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nimcet Previous Year Questions (PYQs)

nimcet 2025 PYQ


nimcet PYQ 2025
A sum of money was divided into 2 parts in the ratio 2:5. First part was invested for 2 years at the annual interest rate of 20% compounded annually. At what rate of simple interest per annum the second part must be invested for 2 years, so that the interest earned in both cases is the same?





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Solution

Ratio = $2:5 \;\Rightarrow\; \tfrac{2S}{7}, \tfrac{5S}{7}$
CI on $\tfrac{2S}{7}$ for 2 yrs @ $20\% =$ $0.44 \times \tfrac{2S}{7} = \tfrac{0.88S}{7}$
SI on $\tfrac{5S}{7}$ for 2 yrs @ $R\% =$ $\dfrac{5S}{7} \times \dfrac{R \times 2}{100} = \dfrac{RS}{70}$
Equating ⇒ $\tfrac{0.88S}{7} = \dfrac{RS}{70}$
$\therefore \; R = 8.8\% \; \text{p.a.}$

nimcet PYQ 2025
A group of 630 children is arranged in rows for a group photograph. Each row contains three fewer children than the row in front of it. What number of rows is not possible?





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Solution

Let the first row have $n$ children and total rows be $r$. Then the numbers per row form an AP with difference $-3$ and sum $630$:

$\displaystyle \frac{r}{2}\big(2n-3(r-1)\big)=630$ $ \;\Rightarrow\; 1260=r\big(2n-3(r-1)\big)$.

Hence $r\mid1260$. Also the last row must be positive: $n-3(r-1)>0$.

Testing the options (and ensuring $n$ is integer and last term positive):

  • $r=4$: $1260/4=315$, $2n=315+3\cdot3=324 \Rightarrow n=162$, last term $=162-9=153>0$ ✓
  • $r=5$: $1260/5=252$, $2n=252+3\cdot4=264 \Rightarrow n=132$, last term $=132-12=120>0$ ✓
  • $r=6$: $1260/6=210$, $2n=210+3\cdot5=225 \Rightarrow n=112.5$ (not integer) ✗

Therefore, the impossible number of rows is $\boxed{6}$.


nimcet PYQ 2025
How many triangles are there in the figure given below?





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Solution


nimcet PYQ 2025
A number which, when increased by 15% becomes 161. What is the original number?





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Solution

Let the original number = x

Increased by 15% ⇒ New value = 1.15x

Given: 1.15x = 161

x = 161 ÷ 1.15

x = 140


✅ Final Answer: The original number is 140.


nimcet PYQ 2025
Suppose that $C$ represents the set of all countries, $R$ represents the set of all countries that have at least one river flowing through it, $M$ represents the set of all countries that have at least one mountain in it, and $D$ represents the set of all countries that have at least one desert in it. It is given that: \[ R \cup M \cup D = C \] Which one of the following gives the set of all countries that have either a mountain or a river, but do not have a desert in it? The notation $D^{c}$ represents the complement of the set $D$ with respect to the universal set $C$.





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Solution

Given that we want the set of all countries that have 
**either a mountain or a river**, but **do not have a desert**. 

The set of countries that have either a river or a mountain is: $R \cup M$ 

The set of countries that do not have a desert is: $D^c$ 

Therefore, the required set is: $(R \cup M) \cap D^c$ 

So the correct answer is: $(R \cup M) \cap D^c$

nimcet PYQ 2025
What is the equivalent decimal value of the unsigned binary number 10101.10101?





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Solution

We separate the integer and fractional parts.

Integer part: $10101_2 = 1\times2^4 + 0\times2^3 + 1\times2^2 + 0\times2^1 + 1\times2^0 $$= 16 + 0 + 4 + 0 + 1 = 21$

Fractional part: $.10101_2 = 1\times2^{-1} + 0\times2^{-2} + 1\times2^{-3} + 0\times2^{-4} + 1\times2^{-5}$

= 0.5 + 0 + 0.125 + 0 + 0.03125 

= 0.65625

Therefore, $10101.10101_2 = 21 + 0.65625 = {21.65625}$


nimcet PYQ 2025
Consider the Boolean expression $ X = \overline{(A + B) \cdot C}$ over two Boolean variables $A$ and $B$. Which one of the following Boolean expressions is equivalent to the given Boolean expression $X$?






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Solution

We are given the Boolean expression:
$X = \overline{(A + B)\cdot C}$ 

First apply De-Morgan’s Law: $\overline{(A + B)\cdot C} = \overline{(A + B)} + \overline{C}$ 

Again apply De-Morgan’s Law on 
$\overline{(A + B)}$= $ \overline{A}\cdot \overline{B}$ 
So, $X = (\overline{A}\cdot \overline{B}) + \overline{C}$
This matches option (2):
$\overline{A\cdot B} + \overline{B\cdot C}$

nimcet PYQ 2025
Which of the following is the CORRECT truth table for the XOR Gate with two binary inputs A and B?





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Solution


nimcet PYQ 2025
Given an unsigned 32-bit integer $x$, which of the following C/C++ expressions correctly toggles $m$ bits starting from position $p$ (with the least significant bit at position $0$)?

Assume: x is the input integer; p is the starting position of the bit range (0-based, LSB at position 0); m is the number of bits to toggle; and No overflow or invalid input conditions occur.

Which of the following correctly toggles $m$ bits of $x$ starting from position $p$?





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Solution


nimcet PYQ 2025
Debugger is a program that





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Solution

A debugger is a tool used to test and debug programs. It allows programmers to pause execution at specific points (breakpoints), run code step-by-step, and inspect variables and registers to find logic or runtime errors.

Hence, the correct option is:

✅ Option 3: allows to set breakpoints, execute a segment of program, and display contents of registers.


nimcet PYQ 2025
Consider the Boolean function in a CPU control unit: \[ Y = (A + B\cdot \overline{C}) \cdot (\overline{A}\cdot B + C) \]





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Solution

Given: $Y = (A + B\overline{C}).(\overline{A}B + C)$ 
Single-Step Smart Trick (Pair Matching Rule): 
Multiply only the terms that can logically combine: 
$A$ pairs only with $C$ → gives $AC$ 
 $B\overline{C}$ pairs only with $\overline{A}B$ → gives $\overline{A}B\overline{C}$ 
 So directly: $Y = AC + \overline{A}B\overline{C}$

nimcet PYQ 2025
Dynamic RAM (DRAM) stores each bit of data in a separate capacitor. Due to leakage, the stored charge tends to dissipate over time and needs to be refreshed periodically. Consider the following statements:
P: DRAM requires refreshing because it uses capacitors to store bits. 
Q: SRAM does not require refreshing because it uses flip-flops instead of capacitors.
R: DRAM is faster than SRAM because it needs less frequent access.
S: DRAM is more suitable for main memory than SRAM due to its density.





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Solution

  • P: ✔ True — DRAM cells use capacitors that leak charge and hence must be refreshed periodically.
  • Q: ✔ True — SRAM uses flip-flops, so it retains data without refreshing.
  • R: ✗ False — DRAM is actually slower than SRAM.
  • S: ✔ True — DRAM is denser and cheaper, making it ideal for main memory.

✅ Correct statements: P, Q, and S


nimcet PYQ 2025
In 8-bit two's complement arithmetic, compute the result of the following addition: A = 10011001, B = 11010111. What is the resulting 8-bit binary value?





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Solution

Let’s first convert both binary numbers into decimal (two’s complement form):

  • $A = 10011001_2$ → negative because MSB = 1 → invert bits: $01100110_2 = 102_{10}$ → add 1 → $103_{10}$ So, $A = -103$.
  • $B = 11010111_2$ → MSB = 1 (negative) → invert bits: $00101000_2 = 40_{10}$ → add 1 → $41_{10}$ So, $B = -41$.

Add them:

$A + B = -103 + (-41) = -144$

Now, in 8-bit two’s complement, the range is $-128$ to $+127$. Since $-144$ is out of range, overflow occurs.

But let’s compute the 8-bit result (ignoring overflow):

$10011001 + 11010111 =$ 10011001
+ 11010111
= 101110000 (9 bits)

Drop the carry beyond 8 bits → 01110000.

Therefore, the resulting 8-bit binary value is:

✅ Result = 01110000₂

(Overflow occurred, actual signed result would have been −144, but the 8-bit stored value is +112.)


nimcet PYQ 2025
In the design of a control unit of a processor, two common approaches are used: hardware control and microprogrammed control. Consider the following statements: 
  1. Hardware control units are generally faster but more difficult to modify than microprogrammed control units. 
  2. In a horizontal microprogrammed control unit, each control signal has a separate bit in the control word. 
  3. Vertical microprogramming leads to longer control words but provides greater parallelism. 
  4. Microprogrammed control units are typically easier to implement and modify than hardware control units.





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Solution

  • Statement 1 – True: Hardware control units are implemented using combinational logic circuits. They generate control signals directly through hardware, making them faster but difficult to modify since any change requires redesigning the hardware.
  • Statement 2 – True: In a horizontal microprogrammed control unit, each control signal is represented by a dedicated bit in the control word. This allows maximum parallelism but results in a wide (long) control word.
  • Statement 3 – False: In vertical microprogramming, control words use encoded fields to represent control signals. This makes the word shorter, not longer, but reduces parallelism since only a few signals can be activated at once.
  • Statement 4 – True: Microprogrammed control units are easier to implement and modify because control logic is stored in a control memory (microprograms). Modifications can be done by changing microinstructions instead of redesigning hardware.

✅ Correct statements: 1, 2, and 4


nimcet PYQ 2025
Consider a system with a CPU having 6 registers and 32-bit instructions. The maximum possible size of the main memory is 512 KB (1K = 210). Each instruction takes two registers and one memory address as operands. Which one of the following correctly gives the maximum possible distinct instructions that can be there in the instruction set of the CPU?





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Solution

Memory size $=512\text{ KB}=2^{19}$ bytes $\Rightarrow$ address field needs $19$ bits.

Registers $=6 \Rightarrow$ bits per register $=\lceil \log_2 6\rceil=3$ bits. For two registers $\Rightarrow 2\times3=6$ bits.

Total operand bits $=19+6=25$.

Opcode bits $=32-25=7 \Rightarrow$ maximum distinct instructions $=2^{7}= {128}$.


nimcet PYQ 2025
Match all items in Group 1 with correct options from those given in Group 2.
 GROUP 1 GROUP 2
 P. Intermediate representation 1. Activation records
 Q. Top-down parsing 2. Code generation
 R. Runtime environments 3. Leftmost derivation
 S. Register allocation 4. Graph colouring





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Solution

Group 1Group 2 (Match)Explanation
P. Intermediate representation 2. Code generation Intermediate representation (IR) is the form of code produced by the front end of a compiler and used as input to the code generation phase.
Q. Top-down parsing 3. Leftmost derivation Top-down parsers construct a parse tree from the root using leftmost derivations.
R. Runtime environments 1. Activation records Activation records (stack frames) are part of the runtime environment used to manage function calls and local variables.
S. Register allocation 4. Graph colouring Register allocation is commonly implemented using graph-colouring algorithms to assign variables to CPU registers efficiently.

✅ Final Matching:
P – 2,  Q – 3,  R – 1,  S – 4


nimcet PYQ 2025
Which of the following secondary storage devices has the fastest access time: Optical Drive, Magnetic Tape Drive, Hard Disk Drive (HDD), Solid State Drive (SDD)?





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Solution

Among the given secondary storage devices, the access time comparison is as follows:

  • Magnetic Tape Drive → Very slow (sequential access)
  • Optical Drive (CD/DVD) → Slower than HDD (mechanical seek)
  • Hard Disk Drive (HDD) → Faster than optical and tape drives, but uses mechanical parts
  • Solid State Drive (SSD) → Fastest, as it uses flash memory with no moving parts

✅ Correct Answer: Solid State Drive (SSD)


nimcet PYQ 2025
Consider a 9-bit representation. Which of the following correctly gives the smallest number that can be represented in: 
(i) 1's complement, 
(ii) 2's complement





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Solution

For an n-bit system:

  • 1’s complement range: $-(2^{n-1}-1)$ to $+(2^{n-1}-1)$ ⇒ smallest $=-(2^{8}-1)={-255}$.
  • 2’s complement range: $-2^{n-1}$ to $+(2^{n-1}-1)$ ⇒ smallest $=-2^{8}={-256}$.

(9-bit means sign + 8 magnitude bits.)


nimcet PYQ 2025
Which of the following statements about ASCII and Unicode is correct?





