CSAT (GS Paper II) 2022

Answer: (a) केवल 1 और 2

The passage says current production models rely heavily on fossil fuels and that this is unsustainable because the resources are finite. Recycling appears only as a remedy some developed countries are using, never as a cause.

• ✓ 1. “Current production models rely heavily on fossil fuels. We now know that this is unsustainable…” — heavy dependence on fossil fuels is named as the problem.
• ✓ 2. The reason given is “because the resources are finite” — that is, resources are available only in limited amounts.
• ✗ 3. Expanding recycling is how some developed countries “alleviate the worst effects”. It eases the problem; it does not cause it.

Remember · Separate cause from remedy: whatever a passage lists as a fix for a problem cannot also be its cause.

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Answer: (b) केवल 1 और 2

The passage names what holds developing countries back — “the high costs and low availability of clean energy sources” — and cites public transport and infrastructure as steps that ease the damage. Real support therefore means removing these material barriers; merely encouraging a change of habits does not.

• ✓ 1. It answers the exact hurdle the passage names: developing countries are “hampered by the high costs and low availability of clean energy sources”.
• ✓ 2. The passage cites “investing in public transport and infrastructure” as a way to alleviate the worst effects; cheap loans let developing countries, held back by cost, do the same.
• ✗ 3. The passage does not say developing countries lack the will to change — it says they are hampered by cost and availability. Encouragement does not touch that barrier, so it is not the support the passage calls for.

Remember · In ‘how can X help Y’ items, match the help to the obstacle the passage names; exhortation is rarely the support meant.

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Answer: (c) केवल 1 और 3

The passage predicts “soil erosion by wind and water” and describes land left “without any water in the streams and springs except in the rainy season”, so soil and water are both lost. It says nothing about land becoming scarce for people.

• ✓ 1. “Soil erosion by wind and water will increase”, leaving “brown earth, bare of vegetation” — the soil resource is depleted.
• ✗ 2. The land does not disappear; it degrades so that agriculture becomes impossible. A shortage of land for the common man is not mentioned.
• ✓ 3. Former perennial streams and springs now have no water “except in the rainy season”, and agriculture “will gradually become impossible”.

Remember · Pick only the consequences the passage actually describes; a plausible real-world effect missing from the text is out.

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Answer: (d) 225

Each term is the previous one multiplied by a factor that grows by 0.5 each time: ×0.5, ×1, ×1.5, ×2, ×2.5. The next factor is ×3, so X = 75 × 3 = 225.

1. Divide each term by the one before it: 10 ÷ 20 = 0.5, 10 ÷ 10 = 1, 15 ÷ 10 = 1.5, 30 ÷ 15 = 2, 75 ÷ 30 = 2.5.
2. The multiplier rises by 0.5 each time, so the next multiplier is 3.
3. X = 75 × 3 = 225.
4. Check: 225 ÷ 75 = 3 continues the pattern 0.5, 1, 1.5, 2, 2.5, 3.

Remember · When differences look irregular, try the ratio of consecutive terms; a steadily growing multiplier is a common pattern.

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Answer: (a) 8

Work through the conditions: G must be 9, E must be 5, the even digits 2, 4, 8 fill B, D, F, and the tests for 3 and 4 fix B = 4 and D = 2. The middle digits C, D, E then add up to 8 (or 14, which is not offered).

1. The digits 1, 2, 4, 5, 7, 8, 9 add up to 36, so the full number is divisible by 9 in any order.
2. ABCDEF is divisible by 6, so F is even and A + B + C + D + E + F is a multiple of 3. That sum is 36 − G, so G is a multiple of 3: G = 9.
3. ABCDE is divisible by 5 and 0 is not available, so E = 5.
4. AB divisible by 2 makes B even; ABCD divisible by 4 makes D even. So B, D, F are the even digits 2, 4, 8, and A, C are 1 and 7 in some order.
5. ABC divisible by 3: A + B + C = 1 + 7 + B = 8 + B must be a multiple of 3, so B = 4.
6. ABCD divisible by 4: the last two digits CD must be 12 or 72 (18 and 78 fail), so D = 2 and F = 8.
7. Middle three digits C + D + E = C + 2 + 5: this is 8 when C = 1 and 14 when C = 7. Only 8 is offered.
8. Check: 7412589 — 741258 ÷ 6 = 123543, 74125 ends in 5, 7412 ÷ 4 = 1853, 741 ÷ 3 = 247, 74 is even.

Remember · Divisibility chains: fix the digit forced by 5, then the even places, then use digit-sum and last-two-digit tests.

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Answer: (a) 100 m

After the first turn the two run in opposite directions, so they end up 60 + 40 = 100 m apart on one line. The next turns send both of them 50 m in the same direction, which does not change the gap.

1. Take the common first run as 50 m north, ending at a point O.
2. X turns right (east) and runs 60 m; Y turns left (west) and runs 40 m. They are now 60 + 40 = 100 m apart on an east–west line.
3. X, facing east, turns left and so runs north 50 m; Y, facing west, turns right and so also runs north 50 m.
4. Both moved 50 m north, so they are still 100 m apart.

Remember · Equal moves in the same direction never change the distance between two people; only the unequal or opposite legs count.

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Answer: (d) 7

Count odd days from the end of 2000: 31 December 2098 falls on a Wednesday, and January to May 2099 add 4 more odd days, so 31 May 2099 is a Sunday. One week later, 7 June 2099, is also a Sunday.

1. 2000 years contain 0 odd days, so 31 December 2000 was a Sunday.
2. From 2001 to 2098 there are 98 years, of which 24 are leap years (2004, 2008, …, 2096). Odd days = 98 + 24 = 122, and 122 ÷ 7 leaves 3.
3. So 31 December 2098 is Sunday + 3 = Wednesday.
4. 1 January to 31 May 2099 (2099 is not a leap year): 31 + 28 + 31 + 30 + 31 = 151 days; 151 ÷ 7 leaves 4.
5. 31 May 2099 is Wednesday + 4 = Sunday, so 1 June 2099 is a Monday and 7 June 2099 is a Sunday.
6. Check: 5 June 2022 (the day of this exam) was a Sunday. Up to 5 June 2099 there are 77 years and 19 leap days: 96 days, which leaves 5 on division by 7. So 5 June 2099 is a Friday and 7 June a Sunday.

Remember · Calendar items: 400 years give 0 odd days; add ordinary years plus leap years, reduce mod 7, then add month days.

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Answer: (d) 1, 2 और 3

Let a, b, c be the numbers of ₹50, ₹20 and ₹10 notes. The two totals reduce to 4a + b = 134, and none of the three statements gives whole, non-negative numbers of notes, so all three are incorrect.

1. Let the numbers of ₹50, ₹20 and ₹10 notes be a, b and c. Then a + b + c = 50 and 50a + 20b + 10c = 1840, that is 5a + 2b + c = 184.
2. Subtract the first equation from the second: 4a + b = 134.
3. Statement 1: a = 25 gives b = 134 − 100 = 34, and then c = 50 − 25 − 34 = −9. Impossible.
4. Statement 2: b = 35 gives 4a = 99, which is not a whole number. Impossible.
5. Statement 3: c = 20 gives a + b = 30; with 4a + b = 134 this means 3a = 104, not a whole number. Impossible.
6. Check: a valid payment does exist, e.g. a = 30, b = 14, c = 6 — that is 50 notes and 1500 + 280 + 60 = ₹1,840 — so the bill itself is fine; only the three statements fail.

• ✗ 1. a = 25 forces c = −9, a negative number of ₹10 notes.
• ✗ 2. b = 35 forces 4a = 99, so a is not a whole number.
• ✗ 3. c = 20 forces 3a = 104, so a is not a whole number.

Remember · With a count total and a value total, subtract to get one clean equation, then test each statement for whole, non-negative answers.

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Answer: (b) 3²¹

4¹⁸ and 8¹² both equal 2³⁶, which is smaller than 2⁴⁰. And 3²¹ = 2187³ is smaller than 2³⁶ = 4096³, so 3²¹ is the smallest of the four.

