EquationsMCQMTP Nov 21Question 998 of 221
All Questions

The value of 'k' for system of equations kx+2y=5\displaystyle kx+2y = 5 and 3x+y=1\displaystyle 3x+y = 1 has no solution is

Options

A6\displaystyle 6
B2/3\displaystyle 2/3
C3/2\displaystyle 3/2
D3\displaystyle 3
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — 6\displaystyle 6

All Options:

  • A6\displaystyle 6
  • B2/3\displaystyle 2/3
  • C3/2\displaystyle 3/2
  • D3\displaystyle 3

Ad

Detailed Solution & Explanation

For a system of linear equations a1x+b1y=c1\displaystyle a_1x + b_1y = c_1 and a2x+b2y=c2\displaystyle a_2x + b_2y = c_2 to have no solution (which means the lines are parallel and do not intersect):
a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}
Given the system of equations:
kx+2y=5  ⟹  a1=k,b1=2,c1=5kx + 2y = 5 \implies a_1 = k, \quad b_1 = 2, \quad c_1 = 5
3x+y=1  ⟹  a2=3,b2=1,c2=13x + y = 1 \implies a_2 = 3, \quad b_2 = 1, \quad c_2 = 1
Substituting these coefficients into the condition:
k3=21≠51\frac{k}{3} = \frac{2}{1} \neq \frac{5}{1}
From the first equality:
k3=2  ⟹  k=6\frac{k}{3} = 2 \implies k = 6
Also, since 2≠5\displaystyle 2 \neq 5, the condition b1b2≠c1c2\displaystyle \frac{b_1}{b_2} \neq \frac{c_1}{c_2} holds true.
Thus, the value of k\displaystyle k for which the system has no solution is 6\displaystyle 6, which corresponds to Option A. (Note: The raw JSON specifies Option C, which is 3/2\displaystyle 3/2, but mathematically Option A is the correct answer).
Hence, **Option A** 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:

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 = \frac{1}{5-2\sqrt{6}}

The cost of 2\displaystyle 2 oranges and 3\displaystyle 3 apples is 28\displaystyle 28. If the cost of an apple is doubled then the cost of 3\displaystyle 3 oranges and 5\displaystyle 5 apples is 75\displaystyle 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 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

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