NIMCET 2012 — Mathematics PYQ

Q: How do I solve NIMCET 2012 Mathematics questions?

Practice NIMCET 2012 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of NIMCET Mathematics questions?

NIMCET Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are NIMCET previous year papers available for free?

Yes. Mathem Solvex provides free access to NIMCET previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

NIMCET 2012 — Mathematics PYQ

NIMCET | Mathematics | 2012

The number of values of $k$ for which the system of equations $(k+1)x + 8y = 4k$ and $kx + (k+3)y = 3k-1$ has infinitely many solutions, is:

Choose the correct answer:

Explanation

Solution

Condition:

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

Steps:

\frac{k+1}{k} = \frac{8}{k+3} = \frac{4k}{3k-1}

1. Solving $\frac{k+1}{k} = \frac{8}{k+3}$ :

(k+1)(k+3) = 8k

k^2 + 4k + 3 = 8k

k^2 - 4k + 3 = 0

(k-1)(k-3) = 0

k = 1, 3

2. Checking $k=1$ in $\frac{8}{k+3} = \frac{4k}{3k-1}$ :

\frac{8}{1+3} = \frac{4(1)}{3(1)-1} \implies \frac{8}{4} = \frac{4}{2} \implies 2 = 2 \text{ (Satisfied)}

3. Checking $k=3$ in $\frac{8}{k+3} = \frac{4k}{3k-1}$ :

\frac{8}{3+3} = \frac{4(3)}{3(3)-1} \implies \frac{8}{6} = \frac{12}{8} \implies \frac{4}{3} \neq \frac{3}{2} \text{ (Not Satisfied)}

Conclusion:

Only $k=1$ is valid. So, the number of values is $1$ .

Correct Option:

(b) 1