is 50 years. *(Note: The input key of 'b' (52) is incorrect. The mathematically correct answer is 50, option D.)* **Correct Option: (d)**

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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