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Solution

Let the first term be $a$ and common difference $d$. $ a + 19d = 30 $ $ a + 29d = 20 $ Subtract → $10d = -10 \Rightarrow d = -1$ Then $a = 49$ 10th term = $a + 9d = 49 - 9 = 40$

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Solution

Sum $S_n = 2n(n-1)$ → $a = 2$, $d = 4$ Sum of squares = $\sum (a + (r-1)d)^2$ $= n[a^2 + (n-1)(a+d)^2 + …]$ Simplifying gives $\dfrac{8n(n+1)(2n+1)}{3}$

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Solution

Let $a=\log_2(5·2^x+1)$, $b=\log_4(2^{1-x}+1)$, $c=1$ For A.P.: $2b=a+c$ Simplify using $\log_4 y = \frac{1}{2}\log_2 y$ After algebra → $x = 1 - \log_2 5$

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Solution

$S_m/S_n = m^2/n^2$ We know $S_n = \dfrac{n}{2}[2a+(n-1)d]$ $\Rightarrow \dfrac{m[2a+(m-1)d]}{n[2a+(n-1)d]} = \dfrac{m^2}{n^2}$ Simplify → $\dfrac{2a+(m-1)d}{2a+(n-1)d} = \dfrac{m}{n}$ $\Rightarrow \text{ratio of } t_m : t_n = (2m-1):(2n-1)$

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Solution

$\log$ both sides: $\log y = \frac{1}{3}\log 9 + \frac{1}{9}\log 9 + …$ Geometric series sum = $\frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}$ $\Rightarrow \log y = \frac{1}{2}\log 9 = \log 3$ $\Rightarrow y = 3$

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Solution

$\alpha+\beta=-p$, $\alpha\beta=q$ $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = p^2 - 2q$ So $\alpha^2 + \alpha\beta + \beta^2 = p^2 - q$

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Solution

Let roots be $r$ and $r+1$ Then $b = r+(r+1) = 2r+1$, $c = r(r+1)$ $\Rightarrow b^2 - 4c = (2r+1)^2 - 4r(r+1) = 1$

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Solution

Each term is non-negative. Sum of squares = 0 ⇒ each = 0 Impossible since $x$ cannot be simultaneously 1, 2, and 3. So, no real root.

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Solution

Let the common root be $x$. From first: $x^2 + 2x + 3\lambda = 0$ From second: $2x^2 + 3x + 5\lambda = 0$ Multiply (1) by 2 and subtract: $(2x^2 + 4x + 6\lambda) - (2x^2 + 3x + 5\lambda) = 0$ $\Rightarrow x + \lambda = 0 \Rightarrow x = -\lambda$

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Solution

${}^nP_r = \dfrac{n!}{(n-r)!}$ and ${}^nC_{r-1} = \dfrac{n!}{(r-1)!(n-r+1)!}$ Equating: $\dfrac{n!}{(n-r)!} = \dfrac{n!}{(r-1)!(n-r+1)!}$ Simplify → $(r-1)! = (n-r+1)!/(n-r)! = (n-r+1)$ $\Rightarrow r-1 = n-r+1 \Rightarrow n = 2r - 2$ Try small integers: for $r=2$, $n=3$ works. Hence $(n, r) = (3, 2)$

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Solution

BANANA → letters: 3 A’s, 2 N’s, 1 B Total = $\dfrac{6}{321} = 60$ Adjacent N’s → treat NN as one unit → $\dfrac{5}{3} = 20$ Required = 60 − 20 = 40

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Solution

Using binomial relation and telescoping pattern, the series reduces to $\dfrac{1}{(n+1)(n+2)}$

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Solution

Use $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. Evaluating the determinant gives value $= -3$.

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Solution

Since $|A^2| = |A|^2 = 0$, we have $|A| = 0$. $\therefore |A| = x^2 - 4 = 0 \Rightarrow x = \pm 2.$

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Solution

Using cross and dot product conditions, $\vec{B} = \hat{k}$.

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Solution

Possible outcomes → $(1,4),(2,3),(3,2),(4,1)$ → $4$ cases. Total $=36$ outcomes. Probability $=\dfrac{4}{36}=\dfrac{1}{9}.$

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Solution

Since $A$ and $B$ are disjoint, elements common with $B'$ are only from $A$. Hence $(A \cup B)\cap B' = A.$

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Solution

$(A-B)$ means elements in $A$ but not in $B$. Subtracting $A$ again gives an empty set.

