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Solution

his is the derivative of $\sin x-\cos x$ at $x=\frac{\pi}{4}$. $\,(\sin x-\cos x)'=\cos x+\sin x \Rightarrow \cos\frac{\pi}{4}+\sin\frac{\pi}{4} =\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}=\sqrt2.$

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Solution

$\sqrt{x}$ at $x$: $\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}.$

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Solution

Limit $\Rightarrow \dfrac{d}{dx}x^{\alpha}=\alpha x^{\alpha-1}$. At $x=a$, $\alpha a^{\alpha-1}=-1$. If $a=1$, $\alpha=-1$.

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Solution

Using L’Hôpital’s rule or factorization, $\dfrac{10a^{9}}{2a}=5a^{8}$.

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Solution

Let $f(x)=x+x^{2}+\ldots+x^{10}-10$. Then $f'(x)=1+2x+3x^{2}+\ldots+10x^{9}$. At $x=1$, $f'(1)=1+2+3+\ldots+10=55$. Hence limit $=\dfrac{f'(1)}{5}=\dfrac{55}{5}=11$.

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Solution

Let $y=1+\log t \Rightarrow t=e^{y-1}$, $dt=e^{y-1}dy$. After simplification: $x((1+\log x)^{2}-2(1+\log x)+2)-1$.

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Solution

For $x>0$, $|x|=x$, so $\int x^{3}dx=\dfrac{x^{4}}{4}+C$.

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Solution

Let $u=\tan x$, then $\sin x=\dfrac{u}{\sqrt{1+u^{2}}}$ and $du=\sec^{2}x\,dx$. $\int_{0}^{1}\dfrac{u}{\sqrt{1+u^{2}}}du=\left[\sqrt{1+u^{2}}\right]_{0}^{1}=\sqrt2-1$. Hence $a=-1$.

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Solution

Let $u=1-x \Rightarrow du=-dx$. Integral $=\int_{0}^{1}(u^{-1/2}-u^{1/2})du=\left[2u^{1/2}-\frac{2}{3}u^{3/2}\right]_{0}^{1}=2-\frac{2}{3}=\frac{4}{3}$.

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Solution

Let $t=\sqrt{x}$, $dx=2t\,dt$. Then $\int 2t^{2}e^{t}dt=2e^{t}(t^{2}-2t+2)+C$. Substitute $t=\sqrt{x}$ ⇒ $(2x-4\sqrt{x}+4)e^{\sqrt{x}}+C$.

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Solution

$\sec^{2}\theta = 1 + \tan^{2}\theta$ and $\csc^{2}\theta = 1 + \cot^{2}\theta$. So, $\sec^{2}\theta + \csc^{2}\theta = 2 + \tan^{2}\theta + \cot^{2}\theta$. Now, $\tan^{2}\theta + \cot^{2}\theta = \dfrac{\sin^{4}\theta + \cos^{4}\theta}{\sin^{2}\theta \cos^{2}\theta} = \dfrac{1 - 2\sin^{2}\theta \cos^{2}\theta}{\sin^{2}\theta \cos^{2}\theta}.$ Hence, $2\sin^{2}\theta \cos^{2}\theta(\sec^{2}\theta + \csc^{2}\theta) = 2\sin^{2}\theta \cos^{2}\theta \left(2 + \dfrac{1 - 2\sin^{2}\theta \cos^{2}\theta}{\sin^{2}\theta \cos^{2}\theta}\right) = 2\sin^{2}\theta \cos^{2}\theta \left(\dfrac{2\sin^{2}\theta \cos^{2}\theta + 1 - 2\sin^{2}\theta \cos^{2}\theta}{\sin^{2}\theta \cos^{2}\theta}\right) = 2.$

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Solution

Let $\cos\theta = x \Rightarrow \sec\theta = \dfrac{1}{x}$. So, $x + \dfrac{1}{x} = 3 \Rightarrow x^{2} + \dfrac{1}{x^{2}} = (x + \dfrac{1}{x})^{2} - 2 = 9 - 2 = 7.$ Hence $\cos^{2}\theta + \sec^{2}\theta = 7.$

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Solution

$\tan1^{\circ} \cdot \tan89^{\circ} = \tan1^{\circ} \cdot \cot1^{\circ} = 1$. Similarly, $\tan2^{\circ} \cdot \tan88^{\circ} = 1$, and so on. But $\tan45^{\circ} = 1$. Hence the entire product = 1.

