CSAT (GS Paper II) 2026

Answer: (a) Select this option if the question can be answered using one of these statements alone, but cannot be answered using other statement

There are only ten ways to pick three different coins, so list their totals. Every total whose digits add up to more than 5 includes the 5-coin, so Statement II settles it; Statement I leaves 32 (2 + 10 + 20) as an exception.

1. Three different coins from {1, 2, 5, 10, 20} can be chosen in 10 ways.
2. With the 5-coin: 1+2+5 = 8, 1+5+10 = 16, 1+5+20 = 26, 2+5+10 = 17, 2+5+20 = 27, 5+10+20 = 35.
3. Without the 5-coin: 1+2+10 = 13, 1+2+20 = 23, 1+10+20 = 31, 2+10+20 = 32.
4. Statement I (m not prime): m can be 8, 16, 26, 27, 35 (5-coin present) or 32 (5-coin absent). Not sufficient.
5. Statement II (digit sum > 5): 8, 16, 26, 17, 27, 35 have digit sums 8, 7, 8, 8, 9, 8; the totals without the 5-coin have digit sums 4, 5, 4, 5. So m always includes the 5-coin. Sufficient.
6. One statement alone works and the other does not.

• ✗ Statement I Not sufficient alone: non-prime totals include 32 = 2 + 10 + 20, which has no 5-coin, as well as 8, 16, 26, 27 and 35, which do.
• ✓ Statement II Sufficient alone: every total with digit sum above 5 (8, 16, 17, 26, 27, 35) uses the 5-coin; totals without it (13, 23, 31, 32) have digit sums of only 4 or 5.

Remember · In data sufficiency with a small finite set, list every case first, then test each statement against the list.

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Answer: (d) Select this option if the question cannot be answered even using any of the statements

Both statements together only trap y between x² and √x, and for 0 < x < 1 the number x itself lies inside that gap. So y can sit on either side of x and the question stays open.

1. Statement I: x = 0.9, y = 0.85 fits (0.81 < 0.85 < 1) and gives x > y; x = 0.1, y = 0.5 also fits and gives y > x. Not sufficient.
2. Statement II: √x < 1 means 0 ≤ x < 1. x = 0.25, y = 0.4 fits (0.4 < 0.5) with y > x; x = 0.25, y = 0.1 fits with y < x. Not sufficient.
3. Together: x² < y < √x with 0 < x < 1. Take x = 0.25: y must lie between 0.0625 and 0.5.
4. y = 0.4 gives y > x, y = 0.1 gives y < x. Still not decided.
5. Check: for 0 < x < 1, x² < x < √x, so x always lies inside the allowed range of y.

• ✗ Statement I Not sufficient alone: x = 0.9, y = 0.85 and x = 0.1, y = 0.5 both satisfy it but order x and y differently.
• ✗ Statement II Not sufficient alone: with x = 0.25, both y = 0.4 and y = 0.1 are below √x = 0.5, yet one is above x and one below.

Remember · For numbers between 0 and 1, x² < x < √x. A range that straddles x never decides which is bigger.

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Answer: (c) Select this option if the question can be answered using both the statements together, but cannot be answered using either statement alone

Statement I says at least one of x and y is even; Statement II says x and y have the same parity. Only together do they force both to be even, which answers the question.

1. Statement I: x²y² = (xy)² is even, so xy is even, so at least one of x, y is even. x could be odd (x = 1, y = 2). Not sufficient.
2. Statement II: 1 + x² + y² is odd, so x² + y² is even, so x and y are both even or both odd. Not sufficient.
3. Together: they share parity and at least one is even, so both are even. x is even — answered.
4. Check: x = 1, y = 1 satisfies II but not I; x = 2, y = 2 satisfies both.

• ✗ Statement I Not sufficient alone: x = 1, y = 2 gives x²y² = 4 (even) with x odd, while x = 2, y = 1 gives x even.
• ✗ Statement II Not sufficient alone: it only says x and y have the same parity; x = y = 1 and x = y = 2 both fit.

Remember · Parity items: turn each statement into a plain rule (at least one even / same parity), then combine the rules.

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Answer: (d) Select this option if the question cannot be answered even using any of the statements

Each statement only tells us that T cannot mix odd and even numbers. An all-odd T and an all-even T satisfy both statements, so whether T comes wholly from Y cannot be decided.

1. Statement I: the sum of two numbers is even only when both have the same parity. So T is all odd or all even. Not sufficient.
2. Statement II: if T is all odd, p − 1 is even, so (p − 1)q is even. If T is all even, q is even, so (p − 1)q is even. Both cases fit.
3. A mixed T fails II: an even p and an odd q give (odd) × (odd) = odd. So II also means 'all odd or all even'. Not sufficient.
4. Together: an all-odd T (from X) and an all-even T (from Y) satisfy both statements. Still not answered.

• ✗ Statement I Not sufficient alone: T = {1, 3} and T = {2, 4} both have even pairwise sums.
• ✗ Statement II Not sufficient alone: for T = {1, 3}, p − 1 is always even; for T = {2, 4}, q is always even. Both give (p − 1)q even.

Remember · When every statement is satisfied by two opposite cases, even combined, mark 'cannot be answered'.

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Answer: (a) Select this option if the question can be answered using one of these statements alone, but cannot be answered using other statement

Factorise both numbers: 116 = 2² × 29 and 261 = 3² × 29. The first leaves only y = 2 and x = 29, a prime; the second allows x = 3, 9, 29 or 87, so it cannot decide.

1. 116 = 2² × 29. With y > 1, the only square factor is y² = 4, so y = 2 and x = 29.
2. 29 is prime, so Statement I alone answers the question (yes).
3. 261 = 3² × 29. With x, z > 1, x can be 3, 9, 29 or 87.
4. 3 and 29 are prime but 9 and 87 are not, so Statement II alone cannot answer.
5. One statement alone works and the other does not.

• ✓ Statement I Sufficient alone: 116 = 4 × 29 forces y = 2 and x = 29, which is prime.
• ✗ Statement II Not sufficient alone: 261 = 9 × 29 allows x = 3 or 29 (prime) and x = 9 or 87 (not prime).

Remember · Prime factorisation first; then ask which factor splits are allowed by the given conditions (here, every variable > 1).

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Answer: (a) 1 and 2 only

Statements 1 and 2 restate what the passage says about the seven Angans and the temple-inspired zones full of regional art. Statement 3 swaps the passage's 'age-old techniques' for 'contemporary craftsmanship', and statement 4 overreaches with 'all regions'.

• ✓ 1. The passage says India's diverse regions were represented by seven 'Sanskriti Angans' in Kalagram.
• ✓ 2. The thematic zones were 'inspired by iconic temples' and were 'treasure troves of regional artistry', so the regional art was presented through temple-inspired zones.
• ✗ 3. The artisans worked with 'age-old techniques'; calling it 'contemporary craftsmanship' changes what the passage says.
• ✗ 4. Only a few examples (Bengal, Assam, Tamil Nadu, Madhya Pradesh) are named; 'all regions of India' is an overstatement.

Remember · Check each qualifying word against the passage: 'all', 'contemporary', 'only' often turn a near-copy into a wrong statement.

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Answer: (a) Paintings from four States of India have been mentioned.

Four States are mentioned, but only two of them for paintings: Bengal's Pattachitra and Tamil Nadu's Thanjavur. Assam is named for bamboo crafts and Madhya Pradesh for tribal sculpture, so statement (a) is the incorrect one.

• ✗ (a) Only Pattachitra (Bengal) and Thanjavur (Tamil Nadu) are paintings; bamboo crafts and tribal sculpture are not. So this is the statement that is not correct.
• ✓ (b) The passage says stepping through the grand portal 'was like entering another world'.
• ✓ (c) The art forms were showcased 'in these living galleries', i.e., the Angans.
• ✓ (d) The Angans are called 'thematic zones, inspired by iconic temples' such as Dakshineshwar Kali Temple and Brahma Mandir.