1. Write the powers of 2 in base 2: 4¹⁸ = 2³⁶ and 8¹² = 2³⁶. Among 2⁴⁰, 4¹⁸ and 8¹² the smallest is 2³⁶.
2. Compare 3²¹ with 2³⁶ using the common exponent 3: 3²¹ = (3⁷)³ = 2187³ and 2³⁶ = (2¹²)³ = 4096³.
3. 2187 < 4096, so 3²¹ < 2³⁶ < 2⁴⁰.
4. Check: 21 × log 3 ≈ 10.02 while 36 × log 2 ≈ 10.84, so 3²¹ (about 1.05 × 10¹⁰) is the smallest.

Remember · To compare powers, rewrite them with a common base or a common exponent, then compare only what differs.

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Answer: (c) 2

The first-row number N must keep 3N below 1000, so it starts with 2 or 3 and uses three of 2, 3, 7, 9. Testing the eight such numbers leaves only 273 (546, 819) and 327 (654, 981).

1. The third row is 3N, a 3-digit number, so N ≤ 333. With digits from 2, 3, 7, 9 (no repeats), N must start with 2 or 3.
2. Candidates: 237, 239, 273, 279, 293, 297, 327 and 329.
3. 237 → 474 (repeats 4); 239 → 478, 717 (repeats 7); 279 → 558 (repeats 5); 293 → 586, 879 (repeats 8 and 9); 297 → 594 (repeats 9); 329 → 658, 987 (repeats 8).
4. 273 → 546 → 819 uses each of 1–9 once; 327 → 654 → 981 also uses each of 1–9 once.
5. So 2 arrangements are possible.
6. Check: 273 × 2 = 546 and 273 × 3 = 819; 327 × 2 = 654 and 327 × 3 = 981.

Remember · Bound the search first (here 3N < 1000, so N ≤ 333); a short, systematic list beats guessing.

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Answer: (c) हमें कोई ऐसी पद्धति सोच निकालने की जरूरत है, जिससे सार्वजनिक पद से अयोग्यता को दूर रखा जा सके।

The author contrasts our insistence on trained shoe-makers and physicians with our acceptance of anyone who can win votes as fit to run the State. The closing question asks for “the wisest and the best”, so the message is that incompetence must be kept out of public office.

• ✗ (a) This is the presumption the author attacks (“we presume that everyone who knows how to get votes knows how to administer a State”). It describes the problem, not the message.
• ✗ (b) Too narrow and specific. The author asks for “the wisest and the best”; the physician analogy is about proven competence, not a rule that politicians must come from administrative training.
• ✓ (c) “Should we not look for the service and guidance of the wisest and the best?” implies that we must find a way to keep the incompetent from governing.
• ✗ (d) Reverses the author’s view — the passage questions exactly the idea that winning votes proves fitness to govern.

Remember · In ‘message of the author’ items, the closing rhetorical question usually carries the message; options restating the criticised view are traps.

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Answer: (d) यह गरीबी को अपर्याप्त आय या अपर्याप्त क्रयशक्ति के रूप में प्रस्तुत करने पर आधारित है।

The passage’s criticism of how the poor are counted is that it “exclusively relies on the notion of poverty as insufficient income or insufficient purchasing power”. Of the four options, only this idea comes from the passage; the others bring in reasons it never mentions.

• ✗ (a) The passage calls the definition “extremely narrow”, not confused; no confusion about what the line should be is mentioned.
• ✗ (b) Rural–urban differences are not discussed anywhere in the passage.
• ✗ (c) A global standard is never mentioned; this is outside the passage.
• ✓ (d) The counting rests on “the notion of poverty as insufficient income or insufficient purchasing power” — that is, meagre income or buying capacity.

Remember · When every other option imports facts the passage never states, choose the one that paraphrases the passage’s own words.

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Answer: (a) यह अन्य सभी की उपेक्षा कर केवल एक प्रकार के वंचन की बात करती है।

If poverty is about all the deprivations that affect well-being, income poverty captures only one of them: “income poverty is only one aspect of it”. Other deprivations, such as the lack of real opportunity, are left out.

• ✓ (a) “If poverty is ultimately about deprivations affecting human well-being, then income poverty is only one aspect of it” — one kind of deprivation, ignoring others such as lack of real opportunity.
• ✗ (b) Too extreme. The passage links low incomes to “curtailing capabilities”; it never says other deprivations are unrelated to purchasing power.
• ✗ (c) Whether income poverty is temporary or permanent is not discussed.
• ✗ (d) “Only at a point of time” is not in the passage; the author’s point is about the narrow range of what is counted, not about timing.

Remember · Distrust absolute words like ‘nothing to do with’; passages rarely support such extremes.

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Answer: (a) मानव-जीवन में वे सभी वंचनाएँ जो केवल आय की कमी से ही नहीं, बल्कि वास्तविक अवसरों की कमी से उत्पन्न होती हैं

The author says poverty of a life lies “not merely in the impoverished state in which the person actually lives, but also in the lack of real opportunity” to choose other ways of living. That means deprivation from lack of income and from lack of real opportunities together.

• ✓ (a) Matches both halves of the definition: the impoverished state and “the lack of real opportunity given by social constraints as well as personal circumstances”.
• ✗ (b) Covers only the first half (the impoverished state), and the rural–urban angle is not in the passage.
• ✗ (c) Only a fragment: personal circumstances are one source of lost opportunity, and the impoverished state is left out.
• ✗ (d) “Permanently” is added; the passage speaks of restricting choices, never of a permanent restriction.

Remember · For ‘what does the author mean by’ items, find the defining sentence and pick the option covering all its parts — nothing added.

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Answer: (b) 3

By the time X finishes 3 km, Y has run 2 km, so X has gained 1000 m — three full laps of 300 m and a little more. Each full lap gained is one pass, so X passes Y 3 times.

1. The race ends when X, the faster runner, completes 3 km = 3000 m.
2. Speeds are 3 : 2, so in that time Y runs 2/3 × 3000 = 2000 m.
3. X gains 3000 − 2000 = 1000 m on Y.
4. X passes Y each time the lead reaches a whole lap: at 300 m, 600 m and 900 m. 1000 ÷ 300 = 3.33, so there are 3 passes.

Remember · On a circular track, passes = lead gained by the faster runner ÷ track length, counting whole laps only.

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Answer: (d) OFXPMLD

Reversing the alphabet swaps each letter with its mirror, whose position adds up to 27 with its own: L→O, U→F, C→X, K→P, N→M, O→L, W→D, giving OFXPMLD.

1. In the reversed alphabet, the letter at position p becomes the letter at position 27 − p (A ↔ Z, B ↔ Y, …).
2. L (12) → O (15); U (21) → F (6); C (3) → X (24); K (11) → P (16).
3. N (14) → M (13); O (15) → L (12); W (23) → D (4).
4. LUCKNOW → OFXPMLD.
5. Check: the second letter separates the options — U (21) → F (6), which rules out (a) and (b); the last letter W (23) → D (4) rules out (c).

Remember · Opposite letters in the alphabet have positions adding to 27 (A–Z, B–Y, …).

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Answer: (c) 149

Every match has exactly one loser, and every loser is eliminated. The tournament ends when one player is left, so 149 players are eliminated — and that takes exactly 149 matches.

1. Each match produces exactly one loser, who is eliminated.
2. To leave a single winner, 150 − 1 = 149 players must be eliminated.
3. So 149 matches are played.
4. Check: with 4 players, 2 semi-finals + 1 final = 3 = 4 − 1 matches.

Remember · Knockout tournament with n players: always n − 1 matches, whatever the byes.

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Answer: (b) 12

A number divisible by 5 with an odd units digit must end in 5. The other two places are filled from 1, 3, 7, 9 without repetition: 4 × 3 = 12 numbers.

