EquationsMCQMTP Nov 21Question 997 of 221
All Questions

If xy+yz+zx=−1\displaystyle xy + yz + zx = -1, then the value of x+y1+xy+y+z1+yz+z+x1+zx\displaystyle 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

Select Your Answer (Keyboard: A, B, C, D)
For any discrepancies in this question, email contact@cadada.in

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}

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 ≠\displaystyle \neq 0, the first term simplifies to:
x+y1+xy=x+y−z(x+y)=−1z\displaystyle \frac{x+y}{1+xy} = \frac{x+y}{-z(x+y)} = -1z\displaystyle \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 ≠\displaystyle \neq 0, the second term simplifies to:
y+z1+yz=y+z−x(y+z)=−1x\displaystyle \frac{y+z}{1+yz} = \frac{y+z}{-x(y+z)} = -1x\displaystyle \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 ≠\displaystyle \neq 0, the third term simplifies to:
z+x1+zx=z+x−y(z+x)=−1y\displaystyle \frac{z+x}{1+zx} = \frac{z+x}{-y(z+x)} = -1y\displaystyle \frac{1}{y}
Now, summing these three simplified terms:
x+y1+xy+y+z1+yz+z+x1+zx=−1z−1x−1y=−(1x+1y+1z)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} = -1z−1x−1y\displaystyle \frac{1}{z} - \frac{1}{x} - \frac{1}{y} = -\left(1x+1y+1z\displaystyle \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(yz+zx+xyxyz\displaystyle \frac{yz + zx + xy}{xyz}\right)
Substituting xy+yz+zx=−1\displaystyle xy + yz + zx = -1 into the numerator:
−(−1xyz)=1xyz-\left(−1xyz\displaystyle \frac{-1}{xyz}\right) = 1xyz\displaystyle \frac{1}{xyz}
Thus, the value of the expression is 1xyz\displaystyle 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 3x+y+2x−y\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = -1 and 1x+y−1x−y=43\displaystyle 1x+y−1x−y=43\displaystyle \frac{1}{x+y} - \frac{1}{x-y} = \frac{4}{3} then (x,y)\displaystyle (x, y) is:

2x+510+3x+1015=5\displaystyle \frac{2x+5}{10} + \frac{3x+10}{15} = 5, find x\displaystyle x

Find value of x2−10x+1\displaystyle x^2 - 10x + 1 if x=15−26\displaystyle x = 15−26\displaystyle \frac{1}{5-2\sqrt{6}}

The cost of 2\displaystyle 2 oranges and 3\displaystyle 3 apples is $28$. If the cost of an apple is doubled then the cost of $3orangesand\displaystyle oranges and5$ apples is $75$. The original cost of 7\displaystyle 7 oranges and 4\displaystyle 4 apples (in Rs) is:

In a multiple choice question paper consisting of $100questionsof\displaystyle questions of1\displaystyle mark each, a candidate gets $60\% marks. If the candidate attempted all questions and there was a penalty of $0.25$ marks for wrong answers is:

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

Ready to Master Equations?

Practice all 221 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free