GATE General Aptitude previous year questions

Verbosity (using too many words) is the opposite of brevity (using few words). So the pair is a pair of antonyms. The opposite of insolence (rude, disrespectful behaviour) is respect. Innocence, solace and wealth are not opposites of insolence. Answer: (B) Respect.

Concept: Analogy by antonymy. The trap is picking a word that merely sounds related (innocence) instead of checking the exact relation in the first pair.

70 = 2 x 5 x 7. Each digit is between 1 and 9 (a zero would make the product 0). Any other split needs a factor of 10, 14, 35 or 70, which is not a single digit, and using 1 forces two digits multiplying to 70, which is impossible with single digits. So the digits are 2, 5 and 7 in some order, and their sum is 2 + 5 + 7 = 14. Answer: (B) 14.

Concept: Prime factorisation of the product fixes the digits. The trap is forgetting to rule out splits that use the digit 1.

Take log base 10 of both. log x = log₁₀ m · log₁₀ n. log y = log₁₀ n · log₁₀ m. The two logarithms are equal, and log is one-to-one, so x = y for every pair of positive m, n (the condition m > n does not matter). Answer: (C) x = y.

Concept: Identity a^(log b) = b^(log a). The trap is assuming the larger base gives the larger value and picking x < y or x > y.

The sentence is a third conditional: 'If ... had been ..., ... would have ...'. This form describes an unreal past situation, so it implies the condition did not happen: the latest movie was not a commercial success. Hence the actor did not make that money, and nothing says he will sponsor the next movie. Answer: (D) His latest movie was not commercially successful.

Concept: Reading counterfactual conditionals. The trap is treating 'had been' as a statement of fact and choosing (B) or (C).

We part AT a place (at the door). The door belongs to the cabin: the door OF the cabin. A cabin is rented FOR a period: for the night. Sequence: at; of; for. Answer: (A) at; of; for.

Concept: Prepositions of place, possession and duration. The trap is 'in the night', which describes when something happens, not the length of a rental.

Sort them a ≤ b ≤ c ≤ d ≤ e. Median c = 18, sum = 5 x 12 = 60. A single mode of 20 needs 20 to occur at least twice, and only d and e can exceed 18, so d = e = 20. Then a + b = 60 - 18 - 40 = 2 with a ≤ b ≤ 18. Options: (0, 2) or (1, 1). (1, 1) makes 1 appear twice, tying with 20, so the mode is no longer single. Only {0, 2, 18, 20, 20} works. Answer: (B) 1.

Concept: Using mean, median and mode together as constraints. The trap is counting (1, 1, 18, 20, 20), which has two modes.

If game 1 were Pavan vs Tanvi, its Class 4 winner must lose game 2 to a Class 5 student (Class 5 needs 2 wins), leaving Rishi vs Swathi for game 3 - a second same-class game. So game 1 is Rishi vs Swathi. Its winner must play both remaining games (otherwise game 3 is Pavan vs Tanvi) and must win game 2 so Class 5 has 2 wins. If Swathi won game 1 and game 2, girls already have 2 wins - not allowed. So Rishi wins game 1 (boy) and game 2 (boy), and game 3 must be won by a girl: Rishi beats Pavan in game 2 and loses to Tanvi in game 3. Tanvi never lost. Answer: (D) Tanvi.

Concept: Case analysis under elimination rules. The trap is assuming the strongest player (Rishi) stays unbeaten, which would give the boys three wins.

n² + (n+1)² < 1000 forces n ≤ 21. P and R must differ, so the two numbers cannot share a tens digit: that leaves 19 & 20 or 20 & 21 (or 10 & 11 type pairs, which share tens digit 1). 20 & 21 share tens digit 2. So the pair is 19 and 20: 361 + 400 = 761 = XYP, giving P = 1, so PQ = 19, RS = 20, X = 7, Y = 6. Digits 1, 9, 2, 0, 7, 6 are all distinct. Answer: (C) 6.

Concept: Bounding then using the distinct-digit condition. The trap is missing that consecutive numbers in the same decade share the tens digit, which violates distinctness.

Split PQRO along OQ into triangles OPQ and ORQ. Since PQ = RQ, OQ bisects angle PQR, so each triangle has angle 22.5° at Q. In isosceles triangle OPQ (OP = OQ = 10), the angle at P is also 22.5°, so angle POQ = 180 - 45 = 135°. Area of OPQ = ½ · 10 · 10 · sin 135° = 50 · (√2/2) = 25√2. Both triangles are congruent, so area = 2 · 25√2 = 50√2 cm². Answer: (C) 50√2.

Concept: Inscribed angle and isosceles triangles at the centre. The trap is using the central angle POR = 90° and computing a sector or triangle POR instead of the two triangles through Q.

The double sum factorises: (Σ_{i≥0} (1/2)^i)(Σ_{j≥1} (1/3)^j) = [1/(1 - 1/2)] · [(1/3)/(1 - 1/3)] = 2 · (1/2) = 1. Answer: 1.

Concept: Separable double geometric series. The trap is the starting index: j starts at 1, not 0.

A protagonist is the leading character of a story, the one the plot follows. The word for the character who opposes the protagonist is antagonist. Agnostic (one unsure about God), arsonist (one who sets fires) and anarchist (one who rejects authority) are unrelated. Answer: (B) antagonist.

Concept: Vocabulary: antonyms built on the same root. 'Pro-' (for) versus 'anti-' (against) attached to 'agonist' (contender). The distractors all end in -ist to look similar.

Every game eliminates exactly one player. To leave one winner out of 64 players, 63 players must be eliminated, so 63 games are played (32 + 16 + 8 + 4 + 2 + 1 = 63). Answer: (C) 63.

Concept: Counting by elimination: in a knock-out with n players, n - 1 games. The trap is 64 (one game per player) or 32 (only the first round).

Maximum total = 70 credits. Fixed: core 15 + project 20 = 35. Specialization at least 10, so take exactly 10 to leave room. Electives at most 70 - 35 - 10 = 25. Answer: (D) 25.

Concept: Maximising one quantity under constraints: push the total to its upper bound and every other component to its lower bound. The 60-credit minimum is irrelevant to the maximum.

The statement is: teacher in room -> all students stand silently. (A) is its contrapositive, so true. (B) and (D) follow because 'stand silently' includes both 'silent' and 'standing'. (C) says all standing -> teacher in room, which is the converse; students could stand while the teacher is absent. Answer: (C).

Concept: Implication p -> q is equivalent to its contrapositive (not q -> not p) but not to its converse (q -> p). The converse is the classic trap.

The passage says combinatorial counting is 'sometimes used' in modeling physical phenomena, so P ('always invoked') is not supported: P is False. It also says that in such models probabilities are assigned to combinatorial possibilities, which enables computing average values of physical quantities: that is exactly Q, so Q is True. Answer: (B) P is False and Q is True.

Concept: Inference questions punish absolute words. 'Always' in P contradicts 'sometimes' in the passage; Q is a restatement of the last sentences.

H₁₀(S) is 0 up to S = 10 and then S - 10 (slope 1). H₂₀(S) is 0 up to S = 20 and then S - 20. For (A): S ≤ 10 gives 0; 10 < S ≤ 20 gives S - 10, rising from 0 to 10; S > 20 gives (S - 10) - (S - 20) = 10. This matches the plot. (B) falls after 20, (C) is negative, (D) starts rising at 15. Answer: (A) H₁₀(S) - H₂₀(S).

Concept: A ramp-then-flat shape is the difference of two shifted 'hinge' functions: the first switches the slope on, the second cancels it. Same idea as a bull call spread payoff.

Neeraj's round-2 throw is the best overall. Vesely's best throw came in rounds 1-4 (he never improved after round 4) but not in rounds 1-2 (no medal throws there), so round 3 or 4. In round 4 Jakub's throw was the best of the round. If Vesely's best were in round 4, Jakub's round-4 throw would exceed it, putting Jakub ahead of Vesely after four rounds, so Vesely could not be second. Hence Vesely's best was in round 3. Answer: (A) Third.

