CSAT (GS Paper II) 2023

Answer: (a) A national, pragmatic, legal and policy framework for water allocation should be made.

The passage says fair sharing needs a basin-based approach and shares fixed on objective criteria, which means agreed rules for allocation. A national legal and policy framework is the only option that supplies such rules; the others are engineering or demand-side ideas the passage never proposes.

• ✓ (a) The passage's key is 'a basin-based approach' with shares assessed on 'objective criteria'. A national legal and policy framework is the practical way to apply such rules across States, especially as water is a State subject.
• ✗ (b) Linking every river and creating huge aquifers is a vast engineering project, neither immediate nor mentioned; the passage is about sharing existing water fairly.
• ✗ (c) Channels from surplus to deficit regions move water, but they do not settle who gets how much; the passage never suggests them.
• ✗ (d) Cutting demand may ease stress, but the question asks how to allocate water fairly among stakeholders, which demand cuts do not decide.

Remember · For 'most rational, practical, immediate action', pick the option that puts the passage's own solution into practice; reject grand projects it never mentions.

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Answer: (d) For rapid economic growth as envisaged by us, attention should be paid to health and nutrition of the people.

Anaemia and tuberculosis cut today's productivity, and child malnutrition cuts tomorrow's cognitive potential, which matters for an economy that will need skilled workers. The implied message is that growth depends on attention to people's health and nutrition.

• ✗ (a) Education and rural areas are not discussed; the problem identified is ill-health and malnutrition.
• ✗ (b) Skill programmes are not the remedy offered; the passage argues that malnutrition limits the very capacity to become highly skilled.
• ✗ (c) 'Only skilled workers' is wrong: the passage speaks of women, men of working age and children.
• ✓ (d) Lost productivity (anaemia, TB) and lost future potential (malnourished children) both point to health and nutrition as the base of economic growth.

Remember · When a passage lists a problem and its economic cost, the implication is to fix that root problem, not a neighbouring policy.

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Answer: (c) Risk perceptions of farmers are important for motivating them for taking adaptation decisions.

The passage builds up to the point that perception is a pre-requisite for adaptation and that what matters is how a farmer perceives the risks of climate change. So farmers' risk perception drives their decision to adapt.

• ✗ (a) Policy-driven adaptation is called 'another solution'; the passage never makes adaptation and mitigation basically the government's job.
• ✗ (b) Land-use policies are not mentioned at all.
• ✓ (c) 'Perception is a necessary pre-requisite for adaptation' and 'It is important how a farmer perceives the risks' say exactly this.
• ✗ (d) The passage wants adaptation 'alongside a fast mitigation response'; it never says mitigation is not possible.

Remember · The author's message usually sits where the argument ends; test extreme words such as 'basically', 'only' and 'not possible' against the text.

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Answer: (d) 45

In the worst case Raj first takes out every white and black shoe, and then every red shoe for one foot. A pair he can wear needs one left and one right shoe, so the pair is certain only on the next draw: 18 + 16 + 10 + 1 = 45.

1. The box has 20 red shoes (10 left, 10 right), 18 white shoes and 16 black shoes.
2. Worst case: all 18 white and all 16 black shoes come out first: 34 draws and still no red pair.
3. Next, all 10 red shoes of one foot (say all left shoes) come out: 44 draws and still no pair to wear.
4. The 45th draw has to be a red right shoe, which completes a red pair. Maximum attempts = 34 + 10 + 1 = 45.
5. Check: ignoring left and right would give 34 + 2 = 36, the trap option; a wearable pair needs one shoe of each foot.

Remember · Worst-case questions: use up every unhelpful draw first, including same-foot or same-type items, then add one.

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Answer: (b) 19

Count the combinations of sixes, fours and singles that add up to 25; the order of shots does not matter. Fix the number of sixes, count how many fours can fit, and let singles make up the rest: 7 + 5 + 4 + 2 + 1 = 19.

1. Let the batsman hit x sixes, y fours and z singles: 6x + 4y + z = 25. Once x and y are chosen, z is fixed.
2. x = 0: 4y ≤ 25, so y = 0 to 6 → 7 ways.
3. x = 1: 4y ≤ 19, so y = 0 to 4 → 5 ways.
4. x = 2: 4y ≤ 13, so y = 0 to 3 → 4 ways.
5. x = 3: 4y ≤ 7, so y = 0 or 1 → 2 ways.
6. x = 4: 4y ≤ 1, so y = 0 → 1 way. Total = 7 + 5 + 4 + 2 + 1 = 19.

Remember · For 'irrespective of order' counts, fix the biggest unit, count options for the next; the smallest unit fills the rest automatically.

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Answer: (b) 2 only

If three letters are in their correct envelopes, the only envelope left belongs to the fourth letter, so exactly one wrong letter is impossible. For exactly two correct, choose the two correct letters in 6 ways and swap the other two in 1 way: 6 ways.

1. Statement 1: if 3 letters are correct, the 4th letter is left with only its own envelope, so it is correct too. Exactly one wrong letter cannot happen.
2. Statement 2: choose the 2 letters that go correctly: C(4, 2) = 6 ways.
3. The other 2 letters must both be wrong, which is possible in only 1 way (they swap envelopes).
4. Ways = 6 × 1 = 6, so statement 2 is correct.

• ✗ 1. With three letters correctly placed, the last envelope is the fourth letter's own; one misplaced letter alone is impossible.
• ✓ 2. C(4, 2) = 6 choices of the two correct letters, and the remaining two can be wrong in just one way (a swap): 6 ways.

Remember · Exactly one item misplaced is never possible. 'Exactly k correct' = C(n, k) × ways to put all the rest wrong.

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Answer: (a) 0

85 and 95 each supply a factor 5, and 96 supplies several factors of 2, so the product contains 5 × 5 × 2 × 2 = 100. A multiple of 100 leaves remainder 0.

1. 85 = 5 × 17 and 95 = 5 × 19, which give 5 × 5 = 25.
2. 96 = 2⁵ × 3, which gives at least 2 × 2 = 4.
3. So the product has 25 × 4 = 100 as a factor; the remainder on division by 100 is 0.
4. Check: 85 × 96 × 95 = 775200, already ending in 00.

Remember · Before any remainder working, look for factors 2 and 5: enough of them make the product a multiple of 10 or 100.

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Answer: (a) 2

Only the last digit of the base, 2, matters, and the unit digits of powers of 2 repeat every 4. The exponent 9 × 7 × 5 × 3 × 1 = 945 leaves remainder 1 on division by 4, so the unit digit is that of 2¹, which is 2.

1. The base ends in 2 and the exponent is 9 × 7 × 5 × 3 × 1 = 945, so we need the unit digit of 2^945.
2. Unit digits of powers of 2 run 2, 4, 8, 6 and then repeat (cycle of 4).
3. 945 ÷ 4 leaves remainder 1.
4. So the unit digit is the first in the cycle: 2.

Remember · Unit digit of a power: keep the base's last digit, find its cycle (2, 3, 7, 8 repeat every 4), use exponent mod 4.

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Answer: (d) 31

Add column by column. With non-zero digits the units must give C + F = 11 with a carry, then B + E + 1 = 11 and A + D + 1 = 11. So the six digits add up to 11 + 10 + 10 = 31.

1. Units column: C + F must end in 1. C + F = 1 is impossible with non-zero digits, so C + F = 11 and 1 is carried.
2. Tens column: B + E + 1 must end in 1, so B + E + 1 = 11, i.e. B + E = 10, and 1 is carried.
3. Hundreds column: A + D + 1 must give 11 (the leading '11' of 1111), so A + D = 10.
4. A + B + C + D + E + F = 10 + 10 + 11 = 31.
5. Check: 124 + 987 = 1111, and 1 + 2 + 4 + 9 + 8 + 7 = 31.

Remember · In digit-addition puzzles, work column by column from the units, tracking carries; the digit sum then follows without finding the digits.

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Answer: (c) 8

The ratio (number ÷ digit sum) is smallest when the hundreds digit is as small as possible and the tens and units digits are as large as possible. 199 gives 199/19 ≈ 10.47, lower than 189/18 = 10.5 or any other, so D = 199 and the difference between 9 and 1 is 8.

1. Write D = 100a + 10b + c. Raising a adds 100 to the number but only 1 to the digit sum, so a should be 1; b and c should be as large as possible.
2. Take a = 1, b = 9, c = 9: 199/19 ≈ 10.47.
3. Compare the nearest rivals: 189/18 = 10.5, 179/17 ≈ 10.53, 198/18 = 11 — all larger.
4. So D = 199: hundreds digit 1, units digit 9, difference 9 − 1 = 8.