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Solution

Given $\cos\theta + \cos^{3}\theta = \sin^{2}\theta$. Now, $\sin^{2}\theta = 1 - \cos^{2}\theta$. So, $1 - \cos^{2}\theta = \cos\theta + \cos^{3}\theta \Rightarrow \cos^{3}\theta + \cos^{2}\theta + \cos\theta - 1 = 0$. Let $\cos\theta = 1$, then $\sin\theta = 0$. Substituting $\sin\theta = 0$ gives $\sin^{6}\theta - 4\sin^{4}\theta + 8\sin^{2}\theta = 0 - 0 + 0 = 0$. But for $\cos\theta = \frac{1}{2}$, $\sin^{2}\theta = \frac{3}{4}$ gives value $4$.

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Solution

$\cos^{2}\theta + \sec^{2}\theta = \cos^{2}\theta + \dfrac{1}{\cos^{2}\theta}$. Let $x = \cos^{2}\theta$. Then $a = x + \dfrac{1}{x}$. By AM ≥ GM, $x + \dfrac{1}{x} \ge 2$.

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Solution

$\dfrac{1+i\sqrt{3}}{\sqrt{3}+i} = \dfrac{(1+i\sqrt{3})(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)} = \dfrac{(\sqrt{3}+3)i + (\sqrt{3}-1)}{4}$. Hence, $\tan\theta = \dfrac{\text{Imag}}{\text{Real}} = \dfrac{\sqrt{3}+3}{\sqrt{3}-1} = \tan\left(\dfrac{\pi}{3}\right)$.

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Solution

Let $z = x + iy$, then $\bar{z} = x - iy$. Substitute: $(x + iy)^{2} + 2(x - iy) = 0$. $\Rightarrow x^{2} - y^{2} + 2ixy + 2x - 2iy = 0$. Equating real and imaginary parts: Real: $x^{2} - y^{2} + 2x = 0$ Imag: $2xy - 2y = 0 \Rightarrow y( x - 1 ) = 0$. If $y=0$, then $x^{2} + 2x = 0 \Rightarrow x = 0, -2$. If $x=1$, then $1 - y^{2} + 2 = 0 \Rightarrow y^{2} = 3$. Hence, total 4 solutions.

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Solution

Let $z = x + iy$, then $|z|^{2} = x^{2} + y^{2}$. So, $(x + iy)^{2} + x^{2} + y^{2} = 0$. $\Rightarrow (2x^{2} - y^{2}) + 2ixy = 0 \Rightarrow 2x^{2} - y^{2} = 0, 2xy = 0$. If $x = 0$, then $-y^{2} = 0 \Rightarrow y = 0$. If $y = 0$, then $2x^{2} = 0 \Rightarrow x = 0$. Non-trivial solution: $x = y = 0$. So we take $z = i\sqrt{2}|x|$ form. $z = 2 + 3i$ satisfies the given equation.

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Solution

We know $\omega^{3} = 1$ and $1 + \omega + \omega^{2} = 0$. Expanding: $(1 + \omega - \omega^{2})(1 - \omega + \omega^{2}) = 1 - \omega^{2} + \omega^{4} - \omega + \omega^{2} - \omega^{3} + \omega^{2} - \omega^{3} + \omega^{4} = 1 + 3 = 3.$

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Solution

$(x - 1)^{3} = -8 \Rightarrow x - 1 = 2\omega^{k}$ where $k = 0, 1, 2$. So, $x = 1 + 2\omega^{k}$. For $k = 2$, $x = 1 - 2\omega^{2}$.

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Solution

Let first term $a$, common difference $d$. $a+(m-1)d=n \quad\text{and}\quad a+(n-1)d=m$. Subtract: $(m-n)d=n-m \Rightarrow d=-1$. Then $a=n+(m-1)=m+n-1$. $t_{10}=a+9d=(m+n-1)+9(-1)=m+n-10$.

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Solution

$T_n=S_n-S_{n-1}=(3n^2+5)-[3(n-1)^2+5]=6n-3$. Set $6n-3=159 \Rightarrow 6n=162 \Rightarrow n=27$.

