CSAT (GS Paper II) 2020

Answer: (d) Increased social sector spending is imperative for large-scale job creation.

The passage is about closing a jobs gap of about 24 million. E-commerce can fill at best half of it, and the author says only demand-driving sectors such as health and education can fill the rest — so spending on these social sectors is what large-scale job creation needs.

• ✗ (a) The passage notes labour leaving agriculture but never calls for curbing rural-to-urban migration.
• ✗ (b) Construction and unregistered manufacturing are called 'not markedly better jobs', but improving their conditions is not the passage's concern; its concern is where new jobs will come from.
• ✗ (c) This reverses the passage: services 'are not generating the additional jobs the country needs'.
• ✓ (d) Health and education are the sectors that 'can comfortably fill the other half' of the jobs gap; that makes more spending on them necessary for jobs on this scale.

Remember · For 'best implied', find where the passage lands (its last line or its solution) and pick the option that restates it.

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Answer: (a) 1 and 2

The author says a right is real only when the State also stops private parties from blocking it. That rests on two unstated premises: the State must stop private violations of privacy (2), and it needs some institutional means to play this active role (1). Nothing suggests the digital economy and privacy are incompatible.

• ✓ 1. The State is asked to 'actively ensure' that private parties cannot block a right. An active guarantee of this kind needs some machinery — institutions — to carry it out.
• ✓ 2. The passage opens with the right to privacy and ends by saying the State's role is to ensure 'private parties are not able to block it' — a substantive right must hold against private force too.
• ✗ 3. The passage never says the digital economy and privacy cannot go together; it only says the digital age brings new realities that shape the focus on privacy.

Remember · An assumption is what the author must take for granted for the argument to work — check whether the conclusion collapses without it.

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Answer: (c) Wetlands need to be reinforced as more than just open sources of water.

The irony the author points to is that water comes from rivers and wetlands, yet policy treats it as a resource divorced from them and engineers supply instead of strengthening ecosystems. The inference is that wetlands must be seen as ecological systems, not merely as taps of water.

• ✗ (a) The Ramsar Convention is outside the passage; the author criticises how policy views water, not the lack of a treaty listing.
• ✗ (b) This is the very approach the author criticises — engineering to augment supply rather than strengthening ecological systems.
• ✓ (c) Policy fails to 'grasp that it is a part of the natural ecosystem'; recognising wetlands as ecosystems rather than just sources of water answers that failure.
• ✗ (d) Pricing, misuse and overuse of water are not discussed at all.

Remember · Name the author's complaint in one line; the right inference is its remedy, not an outside fact or a policy the passage never discusses.

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Answer: (a) Distribute your wealth across different kinds of assets so that your risks would be minimized.

The author opens with asset allocation as the key decision, says no asset is risk-free, and ends by noting that the extreme risks will not all strike at the same time. The practical suggestion is to spread money across different assets so that one failure does not wipe out everything.

• ✓ (a) Asset allocation is 'the most important investment decision', and since the risks 'will not manifest at the same time', spreading wealth across assets lowers overall risk.
• ✗ (b) The author wants informed allocation, not risk-taking as a personality trait; 'necessary component' goes beyond the passage.
• ✗ (c) Handing money to an asset manager is never mentioned; the author even notes our dislike of losing control of investment.
• ✗ (d) This is only the starting fact ('no asset that is risk-free'), not the author's suggestion about what to do.

Remember · When a passage says 'every option is risky, but not all fail together', its advice is diversification.

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Answer: (b) 1 and 3 only

The passage says regulation cannot be framed only for today's single-trait GM crops because multi-trait crops will be the norm — so regulation must keep evolving (1) — and it asks for a comprehensive look that includes socio-economic impacts (3). Public participation and the political executive are never mentioned.

• ✓ 1. Regulation 'cannot be understood solely in the context of the current generation of GM crops' because future crops will differ — it has to keep changing with the technology.
• ✗ 2. The passage does not mention public participation in policy decisions.
• ✓ 3. It asks for a comprehensive look 'including socio-economic impacts'.
• ✗ 4. The passage links domestic regulation to trade policy and treaties, but says nothing about the political executive or what improves effectiveness.

Remember · Accept an assumption only when a phrase of the passage supports it; reject statements that bring in new actors such as 'people' or 'political executive'.

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Answer: (c) Biotechnology’s role is not confined to the current priorities of developing GM crops.

The whole passage widens the frame: future multi-trait crops, socio-economic impacts, climate action and trade obligations all bear on biotechnology. Its crux is that biotechnology's role cannot be seen only through today's GM crops.

• ✗ (a) The precautionary principle is not mentioned; the passage speaks of minimising negative impacts while harnessing the technology.
• ✗ (b) The passage says not using biotechnology in the climate plan 'cannot be an option' — it does not report whether it is used now.
• ✓ (c) Role and regulation 'cannot be understood solely in the context of the current generation of GM crops'; the passage then adds climate, socio-economic and trade dimensions.
• ✗ (d) Negative impacts are to be minimised, but the passage does not claim they are poorly understood.

Remember · The crux restates the passage's central claim; options that take a side detail and add a new judgement are traps.

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Answer: (a) 10

Each trailing zero needs one pair of 2 and 5. Writing the product as 5¹² × 12! shows 14 fives but only 10 twos, so the twos run out first and there are 10 zeroes.

1. The terms after 1 are 5 × 1, 5 × 2, …, 5 × 12, so the product = 5¹² × (1 × 2 × … × 12) = 5¹² × 12!.
2. Fives: 12 from 5¹², plus 2 in 12! (from 5 and 10) = 14.
3. Twos: only from 12! — 6 + 3 + 1 = 10 (multiples of 2, 4 and 8 up to 12).
4. Zeroes = min(10, 14) = 10.
5. Check: 12! = 479001600 has two 5s and ten 2s, as used.

Remember · Trailing zeroes = the smaller of the counts of 2s and 5s. Do not assume 5s are always scarcer.

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Answer: (b) 9

Adding the three rotations puts each digit once in the hundreds, tens and units places, so the sum is 111(X + Y + Z) = 3 × 37 × (X + Y + Z). It is divisible by 3, 37 and (X + Y + Z), but by 9 only if (X + Y + Z) were a multiple of 3 — which it is not.

1. XYZ + YZX + ZXY = (100X + 10Y + Z) + (100Y + 10Z + X) + (100Z + 10X + Y) = 111(X + Y + Z).
2. 111 = 3 × 37, so the sum = 3 × 37 × (X + Y + Z): divisible by 3, 37 and (X + Y + Z).
3. For 9, a second factor 3 must come from (X + Y + Z), which is not a multiple of 3.
4. So the sum is not divisible by 9.
5. Check: 124 + 241 + 412 = 777 = 111 × 7; 777 ÷ 9 is not whole.

Remember · Cyclic rotations of a 3-digit number always add up to 111 × (digit sum) = 3 × 37 × (digit sum).

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Answer: (c) r

Call the common value k. Then each number is k plus or minus a constant, and r = k + 2018 carries the largest addition.

1. Let p − 2016 = q + 2017 = r − 2018 = s + 2019 = k.
2. Then p = k + 2016, q = k − 2017, r = k + 2018, s = k − 2019.
3. The largest is r = k + 2018.

Remember · Set equal expressions to one variable k and compare what is added to k in each.