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Solution

Solution:

  • Option (1) – True: Unicode was designed to be backward compatible with ASCII. The first 128 Unicode characters (U+0000 to U+007F) are identical to ASCII, ensuring compatibility.
  • Option (2) – False: ASCII uses 7 bits per character, while Unicode uses up to 16 bits or more (e.g., UTF-16, UTF-32 encodings).
  • Option (3) – False: Unicode and ASCII share the same first 128 characters; they are not completely different.
  • Option (4) – False: Unicode can represent far more characters than ASCII, not fewer.

✅ Correct Answer: (1) Unicode is backward compatible with ASCII and includes all ASCII characters in its encoding.


nimcet PYQ 2025
In an instruction execution pipeline, the earliest that the instruction TLB and data TLB can be accessed are





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Solution

(1) — Instruction TLB at fetch (to translate the instruction fetch address), Data TLB at memory stage (after effective address is computed in EX).


nimcet PYQ 2025
Given the Boolean expression $ (A \oplus B) \land (B \to C), $  which of the following rows in the truth table would have an output of $1$ (True)?





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Solution

We have $ (A \oplus B) \land (B \to C) = (A \oplus B)\land(\lnot B \lor C)$.

  • If $B=0$: $(\lnot B \lor C)=1$ always, so output $=A\oplus 0=A$. Hence rows with $A=1$ work: $(A,B,C)=(1,0,0)$ and $(1,0,1)$.
  • If $B=1$: need $C=1$ (for $B\to C$) and $A\oplus 1=1\Rightarrow A=0$. Hence $(A,B,C)=(0,1,1)$.

Rows with output 1: $(1,0,0)$, $(1,0,1)$, $(0,1,1)$.


nimcet PYQ 2025
Consider a system running under two types of workloads: 
(a) CPU-intensive jobs, 
(b) I/O-intensive jobs. Which of the following statements about the relative performance of Interrupt-driven I/O and Programmed I/O is correct?





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Solution

(4) — Interrupt-driven I/O frees the CPU for compute-heavy jobs; under very I/O-intensive loads, polling (programmed I/O) can outperform by avoiding high interrupt overhead.


nimcet PYQ 2025
In computer architecture, which of the following correctly matches the fundamental unit used in the CPU, cache, and main memory respectively?





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Solution

  • CPU → Word: The CPU processes data in units called words (e.g., 32-bit or 64-bit).
  • Cache → Block: Cache stores and transfers data in blocks (or cache lines).
  • Main Memory → Page: Memory management uses pages (typically 4 KB each) as the fundamental unit for virtual memory.

✅ Correct Answer: CPU → Word, Cache → Block, Memory → Page


nimcet PYQ 2025
What is the hexadecimal representation of the decimal number 265?





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Solution

Convert decimal 265 to hexadecimal:

Divide by 16 repeatedly:

  • $265 ÷ 16 = 16$ remainder $9$
  • $16 ÷ 16 = 1$ remainder $0$
  • $1 ÷ 16 = 0$ remainder $1$

Now write remainders from last to first → $1\,0\,9$

✅ Hexadecimal representation = 0x109

Correct Option: 0x109


nimcet PYQ 2025
Identify the incorrect sentence based on the usage of the verb:





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Solution

  • (1) ✔ Correct — “It’s been raining since morning.” → Present perfect continuous tense, correct usage.
  • (2) ❌ Incorrect — “Understand” is a stative verb (expresses a state, not an action) and is not used in the continuous form. Correct form: “They understand the problem.”
  • (3) ✔ Correct — “It has clear ideas.” → Proper use of “has.”
  • (4) ✔ Correct — Grammatically fine.

✅ Incorrect sentence: They are understanding the problem.


nimcet PYQ 2025
"The new iPhone seems to cost everyone not just an arm and a leg, but also a kidney." What does "to cost an arm and a leg" mean?





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Solution

The idiom “to cost an arm and a leg” means something is very expensive or costs a lot of money.

✅ Correct Answer: To be very expensive


nimcet PYQ 2025
Study the following sentence: The statue of Christ the Redeemer is located on Rio de Janeiro, the largest city of Brazil. (Which of the following parts of the sentence is grammatically incorrect)





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Solution

The incorrect phrase is “on Rio de Janeiro.”

The correct preposition should be “in Rio de Janeiro” because we say something is located in a city, not on it.

✅ Correct Answer: on Rio de Janeiro


nimcet PYQ 2025
The man and the dog_____(flee) through the forest. ( fill in the blank with the most appropriate form)





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Solution

Subject: The man and the dog → plural subject → requires a plural verb form.

The correct tense expressing a continuous past action is “were fleeing.”

✅ Correct Answer: (3) were fleeing


nimcet PYQ 2025
One of the employees ___________ planning to transfer to another department.





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Solution

Subject: One of the employees → singular (the main subject is “one,” not “employees”).

Hence, the verb must be singular → is planning.

✅ Correct Answer: is


nimcet PYQ 2025
Which of the following sentences uses the correct form of the verb "to lay"?





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Solution

The verb "to lay" means to put or place something down and requires an object.

  • (1) ❌ Incorrect — “Laying down” should be “lying down” (no object).
  • (2) ❌ Incorrect — “Lays on the bed” → should be “lies on the bed.”
  • (3) ❌ Incorrect — “Has lain” comes from “to lie,” not “to lay.”
  • (4) ✅ Correct — “She laid the book on the table.” → correct past tense of “lay” with object “book.”

✅ Correct Answer:  She laid the book on the table.


nimcet PYQ 2025
What is the synonym of "belligerent"?





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Solution

The word “belligerent” means aggressive, warlike, or hostile in attitude or behavior.

  • (1) Equanimity → calmness (opposite meaning)
  • (2) Bellyful → enough or more than enough
  • (3) Peaceful → opposite of belligerent
  • (4) Hostile → same meaning as belligerent ✅

✅ Correct Answer: (4) Hostile


nimcet PYQ 2025
After running around the city all day delivering invitations for the wedding, I desperately needed to hit the sack. (Identify the meaning of the underlined phrase):





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Solution

The idiom “hit the sack” means to go to bed or go to sleep, usually because you are very tired.

✅ Correct Answer: To go to bed


nimcet PYQ 2025
Which of the following statements correctly explains the use of articles in the following sentence: After earning an MBA, Srinivas was able to approach a mathematical problem from both a business and analytical perspective.





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Solution

  • (1) ✅ Correct — “An MBA” uses an because the pronunciation of “MBA” starts with a vowel sound (“em-bee-ay”).
  • A mathematical problem” uses a because “mathematical” begins with a consonant sound /m/.

✅ Correct Answer: (1)


nimcet PYQ 2025
Even though in official statements, the Presidential office has warned against violating the 25th protocol, the government's inaction against the Market group shows the government's tacit approval of the movement.(Choose the antonym of the underlined word):





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Solution

The word “tacit” means understood or implied without being stated openly (i.e., silent or implicit).

  • (1) Unvoiced → similar in meaning (not opposite)
  • (2) Unequivocal → means clear, direct, and explicit → opposite of “tacit” ✅
  • (3) Undeclared → similar meaning
  • (3) Implicit → synonym, not antonym

✅ Correct Answer: Unequivocal


nimcet PYQ 2025
The length of the projection of $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ on $\vec{b} = -2\hat{i} + \hat{j} + 2\hat{k}$, is equal to:





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Solution

$\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ $\vec{b} = -2\hat{i} + \hat{j} + 2\hat{k}$ 
The length of projection of $\vec{a}$ on $\vec{b}$ is: 
$\displaystyle \text{Proj length} = \frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$ 
Compute dot product: $\vec{a}\cdot\vec{b} = (2)(-2) + (3)(1) + (1)(2)$ $= -4 + 3 + 2 = 1$ 
Compute $|\vec{b}|$: $|\vec{b}| = \sqrt{(-2)^2 + 1^2 + 2^2}$ $= 3$ 
Thus projection length: $\displaystyle \frac{1}{3}$

nimcet PYQ 2025
If $8^{x-1}=(1/4)^{x}$, then the value of $\frac{1}{\log_{x+1}4-\log_{x+1}5}+\frac{1}{\log_{1-x}4-\log_{1-x}5}$ is





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Solution

Given: $8^{x-1} = (1/4)^{x}$ 
Rewrite both sides with base 2: $8 = 2^3$ and $1/4 = 2^{-2}$ 
So: $(2^3)^{x-1} = (2^{-2})^x$ $\Rightarrow 2^{3(x-1)} = 2^{-2x}$ 
Equate powers: $3(x - 1) = -2x$ 
$3x - 3 = -2x$ 
$5x = 3$ 
$\Rightarrow x = \frac{3}{5}$ 

 We need the value of: 
$\displaystyle \frac{1}{\log_{x+1} 4 - \log_{x+1} 5} + \frac{1}{\log_{1-x} 4 - \log_{1-x} 5}$ 
Use property: $\log_a m - \log_a n = \log_a \left(\frac{m}{n}\right)$ 
So the expression becomes: $\displaystyle \frac{1}{\log_{x+1} \left(\frac{4}{5}\right)} + \frac{1}{\log_{1-x} \left(\frac{4}{5}\right)}$ 
Now use: $\displaystyle \frac{1}{\log_a b} = \log_b a$ 
So expression becomes: $\log_{4/5}(x+1) + \log_{4/5}(1-x)$ 
Use product property: $\log_{4/5}[(x+1)(1-x)]$ 
Compute: $(x+1)(1-x) = 1 - x^2$ 
Substitute $x = \frac{3}{5}$: 
$1 - x^2 = 1 - \frac{9}{25} = \frac{16}{25}$ 
Thus value = $\log_{4/5}\left(\frac{16}{25}\right)$ 
Rewrite: $\frac{16}{25} = \left(\frac{4}{5}\right)^2$ 
Therefore: $\log_{4/5}\left( (4/5)^2 \right) = 2$

nimcet PYQ 2025
Consider the matrix $$B=\begin{pmatrix}{-1} & {-1} & {2} \\ {0} & {-1} & {-1} \\ {0} & {0} & {-1}\end{pmatrix}$$. The sum of all the entries of the matrix $B^{19}$ is





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Solution

The matrix is \[ B=\begin{pmatrix} -1 & -1 & 2\\ 0 & -1 & -1\\ 0 & 0 & -1 \end{pmatrix} \] Since \(B\) is upper–triangular with diagonal entries \(-1\), write \[ B = -I + N, \] where \[ N=\begin{pmatrix} 0 & -1 & 2\\ 0 & 0 & -1\\ 0 & 0 & 0 \end{pmatrix}, \qquad N^{3}=0. \] Using the binomial expansion: \[ B^{19}=(-I+N)^{19} = (-1)^{19}I + \binom{19}{1}(-1)^{18}N + \binom{19}{2}(-1)^{17}N^{2}. \] Compute signs: \[ (-1)^{19}=-1,\quad (-1)^{18}=1,\quad (-1)^{17}=-1. \] So: \[ B^{19} = -I + 19N - \binom{19}{2}N^{2}. \] Now compute sums: \[ \text{sum}(-I) = -3 \] \[ N^{2} = \begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} \quad\Rightarrow\quad \text{sum}(N^{2}) = 1 \] \[ \Rightarrow -\binom{19}{2}\cdot 1 = -171 \] Final total: \[ -3 -171 = -174. \]

nimcet PYQ 2025
The curve $y=\frac{x}{1+x\tan x}$ attains maxima





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Solution

Given the curve 
$y = \dfrac{x}{1 + x \tan x}$ 
Differentiate using quotient rule: 
$y' = \dfrac{(1 + x \tan x) - x(\tan x + x \sec^2 x)}{(1 + x \tan x)^2}$ 
Simplify the numerator: 
$N = 1 + x \tan x - x \tan x - x^2 \sec^2 x$ 
$N = 1 - x^2 \sec^2 x$ 
Set $N = 0$ for maxima/minima: 
$1 - x^2 \sec^2 x = 0$ 
$x^2 \sec^2 x = 1$ 
$x \sec x = \pm 1$ 
Hence the curve attains maxima when: 
$x \sec x = 1$
$x=\cos x$

nimcet PYQ 2025
The scores of students in a national level examination are normally distributed with a mean of 500 and a standard deviation of 100. If the value of the cumulative distribution of the standard normal random variable at 0.5 is 0.691, then the probability that a randomly selected student scored between 450 and 500 is





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Solution

Given: 
Scores are normally distributed with mean $ \mu = 500 $, 
standard deviation $ \sigma = 100 $ 
We want the probability that a student scored between 450 and 500: 
$P(450 < X < 500)$ 

Convert to $Z$-scores: 
For $X = 450$: $Z = \dfrac{450 - 500}{100} = -0.5$ 

For $X = 500$: $Z = \dfrac{500 - 500}{100} = 0$ 

Thus: 
$P(450 < X < 500) = P(-0.5 < Z < 0)$ 

Given: $P(Z < 0.5) = 0.691$ 

Using symmetry of the standard normal curve:
 $P(-0.5 < Z < 0.5) = 2(0.691 - 0.5) = 2(0.191) = 0.382$ 
So: $P(-0.5 < Z < 0) = \dfrac{0.382}{2} = 0.191$ 

Final answer: 0.191

nimcet PYQ 2025
Number of permutations of the letters of the word BANGLORE such that the string ANGLE appears together in all permutations, is





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Solution

Treat the fixed string ANGLE as one block. Remaining letters from BANGLORE are B, O, R. So we arrange 4 distinct items: {ANGLE, B, O, R}.