1. Divisible by 5 means the units digit is 0 or 5; it must be odd, so it is 5.
2. The hundreds digit can be any of the remaining odd digits 1, 3, 7, 9: 4 choices.
3. The tens digit can be any of the 3 odd digits still unused: 3 choices.
4. Total = 4 × 3 = 12.

Remember · Fill the restricted place first (here the units digit), then count the choices for the free places.

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Answer: (d) कथन–1 और कथन–2, दोनों, प्रश्न का उत्तर देने के लिए पर्याप्त नहीं हैं

x = 1 and x = 1/3 both satisfy both statements, but one is an integer and the other is not. So the statements cannot settle the question even when used together.

1. Statement–1 alone: x = 1 gives x/3 = 1/3 (not an integer) and x is an integer; x = 1/2 gives x/3 = 1/6 and x is not an integer. Not sufficient.
2. Statement–2 alone: x = 1 gives 3x = 3 and x = 1/3 gives 3x = 1; both are integers, yet only one x is an integer. Not sufficient.
3. Together: x = 1 fits both (1/3 is not an integer, 3 is) and so does x = 1/3 (1/9 is not an integer, 1 is). One is an integer, the other is not.
4. So even both statements together are not sufficient.

Remember · Data sufficiency: find two values that satisfy every statement yet give different answers — one such pair proves insufficiency.

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Answer: (b) 10%

A 25% rise followed by a 20% fall brings the price exactly back to where it started (1.25 × 0.8 = 1). Only the last 10% rise remains.

1. Let the price be ₹100. After a 25% increase it is ₹125.
2. A 20% decrease: 125 × 0.8 = ₹100.
3. A 10% increase: 100 × 1.1 = ₹110.
4. The resultant increase is ₹10 on ₹100, i.e. 10%.

Remember · Successive percentage changes multiply: +25% then −20% cancel, because 5/4 × 4/5 = 1.

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Answer: (d) न तो 1 और न ही 2

The passage says “much” of the change comes from affluent farmers in the West — not all of it — and that Western techniques “are being adapted in some places”. Both assumptions overstate or contradict the text.

• ✗ 1. “Much of this change is brought about by affluent farmers in the West/Americas” — much, not all. “Confined to” is too strong.
• ✗ 2. Contradicted: “Techniques developed in the West are being adapted in some places to make tropical crops more productive.”

Remember · Assumptions using ‘confined to’, ‘only’ or ‘never’ fail when the passage says ‘much’, ‘some’ or ‘in some places’.

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Answer: (a) केवल 1

The author says technology is of little use unless it is adapted, and that in the developing world this applies “as much to existing farming techniques” — so poor countries must change how they farm. The passage makes no comparison of infrastructure or food waste with developed countries.

• ✓ 1. Extending “the best of today’s agricultural practices”, such as how much fertilizer to apply and when, to smallholders in Africa and Asia would greatly raise food supply — the author takes it for granted that their present techniques need to change.
• ✗ 2. Better roads and storage are recommended for the developing world, but the passage never says developed countries have better infrastructure or waste less food.

Remember · An assumption must be something the author’s argument needs; a comparison the passage never draws is not one.

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Answer: (a) केवल 1

The passage opens with plateauing yields, little unfarmed land left and rising temperatures — the author assumes that feeding people in future will be hard. Corporate farming is never mentioned; the focus is on smallholders and subsistence farmers.

• ✓ 1. Yields have “reached a plateau”, there is not “much unfarmed land left”, and some places “will become unsuitable for farming” — the case for technology rests on food supply becoming a challenge.
• ✗ 2. The passage speaks of extending best practices to “smallholders and subsistence farmers”; corporate farming lies outside the passage.

Remember · Reject any assumption that brings in a new solution (here corporate farming) the author never discusses.

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Answer: (c) 24

In a row of five, A and E can stand only at places 1 and 4 or 2 and 5, in either order: 4 ways. The other three letters then fill the remaining places in 3! = 6 ways, giving 24.

1. Number the places 1 to 5. Exactly two letters between A and E means A and E occupy places (1, 4) or (2, 5): 2 choices of places.
2. A and E can swap within the chosen places: 2 × 2 = 4 ways.
3. B, C and D fill the other three places in 3! = 6 ways.
4. Total = 4 × 6 = 24.

Remember · Place the constrained pair first (positions × internal order), then arrange the remaining items freely.

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Answer: (d) कथन–1 और कथन–2, दोनों, प्रश्न का उत्तर देने के लिए पर्याप्त नहीं हैं

Both statements tell us X, Y and Z are siblings, and Statement–1 shows X and Y are male, but nothing tells us whether Z is male. Z could be X’s brother or sister, so even together the statements are not sufficient.

1. Statement–1: X is a brother of Y and Y is a brother of Z, so X and Y are male and all three are siblings. Z’s gender is not given.
2. Statement–2: X, Y and Z are siblings — again Z’s gender is not given.
3. Together they still do not say whether Z is male, so Z may be X’s brother or X’s sister.
4. Neither statement alone nor both together can answer the question.

Remember · ‘Is Z a brother?’ needs Z’s gender; statements that fix only other people’s gender cannot answer it.

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Answer: (b) 40·4 m

With plants at both ends of the road, 101 plants make 100 equal gaps along 1010 m, so each gap is 10.1 m. Five consecutive plants span 4 gaps: 4 × 10.1 = 40.4 m.

1. 101 plants in a line from one end of the road to the other leave 101 − 1 = 100 gaps.
2. Road length = 1.01 km = 1010 m, so each gap = 1010 ÷ 100 = 10.1 m.
3. 5 consecutive plants have 4 gaps between them: 4 × 10.1 = 40.4 m.
4. Trap: dividing 1010 by 101 (counting plants instead of gaps) gives 10 m and the wrong answer 40 m.

Remember · n objects in a line make n − 1 gaps; k consecutive objects span k − 1 gaps.

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Answer: (c) 15

One can go directly from A to C (3 ways) or go through B (3 × 4 = 12 ways). These are alternatives, so they add: 3 + 12 = 15.

1. Direct routes from A to C: 3.
2. Routes through B: 3 roads from A to B × 4 roads from B to C = 12.
3. Total = 3 + 12 = 15.

Remember · Legs taken one after another multiply; alternative routes add.

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Answer: (d) 52

Work backwards from D’s 10 coins. Each giver handed over half his coins plus 2, so he had 2 × (coins given − 2): C had 16, B had 28 and A had 52.

1. D now has 10 coins (the smallest two-digit number), all received from C.
2. C gave half his coins + 2 = 10, so half = 8 and C had 16 — all received from B.
3. B gave half + 2 = 16, so half = 14 and B had 28 — all received from A.
4. A gave half + 2 = 28, so half = 26 and A had 52.
5. Check: A (52) gives 26 + 2 = 28 to B; B gives 14 + 2 = 16 to C; C gives 8 + 2 = 10 to D.

Remember · For ‘gives half and k more’ chains, reverse each step: original = 2 × (amount given − k).

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Answer: (c) I

The series is built in blocks A | AB | ABC | ABCD | …, the nth block having n letters. The first 13 blocks use 91 places, so the 100th place is the 9th letter of the 14th block, which is I.

1. Split the series into blocks: A | AB | ABC | ABCD | ABCDE | … The nth block runs from A to the nth letter and has n letters.
2. Places used by the first n blocks = 1 + 2 + … + n = n(n + 1)/2. For n = 13 this is 13 × 14 ÷ 2 = 91.
3. So the 100th place is the 100 − 91 = 9th letter of the 14th block (A to N).
4. The 9th letter of the alphabet is I.
5. Check: the 14th block ends at place 91 + 14 = 105, so place 100 does fall inside it.

Remember · Growing-block series: find the largest n with n(n + 1)/2 below the position; the remainder names the letter.

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Answer: (a) 22

C is 12th from the front, so B is 4th or 20th, and A is then 9th, 15th or 25th. With 13 people behind A, the queue has A’s place + 13 people, which is smallest when A is 9th: 22.