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Answer: (a) Zero

Every such number uses the digits 1 to 5 once, so its digit sum is 15. A digit sum of 15 makes every one of these numbers divisible by 3, so none can be prime.

1. Digit sum = 1 + 2 + 3 + 4 + 5 = 15, whatever the order.
2. 15 is divisible by 3, so every such five-digit number is divisible by 3.
3. A number greater than 3 that is divisible by 3 is not prime, so the count is zero.

Remember · Before listing permutations for primes, check the digit sum: if it is a multiple of 3, no arrangement is prime.

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Answer: (b) 3

Write the symbol as a digit x and expand each number by place value. The sum becomes a simple linear equation, 24x + 61 = 100 + 11x, which gives x = 3.

1. Let ⊗ = x. Then ⊗ = x, 1⊗ = 10 + x, 5⊗ = 50 + x, ⊗⊗ = 11x, ⊗1 = 10x + 1, and 1⊗⊗ = 100 + 11x.
2. Left side = x + (10 + x) + (50 + x) + 11x + (10x + 1) = 24x + 61.
3. 24x + 61 = 100 + 11x, so 13x = 39 and x = 3.
4. Check: 3 + 13 + 53 + 33 + 31 = 133 = 1⊗⊗ with ⊗ = 3.

Remember · In symbol-for-digit sums, expand every number by place value and solve the resulting equation; then check it back.

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Answer: (b) 0.25 foot

Laying the sticks end to end, forwards or backwards, measures any whole-number combination of 7·5 and 3·25 feet. The smallest positive length such combinations can give is their HCF, 0·25 foot.

1. Convert to hundredths of a foot: 7·5 = 750/100 and 3·25 = 325/100.
2. HCF(750, 325) = 25 (750 = 2 × 3 × 5³ and 325 = 5² × 13), so HCF of the lengths = 0·25 foot.
3. Any length made by laying the sticks forwards and backwards is a multiple of 0·25 foot, and 0·25 itself can be made.
4. Check: 7 × 3·25 = 22·75 and 3 × 7·5 = 22·5; 22·75 − 22·5 = 0·25 foot.

Remember · The smallest length measurable with two rods (by adding and subtracting whole lengths) is the HCF of their lengths.

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Answer: (c) 38

Subtracting 1 from each term gives 13, 17, 19, 23, 29, 31 — consecutive prime numbers. The next prime is 37, so the next term is 37 + 1 = 38.

1. Subtract 1: 14 → 13, 18 → 17, 20 → 19, 24 → 23, 30 → 29, 32 → 31.
2. 13, 17, 19, 23, 29, 31 are consecutive primes.
3. The next prime after 31 is 37 (33, 35 are composite).
4. Next term = 37 + 1 = 38.

Remember · When a series 'relates to primes', try adding or subtracting 1 or 2 to each term and look for consecutive primes.

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Answer: (b) 7, 8

The full booklet's page total must be just above 195. With 20 pages the total is 210, so the torn leaf carries numbers adding to 15 — pages 7 and 8.

1. Sum of pages 1 to n is n(n + 1)/2. For n = 19 it is 190 (too small); for n = 20 it is 210.
2. Missing sum = 210 − 195 = 15.
3. A leaf carries two consecutive numbers, odd then even: 7 + 8 = 15.
4. Check: n = 21 gives 231 − 195 = 36, which cannot be an odd + next-even pair (such pairs add to an odd number).

Remember · Missing leaf: find the first 1 + 2 + … + n above the given total; the gap is the two page numbers' sum.

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Answer: (c) a, c, c, e

With a, c, c, e filled in, the string splits into four blocks of five letters — ababa, bcbcb, cdcdc, deded — each alternating two neighbouring letters. No other option gives a regular pattern.

1. The completed string has 20 letters; try blocks of 5.
2. Fill a, c, c, e: a b a b a | b c b c b | c d c d c | d e d e d.
3. Each block alternates two consecutive letters (ab, bc, cd, de), starting where the last block ended.
4. Check option (d) b, c, d, e: the first block becomes ababb, which breaks the alternation.

Remember · For letter-gap series, try splitting the full string into equal blocks; the blank count and length often reveal the block size.

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Answer: (c) 3

Working column by column from the units, the carries force B = 7 and E = 5, and C and F must be 4 and 8. That leaves 6 and 9 for A and D, whose difference is 3.

1. Units: C + F ends in 2; both are at least 4, so C + F = 12, carry 1.
2. Tens: B + 2 + 1 ends in 0, so B + 3 = 10 and B = 7, carry 1.
3. Hundreds: 3 + E + 1 ends in 9, so E = 5, carry 0.
4. Thousands: A + D = 15.
5. Digits left from 4–9 after B = 7, E = 5: {4, 6, 8, 9}. C + F = 12 needs {4, 8}; so A and D are 6 and 9 (6 + 9 = 15).
6. Difference = 9 − 6 = 3.
7. Check: 6374 + 9528 = 15902.

Remember · In letter-addition puzzles, start from the units column, track every carry, and use 'different digits' to fix the pairs.

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Answer: (c) S1 and S2 together are not sufficient to answer the Question.

S1 caps the book at 2 × 51 = 102 figures and S2 guarantees at least 51. Together they only place the total between 51 and 102, which may or may not exceed 100.

1. S1: at most 2 figures on each of 51 pages, so at most 102 figures — could be above or below 100.
2. S2: at least 1 figure per page, so at least 51 figures — no upper limit.
3. Together: between 51 and 102. Both 101 and 60 are possible, so the Question cannot be answered.

Remember · In data sufficiency, find the range each statement allows; if the range still straddles the question's cut-off, the data are insufficient.

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Answer: (a) 7.8

The overall English average of 8·8 lies four times nearer the girls' 9 than the boys' 8, so girls outnumber boys 4 : 1. Using the same weights on Hindi gives (4 × 8 + 1 × 7)/5 = 7·8.

1. Let there be g girls and b boys. English: 9g + 8b = 8·8(g + b).
2. So 0·2g = 0·8b, giving g : b = 4 : 1.
3. Hindi: x = (4 × 8 + 1 × 7)/5 = 39/5 = 7·8.
4. Check: English with the same weights = (4 × 9 + 8)/5 = 44/5 = 8·8.

Remember · Use the known overall average to find the group ratio, then apply that ratio to the other subject.

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Answer: (b) R is S’s wife.

Q, a mother-in-law, belongs to the older couple, so P (U's parent) is her husband. Their children are U, T and one married child whose spouse is R. With three males already P, T and S, R must be female and married to S.

1. Q is R's mother-in-law, so Q is a woman of the older generation and R is married to Q's child.
2. Two couples: the older couple and R with Q's child. U is P's daughter, so P is the older couple's other member — P is Q's husband (male).
3. Males: P, T and S = three, so Q, U and R are the three females.
4. R, a female, is married to a son of Q. T is unmarried, so that son is S — R is S's wife.
5. The remaining two, U and T, are the unmarried siblings.
6. Check (a): R is female, so cannot be U's husband; (c): S is married.

Remember · In family puzzles, fix the generations first, then use the male/female count to settle the remaining roles.

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Answer: (c) 12th July is a Sunday if the year is a leap year.

