NIMCET 2015 — Reasoning PYQ

Q: How do I solve NIMCET 2015 Reasoning questions?

Practice NIMCET 2015 Reasoning 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 Reasoning questions?

NIMCET Reasoning 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 2015 — Reasoning PYQ

NIMCET | Reasoning | 2015

In an examination there are 100 questions divided into 3 parts A, B and C and each part should contain atleast one question. Each question in parts A, B and C carry 1, 2 and 3 marks respectively. Part A is for atleast 60% of the total marks and part B should contain 23 questions. How many questions must be set in part C?

Choose the correct answer:

Explanation

Step 1: Formulate the basic equations

Total number of questions:

$x + y + z = 100$
Given that Part $B$ contains $23$ questions:

$y = 23$
Substituting $y$ into the total:

$x + 23 + z = 100 \implies x + z = 77 \implies x = 77 - z$

Step 2: Determine Total Marks

The marks for each part are:

Marks from Part $A = x \times 1 = x$
Marks from Part $B = 23 \times 2 = 46$
Marks from Part $C = z \times 3 = 3z$
Total Marks ( $T$ ):

$T = x + 46 + 3z$

Substitute $x = 77 - z$ into the total marks equation:

T = (77 - z) + 46 + 3z

T = 123 + 2z

Step 3: Apply the condition for Part A

Part $A$ must be at least $60\%$ of the total marks:

x \ge 0.60 \times T

(77 - z) \ge 0.60 \times (123 + 2z)

Now, solve for $z$ :

77 - z \ge 73.8 + 1.2z

77 - 73.8 \ge 1.2z + z

3.2 \ge 2.2z

z \le \frac{3.2}{2.2}

z \le 1.4545...

Step 4: Final Constraint

Since $z$ represents the number of questions, it must be a positive integer. The problem states each part contains atleast one question, so $z \ge 1$ .

Based on our inequality $z \le 1.45$ , the only possible integer value for $z$ is:

z = 1

Conclusion:

There must be 1 question in part $C$ .

Correct Option: (a) 1