EquationsMCQMTP Nov 21Question 997 of 221
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If xy+yz+zx=−1\displaystyle xy + yz + zx = -1, then the value of x+y1+xy+y+z1+yz+z+x1+zx\displaystyle \frac{x+y}{1+xy} + \frac{y+z}{1+yz} + \frac{z+x}{1+zx} is

Options

A1\displaystyle 1
B−1/yz\displaystyle -1/yz
C1xyz\displaystyle \frac{1}{xyz}
D1x+y+z\displaystyle \frac{1}{x+y+z}
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Correct Answer

✅ Option c — 1xyz\displaystyle \frac{1}{xyz}

All Options:

  • A1\displaystyle 1
  • B−1/yz\displaystyle -1/yz
  • C1xyz\displaystyle \frac{1}{xyz}
  • D1x+y+z\displaystyle \frac{1}{x+y+z}

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Detailed Solution & Explanation

Given:
xy+yz+zx=−1xy + yz + zx = -1
We can rewrite the term 1+xy\displaystyle 1+xy by substituting 1=−(xy+yz+zx)\displaystyle 1 = -(xy + yz + zx):
1+xy=−(xy+yz+zx)+xy=−yz−zx=−z(y+x)=−z(x+y)1 + xy = -(xy + yz + zx) + xy = -yz - zx = -z(y + x) = -z(x + y)
Since x+y≠0\displaystyle x+y \neq 0, the first term simplifies to:
x+y1+xy=x+y−z(x+y)=−1z\frac{x+y}{1+xy} = \frac{x+y}{-z(x+y)} = -\frac{1}{z}
Similarly, for the second term, substituting 1=−(xy+yz+zx)\displaystyle 1 = -(xy + yz + zx):
1+yz=−(xy+yz+zx)+yz=−xy−zx=−x(y+z)1 + yz = -(xy + yz + zx) + yz = -xy - zx = -x(y + z)
Since y+z≠0\displaystyle y+z \neq 0, the second term simplifies to:
y+z1+yz=y+z−x(y+z)=−1x\frac{y+z}{1+yz} = \frac{y+z}{-x(y+z)} = -\frac{1}{x}
And for the third term, substituting 1=−(xy+yz+zx)\displaystyle 1 = -(xy + yz + zx):
1+zx=−(xy+yz+zx)+zx=−xy−yz=−y(x+z)1 + zx = -(xy + yz + zx) + zx = -xy - yz = -y(x + z)
Since z+x≠0\displaystyle z+x \neq 0, the third term simplifies to:
z+x1+zx=z+x−y(z+x)=−1y\frac{z+x}{1+zx} = \frac{z+x}{-y(z+x)} = -\frac{1}{y}
Now, summing these three simplified terms:
x+y1+xy+y+z1+yz+z+x1+zx=−1z−1x−1y=−(1x+1y+1z)\frac{x+y}{1+xy} + \frac{y+z}{1+yz} + \frac{z+x}{1+zx} = -\frac{1}{z} - \frac{1}{x} - \frac{1}{y} = -\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)
Combining the fractions into a single term over a common denominator xyz\displaystyle xyz:
−(yz+zx+xyxyz)-\left(\frac{yz + zx + xy}{xyz}\right)
Substituting xy+yz+zx=−1\displaystyle xy + yz + zx = -1 into the numerator:
−(−1xyz)=1xyz-\left(\frac{-1}{xyz}\right) = \frac{1}{xyz}
Thus, the value of the expression is 1xyz\displaystyle \frac{1}{xyz}, which corresponds to Option C. (Note: The provided key in the raw JSON points to Option B, but Option C is the unique, mathematically correct solution).
Hence, **Option C** 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.

More Questions from Equations

If 3x+y+2x−y=−1\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = -1 and 1x+y−1x−y=43\displaystyle \frac{1}{x+y} - \frac{1}{x-y} = \frac{4}{3} then (x,y)\displaystyle (x, y) is:

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In a multiple choice question paper consisting of 100\displaystyle 100 questions of 1\displaystyle 1 mark each, a candidate gets 60%\displaystyle 60\% marks. If the candidate attempted all questions and there was a penalty of 0.25\displaystyle 0.25 marks for wrong answers is:

The values of x\displaystyle x and y\displaystyle y satisfying the equations 3x+y+2x−y=3\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = 3 and 2x+y+3x−y=233\displaystyle \frac{2}{x+y} + \frac{3}{x-y} = \frac{23}{3} given by

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