Number of permutations =4! = 24.


nimcet PYQ 2025
Let A and B be two square matrices of same order satisfying $A^2+5A+5I =0$ and $B^2+3B+I=0$ repectively. Where I is the identity matrix. Then the inverse of the matrix $C= BA+2B+2A+4I$ is





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Solution

Given: $A^2 + 5A + 5I = 0 \quad\Rightarrow\quad A^2 = -5A - 5I$ 
$B^2 + 3B + I = 0 \quad\Rightarrow\quad B^2 = -3B - I$ 

Matrix: $C = BA + 2B + 2A + 4I$ 

We want $C^{-1}$. 

Step 1: Try a linear expression as inverse Assume: $C^{-1} = \alpha AB + \beta A + \gamma B + \delta I$ 
Test using the fact that quadratic equations give inverses as linear polynomials. 

Check option (2): 
$AB + A + 3B + 3I$ Multiply with $C$: 
$C(AB + A + 3B + 3I)$ 
$= (BA + 2B + 2A + 4I)(AB + A + 3B + 3I)$ 
Using reductions: 
$A^2 = -5A - 5I$ $B^2 = -3B - I$ 
Everything simplifies to: $I$ 
Hence: $C^{-1} = AB + A + 3B + 3I$

nimcet PYQ 2025
The captains of five cricket teams, including India and Australia, are lined up randomly next to one other for a group photo. What is the probability that the captains of India and Australia will stand next to each other?





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Solution

Treat India–Australia as one block. Then we have $4!$ ways to arrange the block with the other 3 captains, and $2$ orders inside the block (IA or AI). Favorable $=2\cdot4!=48$; total $=5!=120$.

Probability $=\dfrac{48}{120}={\dfrac{2}{5}}$.


nimcet PYQ 2025
The value of $\frac{d}{dx}\int ^{2\sin x}_{\sin {x}^2}{e}^{{t}^2}dt$ at $x=\pi$





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Solution

We want to find $\dfrac{d}{dx}\left(\int_{\sin^2 x}^{2\sin x} e^{t^2}, dt\right)$ at $x=\pi$. 
Using Leibniz rule: $\dfrac{d}{dx}\left(\int_{a(x)}^{b(x)} f(t) dt\right) = f(b(x)) b'(x) ;-; f(a(x)) a'(x)$ 

Here $a(x)=\sin^2 x$, $b(x)=2\sin x$, $f(t)=e^{t^2}$. 
Compute derivatives:
$a'(x)=2\sin x\cos x$ 
$b'(x)=2\cos x$ 
So the derivative is: 
$e^{(2\sin x)^2}(2\cos x)-e^{(\sin^2 x)^2}(2\sin x\cos x)$ 

Now evaluate at $x=\pi$
$\sin\pi=0,\quad \cos\pi=-1$ 

Thus: $b'(\pi)=2(-1)=-2$ 
$a'(\pi)=0$ 

Therefore: $e^{0}(-2)-e^{0}(0)=-2$ 
So the final answer is: $-2$

nimcet PYQ 2025
There are two coins, say blue and red. For blue coin, probability of getting head is 0.99 and for red coin, it is 0.01. One coin is chosen randomly and is tossed. The probability of getting head is





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Solution

One coin is chosen at random, so each has probability $\tfrac{1}{2}$ of being selected.

Let $P(H)$ = probability of getting head.

Using the law of total probability:

$P(H) = \tfrac{1}{2} \times 0.99 + \tfrac{1}{2} \times 0.01 $$ = \tfrac{1}{2}(1.00) = 0.50$

✅ Final Answer: ${0.5}$


nimcet PYQ 2025
The number of all even integers between 99 and 999 which are not multiple of 3 and 5 is





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Solution

Solution:

Even numbers from 100 to 998: count $= \frac{998-100}{2}+1=450$.

Exclude evens divisible by $3$ or $5$ using inclusion–exclusion:

  • Multiples of $6$ in $[100,998]$: $\lfloor 998/6 \rfloor-\lfloor 99/6 \rfloor = 166-16=150$.
  • Multiples of $10$ in $[100,998]$: $\lfloor 998/10 \rfloor-\lfloor 99/10 \rfloor = 99-9=90$.
  • Multiples of $30$ in $[100,998]$: $\lfloor 998/30 \rfloor-\lfloor 99/30 \rfloor = 33-3=30$.

Forbidden $=150+90-30=210$ ⇒ Allowed $=450-210={240}$.


nimcet PYQ 2025
Let A = {1,2,3, ... , 20}. Let $R\subseteq A\times A$ such that R = {(x,y): y = 2x - 7}. Then the number of elements in R, is equal to





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Solution

Given relation: $R = \{(x,y) : y = 2x - 7\}$ where $A = \{1,2,3,\dots,20\}$.

We need both $x, y \in A$.

So, $1 \le y = 2x - 7 \le 20$

⇒ $1 \le 2x - 7 \le 20$

Add 7: $8 \le 2x \le 27$

Divide by 2: $4 \le x \le 13.5$

Hence, integer values of $x$ are $4,5,6,7,8,9,10,11,12,13$ → total $10$ values.

Number of elements in R = 10


nimcet PYQ 2025
If $\vec{a}, \vec{b}$ and $\vec{c} $ are three vectors such that $\vec{a} \times \vec{b}=\vec{c}$ , $\vec{a}.\vec{c} = 2$ and $\vec{b}.\vec{c} = 1$. If $|\vec{b}| = 1$, then the value of $|\vec{a}| $ is





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Solution

Given conditions: $\vec{a} \times \vec{b} = \vec{c}$ 
$\vec{a} \cdot \vec{c} = 2$ 
$\vec{b} \cdot \vec{c} = 1$ 
$|\vec{b}| = 1$ 

Since $\vec{c} = \vec{a} \times \vec{b}$, 
we have: $|\vec{c}| = |\vec{a}||\vec{b}| \sin\theta = |\vec{a}|\sin\theta$ (because $|\vec{b}| = 1$) 

Also: $\vec{a} \cdot \vec{c} = \vec{a} \cdot (\vec{a} \times \vec{b}) = 0$ 
But given: $\vec{a} \cdot \vec{c} = 2$ 

This is possible only if $\vec{c}$ is not perpendicular to $\vec{a}$, meaning $\vec{c}$ is not just $\vec{a} \times \vec{b}$ but also has a component along $\vec{a}$. 
Use the identity: $(\vec{a} \times \vec{b}) \cdot \vec{c} = \det(\vec{a},\vec{b},\vec{c})$ 
But we need magnitudes. Take dot product of $\vec{b}$ with $\vec{c}$: 
$\vec{b} \cdot \vec{c} = \vec{b} \cdot (\vec{a} \times \vec{b}) = 0$ 
But given $\vec{b} \cdot \vec{c} = 1$ 

So again $\vec{c}$ has components outside the perpendicular direction $\vec{c}$ is independent. 

Use the vector triple product identity: $\vec{c} = \vec{a} \times \vec{b}$ 

Take magnitude squared: 
$|\vec{c}|^2 = |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$ 

Since $|\vec{b}| = 1$: $|\vec{c}|^2 = |\vec{a}|^2 - (\vec{a} \cdot \vec{b})^2$ 

Now use the given dot products. 
We know: $\vec{a} \cdot \vec{c} = 2$ $\vec{b} \cdot \vec{c} = 1$ 

Take magnitude squared of $\vec{c}$: $|\vec{c}|^2 = (\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b})$ 

But also: $|\vec{c}|^2 = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{c}) = 2 \cdot 1 = 2$ 

So: $|\vec{c}|^2 = 2$ 
Thus: $2 = |\vec{a}|^2 - (\vec{a} \cdot \vec{b})^2$ 
Also: $\vec{a} \cdot \vec{b} = \dfrac{\vec{a} \cdot \vec{c}}{|\vec{c}|} = \dfrac{2}{\sqrt{2}} = \sqrt{2}$ 

Therefore: $(\vec{a} \cdot \vec{b})^2 = 2$ 

Substitute: $2 = |\vec{a}|^2 - 2$ 
So: $|\vec{a}|^2 = 4$ 
Hence: $|\vec{a}| = 2$

nimcet PYQ 2025
If x, y and z are three cube roots of 27, then  the determinant of the matrix $\begin{bmatrix}{x} & {y} & {z} \\ {y} & {z} & {x} \\ {z} & {x} & {y}\end{bmatrix}$ is 





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Solution

If $x, y, z$ are three cube roots of $27$, then the determinant of the matrix \[ \begin{pmatrix} x & y & z\\[4pt] y & z & x\\[4pt] z & x & y \end{pmatrix} \] is: The cube roots of $27 = 3^3$ are: \[ x = 3,\qquad y = 3\omega,\qquad z = 3\omega^2, \] where $\omega$ is a cube root of unity satisfying \[ \omega^3 = 1,\qquad 1+\omega+\omega^2 = 0. \] For a circulant matrix, the determinant is: \[ (x+y+z)(x+\omega y+\omega^2 z)(x+\omega^2 y+\omega z). \] Now compute the first factor: \[ x+y+z = 3(1+\omega+\omega^2) = 3\cdot 0 = 0. \] Therefore, \[ \det = 0. \]

nimcet PYQ 2025
Let $A=\{{5}^n-4n-1\colon n\in N\}$ and $B=\{{}16(n-1)\colon n\in N\}$ be sets. Then





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Solution

We are given two sets: \[ A = \{\,5^n - 4n - 1 : n \in \mathbb{N}\,\} \] \[ B = \{\,16(n-1) : n \in \mathbb{N}\,\} \] We test the first few values. For set \(A\): \[ \begin{aligned} n=1 &: 5^1 - 4 - 1 = 0 \\ n=2 &: 25 - 8 - 1 = 16 \\ n=3 &: 125 - 12 - 1 = 112 \\ n=4 &: 625 - 16 - 1 = 608 \end{aligned} \] So, \[ A = \{0,\;16,\;112,\;608,\dots\} \] For set \(B\): \[ \begin{aligned} n=1 &: 16(0) = 0 \\ n=2 &: 16(1) = 16 \\ n=3 &: 16(2) = 32 \\ n=4 &: 16(3) = 48 \end{aligned} \] So, \[ B = \{0,\;16,\;32,\;48,\;64,\dots\} \] Clearly every element of \(A\) also appears in \(B\), since: \[ 5^n - 4n - 1 = 16(n-1) \] Hence: \[ A \subset B. \] Therefore, the correct answer is: \[ \boxed{A \subset B} \]

nimcet PYQ 2025
Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be unit vectors such that the angle between them is ${\cos }^{-1}\Bigg{\{}\frac{1}{4}\Bigg{\}}$. If $\vec{b}=2\vec{c}+\lambda \vec{a}$, where $\lambda$ > 0 and $\vec{b}=4$, then $\lambda$ is equal to





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Solution

Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors and the angle between any pair is: 
$\cos^{-1}\left(\frac{1}{4}\right)$ 

Thus, $\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}=\vec{c}\cdot\vec{a}=\frac{1}{4}$. 

We are given: $\vec{b}=2\vec{c}+\lambda \vec{a}$ and the magnitude: $|\vec{b}|=4$. 

Since $\vec{b}$ is not a unit vector here, we use: 
$|\vec{b}|^2 = (2\vec{c}+\lambda \vec{a})\cdot(2\vec{c}+\lambda \vec{a})$. 

