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 \frac{xy}{x+y} = 210, yzy+z=140\displaystyle \frac{yz}{y+z} = 140, xzx+z=120\displaystyle \frac{xz}{x+z} = 120

Options

A105;210;420\displaystyle 105; 210; 420
B100;205;400\displaystyle 100; 205; 400
C95;215;395\displaystyle 95; 215; 395
DNone of these
For any discrepancies in this question, email contact@cadada.in

Correct Answer

Option dNone of these

All Options:

  • A105;210;420\displaystyle 105; 210; 420
  • B100;205;400\displaystyle 100; 205; 400
  • C95;215;395\displaystyle 95; 215; 395
  • DNone of these

Ad

Detailed Solution & Explanation

The given system of equations is:
xyx+y=210— (Equation 1)\frac{xy}{x+y} = 210 \quad \text{--- (Equation 1)}
yzy+z=140— (Equation 2)\frac{yz}{y+z} = 140 \quad \text{--- (Equation 2)}
xzx+z=120— (Equation 3)\frac{xz}{x+z} = 120 \quad \text{--- (Equation 3)}
Taking the reciprocal of each equation:
x+yxy=1210    1y+1x=1210— (Equation 4)\frac{x+y}{xy} = \frac{1}{210} \implies \frac{1}{y} + \frac{1}{x} = \frac{1}{210} \quad \text{--- (Equation 4)}
y+zyz=1140    1z+1y=1140— (Equation 5)\frac{y+z}{yz} = \frac{1}{140} \implies \frac{1}{z} + \frac{1}{y} = \frac{1}{140} \quad \text{--- (Equation 5)}
x+zxz=1120    1z+1x=1120— (Equation 6)\frac{x+z}{xz} = \frac{1}{120} \implies \frac{1}{z} + \frac{1}{x} = \frac{1}{120} \quad \text{--- (Equation 6)}
Let u=1x\displaystyle u = \frac{1}{x}, v=1y\displaystyle v = \frac{1}{y}, and w=1z\displaystyle w = \frac{1}{z}. The equations become:
u+v=1210— (Equation 7)u + v = \frac{1}{210} \quad \text{--- (Equation 7)}
v+w=1140— (Equation 8)v + w = \frac{1}{140} \quad \text{--- (Equation 8)}
u+w=1120— (Equation 9)u + w = \frac{1}{120} \quad \text{--- (Equation 9)}
Adding Equations 7, 8, and 9:
2(u+v+w)=1210+1140+11202(u + v + w) = \frac{1}{210} + \frac{1}{140} + \frac{1}{120}
Finding a common denominator of 840\displaystyle 840 for the RHS:
2(u+v+w)=4840+6840+7840=178402(u + v + w) = \frac{4}{840} + \frac{6}{840} + \frac{7}{840} = \frac{17}{840}
u+v+w=171680— (Equation 10)u + v + w = \frac{17}{1680} \quad \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)=1716801140=17121680=51680=1336    x=336u = (u+v+w) - (v+w) = \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)=1716801120=17141680=31680=1560    y=560v = (u+v+w) - (u+w) = \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)=1716801210=1781680=91680=3560    z=5603186.67w = (u+v+w) - (u+v) = \frac{17}{1680} - \frac{1}{210} = \frac{17 - 8}{1680} = \frac{9}{1680} = \frac{3}{560} \implies z = \frac{560}{3} \approx 186.67
Thus, the correct solution is x=336\displaystyle x = 336, y=560\displaystyle y = 560, and z186.67\displaystyle z \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.

More Questions from Equations

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