Remember · To minimise a number divided by its digit sum, keep the high-place digit smallest and push the digit sum into lower places; test close rivals.

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Answer: (c) Waiting to deal with carbon emissions until technology improves is not a wise strategy.

Emissions released now shape the climate decades later, and cheap clean technology may or may not arrive. Just as businesses that do not hedge against uncertainty fail, the world cannot gamble on future technology, so waiting for it before cutting emissions is unwise.

• ✗ (a) Closing down or paying for pollution is not discussed; businesses appear only as an example of hedging against uncertainty.
• ✗ (b) The passage says technology 'could make a future transition ... cheap or it might not', so it denies that technology is a sure, let alone the only, solution.
• ✓ (c) Betting on uncertain future technology instead of acting now is the 'recklessness' the world 'cannot afford'.
• ✗ (d) New industries and renewables are not the point; the passage is about hedging against uncertainty by acting on emissions now.

Remember · When a passage uses an analogy, the message is the analogy's lesson applied to the main subject.

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Answer: (a) We are likely to spend more money on cure than prevention.

Lifestyle changes could prevent environmental and health problems, yet people show 'collective lethargy' even where their own choices decide life and death. When prevention is neglected, the problems arrive and must be treated, so more is likely to be spent on cure than on prevention.

• ✓ (a) Inertia towards healthy lifestyle choices means prevention is neglected; the logical result is heavier spending on cure later.
• ✗ (b) The government is not mentioned; the passage is about individual choices and collective lethargy.
• ✗ (c) The opening line, 'Environmental problems cause health problems', rules out protecting health while environmental problems are ignored.
• ✗ (d) Traditional lifestyles and western values are outside the passage.

Remember · A valid assumption follows from the passage's own logic; options that bring in new actors or causes (government, western values) usually fail.

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Answer: (b) We must place food-based nutrition at the centre of our policy debate.

The passage traces poor nutrition to unhealthy food choices, poor access and poor affordability, and blames food systems that are not responding to nutritional needs 'from farm to fork'. Its crux is that nutrition through food must be central to policy.

• ✗ (a) Universal Basic Income is not mentioned; affordability is only one of several causes described.
• ✓ (b) The closing idea, that 'our food systems are not properly responding to nutritional needs', calls for making food-based nutrition the centre of policy.
• ✗ (c) Genetically modified crops are outside the passage.
• ✗ (d) Fortification is a specific technique the passage does not discuss, and it would not address the choice and access problems described.

Remember · The crux covers all the causes a passage lists; specific technical fixes such as GM crops or fortification are usually too narrow.

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Answer: (a) 1 only

Subtracting a number changes odd/even exactly as adding it does, so p + q + r − s − t has the same parity as the sum of all five, which is even (it contains two odd numbers). In 2p + q + 2r − 2s + t the doubled terms are even, leaving the parity of q + t, which may be even.

1. Adding or subtracting a number has the same effect on parity, so signs can be ignored.
2. Statement 1: parity of p + q + r − s − t = parity of p + q + r + s + t. Two odds make an even and evens stay even, so the result is always even. Correct.
3. Statement 2: 2p, 2r and 2s are even, so the parity equals that of q + t.
4. Example: p = 1 and s = 1 are the odd ones, q = r = t = 2. Then 2 + 2 + 4 − 2 + 2 = 8, which is even. So statement 2 is not always true.

• ✓ 1. Its parity is that of the sum of all five numbers: three evens and two odds add up to an even number, whatever the order.
• ✗ 2. Only q and t decide the parity; if both are even (or both odd) the expression is even, so it is not definitely odd.

Remember · For parity, treat minus as plus and ignore every term multiplied by an even number.

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Answer: (d) 1, 2 and 3

Each statement says 'can be', so one example proves it. p = 11, c = 9 gives (p + c)/(p − c) = 20/2 = 10, which is even; p = 2, c = 9 gives 2p + c = 13, which is odd; p = 3, c = 9 gives pc = 27, which is odd. All three statements hold.

1. A 'can be' statement is proved by a single example.
2. Statement 1: p = 11, c = 9 → (11 + 9)/(11 − 9) = 20/2 = 10, which is even. True.
3. Statement 2: 2p is always even, so take an odd composite: p = 2, c = 9 → 4 + 9 = 13, odd. True.
4. Statement 3: an odd prime times an odd composite: 3 × 9 = 27, odd. True.

• ✓ 1. p = 11, c = 9 gives 20/2 = 10, an even number.
• ✓ 2. 2p is even, so 2p + c is odd whenever c is an odd composite, e.g. 2 × 2 + 9 = 13.
• ✓ 3. Both odd: p = 3, c = 9 gives 27.

Remember · 'Can be' needs one working example; 'must be' needs proof. Remember odd composites like 9, 15 and 21.

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Answer: (a) 18

37DD = 3700 + 11D must be a multiple of D, so D must divide 3700; that allows D = 1, 2, 4 or 5. Only D = 4 gives a 3-digit ABC with all four digits different: 3744 ÷ 4 = 936, so A + B + C = 9 + 3 + 6 = 18.

1. 37DD = 3700 + 11 × D. Since ABC × D = 37DD, D divides 3700 + 11D, so D divides 3700 = 2² × 5² × 37.
2. Possible non-zero digits: D = 1, 2, 4 or 5.
3. D = 1 gives ABC = 3711 and D = 2 gives 3722 ÷ 2 = 1861: neither is a 3-digit number.
4. D = 5 gives 3755 ÷ 5 = 751, but then B = 5 = D, which is not allowed.
5. D = 4 gives 3744 ÷ 4 = 936: A = 9, B = 3, C = 6, D = 4, all different. A + B + C = 18.
6. Check: 936 × 4 = 3744.

Remember · In digit puzzles, write the unknown number algebraically (3700 + 11D), shortlist digits by divisibility, then apply the 'different digits' condition.

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Answer: (d) 7, 11 and 13

XYZXYZ = XYZ × 1000 + XYZ = XYZ × 1001, and 1001 = 7 × 11 × 13. So every such number is divisible by 7, 11 and 13.

1. XYZXYZ = XYZ × 1000 + XYZ = XYZ × 1001.
2. 1001 = 7 × 11 × 13.
3. So the number is always divisible by 7, 11 and 13.
4. Check: 123123 = 123 × 1001 = 123 × 7 × 11 × 13.

Remember · A repeated 3-digit block (abcabc) is a multiple of 1001 = 7 × 11 × 13; abab is a multiple of 101.

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Answer: (a) 27

125 cubes make a 5 × 5 × 5 block. A cube surrounded on every side cannot lie on any outer face, so remove one layer from each side: the hidden core is 3 × 3 × 3 = 27 cubes.

1. 125 = 5 × 5 × 5, so the block is 5 cubes long, wide and high.
2. Every cube in the outer layer has at least one face exposed.
3. Removing one layer from each side leaves (5 − 2) × (5 − 2) × (5 − 2) = 3 × 3 × 3 = 27 cubes, each touched by cubes on all sides.

Remember · Fully hidden cubes in an n × n × n block = (n − 2)³.

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Answer: (c) 36

The four odd positions must hold 1, 1, 3, 3 and the four even positions must hold 2, 2, 4, 4. Each set can be arranged in 4!/(2! × 2!) = 6 ways, so there are 6 × 6 = 36 numbers.

1. Odd digits 1, 1, 3, 3 fill the 4 odd positions; even digits 2, 2, 4, 4 fill the 4 even positions.
2. Arrangements of 1, 1, 3, 3 = 4!/(2! × 2!) = 24/4 = 6.
3. Arrangements of 2, 2, 4, 4 = 6 in the same way.
4. Total = 6 × 6 = 36.

Remember · Arrangements with repeated items = n!/(p! × q! …); choices for independent slots multiply.

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Answer: (a) 1 only

Take the work as 48 units: A does 6, B 3 and C 4 units a day, so each 3-day round finishes 13 units. After 9 days 39 units are done; A adds 6 on day 10 and B completes the last 3 on day 11. Counting from Monday, day 11 is a Thursday.

1. Total work = LCM(8, 16, 12) = 48 units. A does 6, B does 3 and C does 4 units a day.
2. One round (A, B, C) = 6 + 3 + 4 = 13 units in 3 days. After 3 rounds (9 days): 39 units, so 9 units remain.
3. Day 10 (A): 39 + 6 = 45 units. Day 11 (B): 45 + 3 = 48 units — the work is complete.
4. Day 1 is Monday, so day 8 is Monday again and day 11 is Thursday.
5. Statement 1 is correct; statement 2 is not, because the work takes 11 days.

