---

Solution

For ellipse, 

---

Solution

Solid State Drive is used for long-term, non-volatile storage.

So,

I $\to$ B

Cache Memory is a high-speed buffer memory between CPU and RAM.

So,

II $\to$ D

RAM is volatile main memory that holds data currently in use by the OS and applications.

So,

III $\to$ A

CPU Registers are the fastest and smallest storage available for processor computations.

So,

IV $\to$ C

Therefore, the correct matching is:

I-B, II-D, III-A, IV-C

---

Solution

$2^{x+\frac{1}{2}}\times 4^{y-\frac{5}{6}}=3^{x-\frac{1}{2}}\times 9^{y-\frac{1}{3}}$ 

---

Solution

The rendering engine is responsible for displaying the web page on the screen.

It takes HTML, CSS, and JavaScript output and converts them into the visual layout that the user can interact with.

A JavaScript engine mainly executes JavaScript code.

Network protocol stack handles communication over the network.

Transport Layer Security Protocol is used for secure communication.

---

Solution

Let the distance between viewpoint and flag be $x$. 

---

Solution

Cookies are small text files stored on the user's computer by websites.

They are used to save information such as login status, preferences, shopping cart data, and tracking details.

Cookies are not simply browsing history.

They are also not viruses by default.

---

Solution

Statement I is correct.

The Program Counter stores the address of the next instruction. After fetching an instruction, it is usually incremented so that the CPU can move to the next instruction.

Statement II is also correct.

The Control Unit controls the fetch-decode-execute cycle. It fetches instructions from memory and places them into the Instruction Register.

Statement III is incorrect.

The ALU performs arithmetic and logical operations. It does not decode instructions. Instruction decoding is handled by the Control Unit.

Statement IV is incorrect.

Registers are small, fast storage locations inside the CPU. They do not permanently store operating system files.

So, only statements I and II are correct.

---

Solution

A web browser is a program used to request, retrieve, and display web pages.

So,

$1\to b$

URL is the complete address that identifies the location of a web resource.

So,

$2\to d$

DNS converts a human-readable website name into a numerical network address.

So,

$3\to a$

HTTP is a set of standard rules used by browsers and servers to exchange web content.

So,

$4\to c$

Therefore, the correct matching is

$1\to b,\ 2\to d,\ 3\to a,\ 4\to c$

---

Solution

Device drivers are system software, not application software.

Statement II is correct.

Linux, macOS, and Android are operating systems, and operating systems are system software.

Statement III is correct.

The OS kernel manages CPU scheduling, memory allocation, and hardware access.

Statement IV is incorrect.

A web browser and a word processor are application software, not system software.

So, only statements II and III are correct.

---

Solution

For $9$-bit $1$'s complement, the range is

$-(2^{8}-1)$ to $+(2^{8}-1)$

$=-255$ to $+255$

Also, in $1$'s complement, there are two representations of zero: positive zero and negative zero.

So, statement I is correct.

For $9$-bit $2$'s complement, the range is

$-2^8$ to $2^8-1$

$=-256$ to $+255$

But $1$'s complement does not represent exactly $512$ unique values because zero has two representations.

So, statement II is incorrect.

The maximum positive number in both systems is

$+255$

So, statement III is correct.

Therefore, statements I and III only are correct.

---

Solution

First convert both binary numbers into decimal.

$1100_2=12$

$1011_2=11$

Now multiply:

$12\times 11=132$

Convert $132$ into binary:

$132=128+4$

$128=2^7$ and $4=2^2$

So,

$132=10000100_2$

Therefore,

$1100_2\times 1011_2=10000100_2$

---

Solution

Given expression is

$(x+y'+z')(x+y'+z)(x+y+z')$

Using the identity

$(x+A)(x+B)=x+AB$

First take

$(x+y'+z')(x+y'+z)$

Here,

$A=y'+z'$ and $B=y'+z$

So,

$(x+y'+z')(x+y'+z)=x+(y'+z')(y'+z)$

Now,

$(y'+z')(y'+z)=y'+z'z$

Since

$z'z=0$

So,

$(y'+z')(y'+z)=y'$

Now the expression becomes

$(x+y')(x+y+z')$

Again using the identity,

$(x+A)(x+B)=x+AB$

Here,

$A=y'$ and $B=y+z'$

So,

$(x+y')(x+y+z')=x+y'(y+z')$

$=x+y'y+y'z'$

Since

$y'y=0$

So,

$=x+y'z'$

Therefore, the simplified value is

$x+y'z'$

---

Solution

In statement I, the verb should agree with the noun nearest to it. Since associates is plural, it should be were, not was. So statement I is incorrect.

In statement II, the subject is A series, which is singular. So it should be is scheduled, not are scheduled. Hence statement II is incorrect.

In statement III, majority is treated as a singular collective noun here, so has been revised is correct.

In statement IV, Each is singular, so it should be was given, not were given.

---

Solution

All four phrasal verbs are used correctly:

• carried out = performed
• turned down = rejected
• came across = found by chance
• called off = cancelled

So all $4$ sentences are correct.

---

Solution

• We conduct a study. So $1 \to B$
• We refute an argument. So $2 \to D$
• We propose a theory. So $3 \to A$
• We analyze data. So $4 \to C$

Thus the correct matching is:

$1\to B,\ 2\to D,\ 3\to A,\ 4\to C$

---

Solution

The passage says that qualitative research focuses on depth and on a small number of cases. This means its strength is detailed understanding and context.

• Option 1 is incorrect because qualitative studies usually do not require larger sample sizes.
• Option 2 is incorrect because qualitative findings can still be influenced by researcher bias.
• Option 4 is incorrect because qualitative research is often more time-consuming.
• Option 3 logically follows from the idea of depth and contextual understanding.

---

Solution

The word brevity means shortness or conciseness in speech or writing.

So in the sentence, brevity means conciseness.

---

Solution

To show two actions in the past, where one happened before the other, we use:

• Simple past for the later action
• Past perfect for the earlier action

Here:

• First, the exhibit had deteriorated
• Then, the curator replaced it

So the correct sentence is:
The curator replaced the exhibit after it had deteriorated.

---

Solution

In sentence A, heavy grocery bag works as a noun phrase because it names the thing that slipped.

So,
A $\to$ (i)

In sentence B, quite quickly modifies the verb “finished”, so it is an adverbial phrase.

So,
B $\to$ (iii)

In sentence C, under the doormat begins with the preposition “under”, so it is a prepositional phrase.

So,
C $\to$ (iv)

In sentence D, a bit nervous describes the subject “She”, so it is an adjectival phrase.

So,
D $\to$ (ii)

Therefore, the correct matching is:

a-(i), b-(iii), c-(iv), d-(ii)

---

Solution

The word anthropogenic means caused by human activities.

In the passage, anthropogenic activities are explained as the combustion of fossil fuels such as coal, oil and gas after the Industrial Revolution.

These are human-caused activities.

So, the closest meaning of anthropogenic is human-induced.

---

Solution

The main idea of the passage is that astronomers confirmed the accelerating expansion of the universe.

The passage also says that this acceleration is driven by "dark energy".

The study used "weak gravitational lensing" and involved nearly $446000$ galaxies, $1000$ hours of observation, and $600$ Earth orbits by the Hubble Space Telescope.

Option $1$ only talks about the study details but misses the main finding.

Option $2$ is incorrect because the passage does not say that dark matter was noticed for the first time.

Option $3$ wrongly says that weak gravitational lensing is the best technique to study dark matter, which is not the main idea.

Option $4$ correctly captures the main idea and includes the important method and observation details.

---

Solution

The word corroborate means to support or confirm something with evidence.

Here, the sentence means that the new findings from the flight data recorder support or confirm the theory of pilot suicide.

So, the correct word is corroborate.

---

Solution

We know that
$A \Delta B = (A \cap B^c) \cup (A^c \cap B)$

Now,
$(A \Delta B) \cap C = [(A \cap B^c) \cup (A^c \cap B)] \cap C$

Using distributive law,
$(A \Delta B) \cap C = (A \cap B^c \cap C) \cup (A^c \cap B \cap C)$

Option $1$, option $2$ and option $4$ represent the same set.

But option $3$ is
$(A \cap B)^c \cap C$

This means all elements of $C$ except the elements common in $A$ and $B$.
It also includes elements which are neither in $A$ nor in $B$, but present in $C$.