Concept: Ordering puzzle: combine 'best in round 4' with 'second after four rounds'. Students often pick Fourth by not noticing Jakub's round-4 throw beats any Vesely throw that round.

Count ordered pairs (first, second) with second a multiple of first. First = 1: 6 pairs; 2: (2,4,6) 3; 3: (3,6) 2; 4: 1; 5: 1; 6: 1. Total 14 of 36. Probability = 14/36 = 7/18. Answer: (C) 7/18.

Concept: Enumerate favourable outcomes of two independent rolls. A number is a multiple of itself, so pairs like (4,4) count; forgetting them gives 5/18 or less.

Expedite, hasten and hurry all mean to make something happen faster. Accelerate means to increase speed, so it fits the group. Retard means to slow down (opposite), provide means to supply, and disable means to make unable. Answer: (A) Accelerate.

Concept: Synonym grouping. Identify the common meaning of the given words (speeding up) and pick the option sharing it. The trap is retard, which belongs to the same topic of speed but is the antonym.

This is a binomial count with n = 4, p = 0.5. P(exactly 3 cloudy) = C(4,3) x (0.5)^3 x (0.5)^1 = 4 x 1/16 = 1/4. Answer: (A) 1/4.

Concept: Binomial probability: choose which of the four days is sunny (4 ways), each specific pattern has probability 1/16. The trap is forgetting the C(4,3) factor and answering 1/16, or confusing with 3/8 which is the value for exactly 2 of 4.

Rs. 100 buys 100/50 = 2 shares of A; Rs. 80 buys 80/80 = 1 share of B. Total cost = 180. Next day: 2 x 55 + 1 x 70 = 110 + 70 = 180. Profit = 180 - 180 = 0. Answer: (A) 0.

Concept: Convert money invested into number of shares, then revalue. The gain of Rs. 10 on A exactly offsets the loss of Rs. 10 on B. The trap is applying the per-share change (+5, -10) directly without scaling by the number of shares.

The statement is: raining -> dance. The only statement logically equivalent to an implication is its contrapositive: not dance -> not raining, i.e. when peacocks are not dancing, it is not raining. (A) and (B) are the converse, (D) is the inverse; neither follows. Answer: (C).

Concept: Tests implication versus converse, inverse and contrapositive. Only the contrapositive is guaranteed. The usual trap is picking the converse or inverse, which sound natural but are not implied.

Water quenches thirst; food satisfies hunger. The relation is 'remedy : need', so P = Thirst and Q = Hunger. (B) mixes an action (drink) with a need; (C) mixes a need with a resulting state. Answer: (A).

Concept: Analogy questions require the same relation on both sides. Water relieves thirst as food relieves hunger. The trap is options where each pair is related to its word but via different relations.

The trip is a Hamiltonian cycle. (i): a 4 x 4 grid has one - go along the top row, snake down and back through columns 2 to 4, then return up column 1. So (i) works. (ii): Q, R and S each have degree 2, so both their edges must be in the cycle: Q-P, Q-T, R-P, R-T, S-P, S-T. That forces P to have 3 cycle edges, impossible (each vertex has exactly 2 in a cycle). So (ii) fails. Answer: (A).

Concept: Hamiltonian cycle reasoning: degree-2 vertices force their edges into the cycle, which quickly exposes contradictions. The trap is confusing this with an Euler circuit (every edge once) and checking degree parity instead.

Check row counts first: (i) has rows 3,2,1,2 - fails. (iii) has rows 3,2,1,2 - fails. (iv) has rows 3,1,1,3 - fails. (ii) has rows 3,1,2,2. Its columns: col1 rows 1,4 = 2; col2 rows 2,4 = 2; col3 rows 1,3 = 2; col4 rows 1,3 = 2. All constraints hold. Answer: (B) (ii).

Concept: Constraint checking (a nonogram): eliminate options by verifying row totals, then confirm column totals. The trap is checking only rows or only columns.

Possible prime sums are 2, 3, 5, 7, 11. Number of ordered outcomes: sum 2 -> 1, sum 3 -> 2, sum 5 -> 4, sum 7 -> 6, sum 11 -> 2. Total = 15 out of 36. Answer: (C) 15/36.

Concept: Counting ordered outcomes of two dice. The trap is forgetting 2 is prime or counting (1,2) and (2,1) as one outcome.

Courage and bravery are synonyms, so the second word of the pair must be a synonym of yearning. Yearning means a strong desire for something, which is exactly what longing means. Yelling (shouting), yawning (opening the mouth when tired) and glaring (staring angrily) only resemble yearning in sound or spelling. Answer: (A) Longing.

Concept: Synonym analogy. The trap is the distractors that start with Y or share letters with yearning; match the meaning relation, not the look of the word.

The interrupting event (it started to rain) is in the simple past. An action that was already in progress before and up to a past event takes the past perfect continuous: had been playing. 'Have been playing' is present perfect continuous and clashes with the past-tense clause; the modal forms change the meaning to something hypothetical. Answer: (B) had been playing.

Concept: Tests the past perfect continuous for an ongoing action interrupted by a past event. The usual trap is picking the present perfect continuous, which cannot pair with a past-tense time clause.

Count the entries that are at most 4, row by row. Row 1 (0, 1, 0, 2): all 4 qualify. Row 2 (4, 7, 3, 3): 4, 3, 3 qualify - 3 pixels. Row 3 (5, 5, 4, 4): the two 4s - 2 pixels. Row 4 (6, 7, 3, 2): 3 and 2 - 2 pixels. Total = 4 + 3 + 2 + 2 = 11. Answer: (C) 11.

Concept: Simple table reading with an inclusive inequality. The trap is treating ≤ 4 as < 4, which drops the four 4s and gives 7, or miscounting a row.

The labelled region 4 lies in the rectangle (1) and the ellipse (3): 1 + 3 = 4, so a region's number is the sum of the numbers of the shapes containing it. P is in all three: 1 + 2 + 3 = 6. Q is in the triangle and the ellipse: 2 + 3 = 5. R is in the rectangle and the triangle: 1 + 2 = 3. Answer: (A) P = 6; Q = 5; R = 3.

Concept: Pattern induction from the one labelled overlap region. The trap is guessing a product or a count rule; check the rule against the given overlap (4 = 1 + 3) before applying it.

New width = 1.1W, length unchanged. Area: L × 1.1W = 1.1LW, a 10% increase for every L and W. Perimeter: 2(L + 1.1W) = 2(L + W) + 0.2W, an increase of 0.2W/(2(L + W)), less than 10%. Diagonal: √(L² + 1.21W²) grows by less than 10% because L is unchanged. Square: only if 1.1W = L, which is not true in general. Answer: (C) Area increases by 10%.

Concept: Area is linear in each side, so scaling one side scales area by the same factor; perimeter and diagonal are not. The trap is assuming every measure scales by 10%.

Match each phrasal verb to its meaning. P (a messy house) - 'clear it up' means tidy it: P-3. Q (unhappy with marks) - 'take it up with your teacher' means raise the matter with someone in authority: Q-4. R (avoids politics) - 'won't bring it up' means won't mention the topic: R-1. S (doesn't know a word) - 'look it up' means find it in a dictionary: S-2. Answer: (B) P - 3; Q - 4; R - 1; S - 2.

Concept: Tests phrasal verbs with 'up': clear up, take up, bring up, look up. The trap is that 'clear it up' can also mean explain, which tempts matching it to S; only 'look it up' fits the dictionary sense.

Both brothers follow the same weight function w(a), with a₁ ≠ a₂ and w(a₁) = w(a₂). If both ages were at most 50, w would be strictly increasing there, so a₁ < a₂ would force w(a₁) < w(a₂). If both were at least 50, w would be strictly decreasing, giving w(a₁) > w(a₂). Either way the weights could not be equal. So the two ages must lie on opposite sides of the peak, and neither can be exactly 50 (the peak value is attained only once). Hence a₁ < 50 < a₂. Answer: (B).