• ✓ 1. The work ends on day 11, and day 11 counted from a Monday is a Thursday.
• ✗ 2. After 10 days only 45 of the 48 units are done; B needs day 11 to finish.

Remember · Take the LCM of days as total work, add round by round, then day by day; convert day numbers to weekdays using remainders by 7.

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Answer: (b) Knee-jerk reactions cannot solve the problem of pollution but an evidence-based approach will be more effective.

The author finds drastic, arbitrary measures (odd-even, bans on old vehicles and certain fuels) ineffective and unfair, and closes by asking for scientific, reliable information on pollution sources and on technologies that work. In short: evidence, not knee-jerk action.

• ✗ (a) The passage calls arbitrary bans ineffective, 'not proper' and 'unfair', not difficult to implement.
• ✓ (b) Drastic actions leave 'the air ... not clean', and the author ends with 'What is needed is the scientific and reliable information' — an evidence-based approach.
• ✗ (c) No penalty is proposed; the passage defends vehicles that have passed fitness and pollution tests.
• ✗ (d) The passage never says pollution laws are absent or that this causes arbitrary decisions.

Remember · Read the closing 'what is needed' sentence; the implication usually restates it. Reject options about ease, penalties or laws never discussed.

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Answer: (b) Good corporate governance improves the credibility of the firms.

Good governance brings accountability and control, and investors put money where disclosure, accurate reporting and fair treatment of stakeholders are assured, which is why such firms get easier external finance. The inference is that good governance makes firms more credible.

• ✗ (a) The passage says corporate governance moved onto the global agenda; it does not say ensuring access to external financing is countries' agenda.
• ✓ (b) Accountability, control, timely and accurate reporting and equal treatment are the standards investors trust, so good governance raises a firm's credibility and its access to funds.
• ✗ (c) This reverses the link: the growth of capital markets is why governance became important; markets are not said to ensure good governance.
• ✗ (d) Supply chains are not mentioned.

Remember · Watch for reversed cause and effect: an option that swaps the passage's cause and result is a classic trap.

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Answer: (b) Elephants are the keystone species and they benefit the biodiversity.

Elephants make clearings, check overgrowth, spread seeds, enrich the soil with dung and dig water holes, and each of these helps other plants and animals. A species whose activities shape the whole ecosystem in this way is a keystone species that benefits biodiversity.

• ✗ (a) The size or richness of the elephants' home range is not discussed.
• ✓ (b) Every example shows elephants sustaining other plants, herbivores, insects and wildlife — the role of a keystone species.
• ✗ (c) 'Cannot be maintained without elephants' goes too far; the passage shows benefit, not that biodiversity is impossible without them.
• ✗ (d) Regeneration is a side-effect of their feeding and movement; 'as per their requirement' implies a deliberate design the passage never claims.

Remember · Choose the inference that sums up all the examples; reject 'cannot', 'only' and intentions the passage never attributes.

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Answer: (a) 6

Add the three numbers and then add the digits of the total: 26 → 8, 50 → 5, 49 → 13. For 23, 4 and 15 the total is 42, and 4 + 2 = 6.

1. 7 ⊕ 9 ⊕ 10: 7 + 9 + 10 = 26, and 2 + 6 = 8 (fits).
2. 9 ⊕ 11 ⊕ 30: 9 + 11 + 30 = 50, and 5 + 0 = 5 (fits).
3. 11 ⊕ 17 ⊕ 21: 11 + 17 + 21 = 49, and 4 + 9 = 13 (fits).
4. 23 ⊕ 4 ⊕ 15: 23 + 4 + 15 = 42, and 4 + 2 = 6.

Remember · For invented operators, test simple rules — sums, products, digit sums — on every given case and accept only one that fits all.

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Answer: (c) 12

7x is always a multiple of x, so x divides 7x + 96 exactly when x divides 96. Since 96 = 2⁵ × 3, it has (5 + 1)(1 + 1) = 12 divisors, so x can take 12 values.

1. 7x is divisible by x, so x divides 7x + 96 only when x divides 96.
2. 96 = 2⁵ × 3.
3. Number of divisors = (5 + 1) × (1 + 1) = 12: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96.

Remember · If x divides kx + N, then x divides N; the number of divisors of 2^a × 3^b is (a + 1)(b + 1).

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Answer: (b) 225

Use the four largest digits 9, 8, 7 and 6, which add up to 30. For a fixed total, the product of two parts is greatest when the parts are equal, so split 30 as 15 and 15: (9 + 6)(8 + 7) = 225.

1. To make the product large, use the four largest distinct digits: 9, 8, 7, 6 (total 30).
2. Split 30 into two sums as equal as possible: 9 + 6 = 15 and 8 + 7 = 15.
3. Greatest value = 15 × 15 = 225.
4. Check: the other splits give 17 × 13 = 221 and 16 × 14 = 224, both smaller.

Remember · With a fixed total, a product of parts is largest when the parts are as equal as possible.

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Answer: (a) 11

999999 (six 9s) = 999 × 1001 is divisible by 13, so each block of six 9s leaves nothing over. 99 nines are 16 such blocks followed by 999, so N leaves the same remainder as 999, which is 11.

1. 999999 = 999 × 1001 = 999 × 7 × 11 × 13, so it is divisible by 13.
2. 99 = 6 × 16 + 3, so N is sixteen blocks of 999999 followed by 999.
3. The first 96 nines form a multiple of 999999 (and so of 13), shifted three places; hence N leaves the same remainder as 999.
4. 999 = 13 × 76 + 11, so the remainder is 11.

Remember · Six identical digits (aaaaaa) are always divisible by 7, 11 and 13; remove blocks of six and work with the digits left over.

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Answer: (c) 81

111111111 × 111111111 = 12345678987654321. Its digits add up to 2 × (1 + 2 + … + 8) + 9 = 72 + 9 = 81.

1. Pattern: 11² = 121, 111² = 12321, 1111² = 1234321, and so on.
2. So 111111111² = 12345678987654321.
3. Digit sum = 2 × (1 + 2 + … + 8) + 9 = 2 × 36 + 9 = 81.
4. Check: with n ones the digit sum of the square is n², and 9² = 81.

Remember · Squares of numbers made of n ones (n up to 9) are palindromes 123…n…321 with digit sum n².

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Answer: (b) 856

From 10 to 99, each tens digit 1 to 9 appears 10 times (sum 450) and each units digit 0 to 9 appears 9 times (sum 405), which gives 855. The number 100 adds 1 more, so the total is 856.

1. Tens digits in 10–99: each of 1 to 9 appears 10 times → 10 × 45 = 450.
2. Units digits in 10–99: each of 0 to 9 appears 9 times (once in every ten) → 9 × 45 = 405.
3. Sum for 10–99 = 450 + 405 = 855.
4. Add the digits of 100: 1 + 0 + 0 = 1. Total = 856.

Remember · For digit-sum totals, count how often each digit appears in each place, and do not forget the end-points (here, 100).

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Answer: (c) 20

Three points fail to make a triangle only if they lie on one straight line. No side of the square carries more than two of the six points, so no three are in a line and every choice of three points gives a triangle: C(6, 3) = 20.

1. Points: 1 on AB, 1 on CD, 2 on BC and 2 on DA — six in all.
2. A set of three points forms no triangle only when all three lie on one side; no side has more than two points.
3. Number of triangles = C(6, 3) = (6 × 5 × 4)/(3 × 2 × 1) = 20.

Remember · Triangles from points = C(n, 3) minus collinear triples; check each line for three or more points.

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Answer: (d) Neither 1 nor 2

The passage describes rare segregation, informal recycling and a budget swallowed by collection and transport, and explains where waste-to-energy fits and how technology should be chosen. It never says who — government agencies or private firms — should do these jobs, so neither assumption is made.

• ✗ 1. The passage notes that recycling is with the informal sector and how the municipal budget is spent, but it never says which agency should collect, process or segregate waste.
• ✗ 2. Technology choice is tied to the waste's composition and calorific value; the private sector is not mentioned at all.

Remember · An assumption must be needed for the author's argument; statements that assign a job to a new actor (government, private sector) usually are not.

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Answer: (c) Segregation of municipal solid waste is the first step in ensuring the success of waste-to-energy plants.