From 12 January to 12 July is 181 days in an ordinary year and 182 days in a leap year. 182 is exactly 26 weeks, so in a leap year 12 July falls on the same day, Sunday.

1. Days from 12 January to 12 July: January 31, February 28 or 29, March 31, April 30, May 31, June 30.
2. Ordinary year: 181 days = 25 weeks + 6 days, so 12 July is Saturday.
3. Leap year: 182 days = 26 weeks exactly, so 12 July is Sunday.
4. 15 July is then Tuesday (ordinary year) or Wednesday (leap year), never Sunday — (a), (b) and (d) fail.

Remember · Count days between the two dates and divide by 7; the remainder is the number of weekdays to move forward.

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Answer: (b) 1, 3 and 5

Because FDI is volatile by nature and its technology gains depend on India's own physical and human capital, the passage implies: do not lean on foreign investment in the long run (1), build up domestic investment to carry employment (3), and invest in education and health so that technology can spread (5).

• ✓ 1. Employment 'cannot be left to' FDI, its volatility brings unstable jobs and wider inequalities, and even its technology benefit is 'not at all certain'.
• ✗ 2. The passage calls volatility 'a necessary consequence' of FDI's search for the highest returns — something built in, not something policy is expected to remove.
• ✓ 3. If employment cannot be entrusted to the more volatile foreign capital, the author must be counting on domestic investment to carry it.
• ✗ 4. The passage never ranks public against private investment; it contrasts foreign with domestic.
• ✓ 5. Technology diffusion may fail because India's 'physical and human capital' is inadequate; building human capital means investing in education and health.

Remember · In multi-statement items, settle the surest statements first; often only one option contains all of them.

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Answer: (a) 1 and 2

The passage's problem is that monsoon water comes in a short burst, unevenly across regions, and much of it runs to the sea while some areas face drought and others flood. Storing it (dams and canals) and moving it from surplus to deficit rivers (inter-linking) answer exactly that.

• ✓ 1. The monsoon flows have a skewed 'spatial distribution' and river waters 'flow unutilized' while millions face droughts and floods — linking rivers moves surplus water to deficit areas.
• ✓ 2. The passage calls holding rainwater in reservoirs and releasing it over the year 'a necessity nobody can afford to overlook'; dams and canals do this.
• ✗ 3. Borewells draw groundwater; they do nothing to capture the monsoon flows the passage says are being lost.
• ✗ 4. The passage is about capturing and storing water, not about legally restricting its use in farming.
• ✗ 5. Which level of government should allocate river water is never raised.

Remember · A practical implication must solve the passage's own problem; drop steps about other resources or about who holds power.

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Answer: (c) 3 only

The author argues that people invest more in education when they can keep its returns, that is, when economic freedom is greater. That supports the general link between economic freedom and human capital (3), but not the absolute claim in 1 or the reversed claim in 2.

• ✗ 1. Lower taxes make investing in education sensible, but 'invariably' is too strong, and the passage speaks of education in general, not 'higher education' alone.
• ✗ 2. The passage runs the other way: economic freedom encourages investment in education. It never says education guarantees economic freedom.
• ✓ 3. 'The return on education increases as the level of economic freedom rises', so people invest more in education — that is, in human capital.

Remember · Watch for absolute words ('invariably', 'ensures') and for reversed cause and effect — both usually make an assumption invalid.

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Answer: (a) Urban local bodies must recover costs through user charges.

The passage lists the heavy costs of supplying water and treating sewage, says financing mechanisms are essential, and ends: water could be free only if cities could bear the whole cost — 'They are not.' The message is that the cost must be recovered, which points to charging users.

• ✓ (a) Cities cannot meet 'the entire cost', so water cannot be delivered free; the financing the first line demands must come from the users of the service.
• ✗ (b) The passage blames a lack of money, not inefficiency of the urban bodies.
• ✗ (c) The author offers a way out — financing mechanisms — so the problem is not presented as unsolvable.
• ✗ (d) Capping city populations is not discussed at all.

Remember · When a passage ends by ruling out 'free', its crucial message is about who should pay.

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Answer: (b) 2 only

Sustainable delivery, as described, involves collecting, treating and piping water and also building sewers and treatment plants — far more than supplying households. The passage does not say only rich cities can manage it; any city can, once financing mechanisms are in place.

• ✗ 1. Riches would allow water to be delivered 'free'; sustainable delivery needs 'financing mechanisms', which cities that are not rich can also put in place.
• ✓ 2. It includes collecting and treating water, a distribution network, sewerage and sewage treatment so that 'no untreated sewage is discharged' into water bodies.

Remember · Test 'only' claims against the passage: if it names another way to reach the goal, the 'only' assumption fails.

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Answer: (c) 2 and 3

Statement 2 is the passage's own mechanism: better relative prices lifted private farm investment by over 50%. Statements 1 and 4 use absolute words ('impossible', 'essential') that the passage does not support, which leaves 2 and 3 as the only workable pair.

• ✗ 1. The passage calls India's transformation 'slow', not impossible, and does not make small farm size the barrier.
• ✓ 2. The relative price scenario moved in favour of agriculture by over 50%, and 'this boosted private investments in agriculture by more than 50%'.
• ✓ 3. Livestock and horticulture are not named, but the passage's lesson is that better returns on farm produce drive investment, growth and poverty reduction; value chains for high-value products are a way to secure such returns. It is the only remaining statement free of an overreach.
• ✗ 4. Global prices helped in one period, but 'essential' goes too far — the passage shows how the benefit came through prices and investment, not that poverty cannot fall any other way.

Remember · When one statement is certain, strike the options with extreme words ('impossible', 'essential'); the survivor is usually the answer.

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Answer: (d) Inclusive agricultural growth is key to reduce poverty in the near future.

With half the workforce in farming and most farms small, the passage shows poverty falling fastest when agriculture grew — higher investment, faster agri-GDP growth and rising real farm wages. Growth that reaches this large farm population is the key to cutting poverty.

• ✗ (a) The poverty fall came through agriculture itself; off-farm employment is not mentioned.
• ✗ (b) Farmer producer companies are outside the passage.
• ✗ (c) Private investment rose as a result of prices; the passage never compares it with public investment or ranks them.
• ✓ (d) Agri-GDP growth of 4·1%, a trade surplus and real farm wages rising 7% a year together 'led to unprecedented fall in poverty' — growth that reached farmers and farm workers.

Remember · For the 'critical message', follow the passage's cause-and-effect chain to its end point and choose the option that names the whole chain.

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Answer: (c) S1 and S2 together are sufficient to answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question.

S1 makes R bigger than P and Q but says nothing about S; S2 rules out S but says nothing about P, Q and R. Together, S is not the largest and R beats P and Q, so R is the largest.

1. S1 alone: R > P and R > Q, but S could exceed R — not sufficient.
2. S2 alone: the largest is one of P, Q, R — not sufficient.
3. Together: S is not the largest and R is above P and Q, so R is the largest.

Remember · Test each statement alone, then together; choose (c) only when neither alone works but the pair does.

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Answer: (d) S1 and S2 together are not sufficient to answer the Question.

Between 10 and 20 the primes are 11, 13, 17 and 19; of these, 13 and 17 both leave remainder 1 on division by 4. Even together the statements leave two possible values of n, so they are not sufficient.