Concept: A strictly monotone function is one-to-one, so equal values at different inputs require the inputs to straddle the turning point. The trap is option (D), thinking the peak must be involved.

Each triangle has two sides equal to r with an included angle of 360°/12 = 30°. Area of one triangle = (1/2) r² sin 30° = r²/4. Area of a square of side r = r², so each square needs r² / (r²/4) = 4 triangles. With 12 triangles, the number of squares = 12/4 = 3. Answer: (A) 3; 4.

Concept: Area bookkeeping with the triangle area formula (1/2)ab sin C. The trap is using the side d in the area, or dividing 12 by 3 because a dodecagon suggests multiples of 3.

Write everything in base 3: 3^(x²) = 3³ × 3^(2x) = 3^(3 + 2x), so x² = 2x + 3, i.e. x² - 2x = 3 (x = 3 or x = -1). The required value is 2^(x²) / 2^(2x) = 2^(x² - 2x) = 2³ for either root. Answer: (C) 2³.

Concept: Equate exponents after a common base, then notice the target only needs x² - 2x, not x itself. The trap is reading (2^x)² as 2^(x²), or solving for x and plugging in only one root.

Average X = (5×20 + 6×40 + 7×30 + 8×10)/100 = (100 + 240 + 210 + 80)/100 = 6.3. Earnings are linear in X, so average earnings = 1000(6.3 - 0.2) = ₹ 6,100. Answer: (A) 6,100.

Concept: Weighted mean from a frequency chart and linearity of the mean. The trap is forgetting to subtract 0.2 (giving 6,300) or averaging the four X values without weights.

Articles follow the sound, not the spelling. "younger" starts with a consonant sound, so "a younger brother" (first mention, not a specific one). "university" starts with the sound "yoo", a consonant sound, so "a university". "honorable" has a silent h and starts with a vowel sound, so "an honorable man". Sequence: a; a; an. Answer: (A).

Concept: Tests choice of a/an by initial sound. The trap is choosing "an university" because u is a vowel letter, or "a honorable" because h is a consonant letter. Pronunciation decides.

"Myopic" literally means near-sighted and figuratively means lacking foresight. The sentence confirms this: the decision gave up long-term stability for short-term gains. Visionary, progressive and innovative are all positive and mean nearly the opposite. The only synonym is shortsighted. Answer: (B).

Concept: Vocabulary in context. The context clue "sacrificed long-term stability" points to lack of foresight; the distractors are positive words that contradict the sentence's critical tone.

Let n be the number of students. Correcting the entry raises the total by 42 - 24 = 18. The average rises by 31.4 - 30.8 = 0.6. So 0.6 n = 18, giving n = 30. Answer: (C).

Concept: Change in total = change in average x count. The trap is using 24 + 42 or one of the averages directly instead of the difference of the corrected and wrong marks.

R is Q's mother and S is Q's daughter, so R is S's grandmother: (1) true. P is Q's brother and S, T are Q's daughters (T is S's sister), so P is their uncle: (2) true. R's children include P (a son) and Q, whose gender is not given, so "only one son" cannot be concluded: (3) not established. Q has at least two daughters, S and T: (4) false. Answer: (A).

Concept: Blood-relation deduction. The trap is assuming Q's gender (to accept or reject statement 3) - only statements that follow necessarily count as true.

Library and Canteen are in the same top row, Library on the west side, so the library is due west of the canteen, not northwest: (A) wrong. The hospital is north of the chemistry lab (same east side, across the cross road): (B) wrong. Physics Lab is in the north-west block and Chemistry Lab in the south-east block, so the chemistry lab is south-east of the physics lab: (C) correct. Classrooms (south-west, bottom) and Canteen (north-east, top) are diagonally opposite: (D) wrong. Answer: (C).

Concept: Reading directions from a map. The usual trap is calling a same-row neighbour "northwest" because it sits in the upper-left quadrant of the map.

(A) "reflected" means thought about; no mirror is mentioned. (B) He planned a book of poems about the things in his pocket, not a poem on epics. (C) The passage says the pocket-knife is "the type of all human tools, the infant of the sword" - stated directly. (D) He says the book would be too long and the age of epics is past, not that epics are inconvenient to write. Answer: (C).

Concept: Literal comprehension. Distractors exploit a pun (reflect / mirror) and twist nearby phrases (epics, too long). Only a statement directly supported by the text is correct.

Let the height of P above QR be h. The distance between ST and QR is h/2, so P is h/2 above ST. Triangle PST is similar to triangle PQR with ratio (h/2)/h = 1/2, so its area is (1/2)² = 1/4 of triangle PQR. The trapezium SQRT is the rest: 3/4 of PQR. Ratio = (1/4)/(3/4) = 1/3. Answer: (A).

Concept: Similar triangles: areas scale with the square of the linear ratio. The trap is answering 1/4 (PST to the whole triangle) or 1/2 (the linear ratio) instead of PST to the trapezium.

Binomial with n = 3, p = 1/6. P(exactly one 6) = C(3,1) x (1/6) x (5/6)² = 3 x 25/216 = 75/216. Answer: (A).

Concept: Binomial probability. The trap is 25/216, which is the probability for one specific position of the 6 - forgetting the factor 3 for which roll shows it.

We choose 3 scoops from 4 flavours with repetition allowed and order ignored: combinations with repetition, C(n + r - 1, r) = C(4 + 3 - 1, 3) = C(6,3) = 20. Check by cases: all three same flavour 4, two of one and one of another 4 x 3 = 12, all different C(4,3) = 4; total 20. Answer: (B).

Concept: Stars and bars (multisets). Traps: 4³ = 64 or 4 x 3 x 2 = 24 count ordered selections; C(4,3) = 4 forbids repetition.

The blank needs a noun meaning firm determination that "never wavered" despite initial hesitation. Ambivalence means mixed feelings, the opposite of what is described. Satisfaction does not fit "to contribute". Revolve is a verb meaning to rotate. Resolve (noun) means firm determination. Answer: (C) resolve.

Concept: Vocabulary in context: the contrast word "Despite" and the phrase "never wavered" point to determination. The trap is the look-alike word revolve, or ambivalence, which echoes the hesitation rather than contrasting with it.

A nest is the home a bird builds. The home of a bee is a hive. A kennel houses a dog, a lair is a wild animal's den and a hammock is a hanging bed. Answer: (C) Hive.

Concept: Animal-to-dwelling analogy. The trap is lair, which is also an animal home but for wild beasts, not bees.

Multiply both sides by e^x: P e^(2x) = Q for every real x. The left side changes with x unless P = 0, and the right side is a constant. So P = 0, and then Q = P e^(2x) = 0. Check: P = Q = 1 gives e^x = e^(-x), true only at x = 0. P/Q = 0 needs Q nonzero, but Q must be 0, so P/Q is undefined. Answer: (A) P = Q = 0.

Concept: An identity holding for all x forces coefficients of independent functions to match; e^x and e^(-x) are linearly independent. The trap is checking a single value like x = 0, which makes option (B) look right.

(A) fails: 2 + 3 = 5 is prime. (B) p1p2 has divisors 1, p1, p2 and p1p2 with p1, p2 at least 2, so it is never prime: true. (C) fails: 3 + 5 + 1 = 9. (D) fails: 3 x 5 + 1 = 16. Answer: (B) p1p2 is not a prime number.

Concept: "For all values" statements are killed by one counterexample, and the prime 2 is the usual counterexample. A product of two primes always has a nontrivial factor.

"Even if I had known" is a counterfactual conditional: it describes a situation contrary to fact. So Ramya did not know that Josephine was in the hospital. Nothing is said about past friendship or the cause of the hospital stay, so (C) and (D) are not supported, and (A) contradicts the counterfactual. Answer: (B).

Concept: Past-perfect "had known" signals an unreal past condition. The trap is reading emotional content (a broken friendship) that the sentence does not state.