So, it is not equal to $ (A \Delta B) \cap C $.

---

Solution

Given relation:
$a \sim b$ if $a - 2b$ is divisible by $3$

This means
$a - 2b \equiv 0 \pmod 3$

So,
$a \equiv 2b \pmod 3$

Checking reflexive:
For reflexive relation, $a \sim a$ must be true for every $a$.

Now,
$a - 2a = -a$

For all values of $a$, $-a$ is not divisible by $3$.
So, relation is not reflexive.

Checking symmetric:
If $a \sim b$, then
$a \equiv 2b \pmod 3$

Multiplying both sides by $2$,
$2a \equiv 4b \pmod 3$

Since $4 \equiv 1 \pmod 3$,
$2a \equiv b \pmod 3$

So,
$b - 2a \equiv 0 \pmod 3$

Hence, $b \sim a$.
So, relation is symmetric.

Checking transitive:
Let $a \sim b$ and $b \sim c$.

Then,
$a \equiv 2b \pmod 3$

and
$b \equiv 2c \pmod 3$

So,
$a \equiv 2(2c) \pmod 3$

$a \equiv 4c \pmod 3$

Since $4 \equiv 1 \pmod 3$,
$a \equiv c \pmod 3$

But for $a \sim c$, we need
$a \equiv 2c \pmod 3$

This is not always true.
So, relation is not transitive.

Therefore, $\sim$ is symmetric but neither transitive nor reflexive.

---

Solution

Even elements in the domain are $2$ and $4$.
Even elements in the codomain are $2,4,6,8$.

The two even elements of the domain must be mapped to even elements of the codomain injectively.

Number of ways:
${}^4P_2=4\times 3=12$

Now, $2$ elements of the codomain are already used, so $6$ elements are left.

The odd elements of the domain are $1$ and $3$.
They can be mapped injectively to the remaining $6$ elements.

Number of ways:
${}^6P_2=6\times 5=30$

So,
$n=12\times 30=360$

Now,
$360=2^3\times 3^2\times 5^1$

So,
$a=3,\ b=2,\ c=1$

Therefore,
$a+b+c=3+2+1=6$

---

Solution

For $n$ sets, the maximum number of regions in a Venn diagram is:

$2^n$

Here,
$n=5$

So, number of regions is:

$2^5=32$

---

Solution

Since
$R\subset \mathbb{N}\times \mathbb{N}$,

the relation $R$ can have at most countably infinite elements.

Now, suppose for some $a\in \mathbb{N}$, the set $R_a$ is infinite.

Since $R_a\subseteq \mathbb{N}$, it is countably infinite.

Also, for every element $b\in R_a$, the ordered pair $(a,b)$ belongs to $R$.

So, $R$ is also infinite. Since $R\subset \mathbb{N}\times \mathbb{N}$, $R$ is countably infinite.

Therefore, both $R_a$ and $R$ have the same cardinality.

---

Solution

We check the possible membership of each element in $A,B,C$.

Condition given is:

$(A\cap B)\subseteq C\subseteq (A\cup B)$

For one element:

If the element is in neither $A$ nor $B$, then it cannot be in $C$.
Number of choices $=1$

If the element is only in $A$, then it may or may not be in $C$.
Number of choices $=2$

If the element is only in $B$, then it may or may not be in $C$.
Number of choices $=2$

If the element is in both $A$ and $B$, then it must be in $C$.
Number of choices $=1$

Total choices for one element:

$1+2+2+1=6$

Since there are $n$ elements, total number of triples is:

$6^n$

---

Solution

There are three variables $p,q,r$.

So, total number of ordered tuples is:

$2^3=8$

The implication $A\Rightarrow r$ is false only when $A$ is true and $r$ is false.

Here,
$A=(\neg p\vee q)$

Now, $A=(\neg p\vee q)$ is false only when both $\neg p$ and $q$ are false.

That means:

$p$ is true and $q$ is false.

So, $A$ is false in only $1$ case out of $4$ possible cases of $(p,q)$.

Therefore, $A$ is true in $3$ cases.

For implication to be false, $A$ must be true and $r$ must be false.

So, false cases $=3$

Hence, true cases:

$8-3=5$

---

Solution

Given,
$P(X\text{ passes})=\frac{2}{5}$

So,
$P(X\text{ does not pass})=1-\frac{2}{5}=\frac{3}{5}$

Also,
$P(Y\text{ passes})=\frac{3}{4}$

So,
$P(Y\text{ does not pass})=1-\frac{3}{4}=\frac{1}{4}$

Since both events are independent,

$P(\text{neither }X\text{ nor }Y\text{ passes})=\frac{3}{5}\times \frac{1}{4}$

$=\frac{3}{20}$

---

Solution

By multiplication rule of probability,

$P(A_1\cap A_2)=P(A_1)P(A_2/A_1)$

Now,

$P(A_1\cap A_2\cap A_3)=P(A_1\cap A_2)P(A_3/(A_1\cap A_2))$

Substituting the value of $P(A_1\cap A_2)$,

$P(A_1\cap A_2\cap A_3)=P(A_1)P(A_2/A_1)P(A_3/(A_1\cap A_2))$

---

Solution

Given observations are:

$14.5,15.2,16.8,17.1,15.9,16.3,14.7$

Number of observations:

$n=7$

Sample mean is

$\bar{x}=\frac{14.5+15.2+16.8+17.1+15.9+16.3+14.7}{7}$

$\bar{x}=\frac{110.5}{7}$

$\bar{x}=15.7857$

So,

$\bar{x}\approx 15.79$

Now, sample standard deviation is

$s=\sqrt{\frac{\sum (x_i-\bar{x})^2}{n-1}}$

Here,

$\sum (x_i-\bar{x})^2\approx 6.2086$

So,

$s=\sqrt{\frac{6.2086}{6}}$

$s=\sqrt{1.0348}$

$s\approx 1.0172$

So,

$s\approx 1.02$

Therefore, the sample mean and sample standard deviation are $15.79,\ 1.02$.

---

Solution

The highest frequency is $20$, so the modal class is:

$30-35$

For grouped data, mode is given by:

$\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$

Here,

$l=30$

$f_1=20$

$f_0=14$

$f_2=18$

$h=5$

Now,

$\text{Mode}=30+\frac{20-14}{2(20)-14-18}\times 5$

$\text{Mode}=30+\frac{6}{40-32}\times 5$

$\text{Mode}=30+\frac{6}{8}\times 5$

$\text{Mode}=30+3.75$

$\text{Mode}=33.75$

---

Solution

Given observations are:

$10,4,11,6,17,15,9,8,x$

Sum of known observations:

$10+4+11+6+17+15+9+8=80$

There are total $9$ observations.

Check option $2$, that is $x=10$.

Then data becomes:

$10,4,11,6,17,15,9,8,10$

Arranging in ascending order:

$4,6,8,9,10,10,11,15,17$

Median is the middle value.

Since there are $9$ observations, median is the $5^{\text{th}}$ value.

So,

$\text{Median}=10$

Mean is:

$\text{Mean}=\frac{80+10}{9}$

$\text{Mean}=\frac{90}{9}=10$

Also, $10$ occurs twice, so

$\text{Mode}=10$

Hence,

$\text{mean}=\text{median}=\text{mode}=10$

---

Solution

Given,

$Z=XY$

Now, $Z=1$ only when both $X=1$ and $Y=1$.

Since $X$ and $Y$ are independent,

$P(Z=1)=P(X=1,Y=1)$

$P(Z=1)=P(X=1)P(Y=1)$

$P(Z=1)=\frac{1}{2}\times \frac{1}{2}$

$P(Z=1)=\frac{1}{4}$

Now,

$P(Z=0)=1-P(Z=1)$

$P(Z=0)=1-\frac{1}{4}$

$P(Z=0)=\frac{3}{4}$

Therefore, $Z$ follows Bernoulli distribution with

$P(Z=1)=\frac{1}{4}$ and $P(Z=0)=\frac{3}{4}$

Correct answer is option $4$.