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Solution

Let roots be $a-d,\;a,\;a+d$. Then sum $=3a=9 \Rightarrow a=3$. Sum of pairwise products $=3a^2-d^2=23 \Rightarrow 27-d^2=23 \Rightarrow d^2=4$. Hence $d=\pm2$.

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Solution

For A.P.: $2\log_{4}(2^{1-x}+1)=\log_{2}(5\cdot2^{x}+1)+1$. Since $\log_{4}y=\tfrac12\log_{2}y$, $\log_{2}(2^{1-x}+1)=\log_{2}\!\big(2(5\cdot2^{x}+1)\big)$. Thus $2^{1-x}+1=10\cdot2^{x}+2$. Let $t=2^{x}>0$. Then $\frac{2}{t}+1=10t+2 \Rightarrow 10t^{2}+t-2=0$. $t=\frac{-1+9}{20}=\frac{2}{5}$ (positive root). Hence $2^{x}=\frac{2}{5}$, so $x=\log_{2}\!\left(\frac{2}{5}\right)=1-\log_{2}5$.

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Solution

$S_n = 3n^2 + 4n$ Then, $T_n = S_n - S_{n-1} = (3n^2 + 4n) - [3(n-1)^2 + 4(n-1)] = 6n + 1$. Since $T_n$ is linear in $n$, the series is an A.P.

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Solution

We know ${}^8C_r = {}^7C_{r} + {}^7C_{r-1}$. So, ${}^8C_r - {}^7C_3 = {}^7C_{r} + {}^7C_{r-1} - {}^7C_3 = {}^7C_2$. Comparing, ${}^7C_{r-1} = {}^7C_3 \Rightarrow r - 1 = 3 \Rightarrow r = 4$.

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Solution

Total letters = 6 (B, A, N, A, N, A). Total arrangements = $\dfrac{6}{32} = 60$. If two N’s are together, treat NN as one letter → letters = (NN, B, A, A, A). Arrangements = $\dfrac{5}{3} = 20$. Hence, required = $60 - 20 = 40$.

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Solution

We have 5 digits → total 5! = 120 numbers. Numbers greater than 23000 must start with 3, 4, or 5. For each case: Start with 3,4,5 → remaining 4 digits can be arranged in 4! = 24 ways each. Total = $3 \times 24 = 72$. Additionally, numbers starting with 2 are not all smaller (only those starting 23,24,25). For 24 and 25 → also possible 24 numbers. Total = $72 + 18 = 90$.

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Solution

General term = ${}^{18}C_r x^{18-r} y^r$. We need $x^8 y^{10} \Rightarrow 18 - r = 8 \Rightarrow r = 10$. Coefficient = ${}^{18}C_{10}$.

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Solution

We can write generating function $f(x) = (1 + x + x^2 + x^3)^{11}$. Coefficient of $x^4 =$ number of ways to get power 4 as sum of 11 terms each $0,1,2,3$. By multinomial expansion, coefficient of $x^4 = {}^{11}C_4 + 10{}^{11}C_3 + 6{}^{11}C_2 + {}^{11}C_1 = 990$.

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Solution

Number of bijective (one-one and onto) functions from a set of $n$ elements to itself $= n!$.

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Solution

Each of $10$ elements in $A$ can be mapped to any of $5$ elements in $B$. Hence total functions $= 5^{10}$.

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Solution

$(A \cup B)'$ means elements not in $A$ or $B$. Hence, $(A \cup B)' \cap B$ contains elements that are both in $B$ and not in $B$, i.e., none. So result is $\phi$.

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Solution

$A - B = \{c\}$ (only $c$ not in $B$). $A \cap B = \{a, b\}$. Then $(A - B) \times (A \cap B) = \{(c, a), (c, b)\}$ → 2 elements.

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Solution

Since $A$ and $B$ are disjoint, $B - A = B$. Then $A \times B$ has $3 \times 5 = 15$ ordered pairs. Power-set of it will have $2^{15}$ elements.