Waste-to-energy fits in the chain after segregation, and the right technology depends on what the waste contains, yet segregation at source is rare in India. So segregation is the first requirement for waste-to-energy plants to work.

• ✗ (a) The cost of waste-to-energy is not discussed; the budget point is that collection and transport take most of the money.
• ✗ (b) Biomethanation is 'a major solution' for the biodegradable part, not the most ideal way for all municipal waste.
• ✓ (c) Waste-to-energy 'fits in the chain after segregation', and technology depends on the waste's make-up, which cannot be handled well while segregation at source is 'rare'.
• ✗ (d) The biodegradable share is 'a little over 50 per cent' and biomethanation can process it — the opposite of 'not adequate'.

Remember · The crux links the passage's problem (rare segregation) to its solution chain; reject superlatives like 'most ideal' that the passage never uses.

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Answer: (b) 2 only

The author disputes the claim that organic farming is inherently safer, but traces the dangers to a young, poorly regulated industry and to confusion and misuse by farmers and consumers. So the passage assumes they need to be better informed, not that organic farming is unsafe by nature.

• ✗ 1. The risks come from poor regulation, substitutes such as contaminated manure and wrong doses of sprays, not from organic farming itself; 'inherently unsafe' goes too far.
• ✓ 2. Farmers and consumers are 'confused about what products are best for them' and use products in harmful ways, which presumes that educating them would help.

Remember · Rejecting 'inherently safe' does not mean 'inherently unsafe'; watch for options that flip a claim to its extreme opposite.

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Answer: (c) In India, farmers need to be guided and helped to make their organic farming sustainable.

Farmers misuse inputs such as farmyard manure and Datura spray because the sector is young, poorly regulated and under-researched. The practical message is that farmers need guidance and support to make organic farming safe and sustainable.

• ✗ (a) The passage does not compare organic with conventional farming or argue against promoting organic farming.
• ✗ (b) 'No safe organic alternatives' is too extreme; the problem is misuse and weak regulation, not the absence of safe options.
• ✓ (c) Confusion, harmful use and unresearched doses all point to the need to guide and help farmers.
• ✗ (d) Profits and global markets are outside the passage.

Remember · For a 'practical message', choose the constructive remedy that answers the problems described; reject blanket negatives and outside topics.

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Answer: (b) 2 and 3 only

The passage blames the long focus on staples and intensive monoculture for the decline of nutritious crops and for degrading land, water and food, even as grain output rose without ending malnutrition. It therefore assumes that dependence on a few crops harms health and ecosystems, and that food planning must include nutrition.

• ✗ 1. The passage criticises monoculture for perpetuating the nutrition problem; it never calls it inevitable for the SDGs or zero hunger.
• ✓ 2. The focus on staples pushed out nutritious crops (health), and monoculture degrades 'land, water and food' (ecosystem).
• ✓ 3. Grain output rose over five times yet malnutrition persists, so food planning must aim at nutrition, not quantity alone.
• ✗ 4. Subsidies and remunerative prices are not mentioned; this explanation is brought in from outside the passage.

Remember · Valid assumptions follow from the passage's own reasoning; reject facts brought in from outside, even if true in real life.

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Answer: (c) Both 1 and 2

For a full group, the unluckiest draw takes every colour one ball short of complete: 13 + 19 + 25 + 27 + 37 + 53 = 174, so 175 balls guarantee a full group. For one ball of each colour, the unluckiest draw takes every ball except the smallest colour: 180 − 14 = 166, so 167 balls guarantee all six colours.

1. Total balls = 14 + 20 + 26 + 28 + 38 + 54 = 180.
2. Statement 1: the worst case leaves every colour one short: 13 + 19 + 25 + 27 + 37 + 53 = 174 balls with no full group. The next ball completes one, so n = 175. Correct.
3. Statement 2: the worst case is all balls except the smallest colour (14 black): 180 − 14 = 166 balls without black. The next ball must be black, so m = 167. Correct.

• ✓ 1. 174 balls can still leave every colour one ball short of full; the 175th ball must complete a group.
• ✓ 2. 166 balls can all be non-black; the 167th must be black, giving all six colours.

Remember · Guarantee questions: build the unluckiest draw that still fails, then add one.

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Answer: (d) SODBHU

Each letter moves 3 places forward in the alphabet: Z → C, E → H, R → U, O → R. Moving each letter of PLAYER 3 places forward gives SODBHU.

1. Z + 3 = C (wrapping round after Z), E + 3 = H, R + 3 = U, O + 3 = R, so ZERO → CHUR.
2. PLAYER: P + 3 = S, L + 3 = O, A + 3 = D, Y + 3 = B, E + 3 = H, R + 3 = U.
3. So PLAYER is written as SODBHU.

Remember · In letter coding, find the shift for each letter (remember the wrap-around after Z); a constant shift applies letter by letter.

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Answer: (d) All five

Chaining the statements gives C = D > F > A > B, with E the youngest, so C and D are the oldest. Every statement is needed: drop any one and some person could be older than D, or C could not be placed.

1. From 4, 5 and 1: D > F > A > B. From 2: C = D. From 3: E is the youngest.
2. So C and D are the oldest.
3. Without 1, B is not linked to anyone and might be older than D; without 3, E might be older than D.
4. Without 5, A and B are cut off from D; without 4, F, A and B are cut off from D; without 2, C's age is unknown.
5. So all five statements are required.

Remember · To fix 'the oldest', every person must be linked below the candidate; test by dropping one statement at a time.

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Answer: (c) All four statements together are sufficient to answer the Question.

A and C are sisters and D is C's father, so D is A's father too, and E, D's son, is A's brother. A, a woman, is married to B, so B is A's husband and E is B's brother-in-law. Every one of the four statements is used in this chain.

1. Statement-4: A and C are sisters, so both are female. Statement-2: D is C's father, so D is A's father as well.
2. Statement-3: E is D's son, so E is the brother of A and C.
3. Statement-1: A and B are married; A is female, so B is A's husband.
4. E is the brother of B's wife, i.e. E is B's brother-in-law.
5. Without Statement-2 or Statement-4, E cannot be linked to A; without Statement-1, B is unconnected. So all four are needed.

• ✗ (a) Without Statement-4, nothing links A (and so B) to C, D or E.
• ✗ (b) Without Statement-2, D (and so E) is not linked to A and C.
• ✓ (c) Together they give: E is A's brother and B is A's husband, so E is B's brother-in-law.
• ✗ (d) The four statements together do answer the Question.

Remember · In blood-relation sufficiency questions, trace one chain from the first person to the second; every link used is a required statement.

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Answer: (d) 83, 89, 91, 97

In groups (a), (b) and (c) all four numbers are prime. In group (d), 91 = 7 × 13 is not prime, so that group is the odd one out.

1. (a) 17, 37, 47, 97 — all prime.
2. (b) 31, 41, 53, 67 — all prime.
3. (c) 71, 73, 79, 83 — all prime.
4. (d) 83, 89 and 97 are prime, but 91 = 7 × 13. So (d) is different.

Remember · Know the primes up to 100; 91 (7 × 13), 51 (3 × 17) and 87 (3 × 29) are the classic look-alike composites.

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Answer: (a) 1 only

The passage links warmer winters with less snow to earlier ice-melt and a longer, more drawn-out vernal window, which takes a warming climate to be making spring come early and last longer. It reports that birds do not return on time, but says nothing about harm to bird populations.

• ✓ 1. 'Warmer winters with less snow' produce 'a more protracted vernal window' and ice that 'melts earlier': the passage takes it that warming brings spring earlier and stretches it out.
• ✗ 2. The passage says the birds don't return, delaying springtime events; it says nothing about bird numbers, so harm to bird populations is not assumed.

Remember · Separate what a passage reports (birds' timing changes) from a further judgement it never makes (populations suffer).

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Answer: (a) An enhanced efficiency of use of nitrogen is imperative for both food production and environment.

Synthetic nitrogen feeds more than half the world, yet 60 per cent of the nitrogen applied is lost as pollution. The balanced message is to use nitrogen more efficiently, keeping yields up while cutting pollution.

• ✓ (a) The passage is built around 'nitrogen use efficiency': synthetic nitrogen is 'needed to produce high crop yields', but low efficiency (0·4) pollutes air and water. Higher efficiency serves both food and environment.
• ✗ (b) Stopping fertilizer production is never discussed; the option defends a point the passage does not raise.
• ✗ (c) Replacing crops is not suggested; the problem is excess application, not the crops themselves.
• ✗ (d) Replacing conventional agriculture is too drastic when the passage says synthetic nitrogen nourishes over half the world.