1. 11 persons are ahead of C, so C is 12th from the front.
2. 7 persons between B and C means B is 8 places from C: 4th or 20th.
3. 4 persons between A and B means A is 5 places from B: if B is 4th, A is 9th; if B is 20th, A is 15th or 25th.
4. 13 persons are behind A, so the queue has (A’s place + 13) persons: 22, 28 or 38.
5. The minimum is 22, with B 4th, A 9th and C 12th.
6. Check: between B (4th) and A (9th) stand places 5–8, i.e. 4 persons; between B (4th) and C (12th) stand places 5–11, i.e. 7 persons; behind A (9th) stand places 10–22, i.e. 13 persons.

Remember · Queue items: ‘k persons between’ means a gap of k + 1 places; list the cases and take the minimum.

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Answer: (c) 1, 3 और 4

The passage says unprepared species go extinct, that this is due to genomes unable to withstand geological vagaries or biological mishaps, and that extinction “has been a major feature”. It never says that all existing species will go extinct.

• ✓ 1. “The set of existing organisms can never be fully prepared for environmental catastrophes… An outcome of this is the extinction of those species which cannot overcome environmental adversity.”
• ✗ 2. Too extreme: extinction is said to strike “those species which cannot overcome” adversity, not all existing species.
• ✓ 3. “This failure to survive… can be attributed to the genomes which are unable to withstand geological vagaries or biological mishaps.”
• ✓ 4. “Extinction of species has been a major feature”, and more than 90% of species that ever lived are now extinct.

Remember · Knock out the ‘all’ statement first: if the passage limits a claim to ‘those which…’, a universal claim fails.

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Answer: (b) विलोपन, पर्यावरणीय और जैविक हमलों की क्रिया है

“Therefore, the so-called Darwinian theory of evolution is not a theory at all” follows directly from the Darwinian explanation that organisms go extinct “as a function of environmental or biological assaults”. That explanation, being simple and convincing, is what makes the author treat evolution as fact rather than theory.

• ✗ (a) The author criticises creationist doctrines; failing to satisfy them is not offered as the reason.
• ✓ (b) The “Therefore” rests on the preceding sentence: organisms go extinct “as a function of environmental or biological assaults for which their inheritance deems them ill-equipped”.
• ✗ (c) That no evidence has refuted Darwin’s ideas comes later, as support for the mechanism of evolution; it is not the reason attached to “not a theory at all”.
• ✗ (d) Reverses the passage — attributing organisms to a creator is the creationist view the author rejects.

Remember · For ‘why does the author say X’, read the sentence just before the ‘Therefore’ — that is the stated reason.

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Answer: (a) केवल 1

The passage implies that only species able to overcome environmental adversity survive. The 90% figure is about species that ever lived and are now extinct, not present species at risk, and the passage never claims Darwin’s theory explains all natural phenomena.

• ✓ 1. Extinction is “the extinction of those species which cannot overcome environmental adversity”; so only those that can overcome it survive and carry on.
• ✗ 2. Misreads the figure: “more than 90% of species that have ever lived on the earth are now extinct” is about the past, not about present species being in danger.
• ✗ 3. Overreach: Darwinian laws are compared with the laws of Copernicus, Galileo, Newton and Einstein that “explain natural phenomena”; the passage does not say Darwin’s theory explains all of them.

Remember · Check what a statistic actually counts; wrong options often shift its time frame or the group it covers.

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Answer: (d) आने वाले वर्षों में भारत में जल प्रबंधन प्रणालियों में नाटकीय बदलाव लाने की आवश्यकता है।

Both changes the passage describes — protein-rich diets that need more water, and more gadgets and vehicles whose energy needs water to generate — end in rising demand for water. The crux is that India must manage water very differently.

• ✗ (a) The passage describes the move to protein-rich diets; it does not argue for persuading people back to cereal diets.
• ✗ (b) Farm productivity and energy capacity are not the focus; the passage points to water as the common need behind both trends.
• ✗ (c) The passage links affluence, not technology, to changing habits, and the change it stresses is the resulting demand for water.
• ✓ (d) Diets that “need significantly more water” and energy whose “generation… needs water” both point to a growing water demand that current practices must be changed to meet.

Remember · For ‘crux’ items, find the idea every sentence leads to — here each trend ends in ‘water’.

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Answer: (d) 694861x

Convert each unit to seconds: a week is 604800 s, a day 86400 s, an hour 3600 s, a minute 60 s. Adding these and the lone second gives 694861 seconds for every x.

1. 1 week = 7 × 24 × 3600 = 604800 s; 1 day = 24 × 3600 = 86400 s; 1 hour = 3600 s; 1 minute = 60 s.
2. Total = (604800 + 86400 + 3600 + 60 + 1)x = 694861x seconds.
3. Check: the x seconds make the coefficient end in 1, and one week alone is 604800x, so only 694861x fits.

Remember · Use the last digit as a filter: the extra ‘x seconds’ makes the total end in 1.

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UPSC dropped this question from evaluation in its final answer key.

UPSC dropped this item from evaluation. The idea is still worth learning: Q and R are P’s parents and P and U are T’s parents, so T is Q’s granddaughter, while Q’s and R’s genders are never stated.

1. R’s son is P and R is Q’s spouse, so Q and R are P’s parents — the first couple.
2. P’s daughter is T and U is T’s mother, so P (male, being R’s son) and U are T’s parents — the second couple. S is U’s other daughter.
3. The genders of Q and R are never given, so (a), (b) and (c) cannot be decided.
4. T is the daughter of P, who is Q’s son, so T is Q’s granddaughter — as in option (d).

Remember · Blood relations: draw the tree with gender marks; any option that needs an unstated gender cannot be definitely correct.

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Answer: (d) कथन–1 और कथन–2, दोनों, प्रश्न का उत्तर देने के लिए पर्याप्त नहीं हैं

Distances from R to P and to Q fix the distance PQ only if we know their directions from R. Without directions, PQ can be anything from 43 − 18 = 25 km to 43 + 18 = 61 km.

1. Statement–1 gives only QR = 18 km and Statement–2 gives only PR = 43 km; neither alone mentions PQ.
2. Together: if P and Q lie on the same side of R in a straight line, PQ = 43 − 18 = 25 km; on opposite sides, PQ = 43 + 18 = 61 km; in other directions, anything in between.
3. PQ is not fixed, so both statements together are still not sufficient.

Remember · Two distances from a common point give only a range (difference to sum) unless directions are known.

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Answer: (d) इन निष्कर्षों में से कोई भी अनुगमित नहीं होता

“All pens are books” and “No chair is a pen” say nothing definite about chairs and books. Conclusions I and IV are only possibilities, II contradicts Statement–2, and III is impossible, so none follows.

1. Draw the circle ‘pens’ inside ‘books’ (All pens are books) and ‘chairs’ outside ‘pens’ (No chair is a pen).
2. ‘Chairs’ may lie wholly inside ‘books’ (outside the pens), partly inside, or wholly outside — the statements do not decide.
3. So Conclusion–I and Conclusion–IV are possible but not certain; Conclusion–II contradicts Statement–2; Conclusion–III would make every pen a chair, which Statement–2 forbids.
4. None of the conclusions follows.

• ✗ I Possible (chairs may lie inside books, away from the pens) but not certain.
• ✗ II Directly contradicts Statement–2: no chair is a pen.
• ✗ III If all books were chairs, the pens (which are books) would be chairs — impossible under Statement–2.
• ✗ IV Possible (chairs may lie wholly outside books) but not certain.

Remember · Syllogisms: a conclusion follows only if it holds in every possible diagram, not just in one.

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Answer: (c) निष्कर्ष–I और निष्कर्ष–III दोनों

The doctors who are teachers are, by the chain of ‘all’ statements, also engineers and scientists. So “some engineers are doctors” and “some scientists are doctors” both follow; “all engineers are doctors” does not.

1. Some doctors are teachers, and all teachers are engineers, so those doctors are engineers: some engineers are doctors (III follows).
2. All engineers are scientists, so the same doctors are scientists: some scientists are doctors (I follows).
3. Nothing says every engineer is a doctor, so II does not follow.