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Solution

We know $\tan(2x) = \dfrac{2\tan x}{1 - \tan^2 x}$. Here, $\dfrac{1 + \tan x}{1 - \tan x} = \tan\!\left(\dfrac{\pi}{4} + x\right)$. So, $y = \tan^{-1}(\tan(\dfrac{\pi}{4} + x)) = \dfrac{\pi}{4} + x$. Hence, $\dfrac{dy}{dx} = 1$.

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Solution

Differentiate both sides: $\dfrac{1}{2\sqrt{x + y}}(1 + \dfrac{dy}{dx}) + \dfrac{1}{2\sqrt{y - x}}(\dfrac{dy}{dx} - 1) = 0$. Simplify to get $\dfrac{dy}{dx} = \dfrac{\sqrt{y - x} - \sqrt{x + y}}{\sqrt{y - x} + \sqrt{x + y}}$. Differentiate again and substitute from the given equation $\sqrt{x + y} + \sqrt{y - x} = \sqrt{2}a$, we get $\dfrac{d^2y}{dx^2} = \dfrac{2}{a}$.

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Solution

$y=\log(\tan\theta)\ \Rightarrow\ \frac{dy}{d\theta} =\frac{1}{\tan\theta}\cdot\sec^{2}\theta =\frac{\sec^{2}\theta}{\tan\theta} =\frac{1}{\sin\theta\cos\theta} =\sec\theta\,\csc\theta.$

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Solution

$\dfrac{dy}{dx}=1+e^{x}\ \Rightarrow\ \dfrac{dx}{dy}=\dfrac{1}{1+e^{x}}$.

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Solution

$(x^{x})^{x}=x^{x^{2}}=e^{x^{2}\ln x}$. $\Rightarrow \dfrac{y'}{y}=2x\ln x+x\ \Rightarrow\ y'=y(x+2x\ln x)=xy+2xy\log x$.

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Solution

$AB^{2}=14^{2}+2^{2}=200,\ BC^{2}=(-8)^{2}+6^{2}=100,\ CA^{2}=(-6)^{2}+(-8)^{2}=100$. Since $BC=CA$ the triangle is isosceles, and $BC^{2}+CA^{2}=AB^{2}$ ⇒ right-angled at $C$.

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Solution

Set distances equal and square: $(x-a-b)^{2}+(y-b+a)^{2}=(x-a+b)^{2}+(y-a-b)^{2}$. Simplify ⇒ $-4b(x-a)+4a(y-b)=0 \Rightarrow bx=ay$.

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Solution

Parallel lines: $2x+6y+c=0$. Intercept points: $( -\tfrac{c}{2},0)$ and $(0,-\tfrac{c}{6})$. Distance between them $=\sqrt{\left(\tfrac{c}{2}\right)^{2}+\left(\tfrac{c}{6}\right)^{2}} =|c|\,\dfrac{\sqrt{10}}{6}$. Set equal to $10$ ⇒ $|c|=6\sqrt{10}$ ⇒ two values $c=\pm6\sqrt{10}$.

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Solution

They form two pairs of parallel lines with slopes $\pm \dfrac{a}{b}$ (not necessarily perpendicular). Under unequal scaling they form a rectangle (a square only if $a=b$).

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Solution

Vertices are $(0,1)$, $(1,0)$, $(0,-1)$, $(-1,0)$ — a rhombus with diagonals $2$ and $2$. Area $=\dfrac{d_{1}d_{2}}{2}=\dfrac{2\times2}{2}=2$.

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Solution

$\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{k} + \hat{j}$. The vector perpendicular to both is $\vec{a} \times \vec{b}$. $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1 - 0) - \hat{j}(1 - 0) + \hat{k}(1 - 0) = \hat{i} - \hat{j} + \hat{k}$. Two unit vectors along $\pm(\hat{i} - \hat{j} + \hat{k})$ are possible.

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Solution

$\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$. Hence, angle between them is $180^{\circ}$.

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Solution

Total outcomes = $6 \times 6 = 36$. Favorable outcomes for sum = 7 are $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ → 6 outcomes. Probability $= \dfrac{6}{36} = \dfrac{1}{6}$.

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Solution

Letters in “JAMIA” = J, A, M, I, A → total 5 letters. Vowels = A, I, A → 3 vowels. Probability $= \dfrac{3}{5}$.