Remember · In 'not correct' items, count and classify details carefully; a right number with the wrong category is a common trap.

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Answer: (b) P and Q together weigh more than the total weight of A and C.

Adding the last two equations after substituting R = P and S = Q gives 3(P + Q) + B + D = 32, and B + D = 11 from the first two. So P + Q = 7, which is more than A + C = 6.

1. From A + B + C + D = 17 and A + C = 6: B + D = 11.
2. Put S = Q: P + 2Q + D = 15. Put R = P: 2P + Q + B = 17.
3. Add them: 3P + 3Q + B + D = 32, so 3(P + Q) = 32 − 11 = 21 and P + Q = 7.
4. (a) B + D = 11 is not less than 7 — false. (b) P + Q = 7 > A + C = 6 — true.
5. P and Q individually are not fixed (for example P = 3, Q = 4 or P = 4, Q = 3 with suitable B, D), so (c) and (d) cannot be asserted.

• ✗ (a) B + D = 11 is more than P + Q = 7.
• ✓ (b) P + Q = 7 exceeds A + C = 6.
• ✗ (c) Only P + Q = 7 is fixed; P can be larger or smaller than Q.
• ✗ (d) Same reason: the data do not fix which of P and Q is heavier.

Remember · When single values can't be found, add the equations to get the combined quantity the options ask about.

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Answer: (d) 12

Glue Q and U together as one block 'QU'. Then arrange four units — QU, U, E, E — where the two E's are identical: 4!/2! = 12.

1. QUEUE has Q, U, U, E, E.
2. Tie Q to a U as a block: the units are QU, U, E, E.
3. Arrangements = 4! ÷ 2! (two identical E's) = 24 ÷ 2 = 12.
4. Check: Q occurs once, so each word has exactly one block position — no word is counted twice.

Remember · 'X always followed by Y' → treat XY as one block, then divide by factorials of repeated letters.

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Answer: (d) 14

All three touch 199′, and after that their landing marks coincide every LCM(4, 6, 5) = 60 feet. Counting 199, 259, … up to 979 gives 14 common marks in the range.

1. LCM of 4, 6 and 5 is 60, so common marks are 199 + 60k.
2. 199 − 60 = 139 is below 195, so the first common mark in range is 199′ itself.
3. 199 + 60k ≤ 1000 gives k ≤ 13.35, so k = 0, 1, …, 13.
4. Number of common marks = 14 (199′, 259′, …, 979′).

Remember · Common landing points of different step lengths repeat every LCM of the steps; count terms of the resulting sequence.

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Answer: (d) 1, 2 and 3

The passage treats sport as a way of life, says it can help children who are inactive and rarely leave the house, and takes sports-quota jobs as a given benefit. Statement 4 contradicts the skills it lists — time management, teamwork, coordination.

• ✓ 1. The passage opens with 'Sport is not just about winning medals…' and ends by calling for sport to be seen 'as a way of life'.
• ✓ 2. Children who are 'inactive and rarely leave the house' are described, and 'Sport can help tackle these problems.'
• ✓ 3. 'getting jobs through a sports quota' is mentioned as one of the things sport already brings; the passage only adds that sport is more than that.
• ✗ 4. The passage lists time management, teamwork and coordination as things learnt from sport — the opposite of this statement.

Remember · A statement that denies something the passage lists as a benefit is invalid, however plausible it sounds.

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Answer: (b) 2 and 3

The passage says sport teaches time management, teamwork and coordination, and that a culture of sport can build a healthier, happier society. It never blames parents; it only reports what parents say about their children.

• ✗ 1. Parents are only quoted as saying their children are inactive; nothing is said about parents failing to encourage sport.
• ✓ 2. 'From sport you can learn time management, fitness, teamwork, coordination, and so much more' — these are life skills.
• ✓ 3. The passage calls sport as a way of life 'a path to building a healthier, happier society', i.e., it can change the character of society.

Remember · Reported speech is not a judgement: 'parents say children are inactive' does not mean 'parents are not encouraging'.

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Answer: (d) 9

Let f forward and b backward jumps: f + b = 31 and 5f − 2b = 15. This gives f = 11 and b = 20, so the difference is 9.

1. f + b = 31 and 5f − 2b = 15.
2. Substitute b = 31 − f: 5f − 62 + 2f = 15, so 7f = 77 and f = 11.
3. b = 31 − 11 = 20.
4. Difference = 20 − 11 = 9 (more backward jumps than forward).
5. Check: 11 × 5 − 20 × 2 = 55 − 40 = 15′ forward.

Remember · Two unknowns, two facts (count and net result): write both equations and eliminate one variable.

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Answer: (a) 5

The clues chain into the block P-Q-R-S-T, leaving three seats for U-V-W with V in the middle. W must be opposite S, which fixes the full circle as P, Q, R, S, T, U, V, W in clockwise order, so Q passes R, S, T, U and V before W.

1. P and R sit on both sides of Q, and R and T on both sides of S: this gives the block P-Q-R-S-T (5 seats).
2. The remaining 3 seats go to U, V, W with V in the middle (U and W both next to V).
3. Number the seats 1 to 8: P1, Q2, R3, S4, T5. The seat opposite S (4) is seat 8, so W8, V7, U6.
4. From P, going the way P → Q → R → S → T meets R before T, so this direction is clockwise.
5. Clockwise from Q: R, S, T, U, V, then W. Q crosses 5 persons.

Remember · In circular seating, first join 'adjacent' clues into blocks; the 'opposite' clue then fixes the rest.

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Answer: (b) 4

Each 2·5′ × 8′ strip must lie with its 8′ side along a 10′ edge and sit in a corner. Two strips cannot share the same 10′ edge without overlapping, so one strip goes to each long edge, at its left or right end: 2 × 2 = 4 patterns.

1. The 8′ side cannot lie along the 6′ edge (8 > 6), so it lies along a 10′ edge and the 2·5′ side along a 6′ edge.
2. Two sides common with two table edges means the strip sits in a corner (it cannot touch two opposite edges, since 2·5 < 6 and 8 < 10).
3. Two strips along the same 10′ edge would overlap (8 + 8 > 10), so one strip lies along each 10′ edge; they do not meet since 2·5 + 2·5 = 5 < 6.
4. Each strip can be at the left or the right end of its edge: 2 × 2 = 4.
5. The table is fixed, so these four layouts (both left, both right, and the two diagonal ones) all look different from above.

Remember · Fix the orientation first (which side fits which edge), then count positions; a fixed table means mirror images count separately.

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Answer: (d) Neither 1 nor 2

Statement 1 looks like a paraphrase but changes two terms: the passage holds output prices (not output cost) steady and speaks of real value added (not real value). Statement 2 turns 'most countries' into 'all economies', which India itself contradicts.

• ✗ 1. The passage's example keeps output prices unchanged while oil (an input) gets cheaper — so the cost of producing output actually falls. It says real value added, not 'real value', stays the same. The statement misstates both terms.
• ✗ 2. The passage says 'Most countries' use double deflation and notes India uses the same deflator for inputs and outputs, so it is not universal.

Remember · Compare near-copy statements word by word; 'price' vs 'cost' and 'most' vs 'all' decide such items.

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Answer: (a) 1 only

The passage shows that using one deflator for inputs and outputs can make a manufacturer look more productive than it is, so the choice of deflation method can flatter performance. Statement 2 inverts the definition: double deflation deflates input and output prices separately.

• ✓ 1. With a single deflator 'it would look as if the manufacturer had become more productive' — the method of deflation can make performance look better than it is.
• ✗ 2. Double deflation is where 'input and output prices are deflated separately'; using a single deflator is the Indian practice the passage criticises.

Remember · Definition-based statements: check the definition in the passage exactly; a reversed definition is always invalid.

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Answer: (d) bb

The string has 15 places, so the repeating block has length 3 or 5. Length 3 clashes (place 2 is b but place 11 is a); length 5 fits every known letter and gives the block 'abbab', so places 7 and 8 are b and b.