1. S1: n = 11, 13, 17 or 19.
2. S2: n = 13 or 17 (the numbers between 10 and 20 that are 1 more than a multiple of 4).
3. Together: 13 and 17 both satisfy S1 and S2, so n is still not fixed.

Remember · In number data sufficiency, list every value that fits; if more than one survives, the data are insufficient.

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Answer: (c) S1 and S2 together are sufficient to answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question.

A product of 21 alone allows 3 and 7, 1 and 21 and many other pairs, and a sum of 10 alone allows endless pairs. Together, the numbers are the roots of t² − 10t + 21 = 0, which are exactly 3 and 7.

1. S1 alone: 3 × 7, 1 × 21, 2 × 10·5 … — not unique.
2. S2 alone: 3 + 7, 4 + 6, 5 + 5 … — not unique.
3. Together: the numbers satisfy t² − 10t + 21 = 0, i.e. (t − 3)(t − 7) = 0, so they are 3 and 7.

Remember · Two numbers are fixed by their sum and product together: they are the roots of t² − (sum)t + (product) = 0.

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Answer: (d) North-West

A house facing East has its back to the West, so he first walks West. A right turn from West points North, and a left turn from North points West again. He ends 50 m West and 50 m North of the start — North-West.

1. House faces East, so walking down its backside means walking West: 25 m West.
2. Right turn while facing West → facing North: 50 m North.
3. Left turn while facing North → facing West: 25 m West.
4. Net displacement: 50 m West and 50 m North, so he is North-West of the starting point.

Remember · Fix the first direction carefully (the back of an east-facing house is West), then apply each turn relative to the current heading.

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Answer: (a) Only Conclusion-I

A number divisible by both 2 and 3 is divisible by their LCM, 6, so Conclusion-I follows. Divisibility by 2 and 3 says nothing about a second factor of 2, so 4 need not divide the numbers — 6 itself is a counter-example.

1. 2 and 3 have no common factor, so a number divisible by both is divisible by 2 × 3 = 6: Conclusion-I follows.
2. Divisibility by 4 needs two factors of 2; the statements guarantee only one.
3. Counter-example: 6 is divisible by 2 and 3 but not by 4, so Conclusion-II does not follow.

Remember · Divisible by a and b means divisible by LCM(a, b) — no more. Test any stronger claim with the LCM itself.

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Answer: (c) Neither Conclusion-I nor Conclusion-II

The statements only tell us that every cat is both a dog and black, so some dogs are black. The other dogs may be black or not: 'all dogs are black' and 'some dogs are not black' are each possible, but neither is forced, so neither follows.

1. Cats lie inside both 'dogs' and 'black', so some dogs (the cats) are black.
2. Diagram 1: dogs outside the cats circle are not black — then Conclusion-I fails.
3. Diagram 2: all dogs lie inside 'black' — then Conclusion-II fails.
4. Each conclusion fails in some valid diagram, so neither follows. (Exactly one of them is true in any given case, but the statements do not tell us which.)

Remember · Draw one diagram where a conclusion fails; if you can, it does not follow. Contradictory conclusions can both fail to follow.

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Answer: (b) 6

Scan the 26 digits once and mark every place where an odd digit is immediately followed by another odd digit. There are six such places: 5-1, 7-3, 3-9, 5-7, 3-1 and 1-5.

1. Pairs in order: 5-1 ✓, 1-4, 4-7, 7-3 ✓, 3-9 ✓, 9-8, 8-5, 5-7 ✓, 7-2, 2-6, 6-3, 3-1 ✓, 1-5 ✓, 5-8 …
2. After 5-8 the odd digits are 3, 5, 3 and 9, and each is followed by an even digit (8, 2, 4, 6).
3. Total odd-odd pairs = 6.

Remember · For adjacency counts, scan pairs left to right in one pass and tick each hit; do not restart counting.

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Answer: (b) 41

G is 11 places to the right of A, so G is 27th from the left. G is also 3 places to the right of V, so V is 24th from the left. V is 18th from the right, so the row has 24 + 18 − 1 = 41 boys.

1. A is 16th from the left; G is 11th to A's right, so G is 16 + 11 = 27th from the left.
2. G is 3rd to V's right, so V is 27 − 3 = 24th from the left.
3. V is 18th from the right, so total = 24 + 18 − 1 = 41.
4. Check: G is then 41 − 27 + 1 = 15th from the right, which is 3 places right of V (18th).

Remember · Convert every position to one end, then total = position from left + position from right − 1.

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Answer: (d) S1 and S3 together are sufficient to answer the Question.

S1 puts D above C and C above both A and B, so the youngest is A or B. S3 settles that pair: A is older than B, so B is the youngest. S2 adds nothing S1 did not already give.

1. S1: D > C > A and D > C > B, so the youngest is A or B — S1 alone is not enough.
2. S2 only repeats that D is the oldest, so S1 and S2 together still leave A or B.
3. S2 and S3: A > B, but C's position is unknown — C could be youngest.
4. S1 and S3: D > C > A > B, so B is the youngest.

Remember · Find what each statement leaves open, then pick the statement that closes exactly that gap.

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Answer: (c) 12

There are 19 numbers below 100 containing the digit 4: 4, 14, 24, 34, 40 to 49, and 54, 64, 74, 84, 94. Seven of them are multiples of 4, which leaves 12.

1. Units digit 4: 4, 14, 24, 34, 44, 54, 64, 74, 84, 94 = 10 numbers.
2. Tens digit 4 (other than 44): 40, 41, 42, 43, 45, 46, 47, 48, 49 = 9 numbers. Total = 19.
3. Multiples of 4 among them: 4, 24, 40, 44, 48, 64, 84 = 7.
4. Required count = 19 − 7 = 12.

Remember · Count the whole set first, then subtract the ones to exclude — fewer chances to miss a number.

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Answer: (a) u = 4v

Doubling the side multiplies a cube's volume by 2³ = 8, while the mass only doubles. So Q's mass per unit volume is 2/8 = 1/4 of P's, i.e. u = 4v.

1. Side of Q = 2 × side of P, so y = 2³ x = 8x.
2. n = 2m.
3. v = n/y = 2m/8x = (1/4)(m/x) = u/4.
4. So u = 4v.

Remember · Scaling a length by k scales volume by k³; compare ratios by writing each in terms of the smaller object.

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Answer: (c) 35 years

The four ages total 80. If each student is s years old, the teacher is s + 20, so 4s + 20 = 80 gives s = 15 and the teacher is 35.

1. Total of the four ages = 4 × 20 = 80.
2. Let each student be s years; the teacher is s + 20.
3. 3s + (s + 20) = 80, so 4s = 60 and s = 15.
4. Teacher = 15 + 20 = 35 years.
5. Check: (35 + 15 + 15 + 15)/4 = 80/4 = 20.

Remember · Turn an average into a total first; then express every person in one unknown.

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Answer: (d) ₹ 3,75,000

A 20% loss means the selling price is 80% of the cost. So the cost is 3,00,000 ÷ 0·8 = ₹ 3,75,000.

1. Selling price = 80% of cost price.
2. Cost price = 3,00,000 ÷ 0·8 = ₹ 3,75,000.
3. Check: 20% of 3,75,000 = 75,000, and 3,75,000 − 75,000 = 3,00,000.

Remember · Loss and profit percentages are on cost price: CP = SP ÷ (1 − loss%).

