EquationsMCQMTP June 2023 Series IIQuestion 1003 of 221
All Questions

Solve for x,y\displaystyle x, y and z\displaystyle z.xyx+y=210\displaystyle xyx+y=210\displaystyle \frac{xy}{x+y} = 210, yzy+z=140\displaystyle yzy+z=140\displaystyle \frac{yz}{y+z} = 140, xzx+z=120\displaystyle xzx+z=120\displaystyle \frac{xz}{x+z} = 120

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option d — None of these

All Options:

  • A$105; 210; 420$
  • B$100; 205; 400$
  • C$95; 215; 395$
  • DNone of these

Detailed Solution & Explanation

The given system of equations is: xyx+y=210— (Equation 1)xyx+y=210\displaystyle \frac{xy}{x+y} = 210 \quad — (Equation 1)\displaystyle \text{--- (Equation 1)} yzy+z=140— (Equation 2)yzy+z=140\displaystyle \frac{yz}{y+z} = 140 \quad — (Equation 2)\displaystyle \text{--- (Equation 2)} xzx+z=120— (Equation 3)xzx+z=120\displaystyle \frac{xz}{x+z} = 120 \quad — (Equation 3)\displaystyle \text{--- (Equation 3)} Taking the reciprocal of each equation: x+yxy=1210  ⟹  1y+1x=1210— (Equation 4)x+yxy=1210\displaystyle \frac{x+y}{xy} = \frac{1}{210} \implies 1y+1x=1210\displaystyle \frac{1}{y} + \frac{1}{x} = \frac{1}{210} \quad — (Equation 4)\displaystyle \text{--- (Equation 4)} y+zyz=1140  ⟹  1z+1y=1140— (Equation 5)y+zyz=1140\displaystyle \frac{y+z}{yz} = \frac{1}{140} \implies 1z+1y=1140\displaystyle \frac{1}{z} + \frac{1}{y} = \frac{1}{140} \quad — (Equation 5)\displaystyle \text{--- (Equation 5)} x+zxz=1120  ⟹  1z+1x=1120— (Equation 6)x+zxz=1120\displaystyle \frac{x+z}{xz} = \frac{1}{120} \implies 1z+1x=1120\displaystyle \frac{1}{z} + \frac{1}{x} = \frac{1}{120} \quad — (Equation 6)\displaystyle \text{--- (Equation 6)} Let u=1x\displaystyle u = 1x\displaystyle \frac{1}{x}, v=1y\displaystyle v = 1y\displaystyle \frac{1}{y}, and w=1z\displaystyle w = 1z\displaystyle \frac{1}{z}. The equations become: u+v=1210— (Equation 7)u + v = 1210\displaystyle \frac{1}{210} \quad — (Equation 7)\displaystyle \text{--- (Equation 7)} v+w=1140— (Equation 8)v + w = 1140\displaystyle \frac{1}{140} \quad — (Equation 8)\displaystyle \text{--- (Equation 8)} u+w=1120— (Equation 9)u + w = 1120\displaystyle \frac{1}{120} \quad — (Equation 9)\displaystyle \text{--- (Equation 9)} Adding Equations 7, 8, and 9: 2(u+v+w)=1210+1140+11202(u + v + w) = 1210+1140+1120\displaystyle \frac{1}{210} + \frac{1}{140} + \frac{1}{120} Finding a common denominator of $840$ for the RHS: 2(u+v+w)=4840+6840+7840=178402(u + v + w) = 4840+6840+7840=17840\displaystyle \frac{4}{840} + \frac{6}{840} + \frac{7}{840} = \frac{17}{840} u+v+w=171680— (Equation 10)u + v + w = 171680\displaystyle \frac{17}{1680} \quad — (Equation 10)\displaystyle \text{--- (Equation 10)} Now we can solve for u\displaystyle u, v\displaystyle v, and w\displaystyle w: - From Equation 10 and Equation 8: u=(u+v+w)−(v+w)=171680−1140=17−121680=51680=1336  ⟹  x=336u = (u+v+w) - (v+w) = 171680−1140=17−121680=51680=1336\displaystyle \frac{17}{1680} - \frac{1}{140} = \frac{17 - 12}{1680} = \frac{5}{1680} = \frac{1}{336} \implies x = 336 - From Equation 10 and Equation 9: v=(u+v+w)−(u+w)=171680−1120=17−141680=31680=1560  ⟹  y=560v = (u+v+w) - (u+w) = 171680−1120=17−141680=31680=1560\displaystyle \frac{17}{1680} - \frac{1}{120} = \frac{17 - 14}{1680} = \frac{3}{1680} = \frac{1}{560} \implies y = 560 - From Equation 10 and Equation 7: w=(u+v+w)−(u+v)=171680−1210=17−81680=91680=3560  ⟹  z=5603≈186.67w = (u+v+w) - (u+v) = 171680−1210=17−81680=91680=3560\displaystyle \frac{17}{1680} - \frac{1}{210} = \frac{17 - 8}{1680} = \frac{9}{1680} = \frac{3}{560} \implies z = 5603\displaystyle \frac{560}{3} ≈\displaystyle \approx 186.67 Thus, the correct solution is x=336\displaystyle x = 336, y=560\displaystyle y = 560, and z≈186.67\displaystyle z ≈\displaystyle \approx 186.67. Since these values do not match Option A, B, or C, the correct option is Option D. Hence, **Option D** 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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