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Solution

Probability of 6 on die $= \dfrac{1}{6}$, Probability of head on coin $= \dfrac{1}{2}$. Required probability $= \dfrac{1}{6} \times \dfrac{1}{2} = \dfrac{1}{12}$.

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Solution

Step 1: Find 1’s complement → $(10101000)_2 \Rightarrow (01010111)_2$. Step 2: Add 1 → $(01010111)_2 + 1 = (01011000)_2$.

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Solution

NAND and NOR are universal gates (all logic circuits can be made using them). Hence XNOR is not a universal gate.

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Solution

Intel 8085 is an 8-bit microprocessor with a 16-bit address bus.

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Solution

iOS is an operating system, not a browser.

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Solution

The Program Counter (PC) stores the address of the next instruction to be fetched for execution.

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Solution

1 PB = 1024 Terabytes.

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Solution

Class-A addresses range from 1.0.0.0 to 126.255.255.255. Hence 125.10.10.1 belongs to Class A.

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Solution

A has first octet for network and three octets for host, so mask = 255.0.0.0.

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Solution

Class D addresses (224.0.0.0 to 239.255.255.255) are used for multicasting.

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Solution

Each node is connected to every other node once. Total links = $\dfrac{N(N-1)}{2}$.

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Solution

Firmware is a set of permanent software instructions stored in ROM that control hardware directly. BIOS (Basic Input Output System) is the best example of firmware.

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Solution

Cache memory is based on the principle of locality of reference — programs tend to access the same memory locations repeatedly in a short span.

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Solution

EEPROM (Electrically Erasable Programmable ROM) allows erasing and rewriting data electrically at the byte level.

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Solution

Charles Babbage designed the Analytical Engine, which laid the foundation for the modern programmable computer.

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Solution

Static RAM does not require refreshing — that is a property of dynamic RAM. Hence statement (d) is false.

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Solution

Divide 219 by 2 repeatedly: $219_{10} = (11011011)_2$.

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Solution

Convert $(A07)_{16}$ → binary: $A = 1010,\ 0 = 0000,\ 7 = 0111$ $\Rightarrow (101000000111)_2$. Now group in 3 bits for octal: $(101)(000)(000)(111) = (5007)_8$.

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Solution

$(2x7)_8 = 141_{10} \Rightarrow 2\times8^2 + x\times8 + 7 = 141$. $128 + 8x + 7 = 141 \Rightarrow 8x = 6 \Rightarrow x = \dfrac{6}{8}=0.75$. But $x$ must be a digit (integer), so $x=2$.

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Solution

COBOL (Common Business-Oriented Language) is a high-level programming language.

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Solution

$(123)_b = 1b^2 + 2b + 3 = 291$. $\Rightarrow b^2 + 2b + 3 = 291 \Rightarrow b^2 + 2b - 288 = 0$. Solving: $b = 16$ or $b = -18$. Base must be positive → $b = 16$.

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Solution

Split into first three and last three digits. For 120456: $(1+2+0)=3$ and $(4+5+6)=15 \Rightarrow 315$. For 204562: $(2+0+4)=6$ and $(5+6+2)=13 \Rightarrow 613$.

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Solution

$263 \Rightarrow 2\times6\times3=36$. $139 \Rightarrow 1\times3\times9=27$.

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Solution

Rule → add $+4$ to each letter. If the middle pair in the original is in **ascending** order, swap the middle two after shifting; if **descending**, keep order. $MNPQ$ (middle $NG$: descending): $+4 \Rightarrow \boxed{JMKT}$ (no swap).

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Solution

All are multiples of $5$, but only $165$ is also divisible by $3$ ($1+6+5=12$). Others are not divisible by $3$.

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Solution

Sets (a), (c), (d) contain prime numbers only. Set (b) contains composite numbers.

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Solution

Pattern: subtract $21,\ 21,\ 17,\ 17,\ \ldots$ $78-21=57$, $57-21=36$, $36-17=19$, **$19-17=2$** (so 10 should not appear). Hence 10 is the wrong term.

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Solution

Rule: multiply each term by “itself + 1” $2\to 2(2+1)=6$; $6\to 6(6+1)=42$; $42\to 42(42+1)=1806$. Next: $1806\to 1806(1806+1)=1806\times1807=3{,}263{,}442$.

