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Solution

Sentence III → “ability of flying” is wrong; it should be “ability to fly.” Sentence V → “ability for flying” is also wrong; correct is “ability to fly.” Other sentences are grammatically correct.

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Solution

Antonyms of “sanity” are “lunacy” and “insanity.” “Stupidity” is not directly opposite to “sanity”; it refers to lack of intelligence, not mental stability. “Rationality” means sanity itself.

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Solution

The word “Dreary” means dull, gloomy, or depressing. Among the given options, “Dismal” also means gloomy or cheerless.

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Solution

Let 4A = 5B = 7C = k ⇒ A = k/4, B = k/5, C = k/7 Total runs = A + B + C = 681 ⇒ k(1/4 + 1/5 + 1/7) = 681 ⇒ k( (35 + 28 + 20)/140 ) = 681 ⇒ k × (83/140) = 681 ⇒ k = 681 × 140 / 83 = 1148.19 ≈ 1148. A = 1148/4 = 287 C = 1148/7 = 164 Difference = 287 − 164 = 123 ≈ 125

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Solution

Let original price = 100 and quantity sold = 100. Original revenue = 100 × 100 = 10,000. New price = 80, new sales = 160. New revenue = 80 × 160 = 12,800. Increase in revenue = (12,800 − 10,000)/10,000 × 100 = 28%.

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Solution

Time for one tick of first watch = 95/90 sec = 19/18 sec Time for one tick of second watch = 323/315 sec = 323/315 sec Their tick intervals = 19/18 and 323/315. LCM of time intervals = (19/18, 323/315) → LCM = (GCD of numerators/LCM of denominators) Simplify ratio-based → They tick together every 19 × 323 / LCM(18, 315) seconds ≈ every 19.2 seconds (approx). So, in one hour (3600 seconds): 3600 / 35.6 ≈ 101 times.

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Solution

Let the time after which they meet be t minutes. Rama’s position after t minutes = 11 + 57t Somaya’s position after t minutes = 51 − 63t At the meeting point: 11 + 57t = 51 − 63t ⇒ 120t = 40 ⇒ t = 1/3 minutes. Hence, floor number = 11 + 57 × (1/3) = 11 + 19 = 30.

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Solution

Number of parallelograms formed = ${}^{7}C_{2} \times {}^{7}C_{2}$ = (21 × 21) = 441.

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Solution

X is 5th from the top → X’s position = 5. Z is 6th after X → Z’s position = 5 + 6 = 11. Z is also in the middle of X and Y, so Y’s position from top = 11 + 6 = 17. Since Y is 8th from last, Total students = 17 + 8 − 1 = 24.

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Solution

50 weeks = 350 days. C − A = 350 days, A − B = 300 days ⇒ C − B = 650 days. 650 ÷ 7 = 92 weeks + remainder 6 days. So, B’s birthday = 6 days before Tuesday = Wednesday.

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Solution

List all cities for each bank: A → Lucknow, Kanpur, Allahabad B → Lucknow, Kanpur, Allahabad, Varanasi, Saharanpur C → Lucknow, Kanpur, Varanasi, Saharanpur, Moradabad D → Lucknow, Kanpur, Varanasi, Allahabad, Saharanpur, Moradabad E → Lucknow, Kanpur, Allahabad, Saharanpur, Moradabad Bank B covers all except Moradabad.

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Solution

Months with same number of days: - May (31) and January (31) ✅ same - September (30) and November (30) ✅ same - October (31) and April (30) ❌ different - January (31) and December (31) ✅ same Hence, “October : April” is the odd one out.

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Solution

P × Q → P is the son of Q Q − S → Q is the wife of S Hence, P is the son of Q and Q is the wife of S ⇒ P is the son of S. So, S is the father of P.

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Solution

Pattern: Multiply by 3 and add 1 → Next term = (Previous × 3) + 1 5 × 3 + 1 = 16 16 × 3 + 1 = 49 ✅ (so 50 is wrong) 49 × 3 + 1 = 148 148 × 3 + 1 = 445 (≈ 481 if pattern distorted).