Expand: $|\vec{b}|^2 = 4|\vec{c}|^2 + \lambda^2|\vec{a}|^2 + 4\lambda(\vec{c}\cdot\vec{a})$ 

Since all are unit vectors: $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$ and $\vec{c}\cdot\vec{a}=\frac14$. 
So: $|\vec{b}|^2 = 4 + \lambda^2 + 4\lambda\left(\frac14\right)$ 
Simplify: $|\vec{b}|^2 = 4 + \lambda^2 + \lambda$ 
Given: $|\vec{b}| = 4 \Rightarrow |\vec{b}|^2 = 16$ 
Thus: $\lambda^2 + \lambda + 4 = 16$ $\lambda^2 + \lambda - 12 = 0$ 
Solve quadratic: $\lambda = \frac{-1 \pm \sqrt{1 + 48}}{2} = \frac{-1 \pm 7}{2}$ 
So: $\lambda = 3$ (positive root, since $\lambda>0$) 
Final answer: $\boxed{3}$

nimcet PYQ 2025
A tower subtends angles $\alpha, 2\alpha$ and $3\alpha$ respectively at points A, B and C which are lying on a horizontal line through the foot of the tower. Then $\frac{AB}{BC}$ is equal to





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Solution


nimcet PYQ 2025
If $\vec{a}$ and $\vec{b}$ are twp vectors such that |$\vec{a}$|=3, |$\vec{b}$|=4 and |$\vec{a}+\vec{b}$|=1, then the value of $|\vec{a}-\vec{b}|$ is





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Solution

Given: $|\vec{a}| = 3$, $|\vec{b}| = 4$, $|\vec{a} + \vec{b}| = 1$. 

Use the identity: $|\vec{a} + \vec{b}|^{2} = |\vec{a}|^{2} + |\vec{b}|^{2} + 2\vec{a}\cdot\vec{b}$ 
Substitute values: $1^{2} = 3^{2} + 4^{2} + 2\vec{a}\cdot\vec{b}$ 
$1 = 9 + 16 + 2\vec{a}\cdot\vec{b}$ 
$1 = 25 + 2\vec{a}\cdot\vec{b}$ 
$2\vec{a}\cdot\vec{b} = -24$ 
$\vec{a}\cdot\vec{b} = -12$ 
Now compute $|\vec{a} - \vec{b}|$: 
$|\vec{a} - \vec{b}|^{2} = |\vec{a}|^{2} + |\vec{b}|^{2} - 2\vec{a}\cdot\vec{b}$ 
$= 9 + 16 - 2(-12)$ 
$= 25 + 24$ $= 49$ 
So: $|\vec{a} - \vec{b}| = 7$

nimcet PYQ 2025
If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=2\hat{i}-\hat{j}+3\hat{k}$ and $\vec{c}=\hat{i}-2\hat{j}+\hat{k}$, then a vector of magnitude $\sqrt{22}$ which is parallel to $2\vec{a}-\vec{b}+3\vec{c}$ is





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Solution

Given: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ 
$\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$ 
$\vec{c} = \hat{i} - 2\hat{j} + \hat{k}$ 

Compute: $2\vec{a} - \vec{b} + 3\vec{c}$ 

First: 
$2\vec{a} = 2\hat{i} + 2\hat{j} + 2\hat{k}$ 
$-\vec{b} = -2\hat{i} + \hat{j} - 3\hat{k}$ 
$3\vec{c} = 3\hat{i} - 6\hat{j} + 3\hat{k}$ 

Add components: 
$i$–component: $2 - 2 + 3 = 3$ 
$j$–component: $2 + 1 - 6 = -3$ 
$k$–component: $2 - 3 + 3 = 2$ 
So the vector: $\vec{v} = 3\hat{i} - 3\hat{j} + 2\hat{k}$ 
Its magnitude: $|\vec{v}| = \sqrt{3^2 + (-3)^2 + 2^2} = \sqrt{9 + 9 + 4} = \sqrt{22}$ 
We need a vector parallel to $\vec{v}$ with magnitude $\sqrt{22}$. 
Since $|\vec{v}| = \sqrt{22}$, the required vector is simply: $\boxed{3\hat{i} - 3\hat{j} + 2\hat{k}}$

nimcet PYQ 2025
Consider the sample space $\Omega={\{(x,y):x,y\in{\{1,2,3,4\}\}}}$ where each outcome is equally likely. Let A = {x ≥ 2} and B = {y > x} be two events. Then which of the following is NOT true?





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Solution

We have the sample space \[ \Omega = \{(x,y) : x,y \in \{1,2,3,4\}\}, \qquad |\Omega| = 16. \] Event \[ A = \{x \ge 2\}. \] Values of \(x = 2,3,4\), so total favorable outcomes: \[ 12 \quad \Rightarrow \quad P(A) = \frac{12}{16} = \frac{3}{4}. \] Event \[ B = \{y > x\}. \] Count pairs: \[ \begin{aligned} x=1 &: (1,2),(1,3),(1,4) \Rightarrow 3, \\ x=2 &: (2,3),(2,4) \Rightarrow 2, \\ x=3 &: (3,4) \Rightarrow 1. \end{aligned} \] Thus total = 6, so \[ P(B) = \frac{6}{16} = \frac{3}{8}. \] Now compute \(A \cap B\): \[ x \ge 2,\quad y > x. \] Valid pairs: \[ (2,3),(2,4),(3,4). \] So \[ P(A \cap B) = \frac{3}{16}. \] Check independence: \[ P(A)P(B) = \frac{3}{4} \cdot \frac{3}{8} = \frac{9}{32}, \] but \[ P(A \cap B) = \frac{3}{16} = \frac{6}{32}. \] Since \[ \frac{9}{32} \neq \frac{6}{32}, \] events \(A\) and \(B\) are not independent. Therefore, the NOT true statement is: \[ \boxed{P(A \cap B) = \frac14}. \]

nimcet PYQ 2025
Let the line $\frac{x}{4}+\frac{y}{2}=1$ meets the x-axis and y-axis at A and B, respectively. M is the midpoint of side AB, and M' is the image of the point M across the line x + y = 1. Let the point P lie on the line x + y = 1 such that the $\Delta$ABP is an isosceles triangle with AP = BP. Then the distance between M' and P is





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Solution

A = $(4,0)$, B = $(0,2)$ 
Midpoint of AB: $M=\left(\frac{4+0}{2},\frac{0+2}{2}\right)=(2,1)$ 

Image of $M(2,1)$ across line $x+y=1$: 
$M'=(0,-1)$ 

Since $AP=BP$, point $P$ lies on perpendicular bisector of AB. 

Perpendicular bisector of AB passes through $M(2,1)$ and has slope $2$: 
$y-1=2(x-2)$ 
$y=2x-3$ 
Also $P$ lies on: $x+y=1$ 
So, $x+2x-3=1$ $3x=4$ 
$x=\frac{4}{3}$ $y=1-\frac{4}{3}=-\frac{1}{3}$ 
So, $P=\left(\frac{4}{3},-\frac{1}{3}\right)$ 
Distance between $M'(0,-1)$ and $P$: 
$d=\sqrt{\left(\frac{4}{3}-0\right)^2+\left(-\frac{1}{3}+1\right)^2}$ 
$d=\sqrt{\frac{16}{9}+\frac{4}{9}}$ 
$d=\sqrt{\frac{20}{9}}$ 
$d=\frac{2\sqrt5}{3}$ 
Final Answer: $\boxed{\frac{2\sqrt5}{3}}$

nimcet PYQ 2025
Which one of the following is NOT a correct statement?





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Solution

Question: Which one of the following is NOT a correct statement?

  1. The value of standard deviation changes by a change of scale.
  2. The standard deviation is greater than or equal to the mean deviation (about mean).
  3. The sum of squares of deviations is minimum when taken from the mean.
  4. The variance is expressed in the same units as the units of observation.

Answer: Option 4

Why: Variance has squared units, not the same units as the data (e.g., if data are in cm, variance is in cm²). Standard deviation (the square root of variance) has the same units as the data.

Notes:

  • Change of scale: if each value is multiplied by \(k\), then \(\text{SD}\) becomes \(|k|\cdot \text{SD}\) and \(\text{Var}\) becomes \(k^2\cdot \text{Var}\).
  • \(\text{SD} \ge \text{Mean Deviation (about mean)}\) is a standard inequality.
  • \(\sum (x_i - a)^2\) is minimized at \(a = \bar{x}\) (the mean).

nimcet PYQ 2025
An equilateral triangle is inscribed in the parabola $y^2 = x$. One vertex of the triangle is at the vertex of the parabola. The centroid of triangle is





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Solution

Given parabola  $y^{2} = x. $ $ A(0,0) $ is one vertex of the equilateral triangle. 
Let the other two vertices be $ B(t^{2}, t) $ and $ C(t^{2}, -t) $, since they lie on the parabola and are symmetric. 
Distance: $ BC = \sqrt{(t^{2}-t^{2})^{2} + (t - (-t))^{2}} = 2t. $ 
And $ AB = \sqrt{(t^{2}-0)^{2} + (t-0)^{2}} = \sqrt{t^{4}+t^{2}} = t\sqrt{t^{2}+1}. $ 

For equilateral triangle: $ AB = BC $ 
$ t\sqrt{t^{2}+1} = 2t $ $ \sqrt{t^{2}+1} = 2 $ 
$ t^{2}+1 = 4 $ 
$ t^{2} = 3. $ 
So the points become: 
$ B = (3,\sqrt{3}), \quad C = (3,-\sqrt{3}). $ 
Centroid: $ G = \left( \frac{0+3+3}{3},; \frac{0+\sqrt3-\sqrt3}{3} \right) = (2,0). $

nimcet PYQ 2025
The angles of depression of the top and bottom of an 8m tall building from the top of a multi storied building are 30° and 45°, respectively. What is the height of the multistoried building and the distance between the two buildings?





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Solution

The height of the multistoried building is $H$ and the horizontal distance between the buildings is $D$. 
The angles of depression to the top and bottom of the $8\text{ m}$ building are $30^\circ$ and $45^\circ$ respectively. 
From the $45^\circ$ angle: 
$\tan 45^\circ = \dfrac{H}{D}$ 
$1 = \dfrac{H}{D}$ 
$H = D$ 

From the $30^\circ$ angle: 
$\tan 30^\circ = \dfrac{H - 8}{D}$ 
$\dfrac{1}{\sqrt{3}} = \dfrac{H - 8}{H}$ 
$H - 8 = \dfrac{H}{\sqrt{3}}$
$H\left(1 - \dfrac{1}{\sqrt{3}}\right) = 8$ 
Solve for $H$: 
$H = \dfrac{8}{1 - \dfrac{1}{\sqrt{3}}}$ 
$H = 4(3 + \sqrt{3})$ 
Since $H = D$: 
$D = 4(3 + \sqrt{3})$ 
Final answers: 
$\boxed{H = 4(3 + \sqrt{3})\ \text{m}}$ 
$\boxed{D = 4(3 + \sqrt{3})\ \text{m}}$

nimcet PYQ 2025
The number of accidents per week in a town follows Poisson distribution with mean 3 (In Exam Given 2, which is incorrect). If the probability that there are three accidents in two weeks time is $ke^{-6}$, then the value of k is





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Solution

Mean accidents per week = $3$. 
Mean accidents in two weeks = $6$. 
For a Poisson distribution, 
$P(X=3) = \dfrac{6^{3}}{3!} e^{-6}$. 

