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The age of a man is four times the sum of the ages of his two sons and after 10 years, his age will be double the sum of their ages. The present age of the man must be

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Options

A56 years
B45 years
C60 years
D64 years
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Correct Answer

Option c60 years

All Options:

  • A56 years
  • B45 years
  • C60 years
  • D64 years

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Detailed Solution & Explanation

Let the present age of the man be M\displaystyle M years and the sum of the present ages of his two sons be S\displaystyle S years.
According to the first condition:
M=4S— (Equation 1)M = 4S \quad \text{--- (Equation 1)}
After 10 years, the man's age will be M+10\displaystyle M + 10 years.
Since there are two sons, each of their ages increases by 10 years, so the sum of their ages increases by 2×10=20\displaystyle 2 \times 10 = 20 years. The sum of the sons' ages after 10 years will be S+20\displaystyle S + 20 years.
According to the second condition:
M+10=2(S+20)— (Equation 2)M + 10 = 2(S + 20) \quad \text{--- (Equation 2)}
Substitute Equation 1 (M=4S\displaystyle M = 4S) into Equation 2:
4S+10=2(S+20)4S + 10 = 2(S + 20)
4S+10=2S+404S + 10 = 2S + 40
4S2S=40104S - 2S = 40 - 10
2S=30    S=152S = 30 \implies S = 15
Now, find the present age of the man M\displaystyle M:
M=4S=4(15)=60 yearsM = 4S = 4(15) = 60\text{ years}
Thus, the present age of the man is 60 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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