IMAGE reversed is EGAMI; shifting each letter forward by one gives FHBNJ. FIELD reversed is DLEIF; shifted by one gives EMFJG, confirming the rule. BEACH reversed is HCAEB; shifted by one gives IDBFC. Answer: (B) IDBFC.

Concept: Coding-decoding: test reversal and letter shifts together, and verify the rule on the second example before applying it.

X: 20 + (-4) = 16, 16 + 10 = 26, 26 + 15 = 41, so X(i) = X(i-1) + I(i). Y: 20 x (-4) = -80, -80 x 10 = -800, -800 x 15 = -12000, so Y(i) = Y(i-1)I(i). Option (D) would give Y(1) = 20 x 20 = 400. Answer: (A).

Concept: Fit a recurrence by testing each candidate on the first step, then confirm on the rest. The trap in (D) is the off-by-one index I(i-1).

Triangle PLS has base PL and its third vertex S lies on MN, the side opposite PL. So its height over PL equals the rectangle's width, and area(PLS) = half of area(PLMN). Viewed inside the square, triangle PLS has base PS = 2 and L on QR, which is parallel to PS at distance 2, so area(PLS) = (1/2)(2)(2) = 2. Hence area(PLMN) = 2 x 2 = 4 cm². Answer: (D) 4.

Concept: A triangle on one side of a parallelogram with its apex on the opposite side has half its area. The same triangle is half the square, so rectangle and square are equal. Trying to find the rectangle's sides directly is the slow trap.

Treat zones as vertices and bridges as edges. Connecting 5 zones needs at least 4 bridges forming a spanning tree, so (A) with 3 bridges is impossible and (D) with 5 is not least. (B) P, Q, S, T joins Z1-Z2, Z1-Z3, Z1-Z5, Z3-Z5: Z4 is never reached. (C) Q, R, T, V joins Z1-Z3, Z2-Z3, Z3-Z5, Z3-Z4: all five zones connected with 4 bridges. Answer: (C).

Concept: Minimum connections for n regions is n - 1, a spanning tree. The trap is counting bridges without checking that every zone, here Z4, is actually reached.

The first triple moves from mildly unwell (sick) to weak (infirm) to near death (moribund). The second triple must move from mildly foolish (silly) to more foolish to utterly foolish (daft). The middle word must be an adjective on the foolishness scale. 'Frown' is a facial expression, 'fawn' means to flatter (or a young deer) and 'vein' is a blood vessel; none describes foolishness. 'Vain' carries the sense of empty, futile or foolishly conceited, which sits between silly and daft. Answer: (D) vain.

Concept: Tests word meaning along an intensity scale. The trap is the near-homophones: 'vein' and 'vain' sound alike, and 'fawn' looks like a plausible adjective, but only 'vain' is an adjective related to foolishness.

A number is divisible by 3 when its digit sum is. The five digits sum to 1+3+4+6+7 = 21. Using four distinct digits means leaving one digit out, so the digit sum is 21 - d, which is divisible by 3 only when the omitted digit d is divisible by 3, i.e. d = 3 or d = 6. That gives two digit sets, {1,4,6,7} and {1,3,4,7}. Each set of 4 distinct digits can be arranged in 4! = 24 ways (no zero, so every arrangement is a valid 4-digit number). Total = 2 × 24 = 48. Answer: (B) 48.

Concept: Divisibility by 3 via digit sum, plus permutations. The trap is trying to enumerate numbers directly instead of noticing that only the omitted digit matters.

Split the series into two geometric series after the leading 2. Powers of 2: 1/2 + 1/4 + 1/8 + 1/16 + ... = (1/2)/(1 - 1/2) = 1. Powers of 3: 1/3 + 1/9 + 1/27 + ... = (1/3)/(1 - 1/3) = 1/2. Sum = 2 + 1 + 1/2 = 7/2. Answer: (B) 7/2.

Concept: Recognising two interleaved geometric progressions. The trap is treating the terms as a single pattern or forgetting the leading 2.

Valid votes = 1,15,000 - 5,000 = 1,10,000. B and C together have 25% + 20% = 45% of the valid votes. 45% of 1,10,000 = 49,500. Answer: (B) 49,500.

Concept: Percentages from a pie chart. The trap is applying the percentages to all votes cast (1,15,000 gives 51,750) instead of only the valid votes.

The passage says that gene mutations resulted in some people developing the ability to digest dairy milk. (A) overreaches: only some people developed it. (B) contradicts the passage. (C) says nothing the passage supports; it never claims the ability is essential. (D) restates the passage: the ability to digest milk came from a mutated gene. Answer: (D).

Concept: Inference from a short passage: choose what the text directly supports. The trap is universal claims such as 'all' or 'essential' that go beyond the text.

Three independent births give 2³ = 8 equally likely outcomes. Exactly two girls and one boy can occur as GGB, GBG or BGG: C(3,2) = 3 outcomes. Probability = 3/8. Answer: (A) 3/8.

Concept: Binomial counting. The trap is answering 1/8 by counting only one ordering of the children.

Let the return of A be r. Person 1 gets 10,000(1 + r) + 40,000(1.15) = 56,000 + 10,000r. Person 2 gets 20,000(1 + r) + 30,000(1.15) = 54,500 + 20,000r. Given Person 1 - Person 2 = 500: 1,500 - 10,000r = 500, so r = 0.10. Answer: (B) 10%.

Concept: Setting up a linear equation from percentage returns. The trap is comparing only the gains and forgetting that both invested the same total (50,000), or mixing up which person holds more of fund A.

From the views, faces adjacent to 5 are 4, 1 (view 1), 6 and 2 (view 3), so 3 is opposite 5. Faces adjacent to 4 are 5, 1 (view 1), 6, 3 (view 2), so 2 is opposite 4. The remaining pair 1 and 6 are opposite. In each net the middle column is a strip of four faces that wraps round the cube, so alternate faces in it are opposite, and the two side flaps (one at the top-left, one at the bottom-right) are opposite each other. (A): column 1,4,6,2 gives 1-6 and 4-2 opposite, flaps 5 and 3 opposite - all three pairs match. (B): column 1,4,2,6 puts 1 opposite 2 - wrong. (C): column 1,3,2,4 puts 1 opposite 2 - wrong. (D): column 1,4,6,3 puts 4 opposite 3 - wrong. Answer: (A).

Concept: Cube nets: first derive opposite-face pairs from the views (a face adjacent to four others is opposite the fifth), then check which net places those pairs opposite. The trap is checking adjacency by eye instead of opposites.

A plane through both apexes contains the common axis. Each cone is cut along its axis into an isosceles triangle whose base is a diameter of the shared circle. The two congruent isosceles triangles share that diameter as base, pointing in opposite directions, so the outline has four equal sides (the slant heights). That quadrilateral is a rhombus. Answer: (A) A rhombus.

Concept: 3D visualisation: an axial section of a cone is an isosceles triangle. The trap is picturing a slanted cut, which would give an ellipse, rather than a cut through both vertices.

Dry, arid and parched describe the same idea (lack of water) with rising intensity. Diet means eating less, fast means not eating for a period, and the next, more intense step is to go without food to the point of suffering: starve. Reject, feast and deny do not continue the intensity scale (feast is the opposite). Answer: (A) starve.

Concept: Intensity analogy: the third word must be the strongest form of the same idea. The trap is picking feast because it is food-related, though it reverses the direction.

Proportionality gives x + y = k(x − y) for a fixed constant k. Then x + y = kx − ky, so y(1 + k) = x(k − 1), hence x/y = (k + 1)/(k − 1). This depends only on the constant k, not on x or y individually. Answer: (D) is a constant.

Concept: Proportionality means a fixed ratio, so solving for x/y removes both variables. The trap is assuming the ratio still varies with x or y.

Sort: 9, 10, 11, 11, 13, 14, 15, 17, 18, 69. There are 10 values, so the median is the mean of the 5th and 6th values: (13 + 14)/2 = 13.5. Answer: (A) 13.5.

Concept: Median of an even-sized sample is the average of the two middle values after sorting. Traps: not sorting first, or computing the mean (18.7), which the outlier 69 inflates.