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Solution

Let $P(0, y)$. $\sqrt{(0+5)^2+(y-4)^2} = \sqrt{(0-3)^2+(y+2)^2}$ $\Rightarrow 25 + y^2 - 8y + 16 = 9 + y^2 + 4y + 4$ $\Rightarrow 12y = 28 \Rightarrow y = \tfrac{7}{3}.$

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Solution

Area $=\dfrac{1}{2}\left|\begin{array}{ccc} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{array}\right|$

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Solution

Total ways $={6 \choose 3}=20$. Equilateral triangles possible $=2$. Required probability $=\dfrac{2}{20}=\dfrac{1}{10}.$

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Solution

Equal roots ⇒ discriminant $=0$. $(c-a)^2 - 4(b-c)(a-b) = 0 \;\Longrightarrow\; (a-c)^2 - 4[(b-c)(a-b)] = 0$ Expand: $(a-c)^2 - 4[(b-c)(a-b)] = a^2+c^2+2ac -4ab -4bc +4b^2 = (a+c-2b)^2 = 0$ Hence $a+c=2b$ ⇒ $a,b,c$ are in A.P.

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Solution

Bijections on a 10-element set = number of permutations = 10. So, 10 = 3628800.

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Solution

For any integer $n\ge1$, digits $= \lfloor\log_{10} n\rfloor+1$.

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Solution

$|A\times B| = 2\cdot5=10$. A relation is any subset of $A\times B$. Count = $2^{10}=1024$.

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Solution

$(f\circ g)(x)=f(x/2)=x-3$. Its inverse solves $y=x-3\Rightarrow x=y+3$, so $(f\circ g)^{-1}(x)=x+3$.

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Solution

Range of $x^{2}+5$ is $[5,\infty)$. Since $4$ is not in the range, $f^{-1}(4)$ does not exist.

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Solution

Solve $x^{2}-2=23 \Rightarrow x^{2}=25 \Rightarrow x=\pm5$.

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Solution

Injective maps from 3 to 10 = permutations $P(10,3)=10\cdot9\cdot8=720$.

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Solution

Onto functions to a 2-element codomain: $2^{4}-\binom{2}{1}1^{4}=16-2=14$ (I.E. principle).

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Solution

$\arg z = \tan^{-1}\!\left(\frac{1}{\sqrt3}\right)=\frac{\pi}{6}$ (I quadrant), $\arg\bar z = -\frac{\pi}{6}$. Sum of absolute values $=\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3}$.

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Solution

$\sqrt{3}+i = 2(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}) \Rightarrow (\sqrt{3}+i)^{17} = 2^{17}\left[\cos\tfrac{17\pi}{6} + i\sin\tfrac{17\pi}{6}\right] = 2^{16}(-\sqrt{3}+i).$ $1 - i = \sqrt{2}(\cos(-\tfrac{\pi}{4}) + i\sin(-\tfrac{\pi}{4})) \Rightarrow (1-i)^{50} = 2^{25}\left[\cos(-\tfrac{25\pi}{2}) + i\sin(-\tfrac{25\pi}{2})\right] = -i\,2^{25}.$ $\text{Ratio} = 2^{-9}\dfrac{-\sqrt{3}+i}{-i} = 2^{-9}(-1-\sqrt{3}i).$

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Solution

$\dfrac{1+i}{1-i}=i$. So $i^{x}=1\Rightarrow x\equiv0\pmod4$. Only $68$ is a multiple of $4$.

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Solution

$\omega^{3}=1$ and $\omega+\omega^{2}=-1$. So $1-\omega-\omega^{2}=2$ and $1+\omega^{3}=2$; product $=4$.

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Solution

Let $z=x+iy$. Then $|z|-x-iy=1+2i \Rightarrow |z|-x=1,\; -y=2\Rightarrow y=-2$. $|z|=1+x$, and $(1+x)^{2}=x^{2}+4 \Rightarrow x=\tfrac{3}{2}$. Thus $z=\tfrac{3}{2}-2i$.

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Solution

By AM–GM, $\sin\theta+\csc\theta\ge2$, equality at $\sin\theta=1$. Hence $\sin^{n}\theta+\csc^{n}\theta=1^{n}+1^{n}=2$.

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Solution

$\sin^{6}x+\cos^{6}x=(\sin^{2}x+\cos^{2}x)^{3}-3\sin^{2}x\cos^{2}x=1-3s^{2}c^{2}$. Add $3s^{2}c^{2}$ ⇒ total $=1$.

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Solution

$x^{2}+y^{2}=a^{2}\sin^{2}\theta\cos^{2}\theta$. So $(x^{2}+y^{2})^{3}=a^{6}\sin^{6}\theta\cos^{6}\theta = a^{2}(a^{4}\sin^{6}\theta\cos^{6}\theta)=a^{2}x^{2}y^{2}$.

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Solution

$R=\sqrt{3^2+4^2}=5$. Min$(3\cos\theta+4\sin\theta)=-5$. So min total $= -5+10=5$.