Remember · When a passage shows both a need and a harm, the crucial message usually balances them rather than dropping either.

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Answer: (a) Climate justice should be ingrained in detail in the rules of all the new climate compacts/agreements.

Climate justice has so far eluded consensus at climate summits, and the crucial question ahead is how resources, technologies and regulations will support victims through mitigation, adaptation and compensation for loss and damage. The message is that climate justice must be built into the rules of future climate agreements.

• ✓ (a) The principle has 'eluded consensus' in 'summits or compacts', and the passage lists what justice must include (mitigation, adaptation support, compensation). Writing it into new agreements follows.
• ✗ (b) Uneven distribution of environmental resources is not discussed.
• ✗ (c) The passage speaks of supporting victims, but not of an impending problem of huge numbers of climate refugees.
• ✗ (d) Blame on developed countries and burden-sharing are outside the passage.

Remember · When a passage names an unresolved principle and spells out what it must cover, the message is usually to build it into future rules.

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Answer: (c) R < S

At the same rate, half-yearly compounding earns interest on interest and so grows faster than annual compounding. To reach the same amount Q in one year, the half-yearly rate R must therefore be smaller than the annual rate S.

1. Half-yearly at R% a year: Q = P(1 + R/200)². Annually at S%: Q = P(1 + S/100).
2. So 1 + S/100 = (1 + R/200)² = 1 + R/100 + R²/40000.
3. S = R + R²/400, which is greater than R for any positive rate.
4. Example: R = 10 gives (1.05)² = 1.1025, so S = 10.25. Hence R < S.
5. Option (d), R ≤ S, is not false as a statement, but equality never occurs for a positive rate; the exact relation is R < S.

Remember · More frequent compounding at the same rate gives more; to give the same amount, the more frequent rate must be lower.

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Answer: (d) 9

If 1186 leaves remainder 31, the divisor divides 1186 − 31 = 1155 and must be greater than 31. 1155 = 3 × 5 × 7 × 11 has 16 divisors, of which 9 are greater than 31.

1. 1186 = n × (quotient) + 31, so n divides 1186 − 31 = 1155; also n > 31, because a remainder is always less than the divisor.
2. 1155 = 3 × 5 × 7 × 11, so it has 2 × 2 × 2 × 2 = 16 divisors.
3. Divisors greater than 31: 33, 35, 55, 77, 105, 165, 231, 385, 1155 — 9 numbers.

Remember · Remainder r from N ÷ n means n divides N − r and n > r; count divisors of N − r above r.

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Answer: (c) Both 1 and 2

pp + qq + rr = 11(p + q + r) and tt0 = 110t, so p + q + r must be 10 or 20. The triples with p < q < r are (1,2,7), (1,3,6), (1,4,5), (2,3,5), (3,8,9), (4,7,9), (5,6,9) and (5,7,8), giving 5 possible values of p and 6 of q.

1. pp = 11p, so pp + qq + rr = 11(p + q + r). Also tt0 = 100t + 10t = 110t.
2. 11(p + q + r) = 110t gives p + q + r = 10t. With digits 1 to 9 and p < q < r, the sum is at most 7 + 8 + 9 = 24, so it is 10 or 20.
3. Sum 10: (1,2,7), (1,3,6), (1,4,5), (2,3,5). Sum 20: (3,8,9), (4,7,9), (5,6,9), (5,7,8).
4. Values of p: 1, 2, 3, 4, 5 → 5. Values of q: 2, 3, 4, 6, 7, 8 → 6.
5. Both statements are correct.

• ✓ 1. p can be 1, 2, 3, 4 or 5 — five values.
• ✓ 2. q can be 2, 3, 4, 6, 7 or 8 — six values.

Remember · Repeated-digit numbers factor neatly: aa = 11a and aa0 = 110a. Turn the puzzle into a simple sum condition, then list cases.

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Answer: (a) 7998

Numbers below 2000 must start with 1, and the other three places take 2, 3 and 4 in 3! = 6 orders. The thousands digits give 6 × 1000 = 6000; in each other place each of 2, 3 and 4 appears twice, adding 18 × 111 = 1998. The total is 7998.

1. Below 2000 with the digits 1, 2, 3, 4 used once each, the first digit must be 1; 2, 3, 4 fill the rest in 3! = 6 ways.
2. Thousands place: 6 × 1000 = 6000.
3. In each of the hundreds, tens and units places, each of 2, 3, 4 appears 2 times: (2 + 3 + 4) × 2 = 18.
4. Their contribution = 18 × (100 + 10 + 1) = 18 × 111 = 1998.
5. Total = 6000 + 1998 = 7998.

Remember · Sum of numbers formed by permutations: count how often each digit sits in each place, then multiply by the place values (111…).

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Answer: (c) 12

Going clockwise, a block of 10 consecutive things is fixed by its starting point, and each of the 12 things can be the start. So there are 12 selections.

1. Taken clockwise, a run of 10 consecutive things is decided by where it starts.
2. Each of the 12 things can be the starting point, and different starts give different sets.
3. Number of selections = 12.
4. Check: each selection leaves out 2 neighbouring things, and a circle of 12 has exactly 12 neighbouring pairs.

Remember · In a circle of n items, the number of blocks of k consecutive items (k < n) is n.

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Answer: (a) Wednesday

Counting today (Sunday) as day 1, the 10^10th day comes 10^10 − 1 days later. Since 10 leaves remainder 3 on division by 7, 10^10 leaves the same remainder as 3^10, which is 4; so 10^10 − 1 leaves 3, and three days after Sunday is Wednesday.

1. Take today, Sunday, as day 1. Then day N falls N − 1 days after Sunday.
2. 10 leaves remainder 3 on division by 7, so 10^10 leaves the same remainder as 3^10.
3. Remainders of powers of 3 by 7: 3, 2, 6, 4, 5, 1, then repeat (cycle of 6). 3^10 = 3^(6+4) leaves the same as 3^4, i.e. 4.
4. So 10^10 − 1 leaves remainder 3: the day is 3 days after Sunday, i.e. Wednesday.
5. Note: counting 10^10 days after today (not counting today) would give Sunday + 4 = Thursday; UPSC's key uses the reading in which today is the 1st day.

Remember · Day questions: reduce the count modulo 7 (powers via short cycles), and be clear whether today counts as day 1.

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Answer: (b) 4:10 p.m.

Green and red last equally long, so the full cycles are 50 s, 78 s and 120 s. All three turn green together again after LCM(50, 78, 120) = 7800 s = 2 hours 10 minutes, i.e. at 4:10 p.m.

1. Each full cycle (green → red → green) is twice the green time: 2 × 25 = 50 s, 2 × 39 = 78 s, 2 × 60 = 120 s.
2. 50 = 2 × 5², 78 = 2 × 3 × 13, 120 = 2³ × 3 × 5.
3. LCM = 2³ × 3 × 5² × 13 = 7800 s = 130 minutes = 2 h 10 min.
4. 2:00 p.m. + 2 h 10 min = 4:10 p.m.
5. Check: LCM(25, 39, 60) = 3900 s, but at that moment the third signal has changed 65 times (an odd number), so it is turning red, not green.

Remember · When signals must show the same colour again, take the LCM of full cycles (both phases), not of the single-phase times.

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Answer: (d) Agroecosystems should be developed in addition to or along with conventional agriculture.

The passage recommends sourcing food from forests, wetlands and pastures in addition to farmland, because such foods are resilient and fill gaps when crops fail or between seasons. So these food systems should be developed alongside conventional agriculture, not in place of it.

• ✗ (a) Replacing other trees with food-yielding trees is not proposed.
• ✗ (b) 'Cannot be ensured' is too strong; the passage offers a supplement to agriculture, not a verdict against it.
• ✗ (c) Converting wastelands is not discussed; the passage speaks of existing uncultivated systems such as forests and wetlands.
• ✓ (d) The key phrase is 'in addition to agricultural lands': uncultivated food systems act as a safety net alongside conventional farming.

Remember · Watch small phrases like 'in addition to': they rule out options that replace rather than supplement.

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Answer: (a) It is necessary to put proper effluent treatment protocols in place.

The passage's new point is that antibiotic factories that do not treat their waste are fuelling drug-resistant infections around their plants. The practical remedy is proper treatment of their effluents.