1. Number the 15 places: 1 ×, 2 b, 3 ×, 4 a, 5 ×, 6 a, 7 ×, 8 ×, 9 a, 10 ×, 11 a, 12 ×, 13 b, 14 a, 15 b.
2. Block length 3 fails: places 2 and 11 should match but are b and a.
3. Block length 5: places 1, 6, 11 → a; 2, 7, 12 → b; 3, 8, 13 → b; 4, 9, 14 → a; 5, 10, 15 → b.
4. Block = abbab, repeated three times: abbab abbab abbab.
5. Places 7 and 8 = 2nd and 3rd letters of the block = b, b.

Remember · For a repeated pattern, try block lengths that divide the string length; letters one block apart must match.

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Answer: (b) 111

Break the right side into primes: 2³⁷⁵ × 3²⁵ × 5¹¹⁴. The number of 10s is the smaller of the powers of 2 and 5, i.e., 114, and n must hold none of them, so m + 3 = 114.

1. 75²⁵ = (3 × 5²)²⁵ = 3²⁵ × 5⁵⁰.
2. 25³² = 5⁶⁴.
3. 32⁷⁵ = (2⁵)⁷⁵ = 2³⁷⁵.
4. Right side = 2³⁷⁵ × 3²⁵ × 5¹¹⁴, which contains exactly 10¹¹⁴ (limited by the 5s).
5. n = 2²⁶¹ × 3²⁵ is not divisible by 10, so 10ᵐ × 10³ = 10¹¹⁴ and m = 111.

Remember · Powers of 10 in a product = min(power of 2, power of 5). Count both, take the smaller.

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Answer: (c) 13.85

The train's length is fixed, so crossing time is inversely proportional to relative speed. Relative speed changes from 96 km/h to 104 km/h, so the new time is 15 × 96/104 ≈ 13·85 seconds.

1. Same direction: relative speed = 100 − 4 = 96 km/h. Opposite direction: 100 + 4 = 104 km/h.
2. Length of T = 96 × (5/18) m/s × 15 s = 400 m.
3. Time in the opposite case = 400 ÷ (104 × 5/18) = 3600/260 ≈ 13·85 s.
4. Check: 15 × 96/104 = 13·846 s.

Remember · Same length, different relative speed: new time = old time × old relative speed ÷ new relative speed.

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Answer: (b) II only

t + 1/t falls until t = 1 and rises after it, so Statement I breaks when x and y are on opposite sides of 1. Statement II simplifies to √(1/y² + 1) < √(1/x² + 1), which always holds because y > x > 0.

1. Statement I: try x = 1/2, y = 1. Left = 0·5 + 2 = 2·5; right = 1 + 1 = 2. 2·5 < 2 is false, so I is not always correct.
2. Statement II: divide inside the root by the square of the denominator: √(1 + y²)/y = √(1/y² + 1) and √(1 + x²)/x = √(1/x² + 1).
3. Since 0 < x < y, 1/y² < 1/x², so √(1/y² + 1) < √(1/x² + 1). II always holds.
4. Check II with numbers: x = 1, y = 1·5 gives 1·202 < 1·414.

• ✗ I Fails for x = 1/2, y = 1: 2·5 is not less than 2. The function t + 1/t decreases on (1/3, 1).
• ✓ II Equivalent to 1/y² < 1/x², true whenever 0 < x < y.

Remember · For 'always correct', hunt for one counter-example; for the other statement, simplify until the inequality is obvious.

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Answer: (b) 5

Measuring can remove water as well as add it: take 100 litres three times (300 litres) and take out 1 litre twice. That is 5 measurements, and 4 cannot work because 298 is 2 short of 300 and no single cylinder holds 2 litres.

1. 298 = 300 − 2 = 100 + 100 + 100 − 1 − 1: three fills of 100 litres and two removals of 1 litre = 5 measurements.
2. Could 4 work? With at most two 100-litre fills, the other two measures add at most 50 litres: 250 < 298.
3. So three 100-litre fills are needed, leaving one measure to remove exactly 2 litres — impossible with 1, 6, 25 or 100.
4. Minimum = 5.
5. Check: the greedy pour-in-only count 2 × 100 + 3 × 25 + 3 × 6 + 5 × 1 = 13 measures is the trap option.

Remember · For 'minimum measurements', aim just above the target with the big measure, then remove the excess.

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Answer: (b) 8

Fix the number of 1 kg weights at 8, 9, 10 or 11 and make up the rest with 2, 5 and 10 kg weights. The counts are 3, 1, 3 and 1, giving 8 ways.

1. Eight 1 kg weights: 12 kg left → 10 + 2; 5 + 5 + 2; six 2s → 3 ways.
2. Nine 1 kg weights: 11 kg left from 2, 5 and 10 kg weights. 11 is odd, so exactly one 5 is needed, plus three 2s → 1 way.
3. Ten 1 kg weights: 10 kg left → 10; 5 + 5; five 2s → 3 ways.
4. Eleven 1 kg weights: 9 kg left, odd again → 5 + 2 + 2 → 1 way.
5. Total = 3 + 1 + 3 + 1 = 8.

Remember · Counting combinations of coins or weights: fix the most restricted item, then list the rest systematically from the largest.

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Answer: (a) 16

With pieces left unmoved, four plane cuts can give at most 15 pieces (1 + 4 + 6 + 4), so 16 is impossible. The other values arise easily from identical pieces after three cuts: 5 and 8 from four equal slabs, 12 from eight small cubes.

1. Identical pieces after three cuts: 4 equal slabs (three parallel cuts), 6 equal blocks (two parallel + one perpendicular), or 8 small cubes (three mutually perpendicular cuts through the centre).
2. Four plane cuts on an unmoved solid give at most 1 + 4 + 6 + 4 = 15 pieces; equivalently, the fourth plane can cross at most 7 of the 8 small cubes, giving 8 + 7 = 15.
3. So x₄ = 16 is not possible.
4. Check the rest: 4 slabs + a fourth parallel cut through one slab = 5; 4 slabs + a perpendicular cut through all four = 8; 8 cubes + a cut parallel to a face through four of them = 12.

• ✗ (a) Not possible: four plane cuts on unmoved pieces give at most 15 pieces.
• ✓ (b) Possible: after three central perpendicular cuts (8 cubes), a fourth cut parallel to one face through one layer splits 4 cubes → 12.
• ✓ (c) Possible: after three parallel cuts (4 equal slabs), a fourth cut perpendicular to them splits all 4 → 8.
• ✓ (d) Possible: after 4 equal slabs, a fourth cut parallel to them inside one slab → 5.

Remember · Unmoved pieces: plane cuts give at most 2, 4, 8 and 15 pieces for one to four cuts.

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Answer: (b) 3

Adding 100 marks raised the average by 4, so the class has 25 students. The second correction removes 25 marks, lowering the average by 1, so the net rise is 4 − 1 = 3.

1. First correction adds 100 − 0 = 100 marks and raises the average by 4, so number of students = 100 ÷ 4 = 25.
2. Second correction: 81 recorded instead of 56 means the total falls by 25.
3. Average falls by 25 ÷ 25 = 1.
4. y − x = 4 − 1 = 3.

Remember · Change in average = change in total ÷ number of items; find the count from the first change.

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Answer: (b) 2 only

The passage stresses that the two viruses are very different in family, epidemiology, severity and the people they affect, which supports not treating them alike. Statement 1 calls them 'similar viruses' and brings in a pandemic the passage never mentions.

• ✗ 1. The passage says they belong to 'two very different virus families with fundamentally different characteristics and epidemiology' — not 'similar viruses with somewhat different epidemiology'. It also says nothing about a pandemic.
• ✓ 2. Different severity, different long-term effects and non-overlapping affected groups add up to different impacts, which is the passage's reason for not equating HMPV with SARS-CoV-2.

Remember · Reject a statement that contradicts the passage's core contrast, even if its second half sounds reasonable.

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Answer: (d) None of the above

The writer's aim is to show that HMPV is a familiar, milder and very different virus from SARS-CoV-2. The passage develops no method of analysis, does not set out similarities, and says nothing about remedies.