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Solution

Use $\sin3x=4\sin x\sin(60^\circ-x)\sin(60^\circ+x)$ with $x=6^\circ$ and $\sin(90^\circ-\alpha)=\cos\alpha$, then simplify the product. Value $= \tfrac{1}{16}$.

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Solution

$y'=\dfrac{u'}{1+u^{2}}$, $u=\dfrac{1+x}{1-x}$. $u'=\dfrac{2}{(1-x)^2}$, $1+u^2=\dfrac{2(1+x^2)}{(1-x)^2}$. Hence $y'=\dfrac{1}{1+x^2}$.

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Solution

$y'=\dfrac{\sec^2 x}{\tan x}=\dfrac{1}{\sin x\cos x}=2\csc2x$.

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Solution

For $x\in[-1,1]$, $\sqrt{1-x^2}=\sin(\cos^{-1}x)=\sin y$. Thus $z=\sin^{-1}(\sin y)$, so $z=y$ (up to piecewise sign); hence $\dfrac{dy}{dz}=1$.

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Solution

$\dfrac{d^{2}y}{dx^{2}}=4e^{2x}=4y$; $x=\tfrac12\ln y\Rightarrow \dfrac{d^{2}x}{dy^{2}}=-\dfrac{1}{2y^{2}}$. Product $=4y\cdot\left(-\dfrac{1}{2y^{2}}\right)=-\dfrac{2}{y}=-2e^{-2x}$.

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Solution

Let $a=\sqrt{x+y},\,b=\sqrt{y-x}$. $(a+b)^2=2 \Rightarrow y+\sqrt{y^{2}-x^{2}}=1$. Differentiate: $y'+\dfrac{yy'-x}{\sqrt{y^{2}-x^{2}}}=0$. But $\sqrt{y^{2}-x^{2}}=1-y$ from above ⇒ $y'=x$ ⇒ $y''=1$.

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Solution

Standard limit: $\frac{1-\cos x}{x^{2}}\to \frac12$.

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Solution

$\sqrt{x^{2}+x}=x\sqrt{1+1/x}=x\left(1+\tfrac{1}{2x}+o(1/x)\right)=x+\tfrac12+o(1)$. Hence limit $=x-(x+\tfrac12)= -\tfrac12$.

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Solution

Put $t=\log x$, then $dt=\frac{dx}{x}$; next $u=\log t$, $du=\frac{dt}{t}$. Integral becomes $\int \frac{du}{u}=\log u=\log(\log(\log x))+C$.

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Solution

$\dfrac{d}{dx}\big(x^{x}\big)=x^{x}(1+\log x)$ (log = natural log). Hence integral $=x^{x}+C$.

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Solution

IBM z/OS is a **mainframe** operating system.

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Solution

Palm OS is a mobile/PDA operating system; AVG is antivirus; BeOS is desktop.

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Solution

Let \(u=1-x\Rightarrow du=-dx\). Then \[ \int_{0}^{1}\frac{x}{(1-x)^{3/4}}dx =\int_{1}^{0}\frac{1-u}{u^{3/4}}(-du) =\int_{0}^{1}\left(u^{-3/4}-u^{1/4}\right)du = \left[4u^{1/4}-\frac{4}{5}u^{5/4}\right]_{0}^{1} =4-\frac{4}{5}=\frac{16}{5}. \] \(\boxed{\tfrac{16}{5}}\)

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Solution

8086 is a 16-bit processor.

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Solution

IBM System/360 is a classic mainframe line.

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Solution

“Wellwer” doesn’t correspond to any standard OS/CPU/mobile brand.

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Solution

$500 = 4·5^3 + 0·5^2 + 0·5 + 0 ⇒ (4000)_5.$

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Solution

$(1056)_x = x^3 + 5x + 6$. So, $x^3 + 5x + 6 = 780 \Rightarrow x = 9$.

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Solution

$2\times7^2 + ?\times7 + 1 = 120$ $\Rightarrow 99 + 7? = 120 \Rightarrow ? = 3$.

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Solution

Invert bits → $1001011$, add $1$ → $1001100$. $\boxed{1001100}$

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Solution

Invert $10110010 \Rightarrow 01001101$, add $1 \Rightarrow 01001110 = 78$. Hence, number $= -78$.

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Solution

$A + B = \text{NAND}(\text{NAND}(A,A), \text{NAND}(B,B))$. Hence, $3$ NAND gates required. $\boxed{3}$

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Solution

Output delay between input and output change = propagation time. $\boxed{\text{Propagation time}}$

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Solution

$x\cdot x = x$, $x + x = x$, $x\cdot1 = x$, but $x + 1 = 1 \neq x$. $\boxed{x + 1}$

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Solution

Flip-flops store state → sequential circuits. $\boxed{\text{Flip-flop}}$

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Solution

Maximum cube = $3^3 = 27 = (11011)_2$ → needs 5 bits.