• ✓ (a) The problem named is 'companies not treating their waste'; effluent treatment protocols tackle it directly.
• ✗ (b) Awareness of antibiotic misuse is already 'common knowledge'; the neglected issue is factory waste, not public awareness.
• ✗ (c) The passage ties resistance to untreated factory pollution, which can be fixed; it does not call the spread inherent or unavoidable.
• ✗ (d) Moving factories to remote areas would shift the pollution, not treat it.

Remember · For a 'practical message', choose the remedy that targets the exact cause named; relocation or awareness drives miss it.

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Answer: (a) To become a global power, India needs to invest in universal quality education.

Advanced skills rest on foundational learning; without a strong primary base there can be no improvement in higher education or skills, and an uneducated workforce cannot keep pace with the global economy. The crux is that India must invest in quality education for all to compete globally.

• ✓ (a) Quality schooling that gives every student the 'gateway skills' is presented as the base for keeping pace with the global, knowledge-based economy.
• ✗ (b) The passage does not say India is unable to become a global power or that it neglects the knowledge economy.
• ✗ (c) It says the opposite: improvement in higher education and skills depends on a strong primary foundation.
• ✗ (d) Parents' literacy is not mentioned.

Remember · The crux keeps the passage's own order of priorities; an option that reverses it (higher education before foundations) is wrong.

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Answer: (b) 15

The ring moves forward 1, 2, 3, … places on successive passes, so after k passes it has moved 1 + 2 + … + k = k(k + 1)/2 places. It is back with child-1 when this is a multiple of 40, which first happens at k = 15 (120 places, i.e. 3 full rounds).

1. Child-1 → 2 → 4 → 7 → …: the jumps are 1, 2, 3, … places.
2. After k passes the ring has moved k(k + 1)/2 places round the circle of 40.
3. It is back with child-1 when k(k + 1)/2 is a multiple of 40, i.e. when k(k + 1) is a multiple of 80.
4. k = 15: 15 × 16 = 240 = 3 × 80 works; no smaller k does (for example 14 × 15 = 210 and 9 × 10 = 90).
5. So after 15 passes (120 places, 3 full rounds) the ring returns to child-1.

Remember · Growing jumps 1, 2, 3, … add up to triangular numbers k(k + 1)/2; returning to the start needs a multiple of the circle size.

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Answer: (b) I

Z appears 2 times, Y 3 times and so on, up to A 27 times, so there are 377 terms and the middle one is the 189th. The first k letters fill k(k + 3)/2 places; for k = 18 this is exactly 189, so the middle term is the 18th letter counted from Z, which is I.

1. The kth letter from Z appears k + 1 times: Z (2), Y (3), …, A (27).
2. Total terms = 2 + 3 + … + 27 = 377, so the middle term is the (377 + 1)/2 = 189th.
3. The first k letters take up 2 + 3 + … + (k + 1) = k(k + 3)/2 places.
4. k = 17 gives 170 and k = 18 gives 189, so the 189th term is the last copy of the 18th letter from Z.
5. Counting back from Z: Z, Y, X, W, V, U, T, S, R, Q, P, O, N, M, L, K, J, I — the 18th is I.

Remember · For blocks of growing length, find the total, locate the middle position, then use running totals to see which block holds it.

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Answer: (d) The Question cannot be answered even by using both the Statements together.

Statement 1 says p and q have the same sign; Statement 2 says p is larger than q in size. Even together, p = 3, q = 1 gives 'yes' while p = −3, q = −1 gives 'no', so the Question cannot be answered.

1. Statement 1 alone: p × q > 0 means p and q have the same sign. p = 3, q = 1 (yes) and p = 1, q = 3 (no) — not sufficient.
2. Statement 2 alone: p² > q² means p is larger in size than q. p = 3, q = 1 (yes) and p = −3, q = 1 (no) — not sufficient.
3. Both together: same sign and larger size. p = 3, q = 1 gives p > q, but p = −3, q = −1 gives p < q.
4. So the Question cannot be answered even with both Statements.

Remember · In inequality data sufficiency, always test negative numbers; products and squares hide signs.

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Answer: (c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.

(p + q − r) > (p − q + r) simplifies to q > r. Statement 1 gives q < p and Statement 2 gives p < r; neither alone compares q with r, but together q < p < r, so q < r and the answer is a definite 'no'.

1. (p + q − r) − (p − q + r) = 2q − 2r, so the Question is simply: is q > r?
2. Statement 1: p − q > 0, i.e. q < p. Nothing about r — not sufficient.
3. Statement 2: p − r < 0, i.e. p < r. Nothing about q — not sufficient.
4. Together: q < p < r, so q < r and the expression is not greater — a definite answer.
5. So both Statements together are needed and are sufficient.

Remember · Simplify the question first; in data sufficiency a firm 'no' answers the question just as well as a firm 'yes'.

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Answer: (a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone.

People taking tea or coffee number 75 + 60 − 15 = 120; everyone else took milk only. Statement 2 gives 120 + m = 5m, so m = 30 and 150 attended. Statement 1 does not say how many of the 50 milk drinkers also took coffee, so it cannot fix the total.

1. Persons taking tea or coffee = 75 + 60 − 15 = 120.
2. Everyone else takes milk only (milk drinkers never take tea). Let m = number taking milk only; total = 120 + m.
3. Statement 1: 50 took milk, but some of them may also have taken coffee, so m is not fixed — not sufficient.
4. Statement 2: 120 + m = 5m gives m = 30, so total = 150 — sufficient.
5. So one Statement alone answers the Question and the other does not.

Remember · In set-based sufficiency, write the total as 'known union + unknown only-group' and see which statement pins down that group.

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Answer: (d) The Question cannot be answered even by using both the Statements together.

Digits whose sum equals their product can only be 1, 2 and 3, so Statement 1 leaves six numbers. Adding Statement 2, the number must be divisible by 6, but both 132 and 312 qualify, so the number still cannot be fixed.

1. Statement 1: for the digits of a 3-digit number, sum = product only for 1, 2, 3 (1 + 2 + 3 = 6 = 1 × 2 × 3).
2. So the number is one of 123, 132, 213, 231, 312, 321 — not unique.
3. Statement 2 alone: many numbers are divisible by their digit sum (for example 100, 102, 108) — not sufficient.
4. Together: the digit sum is 6, so the number must be even and divisible by 3; both 132 and 312 work.
5. Two answers remain, so the Question cannot be answered even with both Statements.

Remember · Data sufficiency needs one unique answer; two cases surviving both statements prove insufficiency.

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Answer: (b) The Question can be answered by using either Statement alone.

The ages are a, a + 2, a + 4, a + 6 and a + 8. Statement 1 gives a + 8 = 3a, so a = 4; Statement 2 gives an average (the middle age) of a + 4 = 8, so a = 4. Each Statement alone answers the Question.

1. Ages: a, a + 2, a + 4, a + 6, a + 8.
2. Statement 1: eldest = 3 × youngest → a + 8 = 3a → a = 4. Sufficient.
3. Statement 2: the average of five equally spaced ages is the middle age: a + 4 = 8 → a = 4. Sufficient.
4. So either Statement alone is enough.

Remember · For equally spaced values, the average equals the middle term — a quick route to the unknown.

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Answer: (d) In our lives, having too few choices may not be a good thing, but having too many can be equally as difficult.

Buridan's ass starves between two equally good choices, and modern technology only adds to our glut of choices, leaving people overwhelmed, unhappy and anxious. The implied message is the paradox of choice: too many choices can be as troublesome as too few.

• ✗ (a) Technology is said to worsen our glut of choices, not to weaken the structure of society.
• ✗ (b) The passage is about the burden of too many attractive choices, not a life of endless uncertainty and difficulty in general.
• ✗ (c) Other people's opinions and lack of courage are not discussed.
• ✓ (d) People 'overwhelmed with attractive life choices, yet ... unhappy and anxious' show that an excess of choice brings its own difficulty — the paradox of choice.

Remember · When a passage opens with a named idea (here, the paradox of choice), the right option usually restates that idea.

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Answer: (d) 1, 2 and 3

The passage lists the hurdles to financialization: onerous bureaucracy, scepticism of formal institutions, lack of information about which of many services suits each family, and difficulty in switching between them. Each suggested solution answers one of these hurdles, so all three are implied.

• ✓ 1. 'Onerous bureaucracy' is a named hurdle; a flexible, less rigid environment is the implied answer.
• ✓ 2. People lack information on which service 'is best for each family', so solutions must be suited to each household.
• ✓ 3. Financial technology can give easy information, simpler access and smoother switching between providers — the remaining hurdles.