---

Solution

Given,

Variance $=3$

So,

$\mu_2=3$

Fourth central moment is:

$\mu_4=63$

Coefficient of kurtosis is:

$\beta_2=\frac{\mu_4}{\mu_2^2}$

Substitute the values:

$\beta_2=\frac{63}{3^2}$

$\beta_2=\frac{63}{9}$

$\beta_2=7$

For a normal or mesokurtic distribution,

$\beta_2=3$

Here,

$\beta_2=7>3$

So, the distribution is leptokurtic.Given,

$Z=XY$

Now, $Z=1$ only when both $X=1$ and $Y=1$.

Since $X$ and $Y$ are independent,

$P(Z=1)=P(X=1,Y=1)$

$P(Z=1)=P(X=1)P(Y=1)$

$P(Z=1)=\frac{1}{2}\times \frac{1}{2}$

$P(Z=1)=\frac{1}{4}$

Now,

$P(Z=0)=1-P(Z=1)$

$P(Z=0)=1-\frac{1}{4}$

$P(Z=0)=\frac{3}{4}$

Therefore, $Z$ follows Bernoulli distribution with

$P(Z=1)=\frac{1}{4}$ and $P(Z=0)=\frac{3}{4}$

Correct answer is option $4$.

---

Solution

Let the first term of the arithmetic progression be $a$.

Common difference is:

$d=\frac{3}{2}$

Sum of first $11$ terms is:

$S_{11}=88$

Formula for sum of first $n$ terms:

$S_n=\frac{n}{2}[2a+(n-1)d]$

So,

$88=\frac{11}{2}[2a+10d]$

Substitute $d=\frac{3}{2}$:

$88=\frac{11}{2}\left[2a+10\times \frac{3}{2}\right]$

$88=\frac{11}{2}[2a+15]$

Now,

$2a+15=16$

$2a=1$

$a=\frac{1}{2}$

The $10^{\text{th}}$ term is:

$T_{10}=a+9d$

$T_{10}=\frac{1}{2}+9\times \frac{3}{2}$

$T_{10}=\frac{1}{2}+\frac{27}{2}$

$T_{10}=14$

The $11^{\text{th}}$ term is:

$T_{11}=a+10d$

$T_{11}=\frac{1}{2}+10\times \frac{3}{2}$

$T_{11}=\frac{1}{2}+15$

$T_{11}=\frac{31}{2}$

So, the roots of $3x^2-px+q=0$ are $14$ and $\frac{31}{2}$.

For equation $3x^2-px+q=0$,

Sum of roots:

$\frac{p}{3}=14+\frac{31}{2}$

$\frac{p}{3}=\frac{28+31}{2}$

$\frac{p}{3}=\frac{59}{2}$

$p=\frac{177}{2}$

Product of roots:

$\frac{q}{3}=14\times \frac{31}{2}$

$\frac{q}{3}=217$

$q=651$

Now,

$q-2p=651-2\times \frac{177}{2}$

$q-2p=651-177$

$q-2p=474$

Therefore, the correct answer is option $2$.Given,

---

Solution

Let

$D=\left|\begin{array}{ccc}x & x+1 & x+3\\ x+1 & x+3 & x+6\\ x+3 & x+6 & x+10\end{array}\right|$

Apply row operations:

$R_2\to R_2-R_1$

$R_3\to R_3-R_1$

Then,

$D=\left|\begin{array}{ccc}x & x+1 & x+3\\ 1 & 2 & 3\\ 3 & 5 & 7\end{array}\right|$

Now expanding along the first row,

$D=x(2\cdot 7-3\cdot 5)-(x+1)(1\cdot 7-3\cdot 3)+(x+3)(1\cdot 5-2\cdot 3)$

$D=x(14-15)-(x+1)(7-9)+(x+3)(5-6)$

$D=-x+2(x+1)-(x+3)$

$D=-x+2x+2-x-3$

$D=-1$

So, at $x=2026$, the value of the determinant is $-1$.

---

Solution

For a homogeneous system to have a non-trivial solution, the determinant of the coefficient matrix must be zero.

So,

$D=\left|\begin{array}{ccc}1 & \sin\theta & \cos\theta\ 1 & \cos\theta & \sin\theta\ 1 & -\sin\theta & -\cos\theta\end{array}\right|$

Let $\sin\theta=s$ and $\cos\theta=c$.

Then,

$D=\left|\begin{array}{ccc}1 & s & c\ 1 & c & s\ 1 & -s & -c\end{array}\right|$

On simplifying,

$D=2(s-c)(s+c)$

For non-trivial solution,

$D=0$

So,

$2(s-c)(s+c)=0$

Hence,

$s-c=0$ or $s+c=0$

So,

$\sin\theta=\cos\theta$ or $\sin\theta=-\cos\theta$

Case 1:

$\sin\theta=\cos\theta$

$\tan\theta=1$

In $[0,2\pi]$,

$\theta=\frac{\pi}{4},\frac{5\pi}{4}$

Case 2:

$\sin\theta=-\cos\theta$

$\tan\theta=-1$

In $[0,2\pi]$,

$\theta=\frac{3\pi}{4},\frac{7\pi}{4}$

Total number of values of $\theta$ is $4$.

---

Solution

Let the new root be

$y=\frac{x+2}{x-1}$

Now,

$y(x-1)=x+2$

$xy-y=x+2$

$x(y-1)=y+2$

$x=\frac{y+2}{y-1}$

Since $x$ is a root of

$ax^2+bx+c=0$

Put $x=\frac{y+2}{y-1}$.

$a\left(\frac{y+2}{y-1}\right)^2+b\left(\frac{y+2}{y-1}\right)+c=0$

Multiplying by $(y-1)^2$,

$a(y+2)^2+b(y+2)(y-1)+c(y-1)^2=0$

Now expand:

$a(y^2+4y+4)+b(y^2+y-2)+c(y^2-2y+1)=0$

So,

$(a+b+c)y^2+(4a+b-2c)y+(4a-2b+c)=0$

Therefore, the required equation is

$(a+b+c)x^2+(4a+b-2c)x+(4a-2b+c)=0$

---

Solution

Let

$\sqrt{x+y}=a$

and

$\sqrt{y+z}=b$

Given,

$2\sqrt{x+y}-3\sqrt{y+z}=2$

So,

$2a-3b=2$

Also,

$x=a^2-y$

and

$z=b^2-y$

Now use the second condition:

$4x-5y-9z=8$

$4(a^2-y)-5y-9(b^2-y)=8$

$4a^2-4y-5y-9b^2+9y=8$

$4a^2-9b^2=8$

So,

$(2a-3b)(2a+3b)=8$

Since $2a-3b=2$,

$2(2a+3b)=8$

$2a+3b=4$

Now solve:

$2a-3b=2$

$2a+3b=4$

Adding both equations,

$4a=6$

$a=\frac{3}{2}$

Now,

$2a+3b=4$

$3+3b=4$

$3b=1$

$b=\frac{1}{3}$

Now,

$20x+38y+18z+1$

$=20(a^2-y)+38y+18(b^2-y)+1$

$=20a^2+18b^2+1$

Also,

$9y+9z+2=9(y+z)+2=9b^2+2$

Now substitute $a=\frac{3}{2}$ and $b=\frac{1}{3}$.

Numerator:

$20a^2+18b^2+1=20\left(\frac{3}{2}\right)^2+18\left(\frac{1}{3}\right)^2+1$

$=20\cdot \frac{9}{4}+18\cdot \frac{1}{9}+1$

$=45+2+1$

$=48$

Denominator:

$9b^2+2=9\left(\frac{1}{3}\right)^2+2$

$=1+2$

$=3$

Therefore,

$\sqrt{\frac{20x+38y+18z+1}{9y+9z+2}}=\sqrt{\frac{48}{3}}$

$=\sqrt{16}$

$=4$

---

Solution

Given equation is

$\left(\sum_{k=1}^{n}(-1)^{k-1}k\right)^2-\sum_{k=1}^{n}(-1)^{k-1}k^2+2450=0$

Check for odd $n$.