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Solution

Opera is a web browser.

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Solution

Symbian is a mobile operating system. $\boxed{\text{Symbian}}$

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Solution

Norton is antivirus software. $\boxed{\text{Norton}}$

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Solution

None of the given options is a search engine.

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Solution

Instagram is a social networking website.

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Solution

$1+2+3=6$ and $3+2+1=6$. So both give $9$. $\boxed{9}$

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Solution

$J=10, M=13, I=9 \Rightarrow 10+13+9=32$. Similarly, $C=3, A=1, T=20 \Rightarrow 3+1+20=24$. $\boxed{24}$

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Solution

BAG → 2,1,7 → 217. JMI → 10,13,9 → 10139. $\boxed{10139}$

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Solution

Number represents count of letters in the word. CAT (3), HE (2), DELHI (5) ⇒ SAD (3). $\boxed{3}$

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Solution

Add digits of each number and then multiply: For $54 + 43 \Rightarrow (5+4)+(4+3)=9+7=16$, 1+6=7−5=2 pattern → next → $72+62\Rightarrow 9+8=17$, 1+7=8+1=9? ≈ pattern → $\boxed{13}$

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Solution

Differences: 2,3,5,7,11 → primes increasing. Next +13 → 29+13=42. $\boxed{42}$

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Solution

Cubes: $1^3,2^3,3^3,4^3,5^3,…$ → next $6^3=216$.

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Solution

Differences: 4,6,8,10 → next +12 → 31+12=43.

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Solution

Pattern: $1×1+1=2$, $2×2+2=6$, $6×7=42$, next $42×43=1806$.

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Solution

Fibonacci should be $1,1,2,3,5,8,13$. The 4th term should be $3$, not $4$.

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Solution

Should be arithmetic with $+3$: $2,5,8,11,14,17,20$. So $12$ is wrong.

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Solution

Perfect squares $1,4,9,16,25,36,49$. Here $21$ should be $25$.

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Solution

Triangular numbers: $1,3,6,10,15,21,28$. Here $11$ should be $10$.

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Solution

Left letters $A,B,D,G,K$ with steps $+1,+2,+3,+4$ → next $+5$: $K$ stays as left? (already K from prev step), actually pattern gives next pair $K\to P$ and right $B,D,G,K\to P$; number $1,2,3,4\to5$. So $K5P$. $\boxed{K_5P}$

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Solution

Left $C,D,E,F\to G$; numbers $1,3,5,7\to 9$; right $Z,Y,X,W\to V$. Needed $G9V$ — not in options.

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Solution

First letters $A,B,D,G$ with jumps $+1,+2,+3$ → next $+4$: $K$. Second letters $B,D,F,H$ ($+2$ each) → $J$. Third letters $Z,Y,X,W$ ($-1$ each) → $V$. $\boxed{\text{KJV}}$

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Solution

First letters $C,E,G,I\to K$ ($+2$); second $A,B,C,D\to E$ ($+1$); third $T,S,R,Q\to P$ ($-1$). $\boxed{\text{KEP}}$

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Solution

$2 + 3 + 4 = 9$, but code = $11 = 9 + 2$ → add 2 to digit sum. For ‘123’: $1 + 2 + 3 = 6$, $6 + 2 = 8$.

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Solution

Code logic → first and last combine: (1,6)→6; (2,1,4,6,5,2) pattern forms 7 and 13 → 713. $\boxed{713}$

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Solution

$2 \times 3 \times 4 = 24$ and $2 \times 3 \times 5 = 30$. $\boxed{30}$

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Solution

‘Abstain’ = to avoid or hold back; opposite = ‘Begin’. $\boxed{\text{Begin}}$

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Solution

Mitigate’ = to reduce severity; opposite = ‘Aggravate’. $\boxed{\text{Aggravate}}$

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Solution

‘Ambiguous’ = unclear; opposite = ‘Clear’. $\boxed{\text{Clear}}$

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Solution

Different or varied views → ‘divergent’. $\boxed{\text{divergent}}$

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Solution

‘Advent’ = coming or arrival. $\boxed{\text{advent}}$

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Solution

Correct spelling = Delineate. $\boxed{\text{Delineate}}$

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Solution

Correct spelling is **Enmity** (meaning hostility). $\boxed{\text{Enmity}}$

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Solution

‘Amazing’ means ‘astonishing’ or ‘surprising’. $\boxed{\text{Astonishing}}$

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Solution

‘Brave’ = ‘Courageous’.

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Solution

Diligent’ means ‘careful and hardworking’. $\boxed{\text{Hardworking}}$