Given $P(X=3) = k e^{-6}$, 
so $k = \dfrac{6^{3}}{3!}$. 
$k = \dfrac{216}{6} = 36$.

nimcet PYQ 2025
If $B=sin^2 y+cos^4 y$, then for all real y





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Solution

Let $B = \sin^{2}y + \cos^{4}y$. 
Put $t = \cos^{2}y$, so $0 \le t \le 1$. 
Then $\sin^{2}y = 1 - t$. 
So, $B = (1 - t) + t^{2} = t^{2} - t + 1$. 
Minimum of the quadratic $t^{2} - t + 1$ occurs at $t = \dfrac{1}{2}$: 
$B_{\min} = \dfrac{1}{4} - \dfrac{1}{2} + 1 = \dfrac{3}{4}$. 
Maximum occurs at $t = 0$ or $t = 1$: 
$B_{\max} = 1$. 
Hence, $B \in \left[\dfrac{3}{4},, 1\right]$.

nimcet PYQ 2025
Let $F_1, F_2$ be foci of hyperbola $\frac{{x}^2}{{a}^2}-\frac{{y}^2}{{b}^2}=1$, a>0, b>0, and let O be the origin. Let M be an arbitrary point on curve C and above X-axis and H be a point on $MF_1$ such that $MF_2\perp{{F}}_1{{F}}_2$, $MF_1\perp{{O}}{{H}}$, $|OH|=\lambda |OF_2|$ with $\lambda \in(2/5, 3/5)$, then the range of the eccentricity $e$ is





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Solution


nimcet PYQ 2025
A circle with its center in the first quadrant touches both the coordinate axes and the line x-y-2=0. Then the area of the circle is





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Solution

A circle touching both coordinate axes has center $(r, r)$ and radius $r$. 
It also touches the line $x - y - 2 = 0$. 
So the distance from $(r, r)$ to the line equals $r$: $\dfrac{|r - r - 2|}{\sqrt{1^{2} + (-1)^{2}}} = r$ 
$\dfrac{2}{\sqrt{2}} = r$ 
$r = \sqrt{2}$ 
Area of the circle: $\pi r^{2} = \pi(\sqrt{2})^{2} = 2\pi$

nimcet PYQ 2025
If $\alpha$ and $\beta$ are the two roots of the quadratic equation $x^2 + ax + b = 0, (ab \ne 0)$ then the quadratic roots whose roots $\frac{1}{\alpha^3+\alpha}$ and $\frac{1}{\beta^3+\beta}$ is





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Solution

Given $x^{2} + ax + b = 0$ with roots $\alpha,\beta$, 
we use $\alpha + \beta = -a$ and $\alpha\beta = b$. 
Required new roots are $\dfrac{1}{\alpha^{3}+\alpha}$ and $\dfrac{1}{\beta^{3}+\beta}$. 
Since $\alpha^{3}+\alpha = \alpha(\alpha^{2}+1)$ and using $\alpha^{2}=-a\alpha-b$ (and same for $\beta$), after simplification the sum and product of new roots become: 
$u+v = \dfrac{a^{3}+a-3ab}{b(b^{2}+1+a^{2}-2b)}$ 
$uv = \dfrac{1}{b(b^{2}+1+a^{2}-2b)}$ 
So the required quadratic is: 
$b(b^{2}+1+a^{2}-2b)x^{2} - (a^{3}+a-3ab)x + 1 = 0$

nimcet PYQ 2025
The maximum value of $\sin x+\sin(x+1)$ is $k \cos \frac{1}{2}$. Then the value of k is 





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Solution

We use the identity
 $\sin A + \sin B = 2 \sin \dfrac{A+B}{2} \cos \dfrac{A-B}{2}$. 
So, $\sin x + \sin(x+1) = 2 \sin\left(x+\dfrac12\right)\cos\dfrac12$. 
Maximum value of $\sin(x+\tfrac12)$ is $1$. 
Therefore maximum of the expression is: $2 \cos\dfrac12$. 
Given maximum $= k \cos\dfrac12$, so $k = 2$. 

nimcet PYQ 2025
Let $\mathbb{R}\rightarrow\mathbb{R}$ be any function defined as $f(x)=\begin{cases}{{x}^{\alpha}\sin \frac{1}{{x}^{\beta}}} & {,x\ne0} \\ {0} & {,x=0}\end{cases}$, $\alpha , \beta \in \mathbb{R}$. Which of the following is true? ($\mathbb{R}$ denotes the set of all real numbers)





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Solution

The function is $f(x) = x^{\alpha}\sin\left(\dfrac{1}{x^{\beta}}\right)$ for $x \ne 0$, and $f(0)=0$. 
To check continuity at $x=0$, consider: $\displaystyle \lim_{x\to 0} x^{\alpha}\sin\left(\frac{1}{x^{\beta}}\right)$. 
Since $|\sin \theta| \le 1$ for all $\theta$ 
$|x^{\alpha}\sin(1/x^{\beta})| \le |x^{\alpha}|$. 

Now, $\displaystyle \lim_{x\to 0} x^{\alpha} = 0 \quad \text{iff } \alpha > 0$. 
Therefore, $\displaystyle \lim_{x\to 0} x^{\alpha}\sin(1/x^{\beta}) = 0 = f(0)$ iff $\alpha > 0$. 
The value of $\beta$ does not matter because $\sin(1/x^{\beta})$ is always bounded. 
Hence: ${f(x)\text{ is continuous at }x=0\text{ for all }\alpha>0\text{ and }\beta\in\mathbb{R}.}$

nimcet PYQ 2025
The number of 3-digit integers that are multiple of 6 which can be formed by using the digits 1,2,3,4,5,6 without repetition is





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Solution

To be a multiple of $6$: 
• last digit must be even → ${2,4,6}$ 
• sum of digits must be divisible by $3$.
For each even last digit, exactly 8 valid choices of the first two digits satisfy divisibility by $3$ (checked by mod – 3 pairing). 
So total numbers: 3×8=24​

nimcet PYQ 2025
The circle $x^2 + y^2+ \alpha x+ \beta y+ \gamma=0$ is the image of the circle $x^2 + y^2- 6x- 10y+ 30=0$ across the line 3x + y = 2. The value of $[\alpha+ \beta+ \gamma]$ is (where [.] represents the floor function.)





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Solution

Given circle: 
$x^{2}+y^{2}-6x-10y+30=0$ 
Center and radius: $C(3,5), \quad r=2$ 

We reflect $C(3,5)$ across the line $3x + y - 2 = 0$. 
Compute reflection factor: $k = \dfrac{2(3\cdot 3 + 5 - 2)}{3^{2} + 1^{2}} = \dfrac{24}{10} = \dfrac{12}{5}$ 

Reflected center: 
$x' = 3 - 3k = 3 - \dfrac{36}{5} = -\dfrac{21}{5}$ 
$y' = 5 - k = 5 - \dfrac{12}{5} = \dfrac{13}{5}$ 
So new center is: 
$C'\left(-\dfrac{21}{5},\dfrac{13}{5}\right)$ 
The image circle is $x^{2}+y^{2}+\alpha x+\beta y+\gamma = 0$ 
Using center formula: $-\dfrac{\alpha}{2} = -\dfrac{21}{5} \Rightarrow \alpha = \dfrac{42}{5}$ 
$-\dfrac{\beta}{2} = \dfrac{13}{5} \Rightarrow \beta = -\dfrac{26}{5}$ 
Using radius $r=2$: $\left(-\dfrac{21}{5}\right)^{2} + \left(\dfrac{13}{5}\right)^{2} - \gamma = 4$ 
$\dfrac{610}{25} - \gamma = 4$ 
$\gamma = \dfrac{102}{5}$ 
Now compute: 
$\alpha + \beta + \gamma = \dfrac{42}{5} - \dfrac{26}{5} + \dfrac{102}{5} = \dfrac{118}{5} = 23.6$ 
Thus, $\boxed{23}$

nimcet PYQ 2025
Let $\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}$, $\vec{b}=\hat{i}+2\hat{j}-\hat{k}$ and $\vec{c}=3\hat{i}+\hat{j}+\lambda \hat{k}$ be the co-terminal edges
of a parallelopiped whose volume is 5 units. Then the value of $\lambda$ is





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Solution

Given $\vec a = \langle 2,-3,4\rangle,\; \vec b = \langle 1,2,-1\rangle,\; \vec c = \langle 3,1,\lambda\rangle$ 

 Volume condition: $|\vec a \cdot (\vec b \times \vec c)| = 5$ 

 Compute cross product using determinant: \[ \vec b \times \vec c = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 3 & 1 & \lambda \end{vmatrix} \] \[ = (2\lambda + 1)\mathbf{i} - (\lambda + 3)\mathbf{j} - 5\mathbf{k} \] So $\vec b \times \vec c = \langle 2\lambda + 1,\; -(\lambda + 3),\; -5\rangle$ 

 Scalar triple product: \[ \vec a \cdot (\vec b \times \vec c) = 2(2\lambda + 1) + (-3)(-(\lambda + 3)) + 4(-5) \] Simplifying: \[ = 4\lambda + 2 + 3(\lambda + 3) - 20 = 7\lambda - 9 \] Volume equation: \[ |7\lambda - 9| = 5 \] Solve: \[ 7\lambda - 9 = 5 \Rightarrow \lambda = 2 \] \[ 7\lambda - 9 = -5 \Rightarrow \lambda = \frac{4}{7} \] Final answer: \[ \boxed{\lambda = 2 \text{ or } \frac{4}{7}} \]

nimcet PYQ 2025
The area enclosed between the curve y = sin x, y = cosx, $0\leq x\leq\frac{\pi}{2}$ is





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nimcet PYQ 2025
Given the equation $x+y =1$, $x^2+y^2 =2$, $x^5 +y^5 =A$. Let N be the number of solution pairs (x,y) to this system of equations. Then AN is equal to





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Solution

From \(x+y=1\) and \(x^2+y^2=2\): $(x+y)^2=x^2+y^2+2xy$
\[ \;\Rightarrow\; 1=2+2xy \;\Rightarrow\; xy=-\tfrac12. \] Thus \(x,y\) are roots of \(t^2 - t - \tfrac12=0\), giving two ordered pairs \((x,y)\). Hence \(N=2\). Let \(p_n=x^n+y^n\). For quadratic roots with \(s_1=x+y=1\) and \(s_2=xy=-\tfrac12\), \[ p_n = s_1 p_{n-1} - s_2 p_{n-2} \quad (n\ge2), \] with \(p_0=2,\; p_1=1\). \[ p_2=1\cdot1-(-\tfrac12)\cdot2=2,\quad\] $$ p_3=1\cdot2-(-\tfrac12)\cdot1=\tfrac52,$$ \[ p_4=\tfrac72,\quad p_5=1\cdot\tfrac72-(-\tfrac12)\cdot\tfrac52=\tfrac{19}{4}. \] Therefore \(A=p_5=\tfrac{19}{4}\). With \(N=2\), \[ AN=\frac{19}{4}\cdot 2=\boxed{\tfrac{19}{2}}. \]
✅ Final Answer: \(AN=\dfrac{19}{2}\)

nimcet PYQ 2025
Number of three digit numbers that can be formed using 0, 1, 2, 3 and 5 where these digits are allowed to repeat any number of times, is equal to





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Solution

Digits available: 0, 1, 2, 3, 5 (total 5 digits). 
Repetition is allowed. 
 To form a three-digit number: 
 1) The first digit cannot be 0. 
 Possible choices = 1, 2, 3, 5 → 4 choices. 
 2) The second digit can be any of the 5 digits → 5 choices. 
 3) The third digit can also be any of the 5 digits → 5 choices. 

 Total three-digit numbers = $4 \times 5 \times 5 = 100$. 
 Final Answer: 100

nimcet PYQ 2025
Let $g:\mathbb{R}\rightarrow \mathbb{R}$ and $h:\mathbb{R}\rightarrow \mathbb{R}$, be two functions such that $h(x) = sgn(g(x))$. Then select which of the following is not true?( $\mathbb{R}$ denotes the set of all real numbers, sgn stands for signum function)





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Solution

Let $g:\mathbb{R}\to\mathbb{R}$ and $h:\mathbb{R}\to\mathbb{R}$ be such that $h(x) = \operatorname{sgn}(g(x))$. 
 Recall: $\operatorname{sgn}(t) = \begin{cases} 1, & t>0\\ 0, & t=0\\ -1, & t<0 \end{cases}$ 
 Check each statement: 

 1) "The domain of $h(x)$ is the same as the domain of $g(x)$." $\Rightarrow$ True, because $\operatorname{sgn}(g(x))$ is defined for every $x$ where $g(x)$ is defined. 

 2) "The domain of continuity of $h(x)$ equals the domain of continuity of $g(x) - \{x\in\mathbb{R} : g(x)=0\}$." 
 At points where $g(x)\neq 0$, $h(x)$ is locally constant ($1$ or $-1$), hence continuous there (provided $g$ itself is continuous). 
 At points where $g(x)=0$, $h(x)$ jumps from $-1$ to $1$, so it is discontinuous. $\Rightarrow$ 
This statement is true. 

 3) "The domain of $h(x)$ is different from the domain of $g(x)$ at the same point." 
 Since for every $x$ in the domain of $g$, $h(x)=\operatorname{sgn}(g(x))$ is defined, the domains are exactly the same; they never differ. 
 $\Rightarrow$ This statement is false. 

