EquationsMCQMTP May 19Question 1017 of 221
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Ten years ago the age of a father was four times his son. Ten years hence the age of the father will be twice that of his son. The present age of the father and the son are

Chapter 2 Diagram

Options

A(50,20)\displaystyle (50, 20)
B(60,20)\displaystyle (60, 20)
C(55,25)\displaystyle (55, 25)
Dnone of these
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Correct Answer

Option a(50,20)\displaystyle (50, 20)

All Options:

  • A(50,20)\displaystyle (50, 20)
  • B(60,20)\displaystyle (60, 20)
  • C(55,25)\displaystyle (55, 25)
  • Dnone of these

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Detailed Solution & Explanation

Let the present age of the father be F\displaystyle F years and the present age of the son be S\displaystyle S years.
According to the first condition (10 years ago):
F10=4(S10)F - 10 = 4(S - 10)
F10=4S40F - 10 = 4S - 40
F4S=30— (Equation 1)F - 4S = -30 \quad \text{--- (Equation 1)}
According to the second condition (10 years hence):
F+10=2(S+10)F + 10 = 2(S + 10)
F+10=2S+20F + 10 = 2S + 20
F2S=10— (Equation 2)F - 2S = 10 \quad \text{--- (Equation 2)}
We can solve this system by subtracting Equation 1 from Equation 2:
(F2S)(F4S)=10(30)(F - 2S) - (F - 4S) = 10 - (-30)
2S=40    S=202S = 40 \implies S = 20
Substitute S=20\displaystyle S = 20 into Equation 2:
F2(20)=10F - 2(20) = 10
F40=10    F=50F - 40 = 10 \implies F = 50
So the present age of the father is 50 and the son is 20, represented as (50,20)\displaystyle (50, 20).
*(Note: The input key of 'd' is incorrect. Option A is (50,20)\displaystyle (50, 20), which is mathematically correct.)*
**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.

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