EquationsPYQ Jan 26Question 4251 of 155
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One year ago, the ratio of ages (in years) of A and B was 5:4. The ratio of their ages, 4 years from now, will be 6:5. What will be the age of A (in years) after 10 years from now?

Options

A36
B18
C19
D26
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Correct Answer

Option a36

All Options:

  • A36
  • B18
  • C19
  • D26

Detailed Solution & Explanation

Let the ages of A and B one year ago be 5k\displaystyle 5k and 4k\displaystyle 4k respectively, where k\displaystyle k is a constant. Thus, their present ages are: Present age of A=5k+1\text{Present age of A} = 5k + 1 Present age of B=4k+1\text{Present age of B} = 4k + 1
Four years from now, their ages will be: Age of A=(5k+1)+4=5k+5\text{Age of A} = (5k + 1) + 4 = 5k + 5 Age of B=(4k+1)+4=4k+5\text{Age of B} = (4k + 1) + 4 = 4k + 5
According to the problem, the ratio of their ages four years from now will be 6:5\displaystyle 6:5: 5k+54k+5=65\frac{5k + 5}{4k + 5} = \frac{6}{5}
Cross-multiplying to solve for k\displaystyle k: 5(5k+5)=6(4k+5)5(5k + 5) = 6(4k + 5) 25k+25=24k+3025k + 25 = 24k + 30 25k24k=302525k - 24k = 30 - 25 k=5k = 5
Now we can determine the present age of A: Present age of A=5k+1=5(5)+1=26 years\text{Present age of A} = 5k + 1 = 5(5) + 1 = 26\text{ years}
We need to find the age of A after 10 years from now: Age of A after 10 years=26+10=36 years\text{Age of A after 10 years} = 26 + 10 = 36\text{ years} Hence, **Option A** is the correct answer.

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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