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Answer: (d) Relying on a homogeneous food source is not desirable.

Ireland depended almost wholly on the potato, and mostly on a single variety, the Lumper. When one fungus struck that genetically frail variety, nearly the whole food supply failed at once. The lesson is the danger of depending on a uniform food source.

• ✗ (a) The potato suited Ireland's 'cold, rain-soaked soils' well; the famine came from disease, not unsuitable soil or climate.
• ✗ (b) The potato did serve as Ireland's staple; the passage faults reliance on one variety, not tubers as such.
• ✗ (c) The passage does not say the fungus was unstoppable; the scale of disaster came from nearly all plants being the same vulnerable variety.
• ✓ (d) The Irish were 'almost wholly dependent' on the potato and planted 'primarily one' variety, whose 'genetic frailty' let one fungus destroy nearly all of it.

Remember · Ask what turned one fungus into a national famine; the 'critical message' is that underlying cause, not the event itself.

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Answer: (b) 2, 3, 4 and 5

The passage describes child malnutrition that kills, stunts and harms learning and earnings. Delayed marriage and pregnancy, early breastfeeding, safe water and sanitation, and vaccination all act directly on a young child's nutrition and health. Who monitors the PDS is an administrative question the passage gives no basis for.

• ✗ 1. The passage says nothing about the PDS failing or about which government should monitor it; shifting control does not follow from the problem described.
• ✓ 2. Very early pregnancies often mean undernourished mothers and low-weight babies; delaying marriage and first pregnancy gives the child a healthier start against the early damage the passage describes.
• ✓ 3. Children suffer from 'not getting enough of the right foods and nutrients'; breastfeeding from birth is the first and most direct source of them.
• ✓ 4. Unsafe water and poor sanitation cause repeated infections that stop a child's body from using food, feeding the stunting the passage describes.
• ✓ 5. Malnutrition causes about half of under-5 deaths; preventing diseases through vaccination protects already weak children and their growth.

Remember · For 'practical implications', keep steps that act directly on the problem described; drop those about which authority should control a scheme.

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Answer: (c) 2 and 3 only

Arhar straw is green, decomposes fast and can be ploughed back, while silica-rich paddy straw is hard to break down and so gets burnt. Two inferences follow: growing this pulse avoids the burning that pollutes (2), and its straw can enrich the soil (3).

• ✗ 1. The yield is compared with 'existing varieties' of arhar, not with paddy; nothing is said about income.
• ✓ 2. Paddy stalks are burnt because they cannot be ploughed back, while arhar straw is ploughed in — so the pulse crop avoids the smoke from stubble burning.
• ✓ 3. Arhar straw 'can be ploughed back into the soil' and 'will decompose very fast', returning organic matter to the soil.
• ✗ 4. The passage says paddy straw is hard to decompose or salvage, not that it has 'no usefulness' at all.
• ✗ 5. Arhar is also combine-harvested yet its straw is not burnt; the cause is paddy straw's 'high silica content'.

Remember · Reject inferences that compare things the passage never compares (income) or use extreme words ('no usefulness', 'main cause').

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Answer: (d) 1, 2 and 3

Why not the tempting option · UPSC's key accepts all three assumptions. Statement 1 is the weakest, because the passage blames the drive for electricity and irrigation rather than 'over-dependence on hydropower'; but statements 2 and 3 both follow from the passage, and only option (d) contains both, so (d) is the best answer available. In the exam, settle the clear statements first and let the options decide a doubtful one.

The author contrasts the international practice of keeping a reservoir below a set level before the monsoon, so that excess rain can be stored and released in a regulated way, with Indian authorities who fill reservoirs early for power and irrigation. Statement 2 restates that practice, statement 3 is what the author's criticism rests on, and statement 1 names the drive for electricity that the passage blames; UPSC's key accepts all three.

• ✓ 1. The passage says authorities fill reservoirs early 'only to ensure greater electricity generation and irrigation'; the push for power output is treated as the reason for the risky practice. This is the loosest of the three statements, but only option (d) contains both 2 and 3, so it must be taken as valid.
• ✓ 2. The accepted practice is to keep the level 'below a certain level before the onset of monsoon season' so that there is 'space to store the excess rainwater'.
• ✓ 3. By leaving no space for excess rain and regulated release, the authorities act as if a dam's flood-cushioning role mattered less than power and irrigation; the author's complaint assumes that role is being under-valued.

Remember · When two statements are clearly valid, let the option grid decide the third: only (d) holds both 2 and 3.

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Answer: (c) Both 1 and 2

The passage says liberalization was shaped by the government's own economic problems rather than by people's priorities or long-term development goals, that its design flaws have been confirmed by experience — jobless growth, persistent poverty, rising inequality and four quiet crises including education — and that these must be resolved for growth to become meaningful development. UPSC's key takes both inferences as following: the State's economic role needs rethinking (1), and the social outcomes show that social-sector policies were neither delivered well nor funded enough (2).

• ✓ 1. The reforms reflected the government's problems, not 'the economic priorities of the people or ... the long-term development objectives', had 'limitations in conception and design', and 'these problems must be resolved' — so the State's economic role in development needs to be rethought and redefined.
• ✓ 2. Persistent poverty, rising inequality and a crisis in education decades into reform are the passage's evidence on the social side; UPSC's key reads them as showing that policies in the social sectors were neither effectively implemented nor adequately funded — the one step of reasoning that 'most rational and logical' allows.

Remember · For 'most rational and logical inference', ask what the passage's stated outcomes and causes, taken together, point to; UPSC allows one step of common-sense reasoning from outcome to cause.

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Answer: (d) Neither 1 nor 2

The passage is critical of how liberalization was designed and lists jobless growth, persistent poverty and inequality as its results. It neither prescribes deeper global integration nor assumes that liberalization will in time reduce poverty and create enough jobs — its evidence points the other way.

• ✗ 1. Integration with the global economy is never proposed; the remedies the passage points to lie in agriculture, infrastructure, industrialization and education.
• ✗ 2. The author reports 'jobless growth, persistent poverty and rising inequality' since liberalization began — the opposite of this assumption.

Remember · An assumption the author's own evidence contradicts cannot be valid.

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Answer: (d) All the options are equally good for the customer.

Each option multiplies the price by the same three factors — 0·9, 0·8 and 1·1 — only in a different order. Multiplication does not depend on order, so the customer pays the same in every case.

1. Take the price as ₹ 100.
2. Option 1: 100 × 0·9 × 0·8 × 1·1 = ₹ 79·2.
3. Option 2: 100 × 0·8 × 0·9 × 1·1 = ₹ 79·2.
4. Option 3: 100 × 1·1 × 0·8 × 0·9 = ₹ 79·2.
5. All three cost the same, so all options are equally good.

Remember · Successive percentage changes are multiplying factors; their order never changes the final result.

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Answer: (d) 246

G, H, I are 7, 8, 9 and 789 × 2 = 1578; D, E, F are 4, 5, 6 and 456 × 2 = 912. So each code is the three letter numbers written together and doubled: ABC = 123 × 2 = 246.

1. GHI → 7, 8, 9 → 789; 789 × 2 = 1578.
2. DEF → 4, 5, 6 → 456; 456 × 2 = 912.
3. ABC → 1, 2, 3 → 123; 123 × 2 = 246.

Remember · In number codes, write the letter positions side by side first and look for a single operation that gives the code.