• ✗ 1. No methodology is proposed; the passage simply states facts about the two viruses.
• ✗ 2. The passage stresses only differences ('fundamentally different characteristics and epidemiology'); it does not set out similarities.
• ✗ 3. Remedies or treatment are not discussed at all.

Remember · For 'intent of the writer', ask what the passage actually does; reject aims it never attempts (methods, remedies).

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Answer: (b) 1, 2 and 3

HMPV has been known and studied for decades, the two viruses differ in severity and outcomes, and the long-term effects are a clear point of difference. Seasonality is not shared: the passage says strong seasonality is seen in HMPV, unlike SARS-CoV-2.

• ✓ 1. People have been exposed to HMPV 'for decades and the virus is well-studied', while SARS-CoV-2 was 'truly novel' in 2019.
• ✓ 2. They 'cause different severity of symptoms'; HMPV causes milder illness with very rare deaths.
• ✓ 3. Severity differs 'particularly over the long term', and HMPV leaves 'no long-term post-viral symptoms'.
• ✗ 4. Seasonality is a difference, not a common factor: 'strong seasonality seen in HMPV, unlike SARS-CoV-2'.

Remember · Check 'common factor' claims against contrast words such as 'unlike', 'in contrast', 'whereas'.

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Answer: (c) 13

The product dominates the sum, so put the two largest values in the product for M and the two smallest for N. M = 5 × 4 + 3 = 23 and N = 2 × 3 + 4 = 10, so M − N = 13.

1. Largest: x × y = 5 × 4 = 20, z = 3 (largest value left): M = 23. (5 × 3 + 4 = 19 is smaller.)
2. Smallest: x × y = 2 × 3 = 6, z = 4 (smallest value left): N = 10. (2 × 4 + 3 = 11 is larger.)
3. M − N = 23 − 10 = 13.

Remember · To maximise or minimise xy + z, decide the product first; the added term matters less.

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Answer: (b) 1/5

From y = kx², k = 50/25 = 2, so y = 98 gives x = 7. From xz = constant = 5 × 7/25 = 7/5, z = (7/5)/7 = 1/5.

1. y = kx²: 50 = k × 25, so k = 2.
2. y = 98: 2x² = 98, x² = 49, x = 7 (x is positive).
3. xz = constant = 5 × 7/25 = 7/5.
4. z = (7/5) ÷ 7 = 1/5.
5. Check: x rose from 5 to 7, so z must fall in the ratio 5 : 7 — (7/25) × (5/7) = 1/5.

Remember · Direct variation: ratio stays constant; inverse variation: product stays constant. Find each constant from the given pair.

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Answer: (d) 2 and 3 only

The passage's recommendations rest on the belief that affordable, local-language AI tools will reach more small farmers, and it notes AI is already being deployed in India for critical uses despite low penetration. It makes no claim that Western farm productivity rose because of AI-driven economies of scale.

• ✗ 1. The passage only says AI tools in the US and Europe 'have started offering precision farming at scale'; it says nothing about productivity having marched ahead because of them.
• ✓ 2. The passage recommends Indian languages-based and affordable AI tools 'for at-scale integration and accessibility' among smallholder farmers.
• ✓ 3. India 'is starting to deploy AI for critical use cases' such as weather forecasting and pest control, though 'penetration is limited'.

Remember · An unstated cause-and-effect claim (X happened because of Y) is invalid unless the passage links the two.

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Answer: (a) 1 and 3 only

Nothing in the passage says tech-savvy farmers will drive future AgTech firms, or that AI has replaced traditional farming in the West. The government advisory route is recommended as helpful, and real-time use of large datasets is already happening in the US and Europe.

• ✗ 1. Not supported: tech-savvy farmers are mentioned only as the small group AI has reached so far.
• ✓ 2. The passage says it 'would be helpful' to disseminate AI-based advisory services through government programmes.
• ✗ 3. The passage says generative AI tools 'have started offering precision farming at scale' — not that they have replaced traditional practices.
• ✓ 4. In the US and Europe, AI tools are already 'integrating large datasets to provide real-time agronomic insights'.

Remember · In 'not correct' items, mark statements that go beyond the passage ('will drive', 'have replaced') as not correct.

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Answer: (d) Neither I nor II

With P = 4 and R = 1, the ranks 2, 3 and 5 remain for Q, S and T, and only two orders fit the conditions: (Q, S, T) = (2, 3, 5) or (3, 5, 2). In each case the 'then' part of the statement is the opposite of what actually happens.

1. P = 4, R = 1, so Q, S, T share ranks 2, 3 and 5.
2. Q ≠ 5 → Q is 2 or 3; S ≠ 2 → S is 3 or 5; T ≠ 3 → T is 2 or 5.
3. Q = 2: then T ≠ 3 forces T = 5 and S = 3.
4. Q = 3: then S ≠ 2 forces S = 5 and T = 2.
5. S = 3 happens only with T = 5, so I is wrong; Q = 3 happens only with T = 2, so II is wrong.

• ✗ I S = 3 occurs only in the order Q = 2, S = 3, T = 5, so T is 5, not 2.
• ✗ II Q = 3 occurs only in the order Q = 3, S = 5, T = 2, so T is 2, not 5.

Remember · List all arrangements that satisfy the restrictions; there are usually very few, then test each 'if…then' against them.

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Answer: (c) Both I and II

For 'possible', one valid layout is enough. The layout Q4, Q5, Q3, Q1, Q2 beside P3, P5, P1, P4, P2 meets every condition, and it has Q3 in the middle and P2 at an end, so both are possible.

1. Number the portions 1 to 5 along the rods. Try Q3 (and so P1) at portion 3, with Q5 next to it at portion 2.
2. Put Q4 at portion 1 with P3 beside it (Q4 must match P3 or P5).
3. Put Q1 at portion 4 with P4 beside it (P4 must match Q1 or Q2), and Q2 at portion 5.
4. The remaining P5 goes to portion 2 and P2 to portion 5.
5. Layout: Q-rod Q4, Q5, Q3, Q1, Q2; P-rod P3, P5, P1, P4, P2. All conditions hold.
6. Here Q3 is in the middle (I possible) and P2 is at an end (II possible).

• ✓ I Possible: in the layout Q4, Q5, Q3, Q1, Q2, Q3 occupies the middle portion and every condition holds.
• ✓ II Possible: in the same layout, P2 sits at portion 5, an extreme end.

Remember · 'Is possible' needs just one example that satisfies every condition; 'must be' needs proof for all cases.

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Answer: (b) I and III only

B cannot be in Y or with A in X, so B joins C in Z. E and G ride alone together, which leaves them car Y; D and F then fill X with A. So cars are X: A, D, F; Y: E, G; Z: B, C — D is not the only one with F, since A is there too.

1. B is not in Y and not in X (A is in X and A, B are apart), so B is in Z with C.
2. E and G travel together with nobody else; X has A and Z has B, C, so E and G take Y.
3. X has A and two others: the only ones left are D and F.
4. Cars: X = A, D, F; Y = E, G; Z = B, C.
5. I true (no car has one person); II false (A also travels with F); III true (Z has only B and C).

• ✓ I Each car carries two or three persons.
• ✗ II F shares car X with both A and D, so D is not the only one.
• ✓ III Car Z carries only B and C.

Remember · Place the person with the most 'not' conditions first; the rest usually falls into place.

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Answer: (a) I only

Flip each rule into its contrapositive: Z correct → X correct → W incorrect → Y correct. So X correct forces Y correct (I holds), and Z correct also forces Y correct, which makes II wrong.

1. Rule 1: X incorrect → Z incorrect; contrapositive: Z correct → X correct.
2. Rule 3: W correct → X incorrect; contrapositive: X correct → W incorrect.
3. Rule 2: Y incorrect → W correct; contrapositive: W incorrect → Y correct.
4. Chain: Z correct → X correct → W incorrect → Y correct.
5. I: X correct → Y correct — true. II: Z correct forces Y correct, so 'not necessary' is false.