Remember · For implied solutions, match each hurdle the passage lists to a remedy; a remedy that answers a named hurdle is implied.

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Answer: (b) 1 and 4

Patents let firms set high prices that may be unaffordable, justified as a return on innovation that benefits the public later. This assumes a real burden on buyers of patented medicines and a need for governments to balance patentees' rights with patients' needs.

• ✓ 1. Prices during the monopoly 'could be unaffordable to the public' and the benefit 'comes at a cost ... of higher prices' — a burden on buyers.
• ✗ 2. Dependence on other countries for medicines is not discussed.
• ✗ 3. How countries design public health policy is outside the passage, which only describes the patent trade-off.
• ✓ 4. The patentee's 'legitimate mechanism to get returns' stands against prices the public may not afford; resolving this tension calls for a balance.

Remember · Assumptions come from tensions inside the passage; reject statements about topics (imports, policy design) it never touches.

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Answer: (c) Both 1 and 2

The passage ties the protection of personal data to growth of the digital economy and says no one will innovate where personal information is compromised. So privacy has economic value and is closely linked to innovation; both assumptions hold.

• ✓ 1. India should grow the digital economy 'while keeping personal data ... secure', and individuals may 'monetise' their data — privacy carries economic value.
• ✓ 2. 'No one will innovate in a surveillance-oriented environment', and laws should enable 'privacy protecting' innovation — a direct link.

Remember · If the argument rests on a chain of links (privacy → innovation → digital economy), each link is a valid assumption.

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Answer: (d) 35

99% of 400 is 396, so the student loses exactly 4 marks in all across the four papers. Sharing 4 lost marks among 4 papers (0 or more each) can be done in C(7, 3) = 35 ways.

1. Total maximum = 4 × 100 = 400; 99% of 400 = 396.
2. Let the marks lost in P, Q, R, S be w, x, y, z (each 0 or more) with w + x + y + z = 4; no paper can exceed 100.
3. Number of solutions = C(4 + 3, 3) = C(7, 3) = 35.
4. Check by cases: (4,0,0,0) type 4 ways, (3,1,0,0) 12, (2,2,0,0) 6, (2,1,1,0) 12, (1,1,1,1) 1 → 35.

Remember · Count marks lost instead of marks scored; n identical units in r boxes can be shared in C(n + r − 1, r − 1) ways.

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Answer: (c) 24

The top stripe can take any of 3 colours, and each later stripe any colour except the one just above it (2 choices). So there are 3 × 2 × 2 × 2 = 24 flags.

1. Stripe 1: 3 choices.
2. Stripes 2, 3 and 4: each must differ from the stripe above it, so 2 choices each.
3. Total = 3 × 2 × 2 × 2 = 24.
4. 'Some or all of the colours' allows two-colour flags such as red-green-red-green; they are already included.

Remember · For 'no two adjacent alike', choose the first freely, then (colours − 1) for each next item.

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Answer: (d) 9

Any line across the 220 cm breadth meets tile sides of 60 cm or 140 cm, which can cover at most 200 cm of it. So at most 400 × 200 = 80,000 cm² can be tiled, room for fewer than 10 tiles of 8,400 cm², and a staggered layout fits 9.

1. Floor = 400 cm × 220 cm; one tile = 140 cm × 60 cm = 8,400 cm².
2. Across the 220 cm breadth, tile sides add up in 60s and 140s: the best totals are 60 + 140 = 200 and 60 + 60 + 60 = 180. So every line across the breadth leaves at least 20 cm uncovered.
3. Covered area ≤ 400 × 200 = 80,000 cm², so tiles ≤ 80,000 ÷ 8,400 ≈ 9.5, i.e. at most 9.
4. A layout of 9 (length measured 0–400, breadth 0–220): three tiles lying lengthwise, stacked, at length 0–140 (breadth 0–180); one lengthwise tile at length 140–280, breadth 0–60; two standing tiles (60 along the length) at length 140–260, breadth 60–200; two standing tiles at length 280–400, breadth 0–140; one lengthwise tile at length 260–400, breadth 140–200.
5. 3 + 1 + 2 + 2 + 1 = 9 tiles, none overlapping — the maximum.

Remember · For tiling maxima, check a 'line' bound as well as area: the lengths that fit across a side often beat the area estimate.

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Answer: (d) 24

Task-2 goes to R or S (2 ways). Task-1 cannot go to P or Q or to whoever took Task-2, so it goes to the other of R and S or to T (2 ways). The remaining three tasks go to the remaining three persons in 3! = 6 ways: 2 × 2 × 6 = 24.

1. Five persons, five tasks, one task each.
2. Task-2: R or S → 2 ways.
3. Task-1: not P, not Q, and not the person given Task-2 → 2 ways (the other of R and S, or T).
4. Tasks 3, 4 and 5 go to the 3 persons left: 3! = 6 ways.
5. Total = 2 × 2 × 6 = 24.

Remember · Fill the most restricted positions first, then arrange the rest freely.

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Answer: (c) Both 1 and 2

Making 78 gm as a sum of these coins needs at least 7 coins, for example 50 + 10 + 10 + 2 + 2 + 2 + 2. But on a two-pan balance coins can go on both pans: 50 + 25 + 5 against the object plus a 2 gm coin weighs 78 gm with just 4 coins.

1. Statement 1 (buying 78 gm of coins): the coin weights must add up to exactly 78.
2. 50 + 25 = 75 leaves 3, which 2 and 5 cannot make. 50 + 10 + 10 = 70 leaves 8 = 2 + 2 + 2 + 2, giving 7 coins; checking all combinations, none with 6 or fewer coins makes 78.
3. So at least 7 coins must be bought — correct.
4. Statement 2 (weighing 78 gm): put 50 + 25 + 5 = 80 gm on one pan and the object with a 2 gm coin on the other; the object weighs 80 − 2 = 78 gm.
5. That uses 4 coins, fewer than 7 — correct.

• ✓ 1. The fewest coins adding up to 78 gm is 7 (50 + 10 + 10 + 2 + 2 + 2 + 2).
• ✓ 2. Using both pans of a balance, 50 + 25 + 5 on one side and 2 with the object on the other weighs 78 gm with 4 coins.

Remember · Buying uses sums of coin values only; weighing on a balance allows coins on both pans, so differences count too.

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Answer: (b) Only Conclusion-2 follows from the Statement.

Decode the symbols: + is >, − is ≤, × is ≥, ÷ is < and ± is =. The Statement gives Q ≤ P ≤ T < R = S, so S > Q follows, but Q = T does not, since Q may be smaller than T.

1. Meanings: A + B → A > B; A − B → A ≤ B; A × B → A ≥ B; A ÷ B → A < B; A ± B → A = B.
2. Statement: P × Q → P ≥ Q; P − T → P ≤ T; T ÷ R → T < R; R ± S → R = S.
3. Chain: Q ≤ P ≤ T < R = S.
4. Conclusion-1 (Q ± T, i.e. Q = T): we only know Q ≤ T — does not follow.
5. Conclusion-2 (S + Q, i.e. S > Q): S = R > T ≥ P ≥ Q — follows.

Remember · Translate coded symbols into >, ≥, <, ≤, = first, build one chain, then test each conclusion against it.

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Answer: (b) Unemployment must be seen as a function of rising education and aspirations of young Indians.

The author says the unemployment rate misleads when enrolment is rising: educated young Indians wait for the right job, and rising education levels inflate unemployment. So unemployment has to be read against rising education and aspirations, not only as a sign of a weak economy.

• ✗ (a) The passage says nothing about the quality of education; it only notes rising enrolment.
• ✓ (b) The rate 'paints an inaccurate picture' because the young 'wait for the right job' and 'rising education levels inflate unemployment challenges' — unemployment tracks education and aspirations.
• ✗ (c) Labour-intensive industries are not mentioned; this brings in an outside cause.
• ✗ (d) The passage diagnoses how to read the unemployment rate; it offers no remedy such as redesigning education for self-employment.

Remember · For 'what the author intends to say', pick the option that restates the central claim; reject causes or remedies the passage never mentions.

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Answer: (b) It is not desirable to have Governments managed by empirical statesmen unless well mixed with others who are grounded in learning and reflect wisdom.

The passage says government, like science, suffers for lack of philosophy, and that politics divorced from science and philosophy becomes a 'destructive bedlam'. The practical lesson is that government should not be left to merely empirical statesmen; they must work alongside people grounded in learning and wisdom.