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Answer: (d) NPJM

Going from ACPQ to BESU, the four letters move forward by 1, 2, 3 and 4 places. Applying the same shifts to MNGI gives N, P, J and M.

1. A → B (+1), C → E (+2), P → S (+3), Q → U (+4).
2. M + 1 = N, N + 2 = P, G + 3 = J, I + 4 = M.
3. The missing term is NPJM.

Remember · In letter analogies, write the shift for each position separately; the shifts often run 1, 2, 3, 4.

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Answer: (c) (1/3)⁻⁴

A negative power of a fraction flips it: (1/a)⁻ⁿ = aⁿ. The four values are 2⁶ = 64, 4³ = 64, 3⁴ = 81 and 6² = 36, so (1/3)⁻⁴ is the largest.

1. (1/2)⁻⁶ = 2⁶ = 64.
2. (1/4)⁻³ = 4³ = 64.
3. (1/3)⁻⁴ = 3⁴ = 81.
4. (1/6)⁻² = 6² = 36.
5. The largest is 81, i.e. (1/3)⁻⁴.

Remember · (1/a) raised to −n equals aⁿ; rewrite every option this way before comparing.

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Answer: (d) 1¾ m

The greatest length that divides both is their HCF. In quarters of a metre the lengths are 14/4 and 35/4, and HCF(14, 35) = 7, so x = 7/4 = 1¾ m.

1. 3½ = 14/4 m and 8¾ = 35/4 m.
2. HCF(14, 35) = 7, so the greatest common length = 7/4 m = 1¾ m.
3. Check: 3½ ÷ 1¾ = 2 and 8¾ ÷ 1¾ = 5, both whole numbers.

Remember · HCF of fractions: put them over a common denominator, take the HCF of the numerators, keep the denominator.

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Answer: (d) 1971–1981

Natural growth rate is birth rate minus death rate. The gaps for the four offered periods are 12·6, 12·5, 22·2 and 22·3, so 1971–1981 is the highest, just ahead of 1961–1971.

1. 1911–1921: 48·1 − 35·5 = 12·6.
2. 1941–1951: 39·9 − 27·4 = 12·5.
3. 1961–1971: 41·1 − 18·9 = 22·2.
4. 1971–1981: 37·1 − 14·8 = 22·3 — the maximum.
5. Check: the periods not offered are lower too (10·1, 14·0, 18·9).

Remember · Natural growth = birth rate − death rate. Compute only the offered periods, and compute the close ones exactly.

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Answer: (b) 14/11

The repeating block 27 has two digits, so 0·2727... = 27/99 = 3/11. Adding the whole part gives 1 + 3/11 = 14/11.

1. Let x = 1·2727...; then 100x = 127·2727...
2. 100x − x = 126, so x = 126/99 = 14/11.
3. Check: 14 ÷ 11 = 1·2727...

Remember · A pure recurring decimal 0·(ab) repeating = ab/99; handle the whole-number part separately.

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Answer: (b) 1022

A number leaving remainder 2 with each divisor is 2 more than a common multiple of 3, 4, 5 and 6. Their LCM is 60, the first four-digit multiple of 60 is 1020, and so the answer is 1022.

1. LCM(3, 4, 5, 6) = 60.
2. Smallest four-digit multiple of 60: 60 × 17 = 1020 (60 × 16 = 960 has three digits).
3. Required number = 1020 + 2 = 1022.
4. Check: 1022 = 3 × 340 + 2 = 4 × 255 + 2 = 5 × 204 + 2 = 6 × 170 + 2.

Remember · Same remainder r for every divisor: the number = k × LCM + r. Find the smallest k that gives the required size.

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Answer: (c) 34 2/7%

Every married man has exactly one wife in the city, so married men equal married women: 40% of men = 30% of women, which makes men : women = 3 : 4. Then 1·2 + 1·2 = 2·4 of every 7 adults are married, i.e. 34 2/7%.

1. Married men = married women, so 0·4M = 0·3W, giving M : W = 3 : 4.
2. Take 3 men and 4 women: married men = 0·4 × 3 = 1·2, married women = 0·3 × 4 = 1·2.
3. Married adults = 2·4 out of 7 = 24/70 = 12/35.
4. 12/35 × 100 = 240/7 = 34 2/7%.

Remember · In marriage-count problems, equate married men and married women first; that fixes the ratio of the two groups.

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Answer: (a) 50

Only the last two digits matter. 62 × 75 = 4650 ends in 50, and multiplying a number ending in 50 by any odd number again ends in 50. The other factors 51, 27 and 35 are all odd, so the remainder is 50.

1. 62 × 75 = 4650, which ends in 50.
2. 50 × (odd number) always ends in 50, since 50 × odd = 100k + 50.
3. 51 × 27 × 35 = 48195 is odd, so the whole product ends in 50.
4. Remainder = 50.

Remember · Remainder on division by 100 = last two digits; pair factors that quickly give a round ending like 50 or 00.

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Answer: (c) ₹ 400

Clearing the fractions by multiplying by 12 turns the ratio into 6 : 9 : 10, a total of 25 parts worth ₹ 100 each. The largest share (10 parts) exceeds the smallest (6 parts) by 4 parts, or ₹ 400.

1. Multiply 1/2 : 3/4 : 5/6 by 12 to get 6 : 9 : 10.
2. Total parts = 25; one part = 2,500 ÷ 25 = ₹ 100.
3. Maximum − minimum = (10 − 6) × 100 = ₹ 400.
4. Check: shares 600 + 900 + 1,000 = 2,500.

Remember · For a ratio of fractions, multiply every term by the LCM of the denominators to get whole-number parts.

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Answer: (b) For any whole number n

For any positive integer n, 10ⁿ + 1 is 1 followed by zeros and a final 1, so its digit sum is 2. For n = 0 it is 1 + 1 = 2, whose digit sum is also 2. So it holds for every whole number 0, 1, 2, …, but not for fractional n, where 10ⁿ + 1 is not a whole number.

1. n = 1, 2, 3 …: 11, 101, 1001 … — digit sum 2.
2. n = 0: 10⁰ + 1 = 2 — digit sum 2.
3. n = 1/2: 10ⁿ + 1 = √10 + 1 ≈ 4·16, not a whole number, so 'any real number' fails.
4. So the statement holds for any whole number n.

Remember · Always test the edge value n = 0 before choosing 'positive integers only'.

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Answer: (c) It remains the same.

Shifting students between groups inside the class changes the group averages but not the class as a whole: the same students with the same total weight remain. So the class average stays the same.

1. Class average = total weight of all students ÷ number of students in the class.
2. Moving students from A and B to C keeps every student in the class.
3. Both the total weight and the head count are unchanged, so the class average is unchanged.

Remember · Ask which total is being averaged; moving members within that total leaves its average unchanged.

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Answer: (a) 16

Choosing 3, 4 or all 5 of the five denominations gives 10 + 5 + 1 = 16 selections. Each denomination is more than the sum of all the smaller ones, so no two selections can give the same amount — all 16 sums are different.