• ✓ I X correct → W incorrect (rule 3) → Y correct (rule 2).
• ✗ II Z correct → X correct (rule 1) → Y correct; so Y is necessarily correct.

Remember · 'If P then Q' equals 'if not Q then not P'. Rewrite each rule this way and chain them.

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Answer: (a) 35

On a circular track, runners meet once for every full round of difference (same direction) or every full round of combined distance (opposite directions). That gives 2 meetings in the first phase and 34 in the second, but the last one happens exactly at the stop, so 35 are counted.

1. Speeds are in the ratio 7 : 5. Phase 1 (same direction): X runs 7 rounds, Y runs 5, so X gains 2 rounds → 2 meetings (the second at the start point when Y turns).
2. Phase 2: X runs 21 − 7 = 14 more rounds. In the same time Y, at double speed (10 per 7 of X), runs 20 rounds the other way.
3. Opposite directions: they meet once per round of combined distance: 14 + 20 = 34 meetings.
4. The 34th meeting is exactly when they stop (X at 21 rounds, Y at 25 — both at the start point), which is not 'before they finally stopped'.
5. Meetings counted = 2 + 34 − 1 = 35.

Remember · Circular track meetings = relative rounds covered (difference if same direction, sum if opposite). Check whether the final meeting falls at the stop.

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Answer: (b) 81

With c correct and w wrong, 5c − 2w = 69 and 4c − w = 84. Solving gives c = 33 and w = 48, so the paper had 81 questions.

1. 5c − 2w = 69 and 4c − w = 84.
2. From the second: w = 4c − 84.
3. Substitute: 5c − 8c + 168 = 69, so 3c = 99 and c = 33.
4. w = 132 − 84 = 48.
5. Total = 33 + 48 = 81.
6. Check: 5 × 33 − 2 × 48 = 165 − 96 = 69.

Remember · Two scoring schemes for the same answers give two equations; solve for right and wrong counts, then add.

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Answer: (b) 2 only

The 2021 amendment defines a serious offence by the maximum sentence (seven years or more) when there is no minimum sentence. Transfer to the adult system is linked to heinous offences, and activists say serious offences do not merit it — the reverse of statement 1.

• ✗ 1. Trial as adults is allowed for heinous offences; for serious offences, activists say they do 'not merit the transfer of a case to the adult criminal justice system'.
• ✓ 2. A serious offence 'has no minimum sentence, but has a maximum sentence of seven years or more' — it is defined by the maximum sentence.

Remember · Keep categories apart: note which label (heinous or serious) the passage attaches to which consequence.

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Answer: (c) Both 1 and 2

A heinous offence under the 2015 Act needs a minimum punishment of seven years, so an offence without a prescribed minimum cannot be heinous. The 2021 amendment treats an offence with no minimum but a maximum of seven years or more as serious; both statements restate the passage.

• ✓ 1. Culpable homicide and causing death by negligence 'are not heinous offences because they do not have a prescribed minimum punishment'.
• ✓ 2. The amended Act 'categorises an offence that has no minimum sentence, but has a maximum sentence of seven years or more as a serious offence'.

Remember · Statements that restate a definition with its conditions intact are correct; check each condition is carried over.

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Answer: (c) 4:58:42 PM

The second drone's one-way trip takes the 3 minutes between the two pictures, so the spot is 180 × 30 = 5400 m away. The first drone left 3 minutes before 5:02, and the sound had taken 5400 ÷ 300 = 18 s to arrive, so the explosion was 3 min 18 s before 5:02 PM.

1. The second drone starts at 5:02 PM and its picture arrives at 5:05 PM, so one trip takes 180 s.
2. Distance = 30 m/s × 180 s = 5400 m.
3. The first drone also needed 180 s, so it left the camp at 4:59:00 PM — the moment the sound arrived.
4. Sound took 5400 ÷ 300 = 18 s, so the explosion happened at 4:59:00 − 18 s = 4:58:42 PM.
5. Check: 4:58:42 + 18 s + 180 s = 5:02:00 PM.

Remember · Work backwards on a timeline: find the distance from the cleanest interval, then subtract each leg's travel time.

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Answer: (c) 8

Any power of 6 ends in 6. Powers of 7 end in 7, 9, 3, 1 in a cycle of four; 307 leaves remainder 3 on division by 4, so 7³⁰⁷ ends in 3. The product ends in the last digit of 6 × 3 = 18, i.e., 8.

1. 6¹²⁹ ends in 6.
2. 7¹, 7², 7³, 7⁴ end in 7, 9, 3, 1; the cycle length is 4.
3. 307 = 4 × 76 + 3, so 7³⁰⁷ ends in 3.
4. 6 × 3 = 18, so the unit digit is 8.

Remember · Unit digits: 0, 1, 5, 6 never change; 2, 3, 7, 8 cycle every 4 — use the exponent's remainder on division by 4.

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Answer: (b) 30% increase

Take the salary as ₹100: saving ₹10, spending ₹90. After the changes, salary is ₹112 and spending ₹99, so saving becomes ₹13 — a 30% rise on ₹10.

1. Let salary = ₹100. Saving = ₹10, expenditure = ₹90.
2. New salary = ₹112; new expenditure = 90 × 1.1 = ₹99.
3. New saving = 112 − 99 = ₹13.
4. Change = (13 − 10)/10 × 100 = 30% increase.

Remember · Percentage problems with several parts: assume a base of 100 and compute each part directly.

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Answer: (d) Q is a sister of the husband of S

S is a mother, so S is female and her spouse R is male — P's son. P's other child Q must then be the daughter, making Q the sister of S's husband R.

1. S is the mother of T, so S is female.
2. R is S's spouse, so R is male.
3. Q and R are P's two children, and P has one son and one daughter. R is the son, so Q is the daughter.
4. Q is R's sister, and R is S's husband: Q is a sister of the husband of S.

Remember · In blood relations, fix genders first from words like mother, son, spouse; then place each person on a family tree.

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Answer: (c) 3

Powers of 2 run 64, 128, 256, 512, 1024. Only 2⁷, 2⁸ and 2⁹ — 128, 256 and 512 — have three digits.

1. 2⁶ = 64 (two digits), 2⁷ = 128, 2⁸ = 256, 2⁹ = 512, 2¹⁰ = 1024 (four digits).
2. Three-digit powers of 2: 128, 256, 512 — that is 3.

Remember · Know powers of 2 up to 2¹⁰ = 1024 by heart; they settle many CSAT items instantly.

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Answer: (c) 1 and 3

The passage names the fuel we burn as the key to clean air and records the Supreme Court's agreement that fiscal measures should keep clean fuel cheaper. It says bans work when people have affordable alternatives — not when will is strong and people are convinced.

• ✓ 1. 'The key source of the battle for clean skies and clear lungs is the fuel we burn'.
• ✗ 2. The passage's lesson from CNG is that 'people need alternatives for a ban to be effective' and they must be affordable; will and conviction are not mentioned.
• ✓ 3. 'The Supreme Court agreed that fiscal measures were needed to keep clean fuel cheaper than dirty fuel.'

Remember · When a statement gives a reason, check it is the passage's reason; a plausible but different reason makes it wrong.

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Answer: (c) 2 only

Only the CNG statement matches the passage. Thermal plants were to clean up or shut down, not shut down summarily, and the coal ban came from the Delhi Government and was extended to the NCR — the Supreme Court banned pet coke.

• ✗ 1. It was agreed that the plants 'would clean up or shut down' — a choice, not a summary shutdown.
• ✓ 2. 'When diesel buses were stopped, CNG supply had to be assured.'
• ✗ 3. The Delhi Government banned coal, later extended to the NCR; the Supreme Court's ban was on pet coke.

Remember · Match each action to the right actor: who banned what is a favourite trap in policy passages.

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Answer: (d) P, Q and R

The statements form a chain of sets: red inside blue, blue inside green, green inside yellow. Anything inside a smaller set is inside every larger set, so all three conclusions follow.

