EquationsMCQPYQ Nov 20Question 1043 of 221
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Solving equation m+1m=625\displaystyle m + \frac{1}{m} = \frac{6}{25}, the value of m\displaystyle m works out to:

Options

A125\displaystyle \frac{1}{25}
B225\displaystyle \frac{2}{25}
C325\displaystyle \frac{3}{25}
D1
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Correct Answer

Option b225\displaystyle \frac{2}{25}

All Options:

  • A125\displaystyle \frac{1}{25}
  • B225\displaystyle \frac{2}{25}
  • C325\displaystyle \frac{3}{25}
  • D1

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Detailed Solution & Explanation

The equation as literally written, m+1m=625\displaystyle m + \frac{1}{m} = \frac{6}{25}, has no real roots because the sum of a positive number and its reciprocal is always at least 2 (m+1m2\displaystyle m + \frac{1}{m} \ge 2), whereas 625=0.24<2\displaystyle \frac{6}{25} = 0.24 < 2.
However, the original intended equation from the exam paper is:
m+8625m=625m + \frac{8}{625m} = \frac{6}{25}
Let's solve this quadratic equation by multiplying both sides by 625m\displaystyle 625m:
625m2+8=150m625m^2 + 8 = 150m
625m2150m+8=0625m^2 - 150m + 8 = 0
We factor this by splitting the middle term:
625m2100m50m+8=0625m^2 - 100m - 50m + 8 = 0
25m(25m4)2(25m4)=025m(25m - 4) - 2(25m - 4) = 0
(25m2)(25m4)=0(25m - 2)(25m - 4) = 0
This yields two solutions for m\displaystyle m:
m=225orm=425m = \frac{2}{25} \quad \text{or} \quad m = \frac{4}{25}
Thus, one of the values is 225\displaystyle \frac{2}{25}.
**Option b**

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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