1. Ways to choose 3, 4 or 5 of 5 denominations: ⁵C₃ + ⁵C₄ + ⁵C₅ = 10 + 5 + 1 = 16.
2. 100 > 50, 200 > 50 + 100, 500 > 50 + 100 + 200 and 2,000 > 850, so different selections always give different totals.
3. Number of different sums = 16.
4. Check: the 16 sums run from ₹ 350 (50 + 100 + 200) to ₹ 2,850 (all five) with no repeats.

Remember · Count selections with nCr, then check whether two selections can give the same total before answering.

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Answer: (c) India lacks the institutional mechanism to deal with the failure of banks.

The author traces shrinking industrial credit to one missing reform: a resolution corporation for banks. Because there is 'no proper system to shut them down', weak banks keep needing capital. The direct inference is that India lacks a mechanism to handle bank failure.

• ✗ (a) Too sweeping: the passage flags shrinking credit to industry, not a banking system unable to support growth at all.
• ✗ (b) The reforms are called 'incomplete' in one specific respect; the passage does not judge the whole economy against expectations.
• ✓ (c) A 'resolution corporation for banks' was never set up, and 'there is no proper system to shut them down'.
• ✗ (d) Foreign investment as a substitute for bank credit is never discussed.

Remember · Pick the inference that states the passage's own diagnosis; reject sweeping verdicts built from one problem.

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Answer: (c) Getting renewable resources to market too soon may be costly.

Renewables are 'still going through their cost curves and learning curves', and business houses hesitate to invest heavily while 'the technology is not yet ready'. The sound inference is that pushing these resources to market before they mature may prove costly.

• ✗ (a) The government is 'strongly committed' and acting on two fronts; the passage makes no prediction that the target will be missed.
• ✗ (b) Importing gas is not mentioned, and the passage does not ask India to stop investing in renewables.
• ✓ (c) Technology still on its cost and learning curves is expensive per unit of output — which is why businesses are 'wary of investing too heavily' now.
• ✗ (d) The passage mentions a push towards a gas-based economy but nothing about exploration.

Remember · Inferences should stay as cautious as the passage; predictions of failure and new policy prescriptions usually overreach.

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Answer: (a) 1 only

The government has 'invested heavily' in renewable energy even though the technology is 'not yet ready' and still costly — the passage's warning is that such early public spending can be inefficient and expensive (1). The emissions target rests on both a gas-based push and renewables, not on gas alone, so 2 fails.

• ✓ 1. The contrast between heavy government investment and business wariness 'at a time when the technology is not yet ready' assumes that public money put into unready technology can be inefficient and costly — the same idea as item 62's answer.
• ✗ 2. Towards the target the government 'has initiated a strong push towards a gas-based economy and has also invested heavily in renewable energy' — both, not gas alone.

Remember · Read 'and also' carefully: a goal pursued through two means cannot be said to rest on one of them.

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Answer: (c) Both 1 and 2

The passage separates genome editing from genome modification: editing finds and changes the relevant part of the plant's own genome, as plants do naturally under attack. So it needs no gene transfer from another plant (1), and it alters chosen genes precisely in a way akin to the natural adaptive process (2).

• ✓ 1. Editing works by 'finding the part of a plant genome' to change and repairing it with molecular scissors; the stated contrast with genome modification implies that no gene is brought in from another plant.
• ✓ 2. The cutting and repair 'occurs naturally when plants are under attack', and laboratories can now alter 'the relevant genes ... very precisely', speeding up this evolutionary process.

Remember · When a passage opens with 'X is different from Y', expect an assumption that captures the difference.

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Answer: (c) Solid waste management should be integrated with the maintenance of drainage and sewerage networks.

The less understood danger, the author says, arises when poor solid waste management combines with poor drainage and sewage dumped into storm drains, producing choked drains that breed mosquitoes. The message is that these systems must be managed together, not in isolation.

• ✗ (a) The passage says untreated sewage is dumped into storm-water drains, but it does not claim India has no separate networks; that detail is not the message.
• ✗ (b) Resources and legal powers of local bodies are not discussed.
• ✓ (c) The harm comes from 'unscientific solid waste management' combining with 'poor drainage and dumping of untreated sewage into drains' — so they must be handled as one system.
• ✗ (d) The consequence described is disease from stagnant water, not drinking water shortage.

Remember · When a passage says harm arises when two failures 'combine', the critical message is usually to tackle them together.

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Answer: (b) Religious freedom under the Constitution is open to State intervention.

Article 25 grants religious freedom 'subject to public order, morality, health and to the other provisions' of Part III, which places other fundamental rights, such as equality before law, above it. So religious freedom is conditional, and the State can step in on these grounds.

• ✗ (a) The 'subject to' rider allows State action on grounds of public order, morality, health and other fundamental rights.
• ✓ (b) The rider sets 'conditions precedent' for protecting religious practices and subordinates religious freedom to the other rights in Part III.
• ✗ (c) Article 25 sits in Part III, which 'assures people certain fundamental rights' — religious freedom is one of them.
• ✗ (d) This reverses the passage, which stresses that religious practices are protected only subject to these conditions.

Remember · Look for the qualifier ('subject to', 'except', 'unless'); the inference usually follows from the limit, not the grant.

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Answer: (d) 6

With D fixed first and I fixed last, only E, L and H are left to arrange in the three middle places. Three distinct letters can be arranged in 3! = 6 ways.

1. Positions 1 and 5 are fixed: D _ _ _ I.
2. E, L and H fill the three middle places in 3! = 3 × 2 × 1 = 6 ways.
3. Check: DELHI, DEHLI, DLEHI, DLHEI, DHELI, DHLEI — six words.

Remember · Fix the letters whose places are given, then arrange only the remaining ones: n distinct letters give n! orders.

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Answer: (c) 16: 9

Each removal takes away 4/20 = 1/5 of whatever A is present, leaving 4/5 of it. After two rounds A = 20 × (4/5)² = 12·8 litres and B = 7·2 litres, a ratio of 16 : 9.

1. After the first replacement: A = 20 − 4 = 16 litres.
2. Second removal takes 1/5 of the mixture, so A falls to 16 × 4/5 = 12·8 litres.
3. B = 20 − 12·8 = 7·2 litres.
4. A : B = 12·8 : 7·2 = 128 : 72 = 16 : 9.

Remember · Repeated replace-and-remove: remaining original = initial × (1 − removed/total)ⁿ.

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Answer: (c) 62

Turn both averages into totals: 50 × 46·4 = 2320 and 60 × 49 = 2940. The last ten innings added 620 runs, an average of 62.

1. Runs after 50 innings = 50 × 46·4 = 2320.
2. New average = 46·4 + 2·6 = 49; runs after 60 innings = 60 × 49 = 2940.
3. Runs in the last 10 innings = 2940 − 2320 = 620.
4. Average of the last 10 = 620 ÷ 10 = 62.

Remember · Averages over different counts: convert each to a total, subtract, and divide by the new count.

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Answer: (b) ₹ 40

A 25% price rise means the same money buys 1/1·25 = 4/5 of the earlier quantity, i.e. 1/5 less. That lost fifth is 6 kg, so ₹ 1,200 used to buy 30 kg, at ₹ 40 a kg.

1. New price = 5/4 of old, so quantity for ₹ 1,200 becomes 4/5 of the old quantity — a fall of 1/5.
2. 1/5 of old quantity = 6 kg, so old quantity = 30 kg.
3. Original price = 1,200 ÷ 30 = ₹ 40 per kg.
4. Check: new price ₹ 50 buys 1,200 ÷ 50 = 24 kg, which is 6 kg less.