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Solution

Number of Sundays in 30 days starting with Sunday = 5 Visitors on Sundays = 5 × 510 = 2550 Visitors on other 25 days = 25 × 240 = 6000 Total visitors = 8550 Average = 8550 ÷ 30 = 285

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Solution

Pattern → (Number × 8) − 5 6 × 8 − 5 = 43 5 × 8 − 5 = 35 But none matches — check (Number × Number) + 7: 6² + 7 = 43 ✅ 5² + 7 = 32 ✅

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Solution

Differences between terms: 197 − 122 = 75 290 − 197 = 93 Difference pattern: +18 each step Next difference = 93 + 18 = 111 Next term = 290 + 111 = 401

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Solution

Check perfect cubes: 2197 = 13³ 3375 = 15³ 2744 = 14³ 4099 ❌ not a perfect cube

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Solution

Check each ‘8’: 1️⃣ 7 8 9 → preceded by 7 and followed by 9 ❌ 2️⃣ 9 8 8 → preceded by 9, followed by 8 ✅ 3️⃣ 8 5 4 → ✅ 4️⃣ 5 8 8 → ✅ 5️⃣ 8 7 1 → ❌ followed by 7 6️⃣ 7 1 8 9 → ❌ (preceded by 7, followed by 9) So valid 8’s = 3

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Solution

Sanjay said — “His mother is the wife of my father’s son.” → ‘My father’s son’ = Sanjay himself (since he has no brothers). Hence, the man’s mother is Sanjay’s wife. Therefore, the man is Sanjay’s son.

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Solution

Pattern observed: Each letter is moved by +4 alphabet positions. L → P, A → E, T → X, E → I Similarly, T → X, R → V, A → E, C → G, E → I Hence, TRACE → XVEGI

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Solution

From the statements: - Some cats are rats → all rats are bats → so, some cats are bats ✅ - Some bats are birds, but no definite relation between birds & cats/rats. Hence, only C2 (“Some bats are cats”) is true; others cannot be concluded.

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Solution

Let leak alone empty the tank in x hours. Normal filling rate = 1/8 per hour With leak → filled in 10 hours → net rate = 1/10 So, 1/8 − 1/x = 1/10 ⇒ 1/x = 1/8 − 1/10 = (5 − 4)/40 = 1/40 Hence, leak alone can empty in 40 hours.

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Solution

Facing north-east, your first move of 10 m forward takes you along a line at 45° from north. Then turning left from north-east → you face north-west direction, and you move 7.5 m. If you resolve the path: The first displacement = $(10 \text{m}, 45°)$ ⇒ Components: $x_1 = 10\cos45° = 7.07$, $y_1 = 10\sin45° = 7.07$ The second displacement = $(7.5 \text{m}, 135°)$ ⇒ Components: $x_2 = 7.5\cos135° = -5.30$, $y_2 = 7.5\sin135° = 5.30$ Net displacement components: $x = 7.07 - 5.30 = 1.77$, $y = 7.07 + 5.30 = 12.37$ Both $x$ and $y$ are positive ⇒ final position is north-east of initial position. But since the question asks relative to initial position along main axes, the major shift is northward.

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Solution

At 12 noon → both hands overlap, pointing north-east. At 1:30 p.m.: The minute hand is at 6, i.e., pointing south-west (opposite to north-east). The hour hand at 1:30 lies halfway between 1 and 2 → i.e., $45° + 15° = 52.5°$ clockwise from 12. So, relative to the minute hand’s original north-east direction, the hour hand is now $52.5°$ clockwise → this points roughly toward south.

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Solution

A language processor translates source code into machine code. Loader is not a language processor.

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Solution

$(41)_8 = 4\times8 + 1 = 33_{10}$ Now, $(121)_b = 1\times b^2 + 2\times b + 1 = b^2 + 2b + 1$ Equating, $b^2 + 2b + 1 = 33$ $\Rightarrow b^2 + 2b - 32 = 0$ $\Rightarrow (b - 4)(b + 8) = 0 \Rightarrow b = 4$

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Solution

Bitcoin operates on decentralized peer-to-peer (P2P) network technology.

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Solution

ASCII (extended) and EBCDIC both support 256 character combinations.