• ✓ I The doctor-teachers are engineers and hence scientists, so some scientists are doctors.
• ✗ II Only some engineers (those who are teacher-doctors) are known to be doctors; ‘all’ goes beyond the statements.
• ✓ III The doctors who are teachers are engineers, so some engineers are doctors.

Remember · A ‘some’ link passes down a chain of ‘all’ links: some A are B and all B are C, so some A are C.

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Answer: (d) उपर्युक्त में से कोई नहीं

The conditions allow more than one seating. In one valid seating C faces G but E does not face F; in another E faces F but C does not face G; and B can never sit beside A. So none of (a), (b) and (c) is definitely correct.

1. A sits between E and D, so E, A, D occupy three adjacent seats; F sits between B and H, so B, F, H occupy three adjacent seats. C and G take the two seats left.
2. A’s two neighbours are E and D, so B can never sit next to A — (a) is never true.
3. Valid seating 1 (clockwise): A, D, G, B, F, H, C, E. B and D are neither next to nor opposite C. Here C is opposite G, but E is opposite B, not F.
4. Valid seating 2 (clockwise): A, D, H, F, B, G, C, E. Again B and D are neither next to nor opposite C. Here E is opposite F, but C is opposite H, not G.
5. (b) fails in seating 2 and (c) fails in seating 1, so none of the statements is definitely correct.

Remember · For ‘definitely correct’, build two valid arrangements; any option that fails in either one is out.

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Answer: (b) हमारे युग की सबसे बड़ी बुराई अत्यधिक कार्य-तनाव है

The passage traces everything — the stunted spiritual life, the craving for idle entertainment, the shallow culture — to “an excessive amount of labour”. The idea it rests on is that overwork, or overstrain, is the great evil of the age.

• ✗ (a) The author objects to excessive labour, not to hard work as such; “should not work hard” is a prescription the passage never makes.
• ✓ (b) “An excessive amount of labour is rule today in every circle of society, with the result that man’s spiritual element cannot thrive” — every other problem flows from this overstrain.
• ✗ (c) “He does not want to think; or he cannot even if he wants to” is a symptom of overwork, not the idea the passage is built on.
• ✗ (d) The spiritual element failing to thrive is a result of excessive labour; the option picks an effect, not the underlying idea.

Remember · For ‘based on the idea’ items, find the cause from which the other statements flow; effects and symptoms are traps.

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Answer: (d) उसे मनोरंजन प्रिय है और वह मानसिक रूप से खाली है

The passage answers this directly: “He seeks not self-improvement, but entertainment which would enable him to be mentally idle.” The overworked man picks amusement that lets his mind rest.

• ✗ (a) His ability is never questioned; the passage says he does not want to think, or cannot after excessive labour.
• ✗ (b) He has “little leisure”, not none; the issue is how he spends it.
• ✗ (c) Materialism is not mentioned; the distraction the passage names is entertainment.
• ✓ (d) He seeks “entertainment which would enable him to be mentally idle and to forget his usual activities”.

Remember · When the passage answers the ‘why’ in one sentence, match that sentence; do not infer a new cause.

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Answer: (b) केवल 2

By the passage’s own definition, the demographic dividend means “the relative ratio of very young and very old will, for a while, be on the decline”. Rising literacy and a stable population growth rate are not stated as automatic results.

• ✗ 1. The passage says the benefits depend on the quality of the workforce and calls for “bringing back the importance of education” — literacy does not rise automatically.
• ✓ 2. This is the definition itself: the ratio of very young and very old will “for a while, be on the decline”.
• ✗ 3. Stabilisation of the population growth rate is not mentioned at all.

Remember · ‘Invariably’ admits only what follows by definition; benefits the passage makes conditional are out.

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Answer: (d) न तो 1 और न ही 2

The passage says the demographic dividend can drive growth and that its benefits depend on education and skills. It never calls either an essential condition for rapid growth, and it does not single out higher education.

• ✗ 1. The dividend “can be a source of energy and an engine of economic growth” — a help, not a precondition; nothing says growth is impossible without it.
• ✗ 2. The passage names education, skills and human capital as what makes the dividend pay off; making higher education an essential condition for all rapid growth goes beyond it.

Remember · ‘Can help’ is not ‘essential condition’; reject inferences that turn a contributing factor into a necessity.

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Answer: (a) ₹2,600

Statements 2 and 3 give P = 13 and Q = 8; then X = 8, Z = 16 and Y = 20. The five bought 65 notebooks, which cost 65 × ₹40 = ₹2,600.

1. From 2 and 3: P + Q = 21 and P − Q = 5, so P = 13 and Q = 8.
2. From 5: X = 13 − 5 = 8. From 1: Z = 8 + 8 = 16. From 4: Y = 28 − 8 = 20.
3. Total notebooks = (P + Q) + (X + Y) + Z = 21 + 28 + 16 = 65.
4. Cost = 65 × ₹40 = ₹2,600.

Remember · Start with the pair of equations that closes on its own (sum and difference), then chain outwards.

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Answer: (d) 5 मिनट तेज है

Of the 90 minutes he was away by the home clock, 65 were spent driving. The return took 0.625 of the outward time, so the outward drive took 40 minutes and he reached the village at 15:10 home time — the village clock, showing 15:15, is 5 minutes fast.

1. By the home clock he was away from 14:30 to 16:00 = 90 minutes; 25 minutes were spent in the village, so driving took 65 minutes.
2. Let the outward drive take t minutes. The return route is 1.25 times as long at twice the speed, so it takes 1.25 ÷ 2 × t = 0.625t.
3. t + 0.625t = 65, so 1.625t = 65 and t = 40 minutes.
4. He reached the village at 14:30 + 40 min = 15:10 by the home clock, when the village clock showed 15:15.
5. So the village clock is 5 minutes fast.
6. Check: he leaves the village at 15:35 (home time); the return takes 0.625 × 40 = 25 minutes, reaching home at 16:00.

Remember · Use only one clock for elapsed time; compare the other clock’s reading with the true time at that moment.

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Answer: (c) 294

If A gets k pens, the others get k/2, k/3, k/4, k/5 and k/6, and each must be even. So k must be a multiple of 4, 6, 8, 10 and 12; the least such k is 120, giving 120 + 60 + 40 + 30 + 24 + 20 = 294 pens.

1. Let A get k pens. Then B, C, D, E and F get k/2, k/3, k/4, k/5 and k/6.
2. Each share must be an even whole number, so k must be divisible by 4, 6, 8, 10 and 12.
3. LCM(4, 6, 8, 10, 12) = 120, so the least k is 120.
4. Shares: 120, 60, 40, 30, 24, 20 — all even. Total = 294.
5. Check: k = 60 would give 147 pens, but then D gets 15, an odd number, so 147 fails.

Remember · ‘k/n must be even’ means k is a multiple of 2n; take the LCM of all such 2n.

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Answer: (d) कथन–1 और कथन–2, दोनों, प्रश्न का उत्तर देने के लिए पर्याप्त नहीं हैं

Together the statements fix A opposite F, B next to F and C opposite B, which settles who is on A’s right (C) but not on A’s left: that seat can hold D or E. So even both statements are not sufficient.

1. Statement–2 alone says nothing about A’s neighbours. Not sufficient.
2. Statement–1 alone: with B–C and D–E opposite, A and F must be opposite each other, but A’s neighbours are still open. Not sufficient.
3. Together: number the seats 1 to 6 clockwise with A at 1, so F is at 4. For people facing the centre, the immediate left is the next seat clockwise. F on B’s immediate left puts B at 3, and C (opposite B) at 6.
4. D and E take seats 2 and 5 in either order. A’s immediate left is seat 2, which can be D or E.
5. So both statements together are still not sufficient. (With the opposite left–right convention the picture is only mirrored, and A’s left is again D or E.)

Remember · In seating data sufficiency, draw the combined arrangement and look for a pair of people who can still swap.