 4) " $h(x)$ is discontinuous at $g(x)=0$." 
 At any $x_0$ where $g(x_0)=0$, the left and right limits of $h(x)$ are $-1$ and $1$, not equal to $h(x_0)=0$. 
 $\Rightarrow$ $h$ is discontinuous there, so this statement is true. 
 Therefore, the statement which is **not true** is: $\boxed{\text{Option 3}}$

nimcet PYQ 2025
An airplane, when 4000m high from the ground, passes vertically above another airplane at an instant when the angles of elevation of the two airplanes from the same point on the ground are 60° and 30°, respectively. Find the vertical distance between the two airplanes.





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Solution

Let the higher airplane be at a height of $4000\,\text{m}$. 
From a point on the ground, the angles of elevation to the two airplanes are $60^\circ$ (upper plane) and $30^\circ$ (lower plane). 

Let the horizontal distance from the observer to the airplanes be $x$. 

For the upper airplane: 
$\tan 60^\circ = \dfrac{4000}{x}$ 
$\sqrt{3} = \dfrac{4000}{x}$ 
$\Rightarrow x = \dfrac{4000}{\sqrt{3}}$. 
 For the lower airplane with height $h$: 
$\tan 30^\circ = \dfrac{h}{x}$ 
$\dfrac{1}{\sqrt{3}} = \dfrac{h}{4000/\sqrt{3}}$ 
 Thus, $h = \dfrac{4000}{3}$. 
 Now the vertical distance between the two airplanes: $4000 - \dfrac{4000}{3} = \dfrac{8000}{3}.$ 
 Final Answer: $\displaystyle \frac{8000}{3}\text{ m}$.

nimcet PYQ 2025
Suppose $t_1, t_2, ...t_5$ are in AP such that $\sum ^{18}_{l=0}{{t}}_{3l+1}=1197$ and ${{t}}_7+{{3}}t_{22}=174$. If $\sum ^9_{l=1}{{{t}}_l}^2=947b$, then the value of $b$ is





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Solution

Let first term $= a$, common difference $= d$, so $t_n = a + (n-1)d$

Step 1: Expand $\displaystyle\sum_{l=0}^{18} t_{3l+1} = 1197$
Terms: $t_1, t_4, t_7, \ldots, t_{55}$ (19 terms, $l = 0$ to $18$)

$t_{3l+1} = a + 3ld$

$\displaystyle\sum_{l=0}^{18}(a + 3ld) $
$= 19a + 3d\cdot\dfrac{18 \times 19}{2} $
$= 19a + 513d = 1197$

$\Rightarrow a + 27d = 63 \quad \cdots(1)$

Step 2: Use $t_7 + 3t_{22} = 174$
$t_7 = a + 6d,\quad t_{22} = a + 21d$

$(a + 6d) + 3(a + 21d) = 174$
$4a + 69d = 174 \quad \cdots(2)$

Step 3: Solve (1) and (2)
From (1): $a = 63 - 27d$

Substitute in (2):
$4(63 - 27d) + 69d = 174$
$252 - 108d + 69d = 174$
$-39d = -78$
$d = 2$

$a = 63 - 54 = 9$

Step 4: Find $\displaystyle\sum_{l=1}^{9} t_l^2$
$t_l = 9 + (l-1) \times 2 = 7 + 2l$

$\displaystyle\sum_{l=1}^{9} t_l^2 = \sum_{l=1}^{9}(7+2l)^2 = \sum_{l=1}^{9}(49 + 28l + 4l^2)$

$= 49(9) + 28\cdot\dfrac{9 \times 10}{2} + 4\cdot\dfrac{9 \times 10 \times 19}{6}$

$= 441 + 1260 + 1140$

$= 2841$

Step 5: Find $b$
$947b = 2841$

$b = \dfrac{2841}{947} = 3$

Answer: $b = \boxed{3}$

nimcet PYQ 2025
Let E and F be two events such that P(E) > 0 and P(F) > 0. Which one of the following is NOT equivalent to the condition that $P(E) =P(E|F)$?





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Solution

We are given the condition: $P(E) = P(E|F)$. 
 Since $P(E|F) = \dfrac{P(E \cap F)}{P(F)}$, the condition becomes: $P(E)P(F) = P(E \cap F)$, which is exactly the definition of independence of $E$ and $F$. 
Now check each option:
1) "E and F are independent" → This is exactly equivalent to $P(E)=P(E|F)$ (TRUE). 

 3) $P(F) = P(F|E)$ → Also true under independence (TRUE). 

 4) $E^c$ and $F$ are independent → Independence is preserved under complements (TRUE). 

 2) $2P(E^c)P(F^c) \ne P(E \cap F^c)$ → This statement has no relation to $P(E)=P(E|F)$ and does NOT follow from independence (NOT equivalent). 

Therefore, the option that is NOT equivalent is: Option 2.

nimcet PYQ 2025
What is the general solution of the equation $\tan \theta + \cot \theta = 2$ ?





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Solution

Given equation: $ \tan\theta + \cot\theta = 2 $. 

Rewrite $\cot\theta$: 
$ \tan\theta + \dfrac{1}{\tan\theta} = 2 $. 
 Let $t = \tan\theta$. 
Then: $ t + \dfrac{1}{t} = 2 $. 
 Multiply by $t$: 
$ t^2 + 1 = 2t $. 
 Rearrange: 
$ t^2 - 2t + 1 = 0 $. 
 $ (t - 1)^2 = 0 $. 
 So: $ t = 1 $. 
 Thus: $ \tan\theta = 1 $. 
 General solution: $ \theta = \dfrac{\pi}{4} + n\pi,\; n \in \mathbb{Z}. $

nimcet PYQ 2025
If $\cos^2(10°)\cos(20°)\cos(40°)\cos(50°) \cos(70°) = \alpha+\frac{\sqrt{3}}{16} \cos(10°)$, then $3\alpha^{-1}$ is equal to





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nimcet PYQ 2025
The slope of the normal line to the curve $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$ at the point (2,-1) is





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Solution

We have the parametric curve: $$ x = t^2 + 3t - 8,\qquad y = 2t^2 - 2t - 5. $$ The point $(2, -1)$ lies on the curve. Find the value of $t$: Solve $$ t^2 + 3t - 8 = 2. $$ So, $$ t^2 + 3t - 10 = 0. $$ Factor: $$ (t+5)(t-2)=0 \quad\Rightarrow\quad t=2 \text{ or } t=-5. $$ Check which gives $y=-1$: For $t=2$: $$ y = 2(2)^2 - 2(2) - 5 = 8 - 4 - 5 = -1. $$ So the correct parameter is: $$ t = 2. $$ Now compute slope of tangent: $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}. $$ Compute derivatives: $$ \frac{dx}{dt} = 2t + 3,\qquad \frac{dy}{dt} = 4t - 2. $$ Thus: $$ \frac{dy}{dx} = \frac{4t - 2}{2t + 3}. $$ Substitute $t=2$: $$ \frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}. $$ Slope of normal $m_N$ is negative reciprocal of tangent slope: $$ m_N = -\frac{1}{\frac{6}{7}} = -\frac{7}{6}. $$ Final Answer: $\displaystyle -\frac{7}{6}$

nimcet PYQ 2025
A tower subtends an angle of 30° at a point on the same level as the foot of the tower. At a second point h meters above the first, the depression of the foot of the tower is 60°. What is the horizontal distance of the tower from the point?





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nimcet PYQ 2025
There are 40 female and 20 male students in a class. If the average heights of female and male students are 5.15 feet and 5.66 feet, respectively, then the average height (in feet) of all the students in the class equals





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Solution

Given: Females = 40 (avg = 5.15 ft), Males = 20 (avg = 5.66 ft). Total students = 60.

Weighted average height = $$\frac{40\times 5.15 + 20\times 5.66}{40+20}$$

Compute: $$\frac{206 + 113.2}{60}=\frac{319.2}{60}=5.32$$

Answer: 5.32 feet


nimcet PYQ 2025
Let $x$ be a positive real number such that $x^{(8\log_5x-24)}=5^{-4}$. Then the product of all possible values of x is =





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Solution

We have $ x^{8\log_5 x - 24} = 5^{-4} $ 
Let $ \log_5 x = t $. 
Then $ x = 5^t $. 
Substitute: $ (5^t)^{8t - 24} = 5^{-4} $ 
$ 5^{t(8t - 24)} = 5^{-4} $ 
So: $ t(8t - 24) = -4 $ 
$ 8t^2 - 24t + 4 = 0 $ 
Divide by 4: 
$ 2t^2 - 6t + 1 = 0 $ 
Solutions: $ t = \frac{3 \pm \sqrt{7}}{2} $ 
Thus: $ x_1 = 5^{\frac{3 + \sqrt{7}}{2}},\ x_2 = 5^{\frac{3 - \sqrt{7}}{2}} $ 
Product: $ x_1 x_2 = 5^{\frac{3 + \sqrt{7}}{2} + \frac{3 - \sqrt{7}}{2}} = 5^{\frac{6}{2}} = 5^3 = 125 $

nimcet PYQ 2025
The obtuse angle between lines 2y = x + 1 and y = 3x + 2 is





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Solution

Given lines: 
$2y = x + 1$ and $y = 3x + 2$. 
 From $2y = x + 1$ 
we get $y = \dfrac{1}{2}x + \dfrac{1}{2}$, 
so slope $m_1 = \dfrac{1}{2}$. 

 From $y = 3x + 2$, 
 slope $m_2 = 3$. 

 Angle between two lines: $ \tan\theta = \left|\dfrac{m_2 - m_1}{1 + m_1 m_2}\right| $ 
 $ \tan\theta = \left|\dfrac{3 - \dfrac{1}{2}}{1 + \dfrac{1}{2}\cdot 3}\right| = \dfrac{\dfrac{5}{2}}{\dfrac{5}{2}} = 1 $ 
So $\theta = 45^\circ$ or $135^\circ$. 
Required obtuse angle $= 135^\circ$.

nimcet PYQ 2025
What is the value of $\lim _{{x}\rightarrow\infty}-(x+1)\Bigg{(}{e}^{\frac{1}{x+1}}-1\Bigg{)}$?





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Solution

We want to evaluate $ \displaystyle \lim_{x \to \infty} -(x+1)\left(e^{\frac{1}{x+1}} - 1\right). $ 

Rewrite it as a product: $ -(x+1)\left(e^{\frac{1}{x+1}} - 1\right) = -\dfrac{e^{\frac{1}{x+1}} - 1}{\frac{1}{x+1}}. $ 

 Now let $ t = \frac{1}{x+1} \Rightarrow t \to 0^+ $ as $ x \to \infty $. 

The expression becomes: $ -\dfrac{e^{t} - 1}{t}. $ 
Now apply L'Hospital’s Rule to the limit: $ \displaystyle \lim_{t \to 0} \dfrac{e^{t} - 1}{t} = \lim_{t \to 0} \dfrac{e^{t}}{1} = 1. $ 
 So the original limit is: $ -1. $ 
 Final Answer: $ -1 $.

nimcet PYQ 2025
The value of $\int ^{\frac{\pi}{2}}_0\frac{(1+2\cos x)}{({2+\cos x)}^2}dx$ lies in the interval





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Solution

We need to evaluate $ \displaystyle \int_{0}^{\pi/2} \frac{1 + 2\cos x}{(2 + \cos x)^2}\, dx $. 
Let $ t = 2 + \cos x $. Then $ dt = -\sin x\, dx $. 
But the integral has no $\sin x$, so rewrite numerator: 
 $ 1 + 2\cos x = (2 + \cos x) - 1 = t - 1 $. 
 Now express $ dx $ using $ \sin^2 x = 1 - \cos^2 x $, but the standard trick is to differentiate: 
 $ \dfrac{d}{dx}\left(\dfrac{1}{2+\cos x}\right) = -\dfrac{-\sin x}{(2+\cos x)^2} = \dfrac{\sin x}{(2+\cos x)^2}. $ We 
use complementary substitution: 
 Let $ x = \frac{\pi}{2} - y $. 
 Then $\cos x = \sin y$ and $\sin x = \cos y$. 
 Integral becomes: 
 $ I = \int_{0}^{\pi/2} \frac{1 + 2\sin y}{(2 + \sin y)^2}\, dy. $ 
 Average the two forms: 
 $ I = \frac{1}{2}\int_{0}^{\pi/2} \left[ \frac{1 + 2\cos x}{(2 + \cos x)^2} + \frac{1 + 2\sin x}{(2 + \sin x)^2} \right] dx. $ 
 Now observe identity: 
 $ \frac{1 + 2\cos x}{(2 + \cos x)^2} + \frac{1 + 2\sin x}{(2 + \sin x)^2} = \frac{d}{dx}\left(\frac{\sin x - \cos x}{(2+\cos x)(2+\sin x)}\right). $ 
 Thus integral becomes a telescoping form and evaluates to: 
 $ I = \left[ \frac{\sin x - \cos x}{(2+\cos x)(2+\sin x)} \right]_{0}^{\pi/2}. $ 
 Now compute: 
 At $ x = \frac{\pi}{2}$: $ \sin x = 1,\;\cos x = 0 $ 
 Expression = $ \dfrac{1 - 0}{(2+0)(2+1)} = \dfrac{1}{6}. $ 
 At $ x = 0$: $ \sin 0 = 0,\;\cos 0 = 1 $ 
 Expression = $ \dfrac{0 - 1}{(2+1)(2+0)} = -\dfrac{1}{6}. $ 
 Therefore: $ I = \dfrac{1}{6} - (-\dfrac{1}{6}) = \dfrac{2}{6} = \dfrac{1}{3}. $ 
 Final Answer: $\dfrac{1}{3} $

nimcet PYQ 2025
Read definition and all four choices carefully, and find the answer that provides the best example of the given definition. Answer the question solely on the basis of the definition given.
Definition: An informal gathering occurs when a group of people get together in a casual, relaxed manner.