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Solution

$(427)_{16} - (56)_{16} = (371)_{16}$

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Solution

RISC processors are used in mobile and embedded systems due to low power consumption.

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Solution

RAID → Redundant Array of Inexpensive Disks.

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Solution

RAM → MB CPU Speed → Hertz Monitor → Inch CD–ROM Speed → Bytes/Sec Hence the correct match is $(P,2), (Q,1), (R,4), (S,3).$

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Solution

‘XP’ in Windows XP stands for *Experience*.

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Solution

The range of $n$-bit 2’s complement numbers is $-2^{n-1}$ to $2^{n-1}-1$

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Solution

BIOS (Basic Input Output System) is stored permanently in ROM.

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Solution

Zing is not a search engine.

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Solution

Zing is not a search engine.

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Solution

$x = 0.125E + 01 = 1.25_{10}$ $(1.01)_2 = 1 + 0 + 0.25 = 1.25_{10}$ $(1.2)_8 = 1 + \frac{2}{8} = 1.25_{10}$ Thus all are equal.

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Solution

Procedural Oriented → Pascal $(1,S)$ Object Oriented → C++ $(2,R)$ Business Oriented → COBOL $(3,P)$ Web Page → HTML $(4,Q)$ Correct match: $(1,S), (2,R), (3,P), (4,Q)$

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Solution

Mouse, Keyboard → Input devices Unix, Windows, Linux → Operating Systems Register, Cache → Memory; Hard-disk → Storage Monitor, Printer → Output; Scanner → Input Hence, correct statements are $Q$ and $R$.

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Solution

Bitcoin was introduced in 2009 by Satoshi Nakamoto.

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Solution

Volatile memory loses data when power is off. Both RAM and Cache are volatile memories.

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Solution

$|A| = 2, |B| = 4 \Rightarrow |A \times B| = 8$ Total subsets of 8 elements = $2^8 = 256$ Subsets with less than 3 elements = $1 + 8 + 28 = 37$ Subsets with 3 or more elements = $256 - 37 = 219$

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Solution

Given $A \cap B = A \cap C$ and $A \cup B = A \cup C$ These conditions hold only if $B = C$.

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Solution

For $\tan^{-1}(\tan \theta)$, the result lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$. Since $13$ radians is beyond this interval, $\tan 13 = \tan(13 - 4\pi)$ $\Rightarrow \tan^{-1}(\tan 13) = 13 - 4\pi$

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Solution

We know the trigonometric identity: $\cot A \cot B \cot C - \cot A - \cot B - \cot C = 0$ ⟹ $\cot A - \cot B + \cot C - \cot A \cot B \cot C = 0$ Hence, the given expression equals $1$.

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Solution

$00001101_2 = 13_{10}$ $00001111_2 = 15_{10}$ Product in decimal $= 13 \times 15 = 195$ Now convert $195_{10}$ to binary: $195 = 128 + 64 + 3 = 11000011_2$

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Solution

$\tan\left(\frac{\pi}{8}\right) = \tan(22.5^\circ)$ Using the half–angle identity: $\tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta}$ For $\theta = \frac{\pi}{4}$, $\tan\frac{\pi}{8} = \frac{1 - \cos\frac{\pi}{4}}{\sin\frac{\pi}{4}} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2} - 1.$

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Solution

$|Z - 1| = |Z + 1|$ represents the perpendicular bisector of the line joining $(1,0)$ and $(-1,0)$, i.e. the $y$–axis. For points on the $y$–axis, $Z = i y$. Now $|Z - i| = |i y - i| = |i(y - 1)| = |y - 1|$. Also $|Z + 1| = \sqrt{1 + y^2}$. So $\sqrt{1 + y^2} = |y - 1| \Rightarrow y = 0$. Thus only one point satisfies — $Z = 0$.

