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

Options

A60 years
B52 years
C51 years
D50 years
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Correct Answer

Option d50 years

All Options:

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

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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 \text{--- (Equation 1)}
5 years ago:
- The person's age was P5\displaystyle P - 5 years.
- The sum of the ages of his two sons was S(2×5)=S10\displaystyle S - (2 \times 5) = S - 10 years (since each son's age decreases by 5 years).
According to the second condition:
P5=3(S10)— (Equation 2)P - 5 = 3(S - 10) \quad \text{--- (Equation 2)}
Substitute Equation 1 (P=2S\displaystyle P = 2S) into Equation 2:
2S5=3(S10)2S - 5 = 3(S - 10)
2S5=3S302S - 5 = 3S - 30
305=3S2S30 - 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\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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