Which situation below is the best example of an informal gathering?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Definition: An informal gathering occurs when a group of people get together in a casual, relaxed manner.

Check each option:

1) "After finding about his promotion, Jeremy and a few coworkers decide to go out for a quick drink after work."  
- Small group of people ✔  
- Casual and relaxed ✔  
- Decided spontaneously, not formal ✔  
→ This fits the definition of an informal gathering.

2) "Whenever she eats at the Italian restaurant, Kamalika seems to run into Peter."  
- This is a chance meeting, not a planned get-together ✖  
→ Not an informal gathering.

3) "The book club meets on the first Thursday evening of every month."  
- Regularly scheduled and organized ✖  
→ More formal, not purely informal.

4) "Mary sends out 25 invitations for the bridal shower she is giving for her sister."  
- Planned event with invitations, quite formal ✖  
→ Not an informal gathering.

Therefore, the best example of an informal gathering is <b>Option 1</b>.

nimcet PYQ 2025
In a certain code language, the word "REASON" is encoded as TDDQSK. Using the same rule, how is the word PRAYER coded?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Given: "REASON" is coded as "TDDQSK". 
 Write positions of letters (A=1, B=2, ..., Z=26): 
 R(18) → T(20): +2 
 E(5) → D(4): −1 
 A(1) → D(4): +3 
 S(19) → Q(17): −2 
 O(15) → S(19): +4 
 N(14) → K(11): −3 
 So the rule is: +2, −1, +3, −2, +4, −3 (alternating + and −, with magnitude increasing). 
 Now apply same rule to "PRAYER": 
 P(16) +2 → R(18) 
 R(18) −1 → Q(17) 
 A(1) +3 → D(4) 
 Y(25) −2 → W(23) 
 E(5) +4 → I(9)
 R(18) −3 → O(15) 
 Hence, "PRAYER" is coded as: RQDWIO.

nimcet PYQ 2025
Four people (P,Q,R & S) are standing in a queue. Q is ahead of P. R is between P and S. If S is at the end of the queue, who is at the front of the queue?





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Solution

S is at the end 
→ order is _ _ _ S 
R is between P and S 
→ the order of those three from front to back must be P, R, S. 

Q is ahead of P, and only four people are there 
→ Q must stand before P. 
 So full order (front → end): 
Q, P, R, S, 
hence Q is at the front.

nimcet PYQ 2025
Statements:
I. The meteorological department predicts deficient rainfall during monsoon season.

II. The GDP growth estimates have been revised downwards by the government.

Choose the correct option:





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

We are given two statements:

Statement I:
The meteorological department predicts deficient rainfall during monsoon season.
→ This is related to weather/climate.

Statement II:
The GDP growth estimates have been revised downwards by the government.
→ This is related to economic performance.

Now check each possibility:


1. Is Statement I the cause and Statement II the effect?

No.
A rainfall prediction cannot cause the government to revise GDP growth estimates directly.
GDP revisions depend on multiple economic indicators, not weather predictions.

So cause–effect relationship is not possible.


2. Are both statements effects of a common cause?

No.

  • Rainfall prediction is an effect of meteorological data.

  • GDP revision is an effect of economic and financial data.

These two sources of information are completely different → no common cause.


3. Are they effects of independent causes?

Yes.

  • Statement I is caused by climate analysis.

  • Statement II is caused by economic analysis.

Each comes from different, unrelated causes.


Correct Answer:

Both the statements I and II are effects of independent causes.


nimcet PYQ 2025
A monkey climbs 30 feet at the beginning of each hour and rests for a while when he slips back 20 feet before he again starts climbing in the beginning of the next hour. If he begins his ascent at 8.00 am., at what time will he first touch a flag at 120 feet from the ground?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

The monkey climbs 30 ft every hour and slips 20 ft after each climb.

Effective gain per hour = 30 - 20 = 10 ft.
So after 9 hours (till 5:00 pm): 9 × 10 = 90 ft.
In the 10th climb, it will reach the remaining $30$ ft and touch 120 ft in that hour.

Final Answer: The monkey first touches the flag at 6:00 pm.

nimcet PYQ 2025
In a remote research outpost, five scientists (Dr. Anand, Dr. Beena, Dr. Chirag, Dr. Dinesh, and Dr. Elango) are studying a rare phenomenon. They communicate through a secure, text-based system. Over the course of an hour, the following exchanges occur: 
  1. Dr. Anand sends a message to Dr. Beena and Dr. Chirag. 
  2. Dr. Beena replies only to Dr. Anand. 
  3. Dr. Anand sends a reminder message to who did not reply. 
  4. Dr. Chirag sends a message to Dr. Dinesh and Dr. Elango. 
  5. While replying to Dr. Chirag, Dr. Dinesh accidentally sends a "Reply All" 
  6. Dr. Elango, having received Dr. Chirag's message, sends a message to Dr. Beena.
At the end of this hour, who has received the most messages directly from other scientists within the group? (Consider only messages sent within the described hour and only between the five scientists mentioned).





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Dr. Chirag received the most.

Quick count of messages received:

  • B: from A (initial) + from E = 2

  • C: from A (initial) + from A (reminder) + from D (reply-all) = 3

  • D: from C = 1

  • E: from C + from D (reply-all) = 2

  • A: from B = 1

Highest is 3 → Dr. Chirag.


nimcet PYQ 2025
The percentage change in area of a rectangle by decreasing its length and increasing its breadth by 5% is





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Assume: Length = 100, Breadth = 100

Original Area = 100 × 100 = 10,000


New length = 95 (5% decrease)
New breadth = 105 (5% increase)

New Area = 95 × 105 = 9975


Change = 9975 – 10000 = –25

Percentage change = (–25 ÷ 10000) × 100 = –0.25%


✅ Area decreases by 0.25%


nimcet PYQ 2025
In a certain code language, 'SUN' is written as 'RWK' and 'MOON' is written as 'LQLR'. Following the same pattern, how will 'STAR' be written?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Pattern: −1, +2, −3, +4 (with wraparound).
S→R (−1), T→V (+2), A→X (−3), R→V (+4).

Answer: RVXV

nimcet PYQ 2025
If a number is increased by 30% and then decreased by 20%, what is the net change?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Let the original number be $100$.

Increased by $30\%$ ⇒ New value = $100 + 30 = 130$.
Decreased by $20\%$ ⇒ New value = $130 - 20\% \text{ of } 130 = 130 - 26 = 104$.

Net change = $104 - 100 = +4$.
Percentage change = $\dfrac{4}{100} \times 100 = 4\%$.

Final Answer: Net increase of 4%

nimcet PYQ 2025
A two digit number is such that the product of its digits is 12. If 36 is added to the number, the digits interchange their places. What is the number?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Let the two-digit number be 10x + y, where x = tens digit, y = units digit.


Step 1: Product condition

x × y = 12 … (1)

Step 2: Interchange condition

If digits interchange, new number = 10y + x

Given: (10x + y) + 36 = (10y + x)

⇒ 9x – 9y = –36

⇒ x – y = –4 … (2)


Step 3: Solve equations

From (2): x = y – 4

Put in (1): (y – 4) × y = 12

⇒ y² – 4y – 12 = 0

⇒ (y – 6)(y + 2) = 0

⇒ y = 6 (since digit can’t be –2)

Then x = y – 4 = 2


Number = 10x + y = 10×2 + 6 = 26


✅ Final Answer: The number is 26.


nimcet PYQ 2025
At an IIM entrepreneurship summit, two young founders - Karan and Deepak - introduced their startup. In their quirky opening, they said: "The product of our ages is 240. And just like in our startup strategy, twice Deepak's age is 4 years more than Karan's age." How old was Deepak two years ago?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Let Karan’s present age = $K$ years
Let Deepak’s present age = $D$ years

Given:
$K \times D = 240$
and $2D = K + 4 \Rightarrow K = 2D - 4$

Substitute in the first equation:
$(2D - 4)D = 240$
$\Rightarrow 2D^2 - 4D - 240 = 0$
$\Rightarrow D^2 - 2D - 120 = 0$

Solving: $D = \dfrac{2 + \sqrt{484}}{2} = \dfrac{2 + 22}{2} = 12$

Deepak’s present age = $12$ years
Two years ago = $12 - 2 = \mathbf{10}$ years

Final Answer: $10$ years

nimcet PYQ 2025
Average Cost (AC) is defined as the sum of Average Fixed Cost (AFC) and Average Variable Cost (AVC), and dumping is defined as selling a product at a price less than AC but more than AVC. A company in India, suddenly, found that the demand for its product "ZOOM" has fallen to 60% of the output produced in that financial year. As a result, the company must sell 40% of the produced output in a foreign market. If it decides to "dump" 40% of its output in a neighbouring country (by reducing the price by 20%), what would be the objective of its 'dumping strategy', among the following? (The company has set a profit margin of 10% of AC while fixing the price of its product for sale in India).
(a) To minimize losses,
(b) to maximize profits,
(c) to contribute to the recovery of fixed costs,
(d) to contribute to the recovery of variable costs





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Solution

  • Selling at a “dumping” price means the product is sold below AC but above AVC.

  • So, the firm covers all variable costs and earns something to help recover fixed costs.

  • This doesn’t maximize profit, it just minimizes loss and helps recover fixed costs.

Therefore, both
(a) To minimize losses and
(c) To contribute to the recovery of fixed costs
are correct.


nimcet PYQ 2025
What is the next number in each of the following 3 sequences?
8    17    33     67     133     ?    (1)
7    23    67     203   607     ?    (2)
6    27    105   423   1689   ?    (3)





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

(1) 8 → 17 → 33 → 67 → 133 → ?
Pattern: ×2 +1, ×2 −1, ×2 +1, ×2 −1, …
Next: 133×2 + 1 = 267

(2) 7 → 23 → 67 → 203 → 607 → ?
Pattern: ×3 +2, ×3 −2, ×3 +2, ×3 −2, …
Next: 607×3 + 2 = 1823

(3) 6 → 27 → 105 → 423 → 1689 → ?
Pattern: ×4 +3, ×4 −3, ×4 +3, ×4 −3, …
Next: 1689×4 + 3 = 6759

Answers: (1) 267, (2) 1823, (3) 6759

nimcet PYQ 2025
Which number replaces the question mark?
3, 6, 18, 72, ?, 432





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Sequence: 3, 6, 18, 72, ?, 432

Pattern (symmetric multipliers): ×2, ×3, ×4, ×3, ×2
3 → 6 : ×2
6 → 18 : ×3
18 → 72: ×4
72 → ? : ×3 → 72 × 3 = 216
216 → 432: ×2 ✅

Answer: 216

nimcet PYQ 2025
A remote island has a unique social structure. Individuals are either 'Truth-tellers' (who always speak the truth) or 'Tricksters' (who always lie). You encounter three inhabitants: X, Y, and Z.

X says: "Y is a Trickster."
Y says: "Exactly one of us is a Truth-teller."
What can you definitively conclude about Z?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Let me quickly re-check the logic:

  1. Let T = Truth-teller, L = Trickster (liar).

X says: “Y is a Trickster.”
Y says: “Exactly one of us (X, Y, Z) is a Truth-teller.”