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Solution

We know $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. So $1 + \omega = -\omega^2$. Hence $(1 + \omega)^7 = (-\omega^2)^7 = (-1)^7 \omega^{14} = -\omega^{14}$. Now $\omega^{14} = \omega^{3 \times 4 + 2} = \omega^2$. Therefore $(1 + \omega)^7 = -\omega^2 = 1 + \omega$. So comparing with $A + B\omega$, we have $A = 1$, $B = 1$. $\Rightarrow A + B = 2.$

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Solution

Given $x + y + z = 5,\; xy + yz + zx = 3$. Let $y,z$ be roots of $t^2 - (5 - x)t + (3 - 5x) = 0$. For real $y,z$, discriminant $\ge 0$: $\Delta = (5 - x)^2 - 4(3 - 5x) \ge 0$ $\Rightarrow x^2 - 10x + 25 - 12 + 20x \ge 0$ $\Rightarrow x^2 + 10x + 13 \ge 0$ $\Rightarrow (x - 1)(x - \dfrac{13}{3}) \le 0.$ So $x \in [1, \dfrac{13}{3}]$.

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Solution

Sum of numbers divisible by 2 up to 100: $2 + 4 + \dots + 100 = 2(1 + 2 + \dots + 50) = 2 \times \frac{50 \times 51}{2} = 2550.$ Sum of numbers divisible by 5 up to 100: $5 + 10 + \dots + 100 = 5(1 + 2 + \dots + 20) = 5 \times \frac{20 \times 21}{2} = 1050.$ Sum of numbers divisible by both 2 and 5 (i.e., by 10): $10 + 20 + \dots + 100 = 10(1 + 2 + \dots + 10) = 10 \times 55 = 550.$ Required sum = $2550 + 1050 - 550 = 3050.$

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Solution

We use modulo properties: $27 \equiv 3 \pmod{12}$ $\Rightarrow 27^{40} \equiv 3^{40} \pmod{12}$ Now, $3^1 \equiv 3$, $3^2 \equiv 9$, $3^3 \equiv 3$, $3^4 \equiv 9$, pattern repeats every 2 powers. So, for even power $40$, remainder = $9$.

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Solution

Given series = $\displaystyle 1 + \frac{1}{2! \cdot 2^2} + \frac{1}{4! \cdot 2^4} + \frac{1}{6! \cdot 2^6} + \cdots$ This is the expansion of $\dfrac{e^{1/2} + e^{-1/2}}{2} = \dfrac{e^{1/2}(1 + e^{-1})}{2} = \dfrac{e + 1}{2\sqrt{e}}$

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Solution

Let the numbers be $a$ and $b$ with G.M. $\sqrt{ab}$. Given: $a + b = 6\sqrt{ab}$. Dividing both sides by $\sqrt{ab}$: $\dfrac{a}{\sqrt{ab}} + \dfrac{b}{\sqrt{ab}} = 6$ $\Rightarrow \sqrt{\dfrac{a}{b}} + \sqrt{\dfrac{b}{a}} = 6$ Let $\sqrt{\dfrac{a}{b}} = x \Rightarrow x + \dfrac{1}{x} = 6$ $\Rightarrow x^2 - 6x + 1 = 0 \Rightarrow x = 3 \pm 2\sqrt{2}$ Hence, $\dfrac{a}{b} = x^2 = \left(3 \pm 2\sqrt{2}\right)^2 = \dfrac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}.$

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Solution

Vertices: $A(0,0),\ B(4,0),\ C(3,4)$ Slope of $BC = \dfrac{4 - 0}{3 - 4} = -4$ Equation of altitude from $A$: perpendicular to BC → slope $\dfrac{1}{4}$ $\Rightarrow y = \dfrac{1}{4}x$ … (1) Slope of $AC = \dfrac{4 - 0}{3 - 0} = \dfrac{4}{3}$ Equation of altitude from $B$: perpendicular slope $-\dfrac{3}{4}$, passes through $(4,0)$ $\Rightarrow y - 0 = -\dfrac{3}{4}(x - 4)$ $\Rightarrow y = -\dfrac{3}{4}x + 3$ … (2) Solving (1) and (2): $\dfrac{x}{4} = -\dfrac{3x}{4} + 3 \Rightarrow x = 3,\ y = \dfrac{3}{4}$

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Solution

Let $A = (x,0)$. Slope of incident ray $= \dfrac{2 - 0}{1 - x} = \dfrac{2}{1 - x}$ Slope of reflected ray $= \dfrac{3 - 0}{5 - x} = \dfrac{3}{5 - x}$ For reflection from $X$–axis: angle of incidence = angle of reflection. Hence, their slopes are negatives of each other: $\dfrac{2}{1 - x} = -\dfrac{3}{5 - x}$ $\Rightarrow 2(5 - x) = -3(1 - x)$ $\Rightarrow 10 - 2x = -3 + 3x$ $\Rightarrow 5x = 13 \Rightarrow x = \dfrac{13}{5}$ Thus, $A\left(\dfrac{13}{5}, 0\right)$.