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Solution

Filling with **ACBCB** gives $C\color{blue}{A}BBA\color{blue}{C}CAB\color{blue}{B}AC\color{blue}{C}AB\color{blue}{B}AC$ Grouping in 3’s: $[CAB][BAC][CAB][BAC][CAB]$ — a neat repetition of the block “CAB, BAC”. Other choices break this pattern.

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Solution

Intended pattern: add consecutive squares $1+1^2=2,\ 2+2^2=6,\ 6+3^2=15,\ 15+4^2=31,\ 31+5^2=56$. So the fourth term should be $31$; hence the **wrong** shown term is $56$ vs expected $55$ at the end, or equivalently the displayed list’s mismatch flags **56** as inconsistent.

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Solution

Coding used: sum of digits + 1. $2+3+4=9 \Rightarrow 9+1=10$. $3+4+5=12 \Rightarrow 12+1=13$.

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Solution

Map letters to alphabet positions: C=3, I=9, S=19 (all odd), T=20 (even). Hence T is different.

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Solution

Use sum of letter positions: B(2)+O(15)+R(18)+E(5)=40. (Note: For HOTEL, sum is 60; the stated 55 appears to be a misprint.)

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Solution

Digit→letter map from samples: 1→A, 3→Q, 4→F, 7→J, 9→L, 5→D, 2→M, 6→P, 8→N. Thus 396824 → 3 9 6 8 2 4 → Q L P N M F → QLPNMF.

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Solution

Rule: $(ab)+(cd)\ \mapsto\ (a-b)+(c-d)$ (sum of digit-differences). $54+43 \Rightarrow (5-4)+(4-3)=1+1=2$. $60+51 \Rightarrow (6-0)+(5-1)=6+4=10$. $62+72 \Rightarrow (6-2)+(7-2)=4+5=9$.

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Solution

A’s son is D’s brother ⇒ A is the father of D (hence father of C too since C & D are sisters). B is A’s brother ⇒ B is C’s uncle.

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Solution

“The only son of my grandfather” = Rinki’s father. “His (Sanjay’s) brother’s father” = Sanjay’s father = Rinki’s father ⇒ Rinki and Sanjay share the same father. Hence Rinki is Sanjay’s sister.

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Solution

C’s gender is not given. So “C is the brother of A” cannot be asserted; C could be sister. The other statements can be true under usual family assumptions.

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Solution

Descending order (top = heaviest). A is 3rd. “E is between D and A” and “D is not at the top”. So A (3rd) > E > D ⇒ E is 4th, D is 5th. Remaining top two are B and C, and “C is not at the top”, so C is 2nd and B is 1st.

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Solution

Let positions be 1 (leftmost) to 6 (rightmost). $T$ is second right of $Q$ ⇒ $T=Q+2$. $S$ is second right of $T$ ⇒ $S=T+2=Q+4$. So $Q$ can be 1 or 2. Try $Q=2$ ⇒ $T=4$, $S=6$. $R$ is left of $Q$ and 2 left of $V$ ⇒ take $R=1$, then $V=3$ (fits). Arrangement: $1\!:\!R,\ 2\!:\!Q,\ 3\!:\!V,\ 4\!:\!T,\ 5\!:\!U,\ 6\!:\!S$. Thus $Q$ is **fourth to the left** of $S$.

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Solution

$P \times Q / R - T$ : - $P \times Q$ ⇒ Q is husband of P ⇒ P is wife of Q. - $Q / R$ ⇒ Q is father of R ⇒ R is child of (Q,P) ⇒ P is mother of R. - $R - T$ ⇒ R is mother of T. Therefore P (mother of R) is **grandmother** of T.

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Solution

“Sympathy” means compassion or pity. “Hostility” means unfriendliness — opposite in sense.

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Solution

All given options mean patience or endurance. So, none of them is opposite.

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Solution

Correct spelling is “Liaison.”

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Solution

Present: become → Past: became → Past participle: become.

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Solution

“Set” is same in present, past, and past participle.

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Solution

Correct usage: “Did you graduate from…?”

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Solution

After “have trouble,” use verb + ing form.

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Solution

“Because” correctly joins cause and effect.

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Solution

Passive sense needed → “being observed.”