Take 5k, 3k and 13k coins. Value: ₹1 coins = 5k, ₹5 coins = 15k, ₹10 coins = 130k. Total = 150k. Share of ₹5 coins = 15k/150k = 10%. Answer: (C) 10%.

Concept: Convert the count ratio to a value ratio before taking percentages. The trap is using the count share 3/21 = 14 2/7%.

The right side is log(9pq), so p² + q² = 9pq. Square: p⁴ + 2p²q² + q⁴ = 81p²q², so p⁴ + q⁴ = 79p²q². Therefore (p⁴ + q⁴)/(p²q²) = 79. Answer: (A) 79.

Concept: Combine logs into one, drop the log, then use (a + b)² = a² + 2ab + b². The trap is answering 81 by forgetting to subtract 2p²q².

Keep one's head down = stay focused and avoid trouble. A batter heads out to bat. Having a head for something = a natural aptitude for it, so a head for batting. A cool head on one's shoulders = calm and sensible. So (i) down, (ii) out, (iii) for, (iv) on. Answer: (C).

Concept: Tests fixed idioms with prepositions: head down, head out, a head for, a head on one's shoulders. The trap is choosing by sound rather than the fixed phrase.

Joining the two 54 cm edges makes a tube of length 54 cm whose circumference is 4 cm, so r = 2/π. Volume = πr²h = π(4/π²)(54) = 216/π. Sheet area = 216 cm², so cube surface 6a² = 216 gives a = 6 and volume 216. Ratio = (216/π)/216 = 1/π. Answer: (A) 1/π.

Concept: When the longer edges are joined, the shorter side becomes the circumference. The trap is using 54 cm as the circumference, which gives a very different radius.

Fat share = 20% + 20% + 5% = 45% of 2,000 kcal = 900 kcal. All fats give 9 kcal/g, so mass = 900/9 = 100 g. Answer: (C) 100.

Concept: Convert the energy share to kcal, then divide by energy density to get grams. Traps: using 4 kcal/g, or forgetting trans fat.

Each fold halves the current long edge. Fold 1: 20 × 8 becomes 10 × 8. Fold 2: long edge 10 becomes 5, giving 5 × 8. Now the long edge is 8, so fold 3 gives 5 × 4. Perimeter = 2(5 + 4) = 18 cm. Answer: (A) 18.

Concept: 'Its long edge' refers to the sheet as it is at each fold, so the folding direction switches once 8 becomes the longer side. The trap is halving 20 three times (2.5 × 8, perimeter 21).

Walk, jog and sprint are the same action at rising intensity. The second series must go from mildly troubled (bothered) to intimidated into losing courage (daunted). 'Fazed' means disturbed or disconcerted - stronger than bothered, weaker than daunted. 'Phased' (done in stages), 'phrased' (worded) and 'fused' (joined) are sound-alike distractors with unrelated meanings. Answer: (C) fazed.

Concept: Intensity analogy combined with a homophone trap: phased, fazed and phrased sound alike, but only 'fazed' belongs to the bothered-daunted scale. Read the meaning, not the sound.

Each wizard mixes all four elements in every possible order, so each tries every permutation of 4 distinct items: 4! = 4 × 3 × 2 × 1 = 24. They work independently, so each makes 24 attempts on his own (the combined 48 is not asked). Answer: (A) 24.

Concept: Order matters, so this is a permutation count 4!. The trap is doubling to 48 for two wizards - the question asks for attempts by each wizard.

Students who like at least one = 10,000 − 1,500 = 8,500. Let O = students liking other branches and C = O/4 liking core branches. By inclusion-exclusion, C + O − 500 = 8,500, so O/4 + O = 9,000, giving 5O/4 = 9,000, O = 7,200 and C = 1,800. Answer: (A) 1,800.

Concept: Two-set inclusion-exclusion: |C ∪ O| = |C| + |O| − |C ∩ O|. Forgetting to account for the 500 who like both gives a wrong total.

The right side is (1/2) ln(xy) = ln √(xy). So ln((x + y)/2) = ln √(xy), giving (x + y)/2 = √(xy). The arithmetic mean equals the geometric mean only when x = y (equivalently (√x − √y)² = 0). Then x/y + y/x = 1 + 1 = 2. Answer: (C) 2.

Concept: Log rules turn the equation into AM = GM, whose equality case forces x = y. Spotting the equality case avoids messy algebra.

Differences of the known terms: 9 − 6 = 3, 14 − 9 = 5, and 41 − 30 = 11. The differences are consecutive odd numbers 3, 5, 7, 9, 11. So x = 14 + 7 = 21, and indeed 30 − 21 = 9. Answer: (B) 21.

Concept: First-difference pattern (consecutive odd numbers). Check that the chosen value also fits the gap to the next term.

S opens the story with a time marker ('Many millennia ago') and introduces the collision. P refers back to it as 'this fortuitous geological event' and describes the rocks rising. R refers to 'this massive edifice' (the risen rock mass) cooling. Q begins with 'Thus' and draws the conclusion. Order: S P R Q. Answer: (C) SPRQ.

Concept: Para-jumble: find the opener with no back-reference, then chain demonstratives ('this event', 'this edifice') to their antecedents; a 'Thus' sentence concludes, it does not open.

Let each selling price be 99. Cost of the first item = 99/1.1 = 90; cost of the second = 99/0.9 = 110. Total cost 200, total selling price 198. Loss = 2 on 200 = 1%. In general the loss is (10/10)² % = 1%. Answer: (C) 1% loss.

Concept: Equal selling prices with equal percentage profit and loss always give a net loss of (x/10)² %. The trap is assuming the profit and loss cancel.

Renewable share in 2007 = 30 + 5 + 5 = 40%. In 2023 = 35 + 20 + 10 = 65%. Since the total is unchanged, the increase is (65 − 40)/40 × 100 = 25/40 × 100 = 62.5%. Answer: (D) 62.5%.

Concept: Percentage increase is relative to the starting value. Reporting the 25 percentage-point rise as 25% is the trap (option A).

Each full straight cut can at most double the number of pieces (pieces are not rearranged), so at least 3 cuts are needed since 2³ = 8. Three mutually perpendicular cuts through the centre, parallel to the three pairs of faces, give 8 identical smaller cubes. Answer: (A) 3.

Concept: Spatial reasoning: three orthogonal planes through the centre split a cube into 8 congruent cubes; fewer than 3 cuts cannot produce 8 pieces.

Row 3 is 3, X, 6, X and row 4 has no crosses. Count the non-X neighbours of each row-4 cell. Cell (4,1): neighbours (3,1), (3,2)=X, (4,2) - count 2. Cell (4,2): (3,1), (3,2)=X, (3,3), (4,1), (4,3) - count 4. Cell (4,3): (3,2)=X, (3,3), (3,4)=X, (4,2), (4,4) - count 3. Cell (4,4): (3,3), (3,4)=X, (4,3) - count 2. Sum = 2 + 4 + 3 + 2 = 11. Answer: (A) 11.

Concept: Neighbour counting with diagonals included. Only row 3 and row 4 matter for the last row; the usual slips are dropping diagonal neighbours or counting the X cells.

The blank must modify the verb "missed" and convey that the train was almost caught but not quite. "nearly" (an adverb meaning almost) fits: we arrived late and nearly missed the train. "near" is an adjective/preposition, "utterly" means completely and does not collocate with missed a train, and "mostly" makes no sense for a single event. Answer: (B) nearly.

Concept: Tests adverb choice in context. The trap is "near", which sounds similar but is not the adverb meaning "almost"; "nearly" is the correct modifier for a verb.

Run the recurrence backwards using Fₙ₋₁ = Fₙ₊₁ - Fₙ. F5 = F7 - F6 = 60 - 37 = 23. F4 = F6 - F5 = 37 - 23 = 14. F3 = F5 - F4 = 23 - 14 = 9. F2 = F4 - F3 = 14 - 9 = 5. F1 = F3 - F2 = 9 - 5 = 4. Check forwards: 4, 5, 9, 14, 23, 37, 60. Answer: (A) 4.