1. Red ⊂ Blue ⊂ Green ⊂ Yellow.
2. P: Blue ⊂ Green ⊂ Yellow — follows.
3. Q: Red ⊂ Blue ⊂ Green — follows.
4. R: Red ⊂ … ⊂ Yellow — follows.

• ✓ P Every blue is green and every green is yellow.
• ✓ Q Every red is blue and every blue is green.
• ✓ R Following the whole chain, every red is yellow.

Remember · 'Every A is B, every B is C' gives 'every A is C'; draw nested circles to see it at once.

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Answer: (b) 19

5 is the tens digit in 50–59 (10 times) and the units digit in 15, 25, …, 95 (9 times). Counting appearances, 55 contributes two, and the total is 19.

1. Tens place: 50, 51, …, 59 → 10 appearances.
2. Units place: 15, 25, 35, 45, 55, 65, 75, 85, 95 → 9 appearances.
3. Total = 10 + 9 = 19.

Remember · Count digit appearances place by place (tens, then units); a number like 55 is counted once in each place.

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Answer: (d) 6

Add the time for each stretch. X takes 12 + 12 + 45 = 69 minutes; Y takes 15 + 12 + 36 = 63 minutes, so Y finishes 6 minutes earlier.

1. X: 1/5 h = 12 min; 2/10 h = 12 min; 3/4 h = 45 min. Total 69 min.
2. Y: 1/4 h = 15 min; 2/10 h = 12 min; 3/5 h = 36 min. Total 63 min.
3. Y is earlier by 69 − 63 = 6 minutes.

Remember · Average speed over a journey is never the average of the speeds; add time = distance ÷ speed for each stretch.

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Answer: (a) 2

Each weighing has three outcomes (left lighter, right lighter, balance), so split the cubes into three groups: 3, 3 and 1. One weighing narrows the light cube to a group of at most three, and a second weighing of one against one finds it.

1. Weigh 3 cubes against 3. If they balance, the 7th cube is the light one.
2. If one side rises, the light cube is among those 3.
3. Weigh 1 against 1 from that group: the lighter pan shows it; if balanced, it is the third cube.
4. So 2 weighings always suffice; 1 cannot, since one weighing separates at most 3 possibilities, not 7.

Remember · A balance gives three outcomes: with k weighings you can find one odd (known-lighter) item among up to 3ᵏ items.

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Answer: (c) The value of human intervention in the workplace affected by AI might be enhanced through redirection towards more fulfilling tasks.

The author's point is hopeful: if AI does change customer service, its effects may be salutary, freeing human agents for creative and rewarding work. Option (c) captures exactly this redirection of human effort.

• ✗ (a) The passage says humans would be freed for other tasks, not left with no work.
• ✗ (b) It picks up only the side remark that 'AI could yet prove different' and frames AI as a job-killer; the author's view is that humans would be redeployed, not made redundant.
• ✓ (c) 'Human agents could be freed up to spend more time on creative and rewarding tasks' — the author's main point, and 'its effects may be salutary'.
• ✗ (d) Earlier waves such as email and voice menus also 'stoked concerns of job losses', so this is not unique to AI.

Remember · For 'point of view', pick the option that carries the author's attitude (here hopeful), not a side detail or a reversed claim.

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Answer: (d) 1 and 2

Each earlier wave of customer-service technology raised job-loss fears that did not come true, and the passage expects workers to move to more rewarding tasks rather than lose jobs. Nothing suggests technology always brings stress — it would mean less time with irate customers.

• ✓ 1. 'Previous waves of customer-service technology … stoked concerns of job losses' — each wave brought such fears.
• ✓ 2. The feared job losses 'fail[ed] to materialise', and the passage expects agents to be 'freed up' for creative and rewarding tasks — new ways of engaging employees.
• ✗ 3. The passage suggests the reverse: agents would 'spend less time listening to irate customers'.

Remember · Words like 'inevitably' need strong support; if the passage points the other way, reject the statement.

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Answer: (c) 3 only

Rai Sahib's care for his tenants is tied to his holding 'the old values', supporting statement 3. The passage mentions selfish zamindars, has cattle from several nearby villages sharing the ground, and says the hot 'loo' blows later in the day — so 1, 2 and 4 fail.

• ✗ 1. The passage contrasts Rai Sahib with 'those selfish Zamindars'; it does not credit all landlords with goodness.
• ✗ 2. 'Cattle from the nearby villages' graze on the common ground — not just one village's cattle.
• ✓ 3. Rai Sahib 'still held to the old values', and that is why he felt a landlord must look after his tenants.
• ✗ 4. It is the reverse: the morning had 'a cool freshness', and the scorching 'loo' would blow 'the rest of the day'.

Remember · Test each detail against the passage's exact words: 'nearby villages' is not 'that village'; 'rest of the day' tells you when it gets hot.

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Answer: (d) 1, 2 and 3

Only statement 4 fits the passage: Rai Sahib kept the grazing land for the villagers' cattle out of concern for his tenants. The passage speaks only of 'this bit of land', contrasts him with greedy landlords, and says farmers offered him a good price.

• ✗ 1. Only 'this bit of land' was reserved for grazing; nothing is said about the rest of his landholdings.
• ✗ 2. He is explicitly contrasted with 'those selfish Zamindars' who would grab the money.
• ✗ 3. It can be ascertained: farmers 'had offered a good price' to lease the land.
• ✓ 4. He refused to lease the grazing land 'for any price' and felt a landlord who did not look after his tenants was 'less than human'.

Remember · In 'not correct' items, reject overgeneralisations (one plot → all landholdings) and claims the passage directly contradicts.

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Answer: (d) 6 km, North

Track each turn: East 4 km, right to South 3 km, left to East 2 km, left to North 3 km. The south and north legs cancel, leaving the person 6 km east of the start, facing North.

1. Start facing East: 4 km East.
2. Turn right → South: 3 km South.
3. Turn left → East: 2 km East.
4. Turn left → North: 3 km North.
5. Net: 4 + 2 = 6 km East; 3 South − 3 North = 0. Distance 6 km, facing North.

Remember · Direction items: write the facing direction after every turn, then net out opposite legs.

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Answer: (d) 47

Build the sequence by adding the last two terms each time: 4, 7, 11, 18, 29, 47. The sixth term is 47.

1. Term 3 = 4 + 7 = 11.
2. Term 4 = 7 + 11 = 18.
3. Term 5 = 11 + 18 = 29.
4. Term 6 = 18 + 29 = 47.

Remember · Short sequences: just write the terms out; count positions carefully (option 29 is the fifth term).

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Answer: (b) 23: 24

Make the female part equal: write B's 7 : 5 as 14 : 10. Then A has 13 + 10 = 23 parts and B has 14 + 10 = 24 parts.

1. A: male : female = 13 : 10.
2. B: 7 : 5 = 14 : 10, so females are 10 parts in both.
3. Totals: A = 23 parts, B = 24 parts.
4. Ratio = 23 : 24.

Remember · To compare two ratios sharing a common quantity, scale them so that the common term is equal.

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Answer: (c) Both I and II

HCF × LCM equals the product of the two numbers, so a × b = a³ means b = a². The numbers are a and a²: their difference a(a − 1) is a product of consecutive integers and hence even, and a² is a perfect square.

1. For any two numbers, HCF × LCM = a × b.
2. a × b = a³ gives b = a² (a ≠ 1, as the numbers are distinct).
3. Difference = a² − a = a(a − 1), a product of two consecutive integers, so it is even.
4. a² is a perfect square.
5. Check: 3 and 9 — HCF 3, LCM 9, product 27 = 3³; difference 6, and 9 is a square.

• ✓ I a² − a = a(a − 1) is always even.
• ✓ II The second number is a², a perfect square.

Remember · HCF × LCM = product of the two numbers — the first thing to use in any HCF–LCM item.