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Answer: (c) कथन–1 और कथन–2, दोनों, प्रश्न का उत्तर देने के लिए पर्याप्त हैं

Statement–1 gives the age gap and Statement–2 gives a future ratio; neither fixes Manisha’s age alone. Together they give (M + 5) : (M + 29) = 3 : 5, so Manisha is 31.

1. Statement–1 alone gives only the age gap of 24 years. Not sufficient.
2. Statement–2 alone gives a ratio but no actual ages. Not sufficient.
3. Together: let Manisha be M years old, so her mother is M + 24. Then (M + 5)/(M + 29) = 3/5.
4. 5M + 25 = 3M + 87, so 2M = 62 and M = 31.
5. Check: in 5 years they will be 36 and 60, and 36 : 60 = 3 : 5.

Remember · Age data sufficiency: a gap alone or a ratio alone is never enough; a gap plus a ratio fixes both ages.

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Answer: (d) कथन–1 और कथन–2, दोनों, प्रश्न का उत्तर देने के लिए पर्याप्त नहीं हैं

With B fixed in the last period, the block A–F–C can still sit in periods 1–3, 2–4 or 3–5, so the third period could hold C, F or A. Even together the statements do not fix it.

1. There are six one-hour periods from 8:00 a.m. to 2:00 p.m., one lecture in each.
2. Statement–1 alone: A, F, C come in that order, but the block can be anywhere. Not sufficient.
3. Statement–2 alone: B is in the 6th period; the 3rd period is still open. Not sufficient.
4. Together: A–F–C can fill periods 1–3, 2–4 or 3–5 (D and E take the rest), so the 3rd period can be C, F or A.
5. So both statements together are not sufficient.

Remember · Scheduling data sufficiency: slide the fixed block through every free position before calling it sufficient.

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Answer: (a) केवल शिक्षित जनसंख्या ही आर्थिक प्रगति के लाभों का सर्वोत्तम उपयोग कर सकती है।

Machines can give people leisure, but leisure is tedious except for those with intelligent interests, and a leisured population can be happy only if it is educated. So only an educated people can truly enjoy the fruits of economic progress.

• ✓ (a) “If a leisured population is to be happy, it must be an educated population” — education is what lets people benefit from the leisure that productivity brings.
• ✗ (b) The passage says productivity “should lead to” leisure; it does not say all development should aim at leisure.
• ✗ (c) Reverses the logic: education is necessary for happy leisure, not a guarantee that more education raises happiness.
• ✗ (d) The passage does not urge using machines to create leisure; it takes leisure as given and asks what makes it happy.

Remember · ‘Must be X if Y is to happen’ states a necessary condition; options turning it into a sufficient cause are wrong.

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Answer: (b) उपहारों से वयस्कों के कम रोमांचित महसूस करने के एक से अधिक कारण हो सकते हैं।

The author offers two possible reasons with “perhaps… or perhaps”: we already have too much, or we have lost the fullness of the joy of giving. So there can be more than one reason why adults are less thrilled by presents.

• ✗ (a) Too extreme: presents bring “less thrill”, not no thrill.
• ✓ (b) “Perhaps it is because we have too much already; or perhaps it is because we have lost the fullness of the joy of giving” — two possible reasons.
• ✗ (c) The author is tentative but does offer reasons; saying he does not know overstates his doubt.
• ✗ (d) The passage speaks of the joy of giving and receiving presents, not of loving or being loved.

Remember · ‘Perhaps… or perhaps…’ signals more than one possible reason; options saying ‘impossible’ or ‘does not know’ overshoot.

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Answer: (a) अतीत और भविष्य के बारे में अत्यधिक चिन्ता करना

Children’s joys are complete because they “do not look too forward nor too far backward”, and the author feels our cares would shrink “if we thought less about them”. He is against excessive worry about the past and the future.

• ✓ (a) The contrast with children, who “do not look too forward nor too far backward”, and the regret that we are “too cautious, too anxious, too sceptical” show what he opposes.
• ✗ (b) Presents are only the opening example; thinking about them is never criticised.
• ✗ (c) Feeling less thrilled is the puzzle the author explores, not the attitude he argues against; ‘new things’ are not mentioned.
• ✗ (d) Losing the fullness of the joy of giving is one tentative explanation for the lesser thrill, not what the passage as a whole argues against.

Remember · For ‘author is against’, read the advice at the end (think less, worry less); it names what he opposes.

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Answer: (a) केवल 1

The six numbers use each digit twice in every place, so x = 222 × (A + B + C). The least digit sum is 1 + 2 + 3 = 6, giving x = 1332, so x is always at least 1332 and can never be a 3-digit number.

1. The six numbers formed by A, B, C place each digit twice in the hundreds, tens and units places, so x = 2 × (A + B + C) × 111 = 222 × (A + B + C).
2. The smallest possible sum of three distinct non-zero digits is 1 + 2 + 3 = 6, so the least x is 222 × 6 = 1332, a 4-digit number. Statement 1 is correct.
3. A 3-digit x would need 222 × (A + B + C) < 1000, i.e. a digit sum of 4 or less — impossible. So x is never 3-digit, and 888 (digit sum 4) cannot occur. Statement 2 is incorrect.
4. Check: 123 + 132 + 213 + 231 + 312 + 321 = 1332.

• ✓ 1. Digit sum 6 is the minimum, giving x = 222 × 6 = 1332.
• ✗ 2. x is always a multiple of 222 with digit sum at least 6, so it is never below 1332; 888 would need digit sum 4.

Remember · Sum of all permutations of three distinct digits = 222 × (sum of the digits).

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Answer: (c) 10

Each set of three digits from 1–7 that are pairwise at least 2 apart gives exactly one PIN (written in decreasing order). There are 10 such sets, so in the worst case 10 attempts are needed.

1. Once three digits are chosen, the decreasing order fixes the PIN, so count the digit sets.
2. The digits must be from 1–7 with every two at least 2 apart. List by the largest digit: 753, 752, 751, 742, 741, 731, 642, 641, 631, 531 — 10 PINs.
3. Quick count: choosing 3 of 7 numbers with no two consecutive = C(7 − 3 + 1, 3) = C(5, 3) = 10.
4. In the worst case the right PIN is the last one tried, so up to 10 attempts are needed.

Remember · Choosing k of n numbers with no two consecutive: C(n − k + 1, k).

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Answer: (a) 24

Eight equidistant points form 4 diameters. The angle in a semicircle is a right angle, so each diameter with any of the other 6 points gives a right-angled triangle: 4 × 6 = 24.

1. With 8 equidistant points, each point has a point exactly opposite it, giving 8 ÷ 2 = 4 diameters.
2. The angle in a semicircle is a right angle, so a diameter with any third point forms a right-angled triangle.
3. Each diameter can pair with any of the other 6 points: 4 × 6 = 24 triangles.

Remember · Angle in a semicircle = 90°: count diameters × remaining points.

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Answer: (d) कोई निष्कर्ष निकालने के लिए आँकड़े अपर्याप्त हैं

The answer depends on how a man’s work rate compares with a woman’s, and that is not given. Equal rates give exactly 30 days, faster men give more than 30 and faster women fewer — so the data are inadequate.

1. Let a man do m units and a woman w units of work a day. Work = 30 × (24m + 12w).
2. Days for 12 men and 24 women = 30(24m + 12w)/(12m + 24w) = 30(2m + w)/(m + 2w).
3. If m = w this is 30 days; if m > w it is more than 30; if m < w it is less than 30.
4. The relation between m and w is not given, so no conclusion can be drawn. Option (c) fails because it leaves out the possibility of exactly 30 days.

Remember · In time-and-work items with two kinds of workers, you need their relative efficiency before you can swap one for the other.

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Answer: (d) 0

1261 = 13 × 97. The product contains 91 = 7 × 13 and also 97, so it is a multiple of 1261 and the remainder is 0.