Concept: A Fibonacci-type recurrence solved in reverse. The usual slip is subtracting in the wrong order or stopping at F2 = 5, which is also an option.

90% of pregnant women received care at least once. Of these, 60% received care from doctors, i.e. 0.9 × 0.6 = 0.54 = 54% of all pregnant women received care at least once from a doctor. 54% is more than half, so (A) follows with certainty. Nothing is said about how many times care was received, so the "at most once" options cannot be inferred. Answer: (A).

Concept: Percent of a percent: 60% applies to the 90% subgroup, not to everyone. The trap is reading 60% as a share of all women, or being distracted by the "at most once" options, which the data do not address.

A smooth 3D object need not be convex (for example a bowl or a dumbbell shape) nor concave everywhere (a closed surface seen from outside cannot be concave everywhere). Smooth also rules out edges and corners, so (D) is wrong. The only statement that is always true is that the surface may be concave in some places and convex in others. Answer: (C).

Concept: Spatial reasoning about curvature. The trap is assuming every smooth object is convex like a sphere; smoothness only excludes edges and corners, not concave regions.

The government pays people to cycle to work as a measure against health problems caused by lack of physical exercise. That measure only makes sense if the government believes cycling is a form of physical exercise, so (D) is a necessary assumption and can be inferred. (A) and (B) predict outcomes that are not guaranteed, and (C) attributes a recommendation to the experts that the passage never mentions. Answer: (D).

Concept: Inference vs. speculation: the correct option is the belief implicit in the stated policy. Options that predict results or invent facts about what experts said are the usual traps.

f(t) - g(t) = 0.01t² - 4t = 0.01t(t - 400). For 0 < t < 400 this is negative, so g(t) > f(t) (e.g. t = 1: g = 4, f = 0.01), and (i) holds. For t > 400 it is positive, so taking T = 400 gives f(t) > g(t) for all t > T, and (ii) holds. Answer: (C) both (i) and (ii) are correct.

Concept: A quadratic eventually overtakes any linear function, but the linear one can be larger for small t. The trap is assuming one function dominates everywhere.

(iii) introduces Mina and her heavy bags - the natural opening. (ii) follows: she pants after climbing the stairs, explained by the heavy bags. (i) then has her reach the terrace and notice "the man", and (iv) describes what "he" was doing, which needs the man to have been introduced in (i). Order: (iii), (ii), (i), (iv). Answer: (D).

Concept: Narrative ordering using pronoun references and cause-effect. "He" in (iv) must follow the introduction of "the man", and the panting needs the heavy bags first.

Fix y = y0. Then f(x) = g(y0) for every x, so f is constant, say c. Similarly fixing x shows g(y) = f(x0) = c for every y. Hence f(x) = g(y) = c, a constant, but not necessarily zero (f = g = 5 works), so (A) is not necessary. (C) contradicts what we proved. (D) would force g(y) = 0, again not necessary. Answer: (B).

Concept: If a function of x equals a function of y for all independent x and y, both must equal the same constant. The trap is picking (A), which is one valid case but not forced.

The first blank needs a noun meaning the price of travel: 'fare'. The second blank needs an adjective meaning reasonable or just: 'fair'. The sentence reads 'The fare is too high for it to be considered fair.' Answer: (D) fare / fair.

Concept: Homophone pair fare (price of a ticket) and fair (just, reasonable). The trap is choosing by sound alone; read each blank for the part of speech it needs.

The curve is a step function, so the area is a sum of rectangles. First piece: width 1/3, height 2, area 2/3. Second piece: width 3/4 - 1/3 = 5/12, height 3, area 5/4. Third piece: width 1/4, height 1, area 1/4. Total = 2/3 + 5/4 + 1/4 = 2/3 + 3/2 = 4/6 + 9/6 = 13/6. Answer: (C) 13/6.

Concept: Area under a piecewise constant function is the sum of width times height for each piece. The usual slip is mis-computing the middle width 3/4 - 1/3 = 5/12.

From the equation, r² + 2r = −6, i.e. r² = −2r − 6. Pair the factors: (r + 2)(r + 3) = r² + 5r + 6 = (−2r − 6) + 5r + 6 = 3r. (r + 4)(r + 5) = r² + 9r + 20 = 7r + 14. Product = 3r(7r + 14) = 21r² + 42r = 21(r² + 2r) = 21 × (−6) = −126. Answer: (D) −126.

Concept: Reduce every r² using the equation instead of solving for the complex roots. The trap is pairing (r+2)(r+5) and (r+3)(r+4), which does not collapse as neatly, or a sign slip giving 126.

With at least one student, 'All students are inquisitive' and 'No student is inquisitive' are contraries: they cannot both be true. Statements 1 and 2 can both be true (all implies some). Statements 2 and 4 can both be true (some are, some are not). Statements 3 and 4 can both be true (if none is, then some are not). Answer: (A) Statement 1 and Statement 3.

Concept: Square of opposition: A (all) and E (none) are contraries, A and O / E and I are contradictories. The existence assumption is what makes 'some' follow from 'all'.

Rotate every plate upright and list the five letters. (A) A, D, D, D, J: J appears once and A appears once, so two letters have odd counts - no palindrome. (B) A, D, R, A, R: counts A2, R2, D1 - one odd letter, so a palindrome exists: RADAR. (C) A, D, Z, E, D: A, Z, E each once - no. (D) A, D, I, L, Y: all single - no. Answer: (B).

Concept: A five-letter palindrome needs every letter to appear an even number of times except one (the middle). Rotating a plate in its plane does not change which letter it is; mirroring would, but is not allowed.

The passage states that in organizations with good work culture, metrics help improve outcomes, while in poor work culture they are counterproductive. (A) contradicts the passage, (C) and (D) contradict it for good work culture. (B) restates the passage. Answer: (B) Metrics are useful in organizations with good work culture.

Concept: An inference must follow from the passage without adding claims. Watch for absolute words like 'always' and 'never', which the passage does not support.

Girls are 60% of the qualified candidates, so boys are 40% of the qualified. Therefore fewer boys qualified than girls. (A) is wrong (60 vs 40), (B) is wrong (65% vs 35% appeared), (C) is wrong (more boys appeared). Answer: (D) The number of boys who qualified the test is less than the number of girls who qualified.

Concept: Keep the two bases separate: 65% is of those who appeared, 60% is of those who qualified. The trap is mixing the percentages across bases.

Condition on the first draw. Green first (probability 3/5): box becomes 2 green, 2 orange, so P(orange next) = 2/4 = 1/2. Orange first (probability 2/5): it is replaced, box stays 3 green, 2 orange, so P(orange next) = 2/5. Total = (3/5)(1/2) + (2/5)(2/5) = 3/10 + 4/25 = 15/50 + 8/50 = 23/50. Answer: (D) 23/50.

Concept: Total probability over the outcome of the first draw. The trap is assuming the box is unchanged, giving 2/5, or forgetting that only green balls are removed.

Among the six points, two joins are parallel only when one is a midline (two midpoints) and the other lies along the third side. Suppose S were a corner. S lies on two sides, so U (not sharing a side with S) must be the midpoint of the side opposite S. QS through corner S cannot be a midline, so PR must be the midline parallel to a side through S; that midline joins the midpoint opposite S (which is U) to another midpoint, so P or R would be U - impossible. Hence S cannot be a corner. Arrangements exist with P at a corner, U at a midpoint and R at a corner, so (A), (C), (D) are not forced. Answer: (B) S cannot be placed at a corner.

Concept: Midpoint theorem: the segment joining two midpoints is parallel to the third side. Test each option by assuming it false and checking for a contradiction.

The first blank needs a subject and verb: "They're" (they are) bringing. The second blank needs a possessive before "books": "their". The third blank needs a place, answering "from where": "there". So the sentence reads "They're bringing their books from there." Answer: (B).