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Answer: (d) 7

For divisibility by 11, the alternating digit sums must differ by a multiple of 11. Here that gives 7 − (x + y) ≡ 0 (mod 11), so x + y leaves remainder 7 when divided by 11.

1. Digits from the right: 0, 9, 7, y, 5, x, 4.
2. Odd places (1st, 3rd, 5th, 7th): 0 + 7 + 5 + 4 = 16.
3. Even places (2nd, 4th, 6th): 9 + y + x.
4. 16 − (9 + x + y) = 7 − (x + y) must be a multiple of 11.
5. So x + y = 7 or 18; either way the remainder on division by 11 is 7.
6. Check: x = 0, y = 7 gives 4057790 = 11 × 368890.

Remember · Divisibility by 11: (sum of digits in odd places) − (sum in even places) must be 0 or a multiple of 11.

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Answer: (c) 1, 2 and 3

The writer calls Grant's public nap 'quietly radical' — a gentle surrender that defies society's performance metrics. It is framed as yielding to limits, not laziness, given his age, and the closing line plays on 'leading man' as a hero of rest rather than action. All three statements read the passage correctly.

• ✓ 1. The nap is 'something quietly radical' and 'an elegant case for surrender' against 'society's relentless performance metrics'.
• ✓ 2. 'Not to laziness, but to limits' — and the passage cites 'the crushing weight of being a 64-year-old man' among possible reasons.
• ✓ 3. The actor becomes 'a leading man, not of action, but of rest' — a play on the film term.

Remember · Literary passages: decode each figure of speech the passage uses (radical, surrender, leading man) and test the statement against it.

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Answer: (b) 3 only

Grant was watching the men's singles quarterfinals, not a semifinal, and the writer praises the nap rather than condemning it. Statement 3 restates the passage: the body's quiet wisdom was placed over society's performance demands.

• ✗ 1. He was 'watching the men's singles quarterfinals', not a semifinal.
• ✗ 2. 'Ace' is indeed a tennis term, but the writer says 'praise be to Grant' — the nap is admired, not viewed with opprobrium (strong disapproval).
• ✓ 3. He chose 'the body's quiet wisdom over society's relentless performance metrics'.

Remember · Check factual details (round, venue, names) exactly, and check the author's tone before accepting words like 'opprobrium'.

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Answer: (a) 3 and 4 only

The fine-tuning description and the order scavenging → hunting → farming → industry both match the passage. Statement 1 turns 'unchanged' income into 'slow' progress, and statement 2 invents stagnating leaders — the passage says followers catch up by adopting the technology.

• ✗ 1. The followers' per capita income is 'unchanged' in the breakaway phase — there is no progress, not slow progress.
• ✗ 2. Followers close the gap because they 'adopt the new technology'; nothing says the leaders stagnate.
• ✓ 3. Participants 'try to extract the remaining benefits from an increasingly exhausted technology' and incomes become similar.
• ✓ 4. The dates give scavenging (3 MYA), hunting (1 MYA), farming (12,000 years ago), industry (200 years ago); each link in the statement fits this order.

Remember · 'Unchanged' is not 'slow', and 'followers adopt' is not 'leaders stagnate'; reject statements that add a cause the passage never gives.

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Answer: (b) 2 and 3

UPSC's key accepts statements 2 and 3. Statement 3 holds because the leaders develop and profit from each new technology before the followers adopt it. Statement 2 paraphrases the passage's opening sentence: the theory is about how countries close their labour-productivity gap with the leader. Statement 1 confuses eras with nations, and statement 4 wrongly adds up dates that lie on one timeline.

• ✗ 1. The model divides economic eras into three phases and economic entities into two categories (leaders and followers); it does not sort nations into phases.
• ✓ 2. The passage opens with the process by which countries 'close their labour-productivity gap with the technology leader'; gap-closing is the heart of the theory, and the key reads the statement as that idea.
• ✓ 3. Leaders gain from each new technology first (their per capita income rises in the breakaway phase); followers reach the same era later, in the catch-up phase.
• ✗ 4. Three million, one million, 12 thousand and 200 years ago are points on one timeline, so the span is about three million years; adding them to 4,012,200 is wrong.

Remember · Read the exact noun each claim is about: the model divides eras (not nations) into phases; dates on one timeline are never added.

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Answer: (a) 1: 3

R needs 50% copper. By alligation, P (20%) and Q (60%) must be mixed in the ratio (60 − 50) : (50 − 20) = 10 : 30 = 1 : 3.

1. Equal copper and zinc means 50% copper in R.
2. P : Q = (60 − 50) : (50 − 20) = 10 : 30 = 1 : 3.
3. Check: 1 kg P + 3 kg Q has 0.2 + 1.8 = 2 kg copper out of 4 kg = 50%.

Remember · Alligation: ratio of cheaper to dearer = (dearer − mean) : (mean − cheaper).

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Answer: (b) 9

A − B = 9 × (difference of digits), so the digit difference must be a multiple of 3: 3 or 6 (9 would need a 0 as the units digit, making B a one-digit number). That gives 6 + 3 = 9 numbers.

1. A = 10a + b, B = 10b + a, so A − B = 9(a − b), with a > b.
2. 9(a − b) is a multiple of 27 when a − b is 3, 6 or 9.
3. B must be two-digit, so b ≥ 1. a − b = 9 would need b = 0 — not allowed.
4. a − b = 3: 41, 52, 63, 74, 85, 96 → 6 numbers.
5. a − b = 6: 71, 82, 93 → 3 numbers.
6. Total = 9. (Allowing b = 0 would give 12 — the trap option.)

Remember · A two-digit number minus its reverse = 9 × (difference of the digits). Remember that the reverse must stay two-digit.

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Answer: (b) TSPO

Letters shift alternately forward and back by one: Z → A (+1), E → D (−1), R → S (+1), O → N (−1). Applying +1, −1, +1, −1 to S, T, O, P gives T, S, P, O.

1. Z + 1 = A, E − 1 = D, R + 1 = S, O − 1 = N.
2. S + 1 = T, T − 1 = S, O + 1 = P, P − 1 = O.
3. STOP → TSPO.

Remember · Coding items: write the shift for each letter position; alternating patterns are common.

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Answer: (a) 20

Every tile is 3′ wide, so the tiles lie in one row and their lengths must add up to exactly 100: 3a + 7b + 11c = 100. Using mostly 3′ tiles gives at most 32 tiles; using mostly 11′ tiles gives at least 12, so x − y = 20.

1. The strip is 3′ wide and every tile is 3′ on one side (a 7′ or 11′ side cannot fit across), so 3a + 7b + 11c = 100.
2. Maximum: 100 = 3 × 33 + 1 is not possible with 3′ tiles alone; 3 × 31 + 7 = 100 gives 32 tiles — the most.
3. Minimum: 11 × 8 + 3 × 4 = 100 gives 12 tiles (so does 11 × 4 + 7 × 8).
4. 11 tiles cannot work: with a + b + c = 11, the equation becomes 8c + 4b = 67, and the left side is even. Fewer tiles fail the same way or fall short in length.
5. x − y = 32 − 12 = 20.

Remember · Covering a strip: turn it into an equation of lengths, then push towards the smallest (max count) or largest (min count) pieces.

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Answer: (a) 200

With original speed v, the actual trip takes 200/v + 400/(v/2) + 200/(v/8) = 2600/v hours, and the alternative takes 200/v + 400/(v/4) + 200/(v/8) = 3400/v hours. The gap 800/v = 4 hours gives v = 200 km/h.

1. Actual: first 200 km at v; next 400 km at v/2; last 200 km at (v/2)/4 = v/8.
2. Time = 200/v + 800/v + 1600/v = 2600/v.
3. Alternative: 200 km at v; 400 km at v/4; last 200 km at (v/4)/2 = v/8.
4. Time = 200/v + 1600/v + 1600/v = 3400/v.
5. 3400/v − 2600/v = 800/v = 4, so v = 200 km/h.
6. Check: 13 h and 17 h — a 4-hour difference.

Remember · Write each scenario's time as distance ÷ speed with the speed in terms of v, then equate the given difference.