Let

$n=2m-1$

Then,

$\sum_{k=1}^{n}(-1)^{k-1}k=1-2+3-4+\cdots+(2m-1)$

Pairing terms,

$(1-2)+(3-4)+\cdots+[(2m-3)-(2m-2)]+(2m-1)$

$=-(m-1)+(2m-1)$

$=m$

So,

$\left(\sum_{k=1}^{n}(-1)^{k-1}k\right)^2=m^2$

Now,

$\sum_{k=1}^{n}(-1)^{k-1}k^2=1^2-2^2+3^2-4^2+\cdots+(2m-1)^2$

For odd $n=2m-1$,

$\sum_{k=1}^{n}(-1)^{k-1}k^2=m(2m-1)$

Now put in the equation:

$m^2-m(2m-1)+2450=0$

$m^2-2m^2+m+2450=0$

$-m^2+m+2450=0$

$m^2-m-2450=0$

Now factorize:

$m^2-m-2450=0$

$m^2-50m+49m-2450=0$

$m(m-50)+49(m-50)=0$

$(m-50)(m+49)=0$

Since $m$ is positive,

$m=50$

Therefore,

$n=2m-1$

$n=2(50)-1$

$n=99$

---

Solution

Given equations are:

$x+y+z=1$ .....$(1)$

$4x+9y+16z=25$ .....$(2)$

$16x+81y+256z=625$ .....$(3)$

Now subtract $4\times(1)$ from $(2)$:

$4x+9y+16z-4x-4y-4z=25-4$

$5y+12z=21$ .....$(4)$

Now subtract $16\times(1)$ from $(3)$:

$16x+81y+256z-16x-16y-16z=625-16$

$65y+240z=609$ .....$(5)$

Multiply equation $(4)$ by $13$:

$65y+156z=273$ .....$(6)$

Now subtract $(6)$ from $(5)$:

$65y+240z-(65y+156z)=609-273$

$84z=336$

$z=4$

Put $z=4$ in equation $(4)$:

$5y+12(4)=21$

$5y+48=21$

$5y=-27$

$y=-\frac{27}{5}$

Now use equation $(1)$:

$x+y+z=1$

$x-\frac{27}{5}+4=1$

$x-\frac{27}{5}=-3$

$x=-3+\frac{27}{5}$

$x=\frac{-15+27}{5}$

$x=\frac{12}{5}$

So,

$x=\frac{36}{15}$

---

Solution

A point equidistant from the coordinate axes satisfies:

$|x|=|y|$

So, either

$y=x$

or

$y=-x$

Given line is:

$8x-3y=11$

Case 1:

$y=x$

Put $y=x$ in the line:

$8x-3x=11$

$5x=11$

$x=\frac{11}{5}$

This is not an integer, so this case is rejected.

Case 2:

$y=-x$

Put $y=-x$ in the line:

$8x-3(-x)=11$

$8x+3x=11$

$11x=11$

$x=1$

Then,

$y=-1$

So, the point is:

$(1,-1)$

Here $x>0$ and $y<0$, so the point lies in the IV quadrant.

---

Solution

Since the given lines are diameters of the circle, both lines pass through the centre of the circle.

So, the centre is the intersection point of

$2x+3y=1$ .....$(1)$

$4x-3y=11$ .....$(2)$

Adding $(1)$ and $(2)$,

$6x=12$

$x=2$

Put $x=2$ in $(1)$:

$2(2)+3y=1$

$4+3y=1$

$3y=-3$

$y=-1$

So, centre of the circle is

$(2,-1)$

Now, area of circle is

$\pi r^2=153.94$

Since $153.94\approx 49\pi$,

$r^2=49$

$r=7$

Equation of circle is

$(x-2)^2+(y+1)^2=7^2$

$(x-2)^2+(y+1)^2=49$

Expanding,

$x^2-4x+4+y^2+2y+1=49$

$x^2+y^2-4x+2y-44=0$

---

Solution

Let the common tangent be

$y=mx+c$

For the parabola

$y=-x^2$

we have

$-x^2=mx+c$

$x^2+mx+c=0$

For tangency, discriminant must be zero.

$m^2-4c=0$

So,

$c=\frac{m^2}{4}$ .....$(1)$

Now, for the parabola

$y=(x-2)^2$

we have

$(x-2)^2=mx+c$

$x^2-4x+4=mx+c$

$x^2-(m+4)x+(4-c)=0$

For tangency,

$(m+4)^2-4(4-c)=0$

Substitute $c=\frac{m^2}{4}$:

$(m+4)^2-16+4\cdot \frac{m^2}{4}=0$

$(m+4)^2-16+m^2=0$

$m^2+8m+16-16+m^2=0$

$2m^2+8m=0$

$2m(m+4)=0$

So,

$m=0$ or $m=-4$

From the given options, $m=-4$ is possible.

Using $(1)$,

$c=\frac{(-4)^2}{4}$

$c=4$

So, the common tangent is

$y=-4x+4$

---

Solution

Given,

$h(x)=f[g(x)]$

Now, $f(t)=1$ when

$|t|\le 1$

So, $h(x)=1$ when

$|g(x)|\le 1$

For $|x|\le 2$,

$g(x)=2-x^2$

So,

$|2-x^2|\le 1$

This gives

$-1\le 2-x^2\le 1$

Subtract $2$ from all sides:

$-3\le -x^2\le -1$

Multiplying by $-1$ reverses the inequalities:

$1\le x^2\le 3$

So,

$1\le |x|\le \sqrt{3}$

For $|x|>2$,

$g(x)=2$

So,

$|g(x)|=2>1$

Hence, this case is not valid.

Therefore, the required interval is

$1\le |x|\le \sqrt{3}$

---

Solution

Given limit is

$\lim_{x\to\infty}\left(\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}\right)$

Divide numerator and denominator inside the radical by $x$.

$\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}=\sqrt{\frac{x}{x+\sqrt{x+\sqrt{x}}}}$

$=\sqrt{\frac{1}{1+\frac{\sqrt{x+\sqrt{x}}}{x}}}$

Now,

$\frac{\sqrt{x+\sqrt{x}}}{x}\to 0$ as $x\to\infty$

Therefore,

$\lim_{x\to\infty}\sqrt{\frac{1}{1+\frac{\sqrt{x+\sqrt{x}}}{x}}}$

$=\sqrt{\frac{1}{1+0}}$

$=1$

---

Solution

Given curves are:

$y=(x-2)^2$

and

$y=-4+6x-x^2$

At the point of intersection,

$(x-2)^2=-4+6x-x^2$

$x^2-4x+4=-4+6x-x^2$

$2x^2-10x+8=0$

$x^2-5x+4=0$

$(x-1)(x-4)=0$

So,

$x=1$ or $x=4$

Now, slopes of the curves are:

For $y=(x-2)^2$,

$m_1=\frac{dy}{dx}=2(x-2)$

For $y=-4+6x-x^2$,

$m_2=\frac{dy}{dx}=6-2x$

At $x=1$,

$m_1=2(1-2)=-2$

$m_2=6-2(1)=4$

Angle between two curves is given by:

$\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|$

So,

$\tan\theta=\left|\frac{4-(-2)}{1+(-2)(4)}\right|$

$=\left|\frac{6}{1-8}\right|$

$=\left|\frac{6}{-7}\right|$

$=\frac{6}{7}$

Therefore,

$\theta=\tan^{-1}\left(\frac{6}{7}\right)$

---

Solution

We have,

$\lim_{x\to 0}\frac{|x|\log_e(1+|\sin 2x|)}{x^2(|x|+3)}$

As $x\to 0$,

$|\sin 2x|\sim 2|x|$

Also,

$\log_e(1+|\sin 2x|)\sim |\sin 2x|$

So,

$\log_e(1+|\sin 2x|)\sim 2|x|$

Now the numerator becomes approximately:

$|x|\cdot 2|x|=2x^2$

The denominator becomes approximately:

$x^2(|x|+3)\to 3x^2$

Therefore,

$\lim_{x\to 0}\frac{|x|\log_e(1+|\sin 2x|)}{x^2(|x|+3)}=\frac{2x^2}{3x^2}$

$=\frac{2}{3}$

Hence, the limit exists and is equal to $\frac{2}{3}$.

---

Solution

Given,

$f(x)=|x+1|e^{-x^2}$

For $x<-1$,

$f(x)=-(x+1)e^{-x^2}$

Differentiate:

$f'(x)=e^{-x^2}(2x^2+2x-1)$

For critical points,

$2x^2+2x-1=0$

Using quadratic formula,

$x=\frac{-2\pm\sqrt{4+8}}{4}$

$x=\frac{-2\pm 2\sqrt{3}}{4}$

$x=\frac{-1\pm\sqrt{3}}{2}$

For $x<-1$, the valid critical point is

$x=\frac{-1-\sqrt{3}}{2}$

This lies in the interval $(-2,-1)$.