Concept: Tests the homophones they're / their / there: contraction of "they are", possessive pronoun, and adverb of place. The trap is choosing by sound; read each blank for its grammatical role.

The word "but" sets up a contrast with "organized" investigations. The blank therefore needs a word meaning the opposite of organized: "meandering" (wandering, without a fixed course). "Systematic" and "consistent" mean organized, and "timely" is unrelated to the contrast. Answer: (A).

Concept: Vocabulary in context with a contrast signal. "But" means the blank opposes "organized". The trap is picking "systematic" because it sounds like a positive word for an investigation.

Side of the square = √d, so the diagonal = √2 · √d = √(2d). This is the circle's diameter, so the radius r = √(2d)/2 and r² = 2d/4 = d/2. Area of the circle = πr² = πd/2. Answer: (D) (1/2)πd.

Concept: Square diagonal = side × √2, and circle area uses the radius, not the diameter. The usual trap is using the diagonal as the radius (giving 2πd) or squaring d itself, which is already an area.

The number is LCM(20, 42, 76) + 7. 20 = 2² · 5, 42 = 2 · 3 · 7, 76 = 2² · 19. LCM = 2² · 3 · 5 · 7 · 19 = 7980. Smallest number = 7980 + 7 = 7987. Check: 7987 = 20 × 399 + 7 = 42 × 190 + 7 = 76 × 105 + 7. Answer: (C) 7987.

Concept: Same remainder for several divisors means number = k × LCM + remainder; the smallest is LCM + remainder. The trap is multiplying the divisors instead of taking the LCM, which overshoots.

Successive ratios: 12/2 = 6, 60/12 = 5, 240/60 = 4, 720/240 = 3, 1440/720 = 2. The next multiplier is 1, so the missing term is 1440 × 1 = 1440, and then 1440 × 0 = 0, which matches the last term. Answer: (B) 1440.

Concept: A multiplicative series with a decreasing multiplier 6, 5, 4, 3, 2, 1, 0. The final 0 confirms the pattern. The trap is assuming the terms keep growing and picking 2880.

Let m men and w women be eligible, m + w = 300. Money given = 750 × (8/9)m + 1000 × (2/3)w = (2000/3)m + (2000/3)w = (2000/3)(m + w) = (2000/3) × 300 = 2,00,000. The split between men and women does not matter. Answer: (B) 2,00,000.

Concept: The per-person effective amounts happen to be equal (750 × 8/9 = 1000 × 2/3), so the total depends only on m + w. The trap is thinking the problem is unsolvable without knowing m and w separately.

p⁻ˣ = 1/q gives pˣ = q. Similarly qʸ = r and rᶻ = p. Substitute: p = rᶻ = (qʸ)ᶻ = q^(yz) = (pˣ)^(yz) = p^(xyz). So p^(xyz) = p¹, giving xyz = 1. Answer: (C) 1.

Concept: Chained exponent substitution: rewrite each relation without negatives and substitute back to the starting base. The trap is getting lost in the reciprocals and choosing -1 or 1/(pqr).

Take 100 invited guests: 60 male, 40 female. Attendees = 80. All 40 females attended, so males attending = 80 - 40 = 40. Ratio of males to females = 40 : 40 = 1 : 1. Answer: (B) 1:1.

Concept: Assume a convenient total (100) for percentage problems. The trap is applying 80% to males alone (48 males, ratio 6:5) instead of to all guests.

Triangle EAB: ∠DEC = 180° - ∠BAD - ∠ABC. Triangle FAD: ∠BFC = 180° - ∠BAD - ∠ADC. Adding: ∠DEC + ∠BFC = 360° - 2∠BAD - (∠ABC + ∠ADC). In quadrilateral ABCD, ∠ABC + ∠ADC = 360° - ∠BAD - ∠BCD. Substituting: ∠DEC + ∠BFC = 360° - 2∠BAD - 360° + ∠BAD + ∠BCD = ∠BCD - ∠BAD. Answer: (A).

Concept: Angle sums in overlapping triangles combined with the quadrilateral angle sum of 360°. The trap is adding instead of subtracting ∠BAD, since it appears in both triangles and once in the quadrilateral.

P(green) = 2/3, P(red) = 1/3. P(k green in 7) = C(7,k)(2/3)^k(1/3)^(7-k), proportional to C(7,k) · 2^k. k=3: 35 × 8 = 280. k=4: 35 × 16 = 560. k=5: 21 × 32 = 672. k=6: 7 × 64 = 448. The largest is k = 5. Answer: (C) Five green faces and two red faces.

Concept: Binomial mode: compare C(n,k)p^k q^(n-k) across k (mode is near (n+1)p = 5.33, so k = 5). The trap is picking six green because green is more likely per roll, ignoring the C(7,k) factor.

'Until' already carries the negative sense of 'up to the time when', so adding 'does not' after it creates a double negative and changes the meaning (options A and B are wrong). The subject 'the minister' is third person singular, so the present-tense verb must be 'meets', not 'meet' (option C is wrong). 'I will not leave the place until the minister meets me' is correct. Answer: (D).

Concept: Tests subject-verb agreement and the use of 'until'. The trap is the double negative 'until ... does not', which sounds natural in some spoken Indian English but is ungrammatical.

A paraphrase restates a text or speech in other words. A paradox is a self-contradictory statement, a paradigm is a model or pattern, and paraffin is a wax or fuel. Answer: (A) paraphrase.

Concept: Vocabulary question built on look-alike words sharing the prefix 'para-'. Only the meaning matters; do not be distracted by the similar spelling.

Archimedes is not literally proposing to move the Earth; he uses an exaggerated image to express the power of the lever. A statement that uses words in a non-literal, imaginative sense is figurative. 'Literal' is the opposite, 'collateral' means secondary or accompanying, and a 'figurine' is a small statue. Answer: (A) figurative.

Concept: Distinguishes figurative from literal language. The distractor 'figurine' looks like 'figurative' but is a noun meaning a statuette.

All three words share 'taga', which must mean 'care'. In each, 'care' comes first in English but 'taga' comes last in the coded word: relf = free, o = ful, fer = less, so the coded word reverses the English order (modifier + taga). In 'aftercare', 'care' is the last English part, so 'taga' must come first, followed by a new code for 'after'. Only 'tagazen' has that form ('zen' = after). Answer: (C) tagazen.

Concept: Coded-language reasoning: identify the common root and, crucially, the word order. The trap is 'zentaga', which keeps 'taga' at the end as in the examples but ignores that the English order is reversed.

64 unit blocks form a 4 x 4 x 4 cube with surface area 6 x 16 = 96. Removing a corner block takes away its 3 exposed unit faces, but exposes 3 new unit faces of the neighbouring blocks (the inner faces of the notch). The net change per corner is 0, so the surface area stays 96. Answer: (D) 96.

Concept: Visualising surface area after removing corner blocks. The trap is subtracting 3 faces per corner (giving 72) and forgetting the 3 newly exposed faces.

Annual units: Elegance 27300+25222+28976+21012 = 102510; Smooth 20009+19392+22429+18229 = 80059; Soft 17602+18445+19544+16595 = 72186; Executive 9999+8942+10234+10109 = 39284. Revenue: Elegance 102510 x 48 = 4,920,480; Smooth 80059 x 63 = 5,043,717; Soft 72186 x 78 = 5,630,508; Executive 39284 x 173 = 6,796,132. Executive has the largest revenue, hence the greatest fraction. Answer: (B) Executive.

Concept: Revenue = price x quantity. The trap is picking Elegance because it sells the most units; the much higher price of Executive outweighs its lower volume.

The passage says 'at least seventeen', so 'exactly seventeen' (A) is wrong. 'If this is not an indication ..., nothing else is' is a rhetorical way of saying the multilingual notes are a very strong indicator, not the only possible one, so (B) overstates it. Nothing is said about space on notes (C). The claim is simply that linguistic plurality strongly evidences India's diversity. Answer: (D).

Concept: Inference questions reward the option that follows without adding information. The traps are absolute words ('exactly', 'only') that go beyond the passage.