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Answer: (a) 5.1

Weighted scores are 5·2, 5·1, 5·2, 4·5 and 5·0 for candidates 1 to 5. The top three are 5·2, 5·2 and 5·1, so a cutoff of 5·1 selects exactly three.

1. Candidate 1: 0·2 × 5 + 0·3 × 4 + 0·5 × 6 = 1 + 1·2 + 3 = 5·2.
2. Candidate 2: 0·8 + 1·8 + 2·5 = 5·1.
3. Candidate 3: 0·6 + 0·6 + 4 = 5·2.
4. Candidate 4: 1·8 + 1·2 + 1·5 = 4·5.
5. Candidate 5: 1·6 + 2·4 + 1 = 5·0.
6. Ranked: 5·2, 5·2, 5·1, 5·0, 4·5. For exactly three selections the cutoff is 5·1 (a cutoff of 5·2 would select only two).

Remember · Weighted score = Σ(weight × score). The cutoff is the lowest score among those who must be selected.

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Answer: (d) 3 only

All three statements belong to the same climate debate, so relationship 2 is false — and every option except (d) contains it. S3 does not argue against S1; it only notes that extreme voices within activism give deniers a target. The open question in S2 about how much humans contribute can explain why both alarmism and denial thrive.

• ✗ 1. S3 does not deny that protecting the environment is vital; it describes extreme views within activism that deniers seize on. That is a comment on the debate, not a counter-argument to S1.
• ✗ 2. S3 is about environmental activism and climate change deniers — the same subject as S1 and S2.
• ✓ 3. The lack of scientific consensus on the human contribution (S2) leaves room for alarmist and extremist positions and for deniers to target them (S3).

Remember · When one relationship is clearly false, strike every option containing it first; often a single option survives.

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Answer: (b) A-1, B-4, C-3, D-2

Match each relationship on two scales: how formal it is and whether messages are firm or open to discussion. A captain gives firm calls in an informal team setting; a judge's court is formal and firm; a Vice Chancellor and Deputy Registrar talk formally but discuss issues; peers talk informally and openly.

1. A — captain and team: a close, informal group, but on-field instructions are firm → 1.
2. B — judge and lawyers in court: formal procedure and binding directions → 4.
3. C — Vice Chancellor and Deputy Registrar: formal official hierarchy, with open discussion of administrative matters → 3.
4. D — peers and coworkers: informal and open-ended → 2.
5. Code: A-1, B-4, C-3, D-2.

Remember · For communication-type matches, place each relationship on two axes — formal/informal and firm/open-ended — then pair.

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Answer: (c) A-3, B-4, C-2, D-1

A memorandum conveys a directive within an organisation; a flyer spreads general information to anyone; Bcc hides recipients, so it informs confidentially; minutes record the decisions of a meeting.

1. Memorandum → to intimate a directive (3).
2. Flyer → to disseminate non-targeted information (4).
3. Bcc (blind carbon copy) → other recipients cannot see it, so to inform confidentially (2).
4. Minutes → to record decisions (1).
5. Code: A-3, B-4, C-2, D-1.

Remember · Office tools by purpose: memo = instruct, minutes = record, Bcc = confidential copy, flyer = mass publicity.

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Answer: (c) 3 only

Encouraging colleagues to understand each other's perspectives addresses the root of cross-cultural friction. Forcing people to practise other groups' cultural markers is coercive and can deepen resentment, and leaving conflicts alone is a failure of leadership, not a precedent.

• ✗ 1. Ordering people to adopt others' cultural practices is intrusive and can itself cause conflict; respect does not require imitation.
• ✗ 2. Letting conflicts drift allows them to fester; it signals passive, not appropriate, leadership.
• ✓ 3. Seeking each other's perspectives builds empathy and understanding across cultures — a constructive, lasting way to resolve conflict.

Remember · In conflict items, prefer actions that build mutual understanding; reject coercion and inaction.

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Answer: (d) A-2, B-4, C-1, D-3

Semantic barriers concern the meaning of words; cognitive barriers concern the mind's capacity to process, as in information overload; organisational barriers arise from structure and systems, such as missing feedback; affective barriers come from emotions such as fear of social stigma.

1. Semantic (meaning of words) → misunderstanding the meaning of a word (2).
2. Cognitive (processing capacity) → information overload (4).
3. Organisational (systems and channels) → lack of feedback (1).
4. Affective (emotions) → fear of social stigma (3).
5. Code: A-2, B-4, C-1, D-3.

Remember · Barriers by root: semantic = words, cognitive = mental processing, organisational = structure/system, affective = feelings.

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Answer: (c) 1 and 2 only

Feasibility of a family resort turns on who will come and what they will spend, so family size and budget are the right heads. The number of earners is intrusive and adds nothing that the budget does not already capture, and food allergies are an operational detail for later, not a feasibility question.

• ✓ 1. A family-oriented resort must plan room sizes and facilities around typical family size.
• ✓ 2. Visitors' budget shows whether the resort's pricing would be viable — central to feasibility.
• ✗ 3. An intrusive personal question; spending capacity is already captured by the budget head.
• ✗ 4. Useful for running a kitchen later, but not for deciding whether the resort is feasible.

Remember · Questionnaire design: keep only questions that serve the stated objective; drop intrusive or premature details.

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Answer: (d) 1 only

A circular sends the same standard message to many people at once — that is why it is cheap. Individual matters such as appraisals and increments go in personal, confidential letters, not circulars.

• ✓ 1. A circular goes to many recipients — staff, customers or members — at once.
• ✗ 2. Circulars carry standard, identical content; customised content belongs in individual letters.
• ✗ 3. Appraisals and increments concern individuals and are confidential, so they are conveyed by personal letters.
• ✗ 4. One message reproduced for many readers makes circulars more cost-effective, not less.

Remember · Circular = one standard message to many; anything individual or confidential needs a personal letter.

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Answer: (a) ₹ 12,000

Profit is shared in the ratio of capital × time. A: 1/3 × 1/3 = 1/9, B: 1/4 × 1/4 = 1/16, C: 5/12 × 1 = 5/12, i.e., 16 : 9 : 60 out of 85. C gets 60/85 of ₹ 17,000 = ₹ 12,000.

1. C's capital = 1 − 1/3 − 1/4 = 5/12.
2. Capital × time: A = 1/9, B = 1/16, C = 5/12.
3. Multiply by 144: A = 16, B = 9, C = 60; total 85.
4. C's share = 17,000 × 60/85 = ₹ 12,000.
5. Check: A gets ₹ 3,200 and B ₹ 1,800; 3,200 + 1,800 + 12,000 = 17,000.

Remember · Partnership: profit share ∝ capital × time. Clear fractions with a common denominator before dividing.

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Answer: (b) 27

Each replacement keeps 9/10 of whatever A is present. After three replacements A = 10 × (0·9)³ = 7·29 litres, so B = 2·71 litres, about 27% of the 10 litres.

1. Each step removes 1/10 of the mixture, so A falls by a factor 9/10 each time.
2. After 3 steps, A = 10 × 0·9³ = 10 × 0·729 = 7·29 L.
3. B = 10 − 7·29 = 2·71 L ≈ 27%.

Remember · Repeated replacement: amount left = original × (1 − removed/total)ⁿ.

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Answer: (d) The question cannot be answered due to insufficient data

Earnings are 5 × deliveries + 2 × kilometres. Even 90 deliveries give only ₹ 450, so the target depends on the distance travelled, which is not given. No option guarantees more than ₹ 500.

1. Earning = 5d + 2k, where d = deliveries and k ≤ 100 km.
2. d = 80, 85, 90 give ₹ 400, ₹ 425, ₹ 450 from deliveries alone.
3. Whether he crosses ₹ 500 depends on k: 90 deliveries need more than 25 km; 80 need more than 50 km.
4. The distance is not given, so none of the numbers definitely meets the target.

Remember · 'Definitely' means it must hold in every allowed case; test the worst case (here, the shortest distance).

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