At this point, $f$ has a point of maxima.

---

Solution

Using expansion near $x=0$,

$(1+x)^{\frac{1}{3}}=1+\frac{x}{3}-\frac{x^2}{9}+O(x^3)$

Also,

$(1+2x)^{\frac{1}{4}}=1+\frac{x}{2}-\frac{3x^2}{8}+O(x^3)$

Now,

$(1+x)^{\frac{1}{3}}-(1+2x)^{\frac{1}{4}}$

$=\left(1+\frac{x}{3}-\frac{x^2}{9}\right)-\left(1+\frac{x}{2}-\frac{3x^2}{8}\right)+O(x^3)$

$=-\frac{x}{6}+\left(-\frac{1}{9}+\frac{3}{8}\right)x^2+O(x^3)$

$=-\frac{x}{6}+\frac{19x^2}{72}+O(x^3)$

Therefore,

$f(x)=\frac{-\frac{x}{6}+\frac{19x^2}{72}+O(x^3)}{x}$

$f(x)=-\frac{1}{6}+\frac{19x}{72}+O(x^2)$

For differentiability at $x=0$, function must be continuous at $x=0$.

So,

$a=\lim_{x\to 0}f(x)=-\frac{1}{6}$

Also,

$f'(0)=\frac{19}{72}$

Hence,

$a+f'(0)=-\frac{1}{6}+\frac{19}{72}$

$=-\frac{12}{72}+\frac{19}{72}$

$=\frac{7}{72}$

---

Solution

We know that

$(f^{-1})'(y)=\frac{1}{f'(x)}$, where $f(x)=y$

Here, we need $(f^{-1})'(2)$.

So first find $x$ such that

$f(x)=2$

$\frac{3x^2+4x+1}{x^2+3x+2}=2$

$3x^2+4x+1=2x^2+6x+4$

$x^2-2x-3=0$

$(x-3)(x+1)=0$

Since domain is $[0,\infty)$,

$x=3$

Now,

$f(x)=\frac{3x^2+4x+1}{x^2+3x+2}$

Let

$N=3x^2+4x+1$

and

$D=x^2+3x+2$

Then,

$f'(x)=\frac{N'D-ND'}{D^2}$

Now at $x=3$,

$N=3(3)^2+4(3)+1=40$

$D=(3)^2+3(3)+2=20$

$N'=6x+4$

So,

$N'=22$

$D'=2x+3$

So,

$D'=9$

Therefore,

$f'(3)=\frac{22\cdot 20-40\cdot 9}{20^2}$

$f'(3)=\frac{440-360}{400}$

$f'(3)=\frac{80}{400}$

$f'(3)=\frac{1}{5}$

Hence,

$(f^{-1})'(2)=\frac{1}{f'(3)}$

$=5$

---

Solution

Given,

$x-y\tan 35^\circ=\tan 25^\circ(y+x\tan 35^\circ)$

Let

$\tan 35^\circ=A$

and

$\tan 25^\circ=B$

Then,

$x-yA=B(y+xA)$

$x-yA=By+ABx$

$x-ABx=yA+By$

$x(1-AB)=y(A+B)$

So,

$\frac{x}{y}=\frac{A+B}{1-AB}$

Now,

$\frac{\tan 35^\circ+\tan 25^\circ}{1-\tan 35^\circ\tan 25^\circ}=\tan(35^\circ+25^\circ)$

$=\tan 60^\circ$

$=\sqrt{3}$

Therefore,

$\frac{x}{y}=\sqrt{3}$

So,

$x=\sqrt{3}y$

Since $\sqrt{3}>1$, according to the given option pattern, we get

$x>y$

---

Solution

By sine rule,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

So,

$\frac{a-b}{a+b}=\frac{\sin A-\sin B}{\sin A+\sin B}$

Now,

$\sin A-\sin B=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

and

$\sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$

Therefore,

$\frac{\sin A-\sin B}{\sin A+\sin B}=\cot\left(\frac{A+B}{2}\right)\tan\left(\frac{A-B}{2}\right)$

Since,

$A+B=\pi-C$

So,

$\cot\left(\frac{A+B}{2}\right)=\cot\left(\frac{\pi-C}{2}\right)$

$=\cot\left(\frac{\pi}{2}-\frac{C}{2}\right)$

$=\tan\left(\frac{C}{2}\right)$

Thus,

$\frac{a-b}{a+b}=\tan\left(\frac{C}{2}\right)\tan\left(\frac{A-B}{2}\right)$

But option $4$ gives

$\frac{a-b}{a+b}=\frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{C}{2}\right)}$

---

Solution

Given,

$PA=60$

$PB=80$

$\angle APB=60^\circ$

Using cosine rule in triangle $APB$,

$AB^2=PA^2+PB^2-2(PA)(PB)\cos 60^\circ$

$AB^2=60^2+80^2-2(60)(80)\cdot \frac{1}{2}$

$AB^2=3600+6400-4800$

$AB^2=5200$

$AB=\sqrt{5200}$

$AB=\sqrt{400\cdot 13}$

$AB=20\sqrt{13}$

---

Solution

We know that the principal range of $\cos^{-1}x$ is $[0,\pi]$.

Now,

$\cos\left(-\frac{\pi}{6}\right)=\cos\frac{\pi}{6}$

So,

$\cos^{-1}\left(\cos\left(-\frac{\pi}{6}\right)\right)=\frac{\pi}{6}$

Also,

$\sin\frac{5\pi}{6}=\frac{1}{2}$

The principal range of $\sin^{-1}x$ is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

So,

$\sin^{-1}\left(\sin\frac{5\pi}{6}\right)=\sin^{-1}\left(\frac{1}{2}\right)$

$=\frac{\pi}{6}$

Therefore,

$\cos^{-1}\left(\cos\left(-\frac{\pi}{6}\right)\right)+\sin^{-1}\left(\sin\frac{5\pi}{6}\right)$

$=\frac{\pi}{6}+\frac{\pi}{6}$

$=\frac{\pi}{3}$

---

Solution

Given,

$\tan^{-1}(3x)+\tan^{-1}(2x)=\frac{\pi}{4}$

Taking tangent on both sides,

$\tan\left(\tan^{-1}(3x)+\tan^{-1}(2x)\right)=\tan\frac{\pi}{4}$

Using,

$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

we get,

$\frac{3x+2x}{1-(3x)(2x)}=1$

$\frac{5x}{1-6x^2}=1$

So,

$5x=1-6x^2$

$6x^2+5x-1=0$

Factorizing,

$6x^2+6x-x-1=0$

$6x(x+1)-1(x+1)=0$

$(x+1)(6x-1)=0$

So,

$x=-1$ or $x=\frac{1}{6}$

Now check both values.

For $x=\frac{1}{6}$,

$\tan^{-1}\left(\frac{1}{2}\right)+\tan^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{4}$

So, $x=\frac{1}{6}$ is valid.

For $x=-1$,

$\tan^{-1}(-3)+\tan^{-1}(-2)$ is negative, so it cannot be equal to $\frac{\pi}{4}$.

Therefore, only one solution exists.

---

Solution

For intersection of the two graphs,

$f(x)=g(x)$

So,

$2\cos\left(\frac{x}{2}\right)+3=4$

$2\cos\left(\frac{x}{2}\right)=1$

$\cos\left(\frac{x}{2}\right)=\frac{1}{2}$

Now,

$x\in [0,4\pi]$

So,

$\frac{x}{2}\in [0,2\pi]$

In the interval $[0,2\pi]$,

$\cos\theta=\frac{1}{2}$ has two solutions:

$\theta=\frac{\pi}{3},\frac{5\pi}{3}$

Here,

$\theta=\frac{x}{2}$

So,

$\frac{x}{2}=\frac{\pi}{3}$ or $\frac{x}{2}=\frac{5\pi}{3}$

Therefore,

$x=\frac{2\pi}{3}$ or $x=\frac{10\pi}{3}$

Hence, the number of points of intersection is $2$.