(ii): Statement III says S loses to P only sometimes, so S sometimes beats P. S cannot be the absolute worst player, so (ii) fails. (i): The chain P > Q > R > S suggests P is strong, but no statement relates P directly to R, and S beats P part of the time. Winning in poker is not transitive, so 'P is likely to beat all three' cannot be logically inferred from the given statements. Neither inference follows. Answer: (D) neither (i) nor (ii).

Concept: Tests whether you assume transitivity where none is given. The trap is chaining P > Q > R > S into conclusions about pairs the statements never compare, and ignoring that S beats P sometimes.

By the factor theorem, (x - c) is a factor iff f(c) = 0. f(1) = 2 + 3 - 5 = 0, so (x - 1) is a factor. Check the others: f(-1) = -2 - 3 - 5 = -10 ≠ 0; f(5/2) is clearly positive and large; x³ + 8 has root x = -2 and f(-2) = -256 - 6 - 5 ≠ 0. Answer: (B) (x - 1).

Concept: Factor theorem: substitute the root of each candidate factor. A quick check of x = 1 (sum of coefficients equals zero) settles it immediately.

Let N = A·e^(-kL). From N(80) = 100 and N(40) = 10000: N(40)/N(80) = e^(40k) = 100, so k = ln 100 / 40. For N = 5000: N(40)/N(L) = 10000/5000 = 2 = e^(k(L - 40)), so L - 40 = ln 2 / k = 40 · ln 2 / ln 100 = 40 × 0.30103 / 2 = 6.02. L = 46.02. Answer: (B) 46.02.

Concept: Exponential decay model fitted from two points, solved with logarithms. The trap is assuming linear interpolation between 40 and 80, which gives a very different load.

The blank names a job he held before becoming Municipal Commissioner. "A security guard" is right because he was one of many guards (indefinite). "The university" is right because a specific, identifiable university is meant. So the phrase is "a security guard at the university". Answer: (B).

Concept: Definite versus indefinite article. A job held among many people takes "a"; a specific place takes "the". The trap is "the security guard", which implies he was the only guard.

"Cope with" means to deal with or endure a difficult situation. "Put up with" means to tolerate or endure. "Put down to" means to attribute, "put up against" means to pit against, and "put in with" is not an idiom with this sense. Answer: (A) put up with.

Concept: Phrasal-verb meaning in context. All options share "put", so only the particle decides the meaning; only "put up with" carries the sense of enduring.

Mock, deride and jeer all mean to ridicule or scorn. Praise means to express approval, the opposite sense. Answer: (C) praise.

Concept: Odd-one-out by meaning: three near-synonyms and one antonym. Knowing that "deride" means ridicule is the only hurdle.

Write letter positions. CADBE = 3,1,4,2,5: steps -2,+3,-2,+3. JHKIL = 10,8,11,9,12: -2,+3,-2,+3. XVYWZ = 24,22,25,23,26: -2,+3,-2,+3. ONPMQ = 15,14,16,13,17: -1,+2,-3,+4. Only ONPMQ breaks the pattern. Answer: (D) ONPMQ.

Concept: Letter-series pattern: convert letters to alphabet positions and compare the differences. Judging by the look of the letters instead of the step sizes is the trap.

α⁻ⁿ + β⁻ⁿ = 1/αⁿ + 1/βⁿ = (αⁿ + βⁿ)/(αβ)ⁿ. So the ratio is (αⁿ + βⁿ) × (αβ)ⁿ/(αⁿ + βⁿ) = (αβ)ⁿ = 4ⁿ. Answer: (B) 4ⁿ.

Concept: Negative powers combine over the common denominator (αβ)ⁿ. Trying to find the actual roots is unnecessary and is the time trap.

Faculty with at least one account = 150 - 30 = 120. By inclusion-exclusion, both = 55 + 85 - 120 = 20. Only Facebook = 55 - 20 = 35. Answer: (A) 35.

Concept: Two-set inclusion-exclusion. The traps are answering 55 (all Facebook users) or not removing the 30 with neither account first.

(i): the author only wonders whether the developments are good; no belief that computers are bad is stated, so (i) cannot be inferred. (ii): the passage says many believe the internet is an unintended consequence, which contradicts (ii). Neither follows. Answer: (D).

Concept: Inference versus over-reading. Raising a question is not asserting a conclusion, and a statement the passage contradicts cannot be inferred.

"All hill-stations have a lake" means at least one lake; it does not limit the number. So a hill-station may have two lakes and (ii) does not follow. Ooty having two lakes is consistent with being a hill-station, so (i) does not follow either. Answer: (D).

Concept: Reading "have a lake" as "exactly one lake" is the trap. Neither conclusion is forced by the premises.

A rectangle is fixed by choosing 2 of the 3 horizontal lines and 2 of the 5 vertical lines. Count = C(3,2) × C(5,2) = 3 × 10 = 30. Answer: (C) 30.

Concept: Rectangles in an m × n grid: C(m+1,2) × C(n+1,2). The trap is counting only the 8 cells or missing multi-cell rectangles.

The graph is an inverted V with its peak at x = 1, so f(x) = k - |x - 1| with k = f(1) = 2. Check: f(-3) = 2 - 4 = -2 and f(3) = 2 - 2 = 0, both match. Answer: (C) f(x) = 2 - |x - 1|.

Concept: Reading an absolute-value graph: the vertex gives the shift and height, and a downward V means a minus sign. Options with + |x - 1| open upward.

To "cope with" conditions is to manage or adjust to them. "Adapt to" means to adjust oneself to new conditions, which is the same sense. "Adopt" means to take up or accept something as one's own, "adept in" means skilled at, and "accept with" is not idiomatic here. Answer: (B) adapt to.

Concept: Vocabulary in context. The trap is the near-homophones adopt / adapt / adept: only "adapt to" carries the sense of adjusting to circumstances.

Let the roots of ax² + bx + c = 0 be p and q, both positive. For x ≥ 0, |x| = x and the equation is ax² + bx + c = 0, whose roots p and q are both positive, so both are valid: 2 roots. For x < 0, |x| = -x and the equation becomes ax² - bx + c = 0, whose roots are -p and -q (replace x by -x). Both are negative, so both are valid: 2 more roots. Total 4 real roots: p, q, -p, -q. Answer: (D) 4 real roots.

Concept: ax² + b|x| + c is an even function of x, so every positive root has a mirror negative root. The trap is solving only the x ≥ 0 case and answering 2.

The sentence narrates completed past events: he fell down and then lost consciousness. "Fell down" is simple past, so the parallel verb joined by "and" should also be simple past: "lost consciousness". The past continuous "was losing consciousness" breaks the sequence. Parts I, II and III are correct. Answer: (D) IV.

Concept: Tense consistency between verbs joined by 'and' in a narrative of completed events. The trap is 'from under the table', which looks odd but is a correct double preposition.

The main clause ends with "done", a past participle, so it needs "could/would have done": an unreal past (third) conditional. The third conditional takes the past perfect in the if-clause: "If she had known ..., she could have done the experiment." Answer: (C) had known, could have.

Concept: Third conditional: if + past perfect, could/would have + past participle. The trap is option D, which puts a modal ('should have known') inside the if-clause.

The terms are n² + 1 for n = 1, 2, 3, ...: 2, 5, 10, 17, 26, 37, 50, 65. Equivalently the differences are 3, 5, 7, 9, 11, 13, 15. The last term should be 50 + 15 = 65, not 64. Answer: (C) 64.

Concept: Recognising n² + 1 (or successive odd differences). The trap is noticing that 64 is a perfect square and assuming it fits.

Number of students = 21 + 17 + 6 = 44 (each row sums to 44). Total marks scored = 21 × 2 + 15 × 3 + 23 × 2 = 42 + 45 + 46 = 133. Average per student = 133 / 44 ≈ 3.02. Answer: (C) 3.02.

Concept: The class size is the row total, and the average is per student over the whole exam. The trap is averaging per question (133/132 or 133/3) or dividing by the number who answered correctly.