1. Factorise the divisor: 1261 = 13 × 97.
2. The product contains 91 = 7 × 13 and the factor 97.
3. So the product is a multiple of 13 × 97 = 1261, and the remainder is 0.

Remember · Before dividing a big product, factorise the divisor and look for its factors inside the product.

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Answer: (c) 1 और 2 दोनों

Two 8 cm squares plus two 4 cm squares use up the whole 20 × 8 sheet, and five 4 × 8 strips cut along their diagonals give 10 equal triangles. Both statements are possible.

1. Statement 1: cut two 8 × 8 squares from the 20 × 8 sheet, leaving a 4 × 8 strip; cut that into two 4 × 4 squares. That is exactly 4 square sheets with nothing left over.
2. Statement 2: cut the sheet into five 4 × 8 rectangles and cut each along a diagonal: 10 triangles, each of area 16 cm² (160 ÷ 10).
3. Both statements are correct.

• ✓ 1. 8 × 8 + 8 × 8 + 4 × 4 + 4 × 4 = 64 + 64 + 16 + 16 = 160 cm², the whole sheet.
• ✓ 2. Five 4 × 8 rectangles halved diagonally give 10 triangles of 16 cm² each.

Remember · ‘It is possible’ needs only one construction; the squares need not be equal unless the item says so.

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Answer: (a) z < x < y

The first condition gives 0.7x = 0.65y, so y = 14x/13, just above x. The second gives 0.6x = 0.65z, so z = 12x/13, just below x. Hence z < x < y.

1. 0.7x + y = 1.65y, so 0.7x = 0.65y and y = 70x/65 = 14x/13.
2. 0.6x + z = 1.65z, so 0.6x = 0.65z and z = 60x/65 = 12x/13.
3. For positive x, 12x/13 < x < 14x/13, so z < x < y.
4. Check with x = 13: y = 14 and z = 12; 9.1 + 14 = 23.1 = 1.65 × 14 and 7.8 + 12 = 19.8 = 1.65 × 12.

Remember · Turn each percentage condition into ‘part of x = part of y’ and compare the resulting multiples of x.

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Answer: (c) ऐसे लोग भी हैं जो अपनी जरूरतों के लिए धन संचित करने में असफल रह जाते हैं।

The passage opens with “The majority of people who fail to accumulate money sufficient for their needs…”, which takes for granted that such people exist. It says nothing about how many people in all fail or succeed.

• ✗ (a) “Majority” refers to those among the people who fail who are easily influenced, not to most people in general.
• ✗ (b) No claim about most people succeeding is made; outside the passage.
• ✓ (c) Speaking of “people who fail to accumulate money sufficient for their needs” implies that such people exist.
• ✗ (d) The passage treats having money sufficient for one’s needs as desirable; it never says there is no need for it.

Remember · ‘Implied’ often means ‘presupposed’: what must be true for the author’s sentence to make sense at all.

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Answer: (a) लोगों को दूसरों की राय से प्रभावित नहीं होना चाहिए।

The passage links failure to being “easily influenced by the opinions of others” and warns that anyone influenced by opinions when deciding “will not succeed in any undertaking”. The main idea is to resist others’ opinions.

• ✓ (a) The whole passage warns against letting newspapers and neighbours “do their thinking” for you.
• ✗ (b) Money is the example, not the message; the passage never urges accumulating as much as possible.
• ✗ (c) The author warns against being influenced by opinions; he does not forbid giving or accepting opinions altogether.
• ✗ (d) Reverses the conditional: being influenced leads to failure, but rejecting all opinions is not said to guarantee success.

Remember · Main idea = the warning the example serves; and ‘if A then fail’ does not mean ‘if not A then succeed’.

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Answer: (c) 1 और 2 दोनों

The social order is the basis of all other rights, and it is itself founded on conventions — so conventions are the source of man’s rights. And if the social order is the basis of all rights, those rights can be exercised only where a social order exists.

• ✓ 1. Rights rest on the social order, and the social order “must therefore be founded on conventions”; so conventions are ultimately the source of man’s rights.
• ✓ 2. The social order is “the basis of all other rights”; without that basis the other rights cannot be exercised.

Remember · Chain the links: if A is the basis of B and C is the basis of A, then C is the source of B.

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Answer: (a) 15000

X got 56% and Y 44% of the votes cast, so the winning margin is 12% of the votes cast: 1440 votes. That makes 12000 votes cast, which is 80% of the list — so the list has 15000 voters.

1. With two candidates and no invalid votes, Y got 100% − 56% = 44% of the votes cast.
2. Margin = 56% − 44% = 12% of the votes cast = 1440, so votes cast = 1440 ÷ 0.12 = 12000.
3. Votes cast are 80% of the voters list, so total voters = 12000 ÷ 0.8 = 15000.
4. Check: 12000 is 80% of 15000; X gets 6720, Y gets 5280, and 6720 − 5280 = 1440.

Remember · In two-candidate elections, the margin is (winner% − loser%) of the votes cast; convert back to the full list last.

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Answer: (c) 1083

A number leaving remainder 3 with each divisor is 3 more than a common multiple of them. LCM(6, 9, 12, 15, 18) = 180, and the first multiple of 180 that gives a number above 1000 is 1080, so the answer is 1083.

1. LCM(6, 9, 12, 15, 18) = 180.
2. The number must be of the form 180k + 3.
3. 180 × 5 + 3 = 903 is below 1000; 180 × 6 + 3 = 1083 is the first above it.
4. Check: 1083 − 3 = 1080 is divisible by 6, 9, 12, 15 and 18.

Remember · Same remainder r for several divisors: the number = k × LCM + r.

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Answer: (d) 9

2430 = 45 × 54, and 54 is 45 with its digits reversed. The difference between them is 54 − 45 = 9.

1. 2430 = 2 × 3⁵ × 5; look for a two-digit factor whose reverse is the other factor.
2. 2430 = 45 × 54, and 54 is 45 reversed.
3. Difference = 54 − 45 = 9.
4. Check: a two-digit number and its reverse differ by 9 × (difference of the digits) = 9 × (5 − 4) = 9.

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

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Answer: (d) 1, 2 और 3

Each statement says ‘can be’, so one example is enough for each: 3 × 9 = 27 is odd, 6 ÷ 2 = 3 is prime, and 2 + 9 = 11 is prime. All three are correct.

1. p = 3, q = 9: p × q = 27 is odd, so statement 1 is correct.
2. p = 2, q = 6: q/p = 3 is prime, so statement 2 is correct.
3. p = 2, q = 9: p + q = 11 is prime, so statement 3 is correct.
4. All three statements are correct.

• ✓ 1. An odd prime times an odd composite is odd, e.g. 3 × 9 = 27.
• ✓ 2. e.g. 6 ÷ 2 = 3 (also 15 ÷ 5 = 3).
• ✓ 3. e.g. 2 + 9 = 11 (also 3 + 4 = 7).

Remember · ‘Can be’ statements need just one valid example; ‘must be’ statements need just one counter-example to fail.

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Answer: (c) 1 और 2 दोनों

The hour and minute hands meet 16 4/11 minutes after 3 o’clock, i.e. between 3:16 and 3:17. The second hand catches the minute hand about 58.98 seconds after 4:58, i.e. before 4:59. Both statements are correct.

1. At 3:00 the hour hand is 90° ahead of the minute hand. The minute hand moves 6° a minute and the hour hand 0.5°, so the minute hand gains 5.5° a minute.
2. It catches up after 90 ÷ 5.5 = 180/11 = 16 4/11 minutes, about 3:16:22 — between 3:16 and 3:17. Statement 1 is correct.
3. t seconds after 4:58, the minute hand is at the 58 + t/60 minute mark and the second hand at the t mark. They meet when t = 58 + t/60, i.e. t = 58 × 60/59 ≈ 58.98 s.
4. That is before 4:59, so statement 2 is also correct.

• ✓ 1. The hands coincide at 3:16 4/11, inside the stated minute.
• ✓ 2. The second hand passes the minute hand about 58.98 seconds after 4:58, before 4:59.