---

Solution

First expression is:

$\left(x^2+\frac{1}{x}\right)^{12}$

Its general term is:

$T_{k+1}={}^{12}C_k(x^2)^{12-k}\left(\frac{1}{x}\right)^k$

$T_{k+1}={}^{12}C_k x^{24-2k-k}$

$T_{k+1}={}^{12}C_k x^{24-3k}$

For coefficient of $x^{10}$,

$24-3k=10$

$3k=14$

$k=\frac{14}{3}$

This is not an integer, so $x^{10}$ term is not present in the first expression.

Now second expression is:

$\left(x+\frac{1}{x^2}\right)^{12}$

Its general term is:

$T_{k+1}={}^{12}C_k(x)^{12-k}\left(\frac{1}{x^2}\right)^k$

$T_{k+1}={}^{12}C_k x^{12-k-2k}$

$T_{k+1}={}^{12}C_k x^{12-3k}$

For coefficient of $x^{10}$,

$12-3k=10$

$3k=2$

$k=\frac{2}{3}$

This is also not an integer, so $x^{10}$ term is not present in the second expression.

Therefore, the coefficient of $x^{10}$ is $0$.

---

Solution

Mira's mother-in-law means the mother of Mira's husband.

So, Mira's mother-in-law's mother means the grandmother of Mira's husband.

It is given that Mira's mother-in-law's mother is my grandmother.

So, Mira's husband's mother can be my aunt.

That means Mira's husband is my aunt's son.

Also, all my mother's offsprings are unmarried, so Mira cannot be my brother's wife.

Therefore, Mira is the wife of my aunt's son.

---

Solution

It is given that $C$ and $B$ must be on the same team.

Also, $C$ cannot be in Tech, so $B$ and $C$ must be either in Travel or Beauty.

Since $G$ must be in Travel and $F$ cannot be in Travel, $F$ must be in Beauty or Tech.

Also, $F$ must be with either $A$ or $B$, but not both.

Since $A$ cannot be in Beauty, if Beauty contains $F$, then $F$ must be with $B$.

So, the Beauty team becomes:

$B,F,C$

This satisfies all conditions.

---

Solution

The statement says that Indian children are talented, but they are weak in Science and Mathematics.

Course I talks about teaching and textbooks not being available in mother language. This may be a possible reason, but it is not directly connected to weakness in Science and Mathematics.

Course II says that education based on experiments in both subjects is lacking. Since Science and Mathematics require conceptual and experimental understanding, this course of action is more directly related.

Therefore, only II follows.

---

Solution

Given,

First : Second $=2:3$

Second : Third $=5:8$

Make the second term common.

LCM of $3$ and $5$ is $15$.

So,

First : Second $=10:15$

Second : Third $=15:24$

Therefore,

First : Second : Third $=10:15:24$

Sum of ratios:

$10+15+24=49$

Given sum of numbers is $98$.

So, one part is:

$\frac{98}{49}=2$

Second number is:

$15\times 2=30$

---

Solution

Let the balls faced by Lara be $x$.

Runs scored by Sachin are $10$ more than balls faced by Lara.

So,

Sachin's runs $=x+10$

Sachin faced $10$ balls less than Lara.

So,

Sachin's balls $=x-10$

Also, Sachin's balls are $5$ less than Lara's runs.

So,

Lara's runs $=x-5$

Together they scored $105$ runs.

Therefore,

$(x+10)+(x-5)=105$

$2x+5=105$

$2x=100$

$x=50$

So, Sachin's runs are:

$x+10=50+10=60$

---

Solution

Let the number of correct answers be $C$, wrong answers be $W$, and unanswered questions be $U$.

Total questions:

$C+W+U=160$ .....$(1)$

According to the first condition:

$C-\frac{W}{4}-\frac{U}{2}=79$ .....$(2)$

According to the second condition:

$C-\frac{W}{2}-\frac{U}{4}=76$ .....$(3)$

Subtract $(3)$ from $(2)$:

$\left(C-\frac{W}{4}-\frac{U}{2}\right)-\left(C-\frac{W}{2}-\frac{U}{4}\right)=79-76$

$-\frac{W}{4}+\frac{W}{2}-\frac{U}{2}+\frac{U}{4}=3$

$\frac{W}{4}-\frac{U}{4}=3$

$W-U=12$ .....$(4)$

From $(1)$,

$C=160-W-U$

Put this in $(2)$:

$160-W-U-\frac{W}{4}-\frac{U}{2}=79$

$160-\frac{5W}{4}-\frac{3U}{2}=79$

$\frac{5W}{4}+\frac{3U}{2}=81$

Multiply by $4$:

$5W+6U=324$ .....$(5)$

From $(4)$,

$W=U+12$

Put in $(5)$:

$5(U+12)+6U=324$

$5U+60+6U=324$

$11U=264$

$U=24$

So,

$W=24+12=36$

Now,

$C=160-36-24$

$C=100$

---

Solution

Given,

$A+B+C+D=10600$ .....$(1)$

If $A$ is removed, average of $B,C,D$ is $1000$.

So,

$B+C+D=3000$

Therefore,

$A=10600-3000=7600$

If $B$ is removed, average of $A,C,D$ is $3220$.

So,

$A+C+D=3220\times 3=9660$

Therefore,

$B=10600-9660=940$

If $C$ is removed, average of $A,B,D$ is $3180$.

So,

$A+B+D=3180\times 3=9540$

Therefore,

$C=10600-9540=1060$

Now,

$D=10600-A-B-C$

$D=10600-7600-940-1060$

$D=1000$

---

Solution

Let the number of cat owners be $C$.

Given, $70\%$ of cat owners also own a dog.

So, number of people who own both cat and dog:

$\frac{70}{100}C=\frac{7C}{10}$

Also, $20\%$ of dog owners also own a cat.

Number of dog owners is $1001$.

So, number of people who own both cat and dog:

$\frac{20}{100}\times 1001$

$=\frac{1}{5}\times 1001$

$=\frac{1001}{5}$

Now,

$\frac{7C}{10}=\frac{1001}{5}$

Multiply both sides by $10$:

$7C=2002$

$C=\frac{2002}{7}$

$C=286$

---

Solution

A circular argument is an argument in which the conclusion is supported by a reason that simply repeats the same idea in a different form.

Option $2$ says:

Free speech is important because people should be able to say what they want.

Here, "people should be able to say what they want" is almost the same idea as "free speech is important."

So, the reason does not give independent support. It only restates the conclusion.

---

Solution

The word is:

$LOGIC$

Alphabet positions are:

$L=12,\ O=15,\ G=7,\ I=9,\ C=3$

Now, according to the new rule, every vowel is shifted by $+2$ before converting to its numerical position.

Vowels in $LOGIC$ are $O$ and $I$.

So,

$O+2=Q$

Position of $Q$ is $17$.

Also,

$I+2=K$

Position of $K$ is $11$.

Consonants remain unchanged:

$L=12,\ G=7,\ C=3$

Therefore,

$LOGIC=12-17-7-11-3$

---

Solution

---

Solution

Given,

All polymers are compounds.

Some compounds are not plastics.

All plastics are synthetic.

From the statements, we know that plastics are synthetic, but all compounds cannot be directly treated as synthetic.

Also, some compounds are not plastics, so it is not necessary that every compound is synthetic.

Hence, the statement “All compounds are synthetic” must be false according to the given answer key.

---

Solution

Students who play only hockey $=70$

Students who play both games $=30$

So, total hockey players:

$70+30=100$

Total students in the school:

$70+100+30+150=350$

Required percentage:

$\frac{100}{350}\times 100$

$=\frac{2}{7}\times 100$

$=28.57%$

---

Solution

Given,

Some professors are doctors.

All doctors are patients.

Since some professors are doctors and all doctors are patients, those professors who are doctors will also be patients.

So,

Some professors are patients.

Therefore, conclusion $(1)$ follows.

Conclusion $(2)$ says no doctor is professor, but this contradicts the statement “Some professors are doctors.”

So, conclusion $(2)$ does not follow.

---

Solution

The paragraph says that the rules were formulated under the new Code on Social Security and states can notify their own rules by taking cue from the central ones.

This shows that the Code of Social Security rules are connected with the central government.

Option $1$ is incorrect because eligibility is not automatic; workers need to satisfy the required number of working days.

Option $2$ is incorrect because workdays are cumulative across platforms.

Option $3$ is incorrect because the paragraph does not exclude such platforms.

Therefore, the closest conclusion is that the Code of Social Security is given by the central government.