Remember · Price up by 1/n means quantity down by 1/(n + 1) for the same spending; here 25% = 1/4, so quantity falls by 1/5.

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Answer: (a) 6

X would finish the whole work in 40 days and Y in 24 days, so together they do 1/40 + 1/24 = 1/15 of it each day. 40% of the work therefore takes 0·4 × 15 = 6 days.

1. X: 20% in 8 days, so 100% in 40 days. Y: 25% in 6 days, so 100% in 24 days.
2. Together per day: 1/40 + 1/24 = 3/120 + 5/120 = 8/120 = 1/15.
3. Whole work together = 15 days; 40% of it = 0·4 × 15 = 6 days.
4. Check: in 6 days X does 6/40 = 15% and Y does 6/24 = 25%; total 40%.

Remember · Convert partial work to full-job time first, add daily rates, then scale to the fraction of work asked.

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Answer: (b) lies between v and 2v km/hr

All four legs cover the same distance, so the slow legs take the most time and pull the average down. Total time = (d/v)(1 + 1/2 + 1/3 + 1/4) = 25d/12v for a distance of 4d, giving an average of 48v/25 = 1·92v.

1. Let XY = d km. Total distance = 4d.
2. Total time = d/v + d/2v + d/3v + d/4v = (d/v)(12 + 6 + 4 + 3)/12 = 25d/12v.
3. Average speed = 4d ÷ (25d/12v) = 48v/25 = 1·92v.
4. 1·92v lies between v and 2v.

Remember · For equal distances at different speeds, average speed is the harmonic mean — always closer to the slower speeds.

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Answer: (b) 2 only

A circle and a square need not meet at all — a small circle can sit inside the square without touching it — so the minimum is 0, not 2. Each side is a straight segment that can cut a circle at most twice, so four sides give at most 8 points, and 8 is actually reachable.

1. A small circle inside the square, or one far away from it, meets the square in 0 points — so the minimum is 0.
2. A straight line meets a circle in at most 2 points; the square has 4 sides, so at most 4 × 2 = 8 points.
3. 8 is reached by a circle with the square's centre whose radius is more than half the side but less than half the diagonal: it crosses each side twice.

• ✗ 1. The minimum is 0 — the circle can lie wholly inside or wholly outside the square without touching it.
• ✓ 2. Four sides × at most 2 crossings each = 8, and a suitably sized concentric circle achieves it.

Remember · For 'minimum' claims, look for the case where the figures do not touch at all; for 'maximum', count lines × 2 for a circle.

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Answer: (d) 3: 1

Half the time over the same distance means the downstream speed is twice the upstream speed. So b + c = 2(b − c), which gives b = 3c: still-water speed : current = 3 : 1.

1. Let the speed in still water be b and the current c.
2. Downstream time is half the upstream time, so downstream speed = 2 × upstream speed.
3. b + c = 2(b − c), so b = 3c.
4. b : c = 3 : 1.
5. Check: b = 3, c = 1 gives speeds 4 and 2 — half the time downstream.

Remember · Same distance: time ratio is the inverse of speed ratio. Downstream = b + c, upstream = b − c.

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Answer: (a) 3

x² − y² = (x − y)(x + y) = 63, so each way of writing 63 as a product of two factors gives one pair. 63 = 1 × 63 = 3 × 21 = 7 × 9, giving (32, 31), (12, 9) and (8, 1).

1. x² − y² = (x − y)(x + y) = 63, with x − y < x + y.
2. Factor pairs of 63: (1, 63), (3, 21), (7, 9) — all odd, so x and y come out whole.
3. (1, 63): x = 32, y = 31. (3, 21): x = 12, y = 9. (7, 9): x = 8, y = 1.
4. 3 pairs.
5. Check: 32² − 31² = 63, 144 − 81 = 63, 64 − 1 = 63.

Remember · Difference of squares: factor the number as (x − y)(x + y) with both factors of the same parity; each factor pair gives one solution.

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Answer: (d) 5/6

Each fraction has the form n/(n + 1). Adding 5 to both parts changes it by 5/[(n + 1)(n + 6)], which shrinks as n grows. So 5/6, the largest, changes least (by 5/66).

1. 2/3 → 7/8: change 7/8 − 2/3 = 5/24 ≈ 0·208.
2. 3/4 → 8/9: change 5/36 ≈ 0·139.
3. 4/5 → 9/10: change 5/50 = 0·1.
4. 5/6 → 10/11: change 5/66 ≈ 0·076 — the smallest.

Remember · Adding the same number to top and bottom pushes a proper fraction towards 1; fractions already closer to 1 move less.

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Answer: (d) 3n + 1

The digits greater than 3 that are divisible by 3 are 6 and 9; 6 is divisible by 6, so n = 9. Then 3n + 1 = 28 is the only option divisible by 4.

1. Digits above 3 divisible by 3: 6 and 9. 6 is divisible by 6, so n = 9.
2. 2n = 18, 3n = 27, 2n + 4 = 22 — none divisible by 4.
3. 3n + 1 = 28 = 4 × 7 — divisible by 4.

Remember · When a condition pins a single digit, find that digit first and then test every option numerically.

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Answer: (c) 100

1 litre = 1 kg means 1 gram of water occupies 1 cm³. Since 1 cm³ = 10 × 10 × 10 = 1000 mm³, 0·1 g occupies 100 mm³.

1. 1 litre = 1000 cm³ weighs 1 kg = 1000 g, so 1 g of water = 1 cm³.
2. 1 cm³ = 10 mm × 10 mm × 10 mm = 1000 mm³.
3. 0·1 g = 0·1 cm³ = 100 mm³.

Remember · Water: 1 g = 1 cm³ = 1 mL; and 1 cm³ = 1000 mm³.

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Answer: (a) 10 kg

Going from full to one-third full removes two-thirds of the water and 20 kg of weight, so the full water weighs 30 kg. The empty vessel is 40 − 30 = 10 kg.

1. Full: vessel + water = 40 kg. One-third full: vessel + water/3 = 20 kg.
2. Subtracting: (2/3) × water = 20 kg, so water = 30 kg.
3. Empty vessel = 40 − 30 = 10 kg.
4. Check: 10 + 30/3 = 20 kg.

Remember · Subtract the two readings: the difference belongs to the water alone, which fixes the water's full weight.

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Answer: (b) 29

Each jump-and-slide gains a net 15 cm, but the final jump does not slide back. The frog needs to be within 30 cm of the top, i.e. at 420 cm, before its last jump; 28 jumps with slides get it there, and the 29th jump takes it out.

1. Well depth = 450 cm; net gain per jump-and-slide = 30 − 15 = 15 cm.
2. The last jump of 30 cm must start from at least 450 − 30 = 420 cm.
3. 420 ÷ 15 = 28 jumps (with slides) to reach 420 cm.
4. The 29th jump lifts it from 420 to 450 cm and out — 29 jumps.
5. Check: after 27 jumps it is at 405 cm; the 28th reaches only 435 cm and slides back to 420 cm.

Remember · Climb-and-slip problems: subtract the last full jump from the height, count net gains to reach that point, then add one jump.

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