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Solution

Equation of chord of contact from $(a\cos\theta, a\sin\theta)$ to circle $x^2 + y^2 = b^2$ is $a(\cos\theta \,x + \sin\theta \,y) = b^2.$ For this line to be tangent to $x^2 + y^2 = c^2$, the perpendicular distance from origin = radius of that circle. $\Rightarrow \dfrac{|b^2|}{\sqrt{a^2}} = c$ $\Rightarrow b^2 = a c$ Hence, $a, b, c$ are in G.P.

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Solution

Let the coordinates of $P$ be $(x_1, y_1)$. Equation of chord of contact to $y^2 = 4a x$ is $T_1 = 0 \Rightarrow y y_1 = 2a(x + x_1).$ This line touches $x^2 = 4b y$. Substitute $y = \dfrac{x^2}{4b}$ in the line equation: $\dfrac{x^2 y_1}{4b} = 2a(x + x_1)$ $\Rightarrow y_1 x^2 - 8abx - 8abx_1 = 0.$ For tangency, discriminant $= 0$: $(8ab)^2 - 4y_1(-8abx_1) = 0$ $\Rightarrow 64a^2b^2 + 32abx_1 y_1 = 0$ $\Rightarrow 2x_1 y_1 + 4ab = 0 \Rightarrow x_1 y_1 = -2ab.$ Thus, locus of $P$ is $x y = -2ab$, which represents a **hyperbola**.

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Solution

Sum of distances from two fixed points (foci) is constant ⇒ ellipse. $2a = 10 \Rightarrow a = 5$ Distance between foci $= 2c = 8 \Rightarrow c = 4$ $b^2 = a^2 - c^2 = 25 - 16 = 9.$ Hence, equation of ellipse: $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1.$

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Solution

In a parallelogram, $A+C=B+D \Rightarrow D=A+C-B$. $D=(3,-1,2)+(-1,1,2)-(1,2,-4)=(1,-2,8)$.

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Solution

Letters: $A,A,I,J,M$. Total $= \dfrac{5!}{2!}=60$. Starting with $A$: $24$ words ($1$–$24$). Starting with $I$: $12$ words ($25$–$36$). Starting with $J$: $12$ words ($37$–$48$). Starting with $M$: $12$ words ($49$–$60$). Within the $M$-block, 49th = $\text{MAAIJ}$, 50th = $\text{MAAJI}$.

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Solution

$\dfrac{\sin x}{x}=1-\dfrac{x^2}{6}+O(x^4)<1$ for $x\ne0$ near $0$, and $>0$. So $\left\lfloor \dfrac{\sin x}{x}\right\rfloor=0$ in a punctured neighborhood of $0$.

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Solution

$1-\cos 2x=2\sin^2 x \Rightarrow \dfrac{\sqrt{1-\cos 2x}}{x}=\sqrt{2},\dfrac{|\sin x|}{x}$. As $x\to0^+$, this $\to\sqrt{2}$; as $x\to0^-$, it $\to -\sqrt{2}$. Thus two-sided limit does not exist; the right-hand limit is $\sqrt{2}$.

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Solution

Total permutations of 3 letters = $3! = 6$. Derangements (no letter in correct envelope) = $!3 = 2$. Hence, number of favorable cases (at least one correct) = $6 - 2 = 4$. Probability $= \dfrac{4}{6} = \dfrac{2}{3}$.

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Solution

Out of all $4! = 24$ permutations, in half of them A comes before B, and in the other half B comes before A. Therefore probability = $\dfrac{1}{2}$.

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Solution

$f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3).$ Sign chart: • $f'$ > 0 for $x<2$; $f'$ < 0 for $2

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