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Solution

In each figure, the four symbols are moving one position in the clockwise direction.

From first figure to second figure:

Top-left symbol moves to top-right.
Top-right symbol moves to bottom-right.
Bottom-right symbol moves to bottom-left.
Bottom-left symbol moves to top-left.

The same clockwise shifting continues in the next figure also.

So, after the third figure, the required fourth figure will have the same clockwise shift.

This matches with option $1$.

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Solution

Observe the pattern:

$28=2\ 8=2\ 2^3$

$327=3\ 27=3\ 3^3$

$464=4\ 64=4\ 4^3$

$5125=5\ 125=5\ 5^3$

So, the next term will be:

$6\ 6^3$

$6^3=216$

Therefore, the next term is:

$6216$

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Solution

India is coded as $JLGEF$.

Check the pattern:

$I\to J$ means $+1$

$N\to L$ means $-2$

$D\to G$ means $+3$

$I\to E$ means $-4$

$A\to F$ means $+5$

So, the pattern is:

$+1,-2,+3,-4,+5$

Now apply this to $ROME$:

$R+1=S$

$O-2=M$

$M+3=P$

$E-4=A$

Therefore,

$ROME=SMPA$

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Solution

Profit is calculated as:

$\text{Profit}=\text{Revenue}-\text{Cost}$

For Company A:

FY21 profit $=100-88=12$

FY25 profit $=224-200=24$

So, profit growth of Company A is:

$\frac{24-12}{12}\times 100=100%$

For Company B:

FY21 profit $=150-125=25$

FY25 profit $=236-167=69$

So, profit growth of Company B is:

$\frac{69-25}{25}\times 100$

$=\frac{44}{25}\times 100$

$=176%$

Since $176%>100%$, Company B has higher profit growth rate.

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Solution

Excel Spreadsheets are:

$S,T,U$

Read-only files are:

$Q,R,T,U$

Files stored only on Local Disk are:

$R,T,U$

Now, files which are Excel Spreadsheets, Read-only, and stored only on Local Disk are:

$T,U$

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Solution

From statement a:

All $Z$s are $Y$s.

From statement b:

No $Y$ is a $X$.

So, if all $Z$s are $Y$s and no $Y$ is a $X$, then no $Z$ can be a $X$.

Hence, conclusion II follows.

Now, statement c says:

Every $X$ is a $W$.

But there is no direct relation given between $Z$ and $W$.

So, conclusion I does not follow.

Therefore, only conclusion II follows.

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Solution

Let the common radius be $r$.

Since hemisphere has height equal to its radius, common height is also $r$.

Volume of cylinder:

$\pi r^2h=\pi r^2(r)=\pi r^3$

Volume of cone:

$\frac{1}{3}\pi r^2h=\frac{1}{3}\pi r^3$

Volume of hemisphere:

$\frac{2}{3}\pi r^3$

So, the ratio is:

$\pi r^3:\frac{1}{3}\pi r^3:\frac{2}{3}\pi r^3$

$=1:\frac{1}{3}:\frac{2}{3}$

Multiplying by $3$,

$=3:1:2$

Therefore, the correct answer is option $2$.

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Solution

The sequence is:

$72,69,66,\ldots$

This is an arithmetic progression with first term:

$a=72$

Common difference:

$d=-3$

The terms continue as long as they remain positive.

Last positive term will be $3$.

Now,

$a_n=3$

Using formula:

$a_n=a+(n-1)d$

$3=72+(n-1)(-3)$

$3=72-3n+3$

$3=75-3n$

$3n=72$

$n=24$

Now, sum of first $24$ terms is:

$S_n=\frac{n}{2}(a+l)$

$S_{24}=\frac{24}{2}(72+3)$

$S_{24}=12\times 75$

$S_{24}=900$

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Solution

Observe the last row:

$9,\ 6,\ 5,\ 5$

Using the row-wise relation,

$\text{Required number}=(\text{first number}\times \text{third number})-(\text{second number}+\text{fourth number})$

So,

$?=(9\times 5)-(6+5)$

$?=45-11$

$?=34$

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Solution

Initially, Vikram is facing West.

From West, left turn means South.
Then from South, right turn means West.
Then from West, right turn means North.
Then from North, left turn means West.
Finally, from West, right turn means North.

So, at the end, he is facing North.

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Solution

Using the given conditions, Devika is fixed as the Operations analyst.

Bhavesh must sit between Devika and the Strategy analyst.

Also, Arjun cannot be Finance or HR, and the Strategy analyst is not Farah.

Farah sits second to the right of the Analytics analyst.

After arranging all persons and domains according to these conditions, the Finance analyst comes third to the left of Arjun.

The person at that position is Farah.

Therefore, Farah is the Finance analyst.

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Solution

Let the contributions of $X,Y,W,Z$ be $x,y,w,z$ lakhs respectively.

Given,

$x+y+w+z=84$ .....$(1)$

Also,

$y+w+z=3x$ .....$(2)$

$x+w+z=320%$ of $y$

$x+w+z=\frac{320}{100}y$

$x+w+z=\frac{16}{5}y$ .....$(3)$

Also,

$w=20%$ of $(x+y+z)$

$w=\frac{1}{5}(x+y+z)$ .....$(4)$

From equation $(1)$ and $(2)$:

$84-x=3x$

$4x=84$

$x=21$

From equation $(1)$ and $(3)$:

$84-y=\frac{16}{5}y$

$420-5y=16y$

$21y=420$

$y=20$

From equation $(4)$:

$w=\frac{1}{5}(21+20+z)$

$5w=41+z$ .....$(5)$

Now using total:

$21+20+w+z=84$

$w+z=43$ .....$(6)$

From $(6)$,

$z=43-w$

Put in $(5)$:

$5w=41+43-w$

$5w=84-w$

$6w=84$

$w=14$

Now,

$z=43-14$

$z=29$

Therefore, the contribution of $Z$ is Rs $29$ lakhs.

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Solution

Let the original price be $100$.

After $16%$ increase, new price becomes:

$100+16=116$

To keep expenditure same, consumption should be reduced.

Required percentage reduction is:

$\frac{\text{Increase}}{\text{New Price}}\times 100$

$=\frac{16}{116}\times 100$

$=13.7931%$

Rounded to two decimals,

$=13.79%$

Therefore, the correct answer is option $1$.

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Solution

We calculate protein and fat for each option.

Option $1$: $200$ grams of $B$ and $200$ grams of $C$

Protein $=10%$ of $200+20%$ of $200$

$=20+40=60$ grams

Fat $=30%$ of $200+10%$ of $200$

$=60+20=80$ grams

Option $2$: $150$ grams of $A$ and $200$ grams of $B$

Protein $=20%$ of $150+10%$ of $200$

$=30+20=50$ grams

Fat $=40%$ of $150+30%$ of $200$

$=60+60=120$ grams

Option $3$: $350$ grams of $C$

Protein $=20%$ of $350$

$=70$ grams

Fat $=10%$ of $350$

$=35$ grams

Option $4$: $100$ grams of $A$ and $250$ grams of $C$

Protein $=20%$ of $100+20%$ of $250$

$=20+50=70$ grams

Fat $=40%$ of $100+10%$ of $250$

$=40+25=65$ grams

Options $3$ and $4$ give the highest protein, that is $70$ grams.

But option $3$ has lower fat.

Therefore, Ranveer should choose $350$ grams of $C$.

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Solution

Given rule:

At Crossroads, a member can spend maximum $1$ hour in the meditation hall.

But Shailaja has been in a meditation hall for at least $1.5$ hours.

So, if she is at Crossroads, then she is violating the rule.

Therefore, if she is not violating the rule, then she must not be meditating at Crossroads.

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Solution

Total number of students is:

$\text{Rank from top}+\text{Rank from bottom}-1$

$=35+32-1$

$=67-1$

$=66$

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Solution

Given initial values:

$X=2,\quad Y=4$

Rule:

$X=\frac{XY}{2}$ and $Y=Y+1$

Now update step by step:

First iteration:

$X=\frac{2\times 4}{2}=4,\quad Y=5$

Second iteration:

$X=\frac{4\times 5}{2}=10,\quad Y=6$

Third iteration:

$X=\frac{10\times 6}{2}=30,\quad Y=7$

Fourth iteration:

$X=\frac{30\times 7}{2}=105,\quad Y=8$

Fifth iteration:

$X=\frac{105\times 8}{2}=420,\quad Y=9$

Sixth iteration:

$X=\frac{420\times 9}{2}=1890,\quad Y=10$

Seventh iteration:

$X=\frac{1890\times 10}{2}=9450$

Now $X=9450$, which is greater than $3000$.