Case 1: Assume X is T.
Then Y must be L.
Y’s statement must be false: “Exactly one of us is a Truth-teller” is false.

  • Truth-tellers so far: X (1).
    If Z were L, then exactly one (X) would be T → Y’s statement would be true, contradiction.
    So Z must be T.
    Configuration: (X T, Y L, Z T) works.

Case 2: Assume X is L.
Then “Y is a Trickster” is false → Y is T.
Now Y’s statement must be true: exactly one of X, Y, Z is T.

  • We already have Y as T, so X and Z must both be L.
    Configuration: (X L, Y T, Z L) works.


Both configurations are consistent:

  • In one, Z is T.

  • In the other, Z is L.

So Z can be either a Truth-teller or a Trickster


nimcet PYQ 2025
Ten years ago, Ram was twice as old as Shyam. Ten years from now, the ratio of their ages will be 3:2. What is Ram's present age?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Let Ram’s present age = $R$ years
Let Shyam’s present age = $S$ years

Ten years ago: $R - 10 = 2(S - 10)$
⇒ $R = 2S - 10$

Ten years from now: $\dfrac{R + 10}{S + 10} = \dfrac{3}{2}$
⇒ $2R + 20 = 3S + 30$
⇒ $2R = 3S + 10$

Substitute $R = 2S - 10$ into it:
$2(2S - 10) = 3S + 10$
⇒ $4S - 20 = 3S + 10$
⇒ $S = 30$

Then $R = 2(30) - 10 = 50$

Final Answer: Ram’s present age = 50 years

nimcet PYQ 2025
Consider the following statements followed by two conclusions. 
Statements: 
1. Some men are great. 
2. Some men are wise 
Conclusions: 
1. Men are either great or wise. 
2. Some men are neither great nor wise.





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Answer: Neither conclusion follows.

Reasoning:
From “Some men are great” and “Some men are wise,” we only know there exists at least one great man and at least one wise man (they may be the same or different).

  • Conclusion 1 (“All men are great or wise”) is not guaranteed—there could be men who are neither.

  • Conclusion 2 (“Some men are neither great nor wise”) is also not guaranteed—it's possible every man is in great or wise (or both).


nimcet PYQ 2025
Complete the next series:
2      8       12
8      40     45
40   240    246
....    ....      ....





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Each row uses a multiplier that increases by 1: 4, 5, 6, 7.
Middle = Left × multiplier; 
Right = Middle + multiplier.
So for row 4: 
Left = previous row’s 
Middle = 240 → Middle = 240×7 = 1680 
Right = 1680+7 = 1687.

nimcet PYQ 2025
Consider the following arguments
"In order to be a teacher one must graduate from college. All poets are poor. Some Mathematicians are poets. No college graduate is poor". Which one of the following is not a valid conclusion regarding the above
argument?

Note: In NIMCET 2025 Original Exam given line was missing.
"Which one of the following is not a valid conclusion regarding the above argument?"





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Answer: (1) Some teachers are not Mathematicians — NOT valid.

Easy check:

  • Teacher ⇒ College Graduate, and no graduate is poorTeachers are not poor → (4) valid.

  • All Poets are poor; graduates aren’t poor ⇒ Poets aren’t teachers → (3) valid.

  • Some Mathematicians are Poets ⇒ they’re poor ⇒ not graduates ⇒ not teachers → (2) valid.

  • (1) doesn’t follow: we’re told nothing about teachers vs mathematicians (and there might even be no teachers). Hence (1) is not a valid conclusion.


nimcet PYQ 2025
A, B, C, D, E, F and G are travelling in three different vehicles: Swift, Creta, and Nexon. There are at least two passengers in each vehicle. Among them, only two are male. There are two engineers, two doctors and three teachers.
  1. C is a lady doctor and she does not travel with A and F, who are sisters.
  2. B, a male engineer, travels with only G, a teacher, in a Swift.
  3. D is a male doctor.
  4. Two persons belonging to the same profession do not travel in the same vehicle.
  5. A is not an engineer and travels in a Creta.
  6. The pair of sisters A and F travels in the same vehicle
What is the profession of F?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

PersonEngineerDoctorTeacher
A
B
C
D
E
F
G
PersonSwiftCretaNexon
A
B
C
D
E
F
G
  • B is an engineer; doctors are C (lady) and D (male) ⇒ only one engineer left among A/E/F/G.

  • A is not an engineer and is a teacher; G is a teacher; we need 3 teachers total ⇒ E can be the 3rd teacher.

  • A and F sit together (in Creta), and no two same professions in one vehicle. Since A is a teacher, F cannot be a teacher; and doctors are already C & D.

  • Hence, F must be the second engineer.


nimcet PYQ 2025
In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Red = 8
Blue = 7
Green = 6
Total balls = 8 + 7 + 6 = 21

Probability (neither red nor green) = Probability (blue)
= Number of blue balls / Total balls
= 7 / 21
= 1 / 3

Final Answer: 1/3

nimcet PYQ 2025
Five friends A,B,C,D and E are sitting in a row. The following conditions hold:
A is sitting next to B.
C is sitting at one of the ends.
D is sitting between A and E.
E is sitting to the right of D.
Who is sitting at the second position?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Given:
A, B, C, D, E are sitting in a row.
1️⃣ A is next to B.
2️⃣ C is at one end.
3️⃣ D is between A and E.
4️⃣ E is to the right of D.

Possible arrangements:
(i) B – A – D – E – C ✅
(ii) C – B – A – D – E ✅

Second position:
➤ In (i): A is 2nd.
➤ In (ii): B is 2nd.

✅ Final Answer: Cannot be determined (could be A or B).

nimcet PYQ 2025
Statements:
I. All cakes are ice cream
II. No ice cream is chocolate

Conclusions:
I. Some cakes are chocolate
II. No cake is chocolateChoose the correct option:





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Only Conclusion II follows. 
Reason: All cakes ⊆ ice cream and no ice cream is chocolate ⇒ no cake is chocolate. 

So I is false, II is true.

nimcet PYQ 2025
If 1 man or 2 women can finish a job in 20 days, how may days will 3 men and women take to finish the same job?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

$1$ man or $2$ women can finish the work in $20$ days.

Work by $1$ man in $1$ day $= \dfrac{1}{20}$
Work by $1$ woman in $1$ day $= \dfrac{1}{40}$

For $3$ men and $1$ woman:
Total work per day $= 3 \times \dfrac{1}{20} + 1 \times \dfrac{1}{40}$
$= \dfrac{6}{40} + \dfrac{1}{40} = \dfrac{7}{40}$

Time to finish the job $= \dfrac{1}{(7/40)} = \dfrac{40}{7}$ days
$\approx 5.71$ days

Final Answer: $ \dfrac{40}{7} $ days

nimcet PYQ 2025
If 1st January 2022 was a Saturday, what day of the week will 1st January 2023 be?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Given: 1st January 2022 was a Saturday.

Year 2022 is not a leap year → it has $365 = 52$ weeks $+ 1$ extra day.

So, the same date next year moves forward by one weekday:
Saturday → Sunday.

Final Answer: 1st January 2023 = Sunday

nimcet PYQ 2025
What comes next in the sequence?

2, 6, 12, 20, 30,





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Sequence: 2, 6, 12, 20, 30

Differences: 4, 6, 8, 10
Next difference = 12

Next term = 30 + 12 = 42

nimcet PYQ 2025
Statements: 
All apples are fruits. 
All fruits are tasty. 

Conclusions: 
I. All apples are tasty. 
II. Some tasty things are apples.





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Answer: Both conclusions follow.

Why:
From “All apples are fruits” and “All fruits are tasty,” it follows that all apples are tasty (I).
And from “All apples are tasty,” we can validly infer the particular statement some tasty things are apples (II), via valid conversion of a universal affirmative (“All A are B ⟹ Some B are A”).


nimcet PYQ 2025
Which word in the following is the "odd man" out?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Odd man out: 2) Pampering 
Why: Promoting, Nurturing, and Incubating all involve fostering growth/development. Pampering means indulging/spoiling, not necessarily aiding growth.

nimcet PYQ 2025
If John's mother's father's granddaughter is Maria, what is Maria's relation to John?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Cannot be determined uniquely.

John’s mother’s father = John’s maternal grandfather.
His granddaughter could be:

  • John’s sister (his mother’s daughter), or

  • John’s maternal cousin (daughter of his mother’s sibling).

So Maria is either John’s sister or his maternal cousin; the statement doesn’t fix which.


nimcet PYQ 2025
A person walks 8 km north, turns right, walks 4 km, turns right again and walks 6 km. Finally, he turns left and walks 3 km. How far is he from the starting point?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

A person walks $8$ km north. 
 Position: $(0,\,8)$ 
 He turns right (east) and walks $4$ km. 
 Position: $(4,\,8)$ 
 He turns right again (south) and walks $6$ km. 
 Position: $(4,\,2)$ 
 He turns left (east) and walks $3$ km. 
 Final position: $(7,\,2)$ 
 Distance from starting point: 
 $ d = \sqrt{(7-0)^2 + (2-0)^2} $ 
 $ d = \sqrt{49 + 4} $ 
 $ d = \sqrt{53} $ 
 $d=7.28$

nimcet PYQ 2025
Mohan introduced Sarita as the mother of his sister's brother. How is Sarita related to Mohan?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Mohan’s sister’s brother = Mohan himself. 
Sarita is introduced as the mother of Mohan, 
So: Sarita is Mohan’s mother.

nimcet PYQ 2025
Which number replaces the question mark?
4, 9, 16, 25, ?, 49





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

22 = 4
32 = 9
42 = 16
52 = 25
62 = 36
72 = 49
Answer = 36

nimcet PYQ 2025
Consider a sequence formed by concatenating the positive integers in increasing order: 12345678910111213 ... Which among the following is the closest to the 2028th digit in this infinite sequence?





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

1-digit numbers: 1 to 9 → 9 digits
2-digit numbers: 10 to 99 → 90 × 2 = 180 digits
Total till 99 → 9 + 180 = 189 digits
Remaining → 2028 − 189 = 1839 digits (in 3-digit range)
Each 3-digit number has 3 digits → 1839 ÷ 3 = 613 numbers (approx.)
Last full number before 2028th digit → 100 + 612 = 712
712 → digits are 7, 1, 2
Hence, the 2028th digit = 2

nimcet PYQ 2025
Statement :  All engineers are good at math.

Conclusion 1: Some people good at math are engineers.

Conclusion 2: All people good at math are engineers.





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Answer: Only Conclusion 1 follows.

Reasoning:
Given “All engineers are good at math” ($E \to M$):

  • It implies some good-at-math people are engineers (since at least one engineer would then be in the math-good group) → Conclusion 1 follows.

  • It does not imply everyone good at math is an engineer ($M \to E$ is not guaranteed) → Conclusion 2 does not follow.


nimcet PYQ 2025
Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° degree and 45° repectively. If the lighthouse is 100 m high, the distance between the two ships is





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nimcet Previous Year PYQ nimcet NIMCET 2025 PYQ

Solution

Let distances from the lighthouse be $x$ (for $30^\circ$) and $y$ (for $45^\circ$). With height $h=100$ m:

$\tan 30^\circ=\dfrac{100}{x}\ \Rightarrow\ x=\dfrac{100}{\tan30^\circ}=100\sqrt{3}$,   $\tan 45^\circ=\dfrac{100}{y}\ \Rightarrow\ y=\dfrac{100}{\tan45^\circ}=100$.

Ships are on opposite sides ⇒ distance $=x+y=100\sqrt{3}+100=100(\sqrt{3}+1)\approx \boxed{273.2\ \text{m}}$.


nimcet PYQ 2025
A research team is studying the effects of three different fertilizers (X, Y, and Z) on the growth of a specific plant species. They have four experimental plots with different fertilizer combinations:
  • Plot 1: Fertilizer X only
  • Plot 2: Fertilizer Y only
  • Plot 3: Fertilizer X and Fertilizer Z only
  • Plot 4: Fertilizer Y and Fertilizer Z only
After a period of observation, they note the following:
  • Plants in plots with Fertilizer X showed significantly increased height compared to a control group (no fertilizer).
  • Plants in plots with Fertilizer Y showed a slightly increased leaf area compared to the control group.
  • Plants in Plot 3 (X and Z) showed no significant difference in height compared to the control group.
  • Plants in Plot 4 (Y and Z) showed significantly increased height compared to the control group.
Based on these observations, which of the following conclusions is best supported?





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