Remember · Clock overlaps: gap ÷ relative speed. Minute hand gains 5.5° per minute on the hour hand.

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Answer: (a) m = n

After the two transfers each container again holds 100 ml. Whatever water went into X must be matched by the same amount of milk left in Y, so X’s milk share equals Y’s water share: m = n = 5/6.

1. After the first transfer: X has 80 ml milk; Y has 100 ml water + 20 ml milk = 120 ml.
2. 20 ml of Y’s mixture carries 20 × 20/120 = 10/3 ml milk and 20 × 100/120 = 50/3 ml water back to X.
3. X: milk = 80 + 10/3 = 250/3 ml out of 100 ml, so m = 5/6.
4. Y: water = 100 − 50/3 = 250/3 ml out of 100 ml, so n = 5/6.
5. So m = n.

Remember · Back-and-forth transfers of equal volume: the milk in the water jar always equals the water in the milk jar.

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Answer: (d) 250/9

The angles of a pie chart add up to 360°, so A takes 360° − 260° = 100°. As a percentage, that is 100/360 × 100 = 250/9 %.

1. Angles in a pie chart add up to 360°. A = 360° − (90° + 50° + 45° + 75°) = 360° − 260° = 100°.
2. Percentage for A = 100/360 × 100 = 250/9 % (about 27.8%).

Remember · Pie chart: percentage = angle ÷ 360 × 100, so every 18° is 5%.

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Answer: (c) यह कभी भी कुछ-न-कुछ मूल्य के बिना नहीं होता

The author writes that “Scientific research is probably never completely valueless” — there is always a chance of some value. The assumption is that it is never without some value, not that it is always very valuable.

• ✗ (a) The author says the results may be of value to an investigator of talent; “never very valuable” is not what he assumes.
• ✗ (b) The passage does not grade research as sometimes very valuable; its claim is only about some value.
• ✓ (c) “Probably never completely valueless” means never without some value.
• ✗ (d) Too strong: the author admits research may seem “silly and insignificant” and claims only some value.

Remember · Match the strength of the passage’s claim: ‘never completely valueless’ is not ‘always very valuable’.

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Answer: (c) किसी प्रबुद्ध अन्वेषक के लिए कोई भी अनुसंधान परिणाम महत्त्व का हो सकता है

“There is always a chance that the results may be of value to the investigator of talent” — so any result can be useful to such an investigator, though none is guaranteed to be.

• ✗ (a) The author stresses that even silly-looking research can help; “not many” goes against the passage.
• ✗ (b) “Always valuable” turns “always a chance” into a certainty.
• ✓ (c) However silly the research, its results “may be of value” to an investigator of talent — they can be, not must be.
• ✗ (d) “Must always be” is again a certainty; the passage speaks only of a chance.

Remember · ‘Always a chance that it may’ means ‘can’, not ‘always is’ or ‘must be’.

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Answer: (d) बाँधों और जलभरों के जरिए जल का सुरक्षित भण्डार (बफर) रखना

The problem is one of timing: about 75% of the rain falls in 120 days and too little in the other 245. The lasting fix is to store the monsoon water — in dams and in recharged aquifers — and use it through the year.

• ✗ (a) Huge concrete tanks and canals across the country are neither practical nor lasting at that scale, and canals move water rather than store it.
• ✗ (b) Changing cropping patterns adjusts demand; it does not manage the skewed availability the passage describes.
• ✗ (c) Interlinking rivers moves water from place to place; the passage’s problem is uneven supply over the year.
• ✓ (d) Buffer stocks in dams and recharged aquifers capture the ‘too much’ of June–September and release it in the ‘too little’ months — exactly the managed, year-round use the passage asks for.

Remember · Match the solution to the exact problem: uneven supply over time needs storage, not transfer or demand changes.

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Answer: (b) 6

The largest power of 3 dividing 15! decides m. The multiples of 3 up to 15 give five 3s and 9 = 3² gives one more, so m can be at most 6.

1. Count the factors of 3 in 15!: the multiples of 3 up to 15 are 3, 6, 9, 12, 15 — five numbers, one 3 each.
2. 9 = 3² contributes one extra 3.
3. Total power of 3 = ⌊15/3⌋ + ⌊15/9⌋ = 5 + 1 = 6, so the maximum m is 6 (n is then a whole number).

Remember · Highest power of prime p in n! = ⌊n/p⌋ + ⌊n/p²⌋ + ….

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Answer: (b) 252

The terms are n² × (n + 1) — that is, n³ + n² — for n = 1, 2, 3, 4, 5: 2, 12, 36, 80, 150. The next term is 6² × 7 = 252.

1. Test n²(n + 1): 1 × 2 = 2, 4 × 3 = 12, 9 × 4 = 36, 16 × 5 = 80, 25 × 6 = 150.
2. Next term: 36 × 7 = 252.
3. Check with differences: 10, 24, 44, 70, 102; second differences 14, 20, 26, 32 rise steadily by 6, as they should for a cubic pattern.

Remember · If terms grow roughly like cubes, test n³ ± n² or n³ ± n before trying differences.

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Answer: (c) 945

The password has three characters: the vowel first, the consonant last and the digit in the middle. That gives 5 × 9 × 21 = 945 passwords.

1. Each password has three characters. It starts with the vowel and ends with the consonant, so the digit sits in the middle.
2. Choices: 5 vowels (A, E, I, O, U) × 9 non-zero digits × 21 consonants.
3. Total = 5 × 9 × 21 = 945.

Remember · Fix the forced positions first; each remaining position multiplies by its number of choices.

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Answer: (d) 81

The 9 cups form a 3 × 3 grid. Choosing the 3 places for tea can be done in C(9, 3) = 84 ways; only the 3 choices that fill a whole row with tea leave a row without coffee. So 84 − 3 = 81.

1. Equal rows and columns with 9 cups means a 3 × 3 grid.
2. Cups of the same drink are alike, so an arrangement is fixed by the 3 places holding tea: C(9, 3) = 84.
3. A row lacks coffee only when all 3 tea cups fill that row: 3 such arrangements.
4. Arrangements with coffee in every row = 84 − 3 = 81.

Remember · ‘At least one’ counts are easiest as total minus the bad cases.

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Answer: (c) केवल तीन

Writing the integers as n − 1, n, n + 1 gives 3n = n(n² − 1), so n = 0, 2 or −2. That produces three triples: (−1, 0, 1), (1, 2, 3) and (−3, −2, −1).

1. Let the integers be n − 1, n, n + 1. Sum = 3n; product = n(n² − 1).
2. 3n = n(n² − 1) gives n = 0 or n² − 1 = 3, i.e. n = 0, 2 or −2.
3. The triples are (−1, 0, 1), (1, 2, 3) and (−3, −2, −1), with sums 0, 6, −6 equal to their products.
4. So there are exactly three possibilities. (Testing only positive integers finds just 1, 2, 3.)

Remember · ‘Integers’ includes zero and negatives; solve the algebra instead of testing only positive numbers.

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Answer: (a) 72

X can be any of the 9 non-zero digits and Y any of the other 8, giving 9 × 8 = 72 different decimals.

1. X can be any of 1–9: 9 choices.
2. Y must be non-zero and different from X: 8 choices.
3. Total = 9 × 8 = 72, and each ordered pair gives a different number.

Remember · Distinct choices from the same pool: n × (n − 1) ordered pairs.

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Answer: (c) 41 kg

A + B + C = 120 and B + D + E = 126. Since F weighs the same as B, the six weights total 120 + 126 = 246 kg, and the average is 41 kg even though B itself is unknown.

1. A + B + C = 3 × 40 = 120 kg and B + D + E = 3 × 42 = 126 kg.
2. Since F = B, A + B + C + D + E + F = (A + B + C) + (F + D + E) = 120 + (B + D + E) = 120 + 126 = 246 kg.
3. Average = 246 ÷ 6 = 41 kg.

Remember · Before answering ‘cannot be determined’, check whether the unknown cancels — here F = B fills the gap exactly.

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