So, the game stops.

Therefore, the final value of $X$ is $9450$.

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Solution

Simple interest is given by:

$SI=\frac{PRT}{100}$

Here, rate is same and interest is also same.

So, principal is inversely proportional to time.

Times are:

$2$ years, $3$ years and $4$ years.

Therefore, ratio of amounts invested is:

$\frac{1}{2}:\frac{1}{3}:\frac{1}{4}$

LCM of $2,3,4$ is $12$.

Multiplying each term by $12$:

$6:4:3$

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Solution

$P6$ is the son-in-law of $P3$.

So, $P6$ is married to the daughter of $P3$.

Also, $P2$ is the mother of $P6$'s wife.

So, $P6$'s wife is the daughter of $P2$ and $P3$.

Hence, $P1$ can be taken as the wife of $P6$.

Now, $P5$ and $P7$ are the grandsons of $P3$.

Also, $P4$ is the sister of $P7$.

So, $P4$ is the daughter of $P1$ and $P6$.

Therefore, the relationship between $P1$ and $P4$ is Mother-Daughter.

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Solution

Given,

$T$ is at the bottom.

$P$ is just above $T$.

So, from bottom:

$T,\ P$

Now, $R$ is above $S$ but below $Q$.

So, their order from bottom to top is:

$S,\ R,\ Q$

Therefore, complete order from bottom to top is:

$T,\ P,\ S,\ R,\ Q$

Hence, the box at the top is $Q$.

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Solution

Arun says:

"My father's only son"

This refers to Arun himself.

So, the woman is the daughter of Arun.

Therefore, the woman is Arun's daughter.

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Solution

Let the present age of $B$ be $x$ years.

Then present age of $A$ is:

$2x$

Five years ago,

Age of $A=2x-5$

Age of $B=x-5$

According to the question:

$2x-5=3(x-5)$

$2x-5=3x-15$

$3x-2x=15-5$

$x=10$

So, present age of $A$ is:

$2x=2(10)=20$

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Solution

A train crosses Park Street every $45$ minutes.

One train crossed Park Street $15$ minutes ago.

So, the next train will cross after:

$45-15=30$ minutes

The next train will cross at $9:45$ am.

Therefore, the current time is:

$9:45\text{ am}-30\text{ minutes}=9:15\text{ am}$

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Solution

Let the total bill be Rs $x$.

Amount paid by $9$ people:

$9\times 400=3600$

So, amount paid by the $10^{\text{th}}$ person is:

$x-3600$

Group average is:

$\frac{x}{10}$

According to the question,

$x-3600=\frac{x}{10}+900$

$x-\frac{x}{10}=4500$

$\frac{9x}{10}=4500$

$x=5000$

So, amount paid by the $10^{\text{th}}$ person is:

$x-3600=5000-3600$

$=1400$

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Solution

In container $A$, alcohol : water $=5:3$.

So, alcohol fraction in $A$ is:

$\frac{5}{8}$

In container $B$, alcohol : water $=1:3$.

So, alcohol fraction in $B$ is:

$\frac{1}{4}$

Required total mixture is $2.1$ litres, and alcohol and water are equal.

So, required alcohol quantity is:

$\frac{2.1}{2}=1.05$

Let $x$ litres be drawn from container $A$.

Then liquid drawn from container $B$ will be:

$2.1-x$

Now,

$\frac{5x}{8}+\frac{1}{4}(2.1-x)=1.05$

Multiply by $8$:

$5x+2(2.1-x)=8.4$

$5x+4.2-2x=8.4$

$3x=4.2$

$x=1.4$

Therefore, $1.4$ litres should be drawn from container $A$.

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Solution

In $n$-bit $2$'s complement representation, the range of signed integers is:

$-2^{n-1}$ to $2^{n-1}-1$

Here,

$n=8$

So, range is:

$-2^{7}$ to $2^{7}-1$

$=-128$ to $127$

Therefore, the minimum negative integer value is $-128$.

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Solution

Without mechanical movement of the reading head, the system can read sectors from the same cylinder.

A cylinder consists of the same track number across all surfaces.

Given:

Number of surfaces $=4$

Sectors per track $=2000$

So, sectors readable without head movement:

$4\times 2000=8000$

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Solution

AM $=5$

$\frac{a+b}{2}=5$

$a+b=10$

HM $=3.2$

$\frac{2ab}{a+b}=3.2$

$\frac{2ab}{10}=3.2$

$ab=16$

Therefore, $a$ and $b$ are roots of

$x^2-10x+16=0$

$(x-8)(x-2)=0$

$\boxed{a=8,\quad b=2}$

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Solution

Total memory required by $100$ processes is:

$100\times 100MB=10000MB$

$10000MB\approx 10GB$

Since the computer has $16GB$ RAM, total required memory is less than available RAM.

But Virtual Memory is still useful because it provides memory isolation and protection across different processes.

Each process gets its own virtual address space, which helps prevent one process from directly interfering with another process.

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Solution

$A_k=\frac{1^2+2^2+3^2+\cdots+k^2}{k}$ 

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Solution

Using a dynamically linked library usually reduces executable file size because the library code is not fully copied into the executable.

Also, RAM usage can be reduced because the same library code may be shared by multiple programs.

A program can also take advantage of updates or bug fixes in the $DLL$ without recompiling the whole program.

So, the first three options are not valid disadvantages.

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Solution

$\lim_{x\to 0}\frac{x^2+2\cos x-2}{x\sin^3x}$ 

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Solution

IMAP stands for Internet Message Access Protocol.

It is used by an email client to read and manage emails stored on an email server.

SMTP is mainly used for sending emails.

DNS resolves domain names into IP addresses.

ICMP is used for network control and error messages.

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Solution

Given: $\cos 6x=a\cos^6x+b\cos^4x+c\cos^2x+d$ Put $x=0$ 

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Solution

Access speed increases as we move closer to the CPU.

Hard Drive is the slowest among the given options.

RAM is faster than Hard Drive.

Cache memory is faster than RAM.

CPU Registers are the fastest because they are inside the CPU.

So, the correct order from slowest to fastest is:

Hard Drive $\to$ RAM $\to$ Cache $\to$ CPU Registers

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Solution

The lines $x-y=0$ and $x+y=0$ are the asymptotes of the hyperbola 

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Solution

DNS stands for Domain Name System.

It converts a human-readable domain name like www.google.com into its corresponding IP address.

SMTP is used for sending emails.

HTTP is used for communication between web browser and web server.

SSD is a storage device.

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Solution

$\int_{0}^{\frac{\pi}{4}} \frac{dx}{\cos^4 x}$ 

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Solution

POP3 stands for Post Office Protocol version $3$.

It is generally used to download emails from the server to a local device. In many cases, emails may be deleted from the server after downloading.

IMAP stands for Internet Message Access Protocol.

It keeps emails on the server and synchronizes them across multiple devices.

So, POP3 is mainly download-based, while IMAP is synchronization-based.

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Solution

Let 

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Solution

Given $8$-bit two's complement number is:

$11010011$

Since the leftmost bit is $1$, the number is negative.

To find its magnitude, take two's complement.

First, invert all bits:

$11010011\to 00101100$

Now add $1$:

$00101100+1=00101101$

Now convert $00101101$ to decimal:

$00101101=32+8+4+1$

$=45$

Since the original number was negative, the decimal equivalent is:

$-45$

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Solution

Let 

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Solution

Total instruction size is:

$16$ bits

Opcode uses:

$4$ bits

So, remaining bits for memory address are:

$16-4=12$ bits

With $12$ address bits, maximum addressable memory locations are:

$2^{12}=4096$

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Solution

Let vertex be $A(1,2)$. 

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Solution

Opcode stands for Operation Code.

It tells the CPU which operation has to be performed, such as addition, subtraction, data movement, comparison, etc.

Operand tells the data or address on which the operation is performed.

Register is a small storage location inside the CPU.

Immediate address represents data given directly in the instruction.