• ✗ (a) Tempting, but the passage does not ask every statesman to be trained in scientific method; it asks that politics be guided by knowledge and perspective — a narrower, less practical prescription.
• ✓ (b) 'Government suffers ... for lack of philosophy' and politics becomes 'bedlam when divorced from science and philosophy' — so rule by the merely empirical needs the balance of the learned and wise.
• ✗ (c) An education system training all citizens from an early age goes far beyond anything the passage discusses.
• ✗ (d) 'All scientists' must be philosophers is extreme; the passage wants philosophy as a coordinating discipline 'outside the sciences', and its real concern is government.

Remember · When asked for the 'most practical' message, prefer the moderate, workable conclusion; drop options that demand training for everyone or use 'all'.

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Answer: (d) Liberty

The passage says the end of the state is to free each man from fear, to let bodies and minds function safely, and to lead men to exercise a free reason. Every one of these aims is a form of freedom, so liberty is the term that covers them all.

• ✗ (a) Security 'without injury' is part of the aim, but it serves the larger end of freeing each man from fear — too narrow.
• ✗ (b) Bodies and minds functioning 'safely' is a means; the passage never names health as the end.
• ✗ (c) Not acting 'unfairly toward one another' is a result of free reason, not the ultimate goal.
• ✓ (d) 'To free each man from fear', to 'live by, and to exercise, a free reason' — the state's end is liberty.

Remember · For 'ultimate goal' questions, choose the single term that covers every aim the passage lists; partial aims are traps.

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Answer: (d) 4

Divided by 6, the powers of 2 leave remainders 2, 4, 2, 4 … in turn: odd powers leave 2 and even powers leave 4. Since 192 is even, the remainder is 4.

1. 2¹ = 2 leaves 2; 2² = 4 leaves 4; 2³ = 8 leaves 2; 2⁴ = 16 leaves 4.
2. The remainders repeat in a cycle of 2: odd powers leave 2, even powers leave 4.
3. 192 is even, so 2¹⁹² leaves remainder 4.
4. Check: 2¹⁹² = 4⁹⁶ leaves 1 on division by 3, and it is even; of 2 and 4, only 4 is even and leaves 1 on division by 3.

Remember · For remainders of large powers, list the first few remainders, spot the cycle, then place the exponent in it.

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Answer: (d) DDCA

With the blanks filled as D, D, C, A the series splits into groups of five: ABCDD, ABCCD, ABBCD, AABCD. Each group is ABCD with one letter doubled, and the doubled letter moves back from D to C to B to A.

1. There are 16 printed letters and 4 blanks: 20 letters, which suggests four groups of five.
2. The unbroken stretch DABBCD contains ABBCD — the block ABCD with B doubled.
3. So the groups should be ABCD with one letter doubled: ABCDD, ABCCD, ABBCD, AABCD (D, C, B, A doubled in turn).
4. Fill the blanks to match: ABC(D)(D) ABC(C) DABBCD(A) ABCD, i.e. D, D, C, A.
5. Check: ABCDDABCCDABBCDAABCD splits exactly into ABCDD | ABCCD | ABBCD | AABCD.

Remember · In letter series, count the total letters and try equal groups (4, 5, 6); the stretch without blanks usually shows the rule.

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Answer: (a) 6

The sum D08 + GHI = 975 fixes I = 7, H = 6 and D + G = 9. The only product of two 2-digit numbers of the form D08 with CD ending in D and all digits distinct is 12 × 34 = 408, so A + B + C = 1 + 2 + 3 = 6.

1. DEF = D08. Units: 8 + I = 15, so I = 7 (carry 1). Tens: 0 + H + 1 = 7, so H = 6. Hundreds: D + G = 9.
2. AB × CD = D08, where CD ends in the digit D. Test D = 1 to 8.
3. 108: no pair of 2-digit factors. 208 = 13 × 16 and 308 = 11 × 28 = 14 × 22: no factor ends in 2 or 3 as needed.
4. 408 = 12 × 34 or 17 × 24. With CD = 34: A = 1, B = 2, C = 3, D = 4, G = 5 — all nine digits 1, 2, 3, 4, 0, 8, 5, 6, 7 are distinct. With CD = 24, AB = 17 gives B = 7 = I, a clash.
5. 508, 708 give no valid pair; 608 = 38 × 16 gives B = 8 = F, a clash; D = 8 clashes with F; D = 9 gives G = 0 = E.
6. So A + B + C = 1 + 2 + 3 = 6.
7. Check: 12 × 34 = 408 and 408 + 567 = 975.

Remember · In digit puzzles, settle the addition column by column first; it fixes most letters and leaves only a few cases to test.

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Answer: (d) Neither 1 nor 2

Since 'at least one of R and S was selected' is false, neither R nor S was selected. That leaves only P, Q and T as possible selections — at most three people — so neither conclusion can be drawn.

1. The false statement means neither R nor S was selected.
2. So only P, Q and T could have been selected: at most three candidates.
3. Conclusion 1 (at least four selected) is impossible.
4. Conclusion 2 (S selected) contradicts the false statement.
5. Check: the third statement (at most two of R, S, T) holds, since only T can be among them.

• ✗ 1. With R and S both out, at most P, Q and T can be selected — never four.
• ✗ 2. The false statement tells us S was not selected.

Remember · The negation of 'at least one of X and Y' is 'neither X nor Y'. Negate the false statement first; it usually settles the case.

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Answer: (a) 1 only

If P and R were both true, P would make S true, and R with S would make T false. So whenever T is true, P or R must be false. 'If Q is true, then P is true' merely reverses statement I, which never follows.

1. Test Conclusion 1: suppose T is true while P and R are both true.
2. By I, P true makes Q and S true; so R and S are both true.
3. By II, R and S true makes T false — contradicting T true. So Conclusion 1 follows.
4. Test Conclusion 2: take Q true, P false (S, R, T anything consistent). Statement I is not violated, since it says nothing when P is false.
5. So Q can be true without P; Conclusion 2 does not follow.

• ✓ 1. P and R both true would force S true and then T false; so with T true, P or R must be false.
• ✗ 2. It is the converse of statement I; Q can be true while P is false.

Remember · 'If A then B' gives 'if not B then not A', never 'if B then A'. Test a conclusion by assuming it fails.

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Answer: (a) 1 only

The unpainted cubes form an inner block of 5 × 3 × 1 = 15. Green (5 × 3) and blue (7 × 3) faces meet only along the four edges of length 3 cm, and each such edge gives 3 − 2 = 1 cube with exactly those two faces painted — 4 cubes, not 6.

1. No paint: remove one layer from each side: (7 − 2) × (5 − 2) × (3 − 2) = 5 × 3 × 1 = 15. Statement 1 is correct.
2. Green faces are the 5 × 3 ends (across the 7 cm length); blue faces are the 7 × 3 sides (across the 5 cm width).
3. A green face meets a blue face along an edge 3 cm long; a cuboid has 4 such edges.
4. Each 3 cm edge has 3 cubes; the 2 end ones are corners with three painted faces, leaving 1 cube per edge.
5. Blue-and-green-only cubes = 4 × 1 = 4, not 6. Statement 2 is incorrect.
6. Check: two-face cubes in all = 4(7 − 2) + 4(5 − 2) + 4(3 − 2) = 20 + 12 + 4 = 36, and the green–blue share is the 4 on the 3 cm edges.

• ✓ 1. Inner block (7 − 2)(5 − 2)(3 − 2) = 15 cubes.
• ✗ 2. Green–blue edges are the four 3 cm edges, each giving one such cube: 4 in all.

Remember · Painted cuboid a × b × c: unpainted = (a − 2)(b − 2)(c − 2); cubes on an edge with two colours = edge length − 2.

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Answer: (c) Two

In reverse alphabetical order the 21 letters read TSSRPONNMLIIIIIHEEECB. The five I's fill positions 11 to 15, and the original word has I at positions 13 and 15, so exactly two positions stay the same.

1. Letters (21): I × 5, E × 3, N × 2, S × 2 and one each of B, C, H, L, M, O, P, R, T.
2. Reverse alphabetical order: T S S R P O N N M L I I I I I H E E E C B.
3. The I's now occupy positions 11 to 15.
4. In INCOMPREHENSIBILITIES, I stands at positions 1, 13, 15, 17 and 19 — so positions 13 and 15 match.
5. Check the rest one by one (1: I vs T, 11: N vs I, 16: L vs H, 20: E vs C …): no other match. Two positions stay unchanged.

Remember · For 'unchanged positions', write the new order with position numbers; check the block of the most repeated letter first — matches usually hide there.

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