NIMCET 2027 2024 Previous Year Questions (PYQs) – Page 3 of 12

Logical Deduction: Syllogism Based Reasoning

Statements:

1. All students are intelligent.
2. No intelligent person is lazy.
3. Some lazy people are poor.

Conclusions:

1. ✅ No student is lazy. (Follows)
2. ✅ Some poor people are not intelligent. (Follows)
3. ❌ All poor people are lazy. (Does not follow)

✅ Final Answer: Only Conclusions 1 and 2 follow.

Given:

• Arjun is 25 years younger than his mother.
• Arjun’s brother was born in 1964 and is 35 years younger than their mother.

We want: Arjun’s birth year.

Step 1: Calculate the mother's birth year:

\[ 1964 + 35 = 1999 \] So, the mother was born in 1999.

Step 2: Arjun is 25 years younger than his mother:

\[ 1999 - 25 = 1974 \] Therefore, Arjun was born in 1974.

✅ Final Answer: Arjun was born in 1974

Given:

• A says: "Yes, B is a truth-teller."
• B says: "No, I am not a truth-teller."
• C says: "B is a liar."

We need to determine: Who is telling the truth?

Step 1: Assume B is a truth-teller:

If **B is a truth-teller**, then B's statement "No, I am not a truth-teller" would be false, which is a contradiction because a truth-teller cannot lie.

So, **B cannot be a truth-teller**.

Step 2: Assume B is a liar:

If **B is a liar**, then B's statement "No, I am not a truth-teller" must be false, meaning B is indeed a truth-teller. But this is a contradiction because B can't be both a liar and a truth-teller.

Thus, **B must be a liar**.

Step 3: A’s statement analysis:

A says, "Yes, B is a truth-teller." Since we concluded that **B is a liar**, A's statement is false.

This means **A is a liar**.

Step 4: C’s statement analysis:

C says, "B is a liar." Since **B is indeed a liar**, C’s statement is true.

Therefore, **C is a truth-teller**.

✅ Final Answer: C is the truth-teller.

Given:

\[ x = 1 + 2^{1/6} + 4^{1/6} + 8^{1/6} + 16^{1/6} + 32^{1/6} \]

Step 1: Write in powers of \( a = 2^{1/6} \)

\[ x = 1 + a + a^2 + a^3 + a^4 + a^5 = 1 + \frac{a(a^5 - 1)}{a - 1} \]

Step 2: Use identity \( a^6 = 2 \Rightarrow a^5 = \frac{2}{a} \)

\[ x = 1 + \frac{2 - a}{a - 1} = \frac{1}{a - 1} \Rightarrow 1 + \frac{1}{x} = a \Rightarrow \left(1 + \frac{1}{x} \right)^{24} = a^{24} \]

Step 3: Final calculation

\[ a = 2^{1/6} \Rightarrow a^{24} = (2^{1/6})^{24} = 2^4 = \boxed{16} \]

✅ Final Answer: $\boxed{16}$

Family Age Puzzle: Arjun's Birth Year

Given:

• Arjun is 25 years younger than his mother.
• His brother was born in 1964 and is 35 years younger than their mother.

Step-by-Step Solution:

• Mother’s birth year = 1964 − 35 = 1929
• Arjun’s birth year = 1929 + 25 = 1954

✅ Final Answer: Arjun was born in 1954.

Step 1: Recognize the series

The exponent is an infinite geometric series: $$ 1 + |\sin x| + |\sin x|^2 + |\sin x|^3 + \cdots $$

This is a geometric series with first term \( a = 1 \), common ratio \( r = |\sin x| \in [0,1] \), so: $$ \text{Sum} = \frac{1}{1 - |\sin x|} $$

Step 2: Rewrite the equation

$$ 5^{\frac{1}{1 - |\sin x|}} = 25 = 5^2 $$

Equating exponents: $$ \frac{1}{1 - |\sin x|} = 2 \Rightarrow 1 - |\sin x| = \frac{1}{2} \Rightarrow |\sin x| = \frac{1}{2} $$

Step 3: Solve for \( x \in (-\pi, \pi) \)

We want all \( x \in (-\pi, \pi) \) such that \( |\sin x| = \frac{1}{2} \)

So \( \sin x = \pm \frac{1}{2} \). Within \( (-\pi, \pi) \), the values of \( x \) satisfying this are:

• $x = \frac{\pi}{6}$
• $x = \frac{5\pi}{6}$
• $x = -\frac{\pi}{6}$
• $x = -\frac{5\pi}{6}$

✅ Final Answer: $\boxed{4}$ solutions

Code Language Puzzle: Decoding Words into Numbers

Rule: Letters are converted to their alphabetical positions and then reversed.

Example:

• HEART → H(8), E(5), A(1), R(18), T(20) → Reversed: T R A E H → 20 18 01 05 08
• LUNGS → L(12), U(21), N(14), G(7), S(19) → Reversed: S G N U L → 19 07 14 21 12

Now for BRAIN:

• B(2), R(18), A(1), I(9), N(14)
• Reversed: N I A R B → 14 09 01 18 02

✅ Final Answer: The last number is 02.

Given System of Equations:

• $x + 2y + 2z = 5$
• $x + 2y + 3z = 6$
• $x + 2y + \lambda z = \mu$

Goal: Find values of $\lambda$ and $\mu$ such that the system has infinitely many solutions

Step 1: Write Augmented Matrix

$ [A|B] = \begin{bmatrix} 1 & 2 & 2 & 5 \\ 1 & 2 & 3 & 6 \\ 1 & 2 & \lambda & \mu \end{bmatrix} $

Step 2: Row operations: Subtract $R_1$ from $R_2$ and $R_3$

$ \Rightarrow \begin{bmatrix} 1 & 2 & 2 & 5 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & \lambda - 2 & \mu - 5 \end{bmatrix} $

Step 3: For infinitely many solutions, rank of coefficient matrix = rank of augmented matrix < number of variables (3)

This happens when the third row becomes all zeros:

$ \lambda - 2 = 0 \quad \text{and} \quad \mu - 5 = 0 $

$\Rightarrow \lambda = 2,\quad \mu = 5$

✅ Final Answer: $\boxed{\lambda = 2,\ \mu = 5}$