UPSC 2019 Prelims Questions with Answers

Twice Sohan's weight is less than the weight of Mohan and less than that of Rohan, so Sohan weighs less than both Mohan and Rohan. Hence Sohan's weight is the least.

Twice Rohan's weight is greater than Mohan's weight and greater than Sohan's weight. This only tells us Rohan is heavier than Sohan; Rohan's relation to Mohan (and Mohan's to Sohan) is not decided by the given statements. Hence whose weight is greatest cannot be determined.

Either C or F scores the highest. E is ranked third, so E is not the least; but when C scores the highest, E scores the least, so C cannot be the highest here - F is first and B is last. D and A are close competitors with D above A. With E third, the arrangement is F, C, E, D, A, B. Hence C gets more marks than E.

B scores the least. If C were the highest, E would be the least, which contradicts B being least, so F is the highest. With F first and B last, C - which is higher than A and below F - can be either second or third depending on where E falls. Hence C's rank is second or third.

F stands second, so F is not the highest; therefore C is the highest and E scores the least. With C first, F second, D above A (close competitors) and A above B, the ranking is C, F, D, A, B, E. B is therefore fifth.

3. The intended number is 42527B63. A number is divisible by 3 when its digit sum is divisible by 3. The fixed digit sum is 4+2+5+2+7+6+3 = 29. Thus 29+B must be divisible by 3. Possible digit values are B=1,4,7. Therefore 3 values are possible.

9. Replace symbols: $ means division, @ means multiplication and # means minus. So 10#5@1$5 becomes 10 - 5 x 1 / 5. Apply multiplication/division before subtraction: 5 x 1 / 5 = 1. Therefore value = 10 - 1 = 9.

10. Since x, y and z are natural numbers, each is at least 1. Put x'=x-1, y'=y-1, z'=z-1. Then x'+y'+z'=3, where x', y', z' are non-negative integers. Number of solutions = C(3+3-1,3-1)=C(5,2)=10.

Classify the colourings by the number of black faces (0 to 6). By rotational symmetry: 0 or 6 black faces - 1 way each; 1 or 5 black faces - 1 way each; 2 or 4 black faces - 2 ways each (adjacent or opposite); 3 black faces - 2 ways (all meeting at a corner, or two opposite faces plus one). Total = 1 + 1 + 2 + 2 + 2 + 1 + 1 = 10.

F is D's father, and B (D's sibling) is also F's child. A is B's daughter, so F is A's grandfather and F's wife E is A's grandmother. C is B's cousin, so C is a child of F's sibling, making E (F's wife) C's aunt. Hence E is the wife of F, the grandmother of A and the aunt of C - statements 1, 2 and 3.

1. Sequence: 1, 5, 7, 3, 5, 7, 4, 3, 5, 7. The 5 at position 2 is followed by 7 and is not preceded by 3. The 5 at position 5 is followed by 7 but is preceded by 3. The 5 at position 9 is followed by 7 but is preceded by 3. Hence only one such 5 exists.

First piece = average of the single-digit odd primes 3, 5, 7 = 5 units. Let the third piece be z. Second piece = 5 + z/3, and the third equals the other two together: z = 5 + 5 + z/3, so 2z/3 = 10 and z = 15. Second piece = 10. Total length = 5 + 10 + 15 = 30 units.

Alternative letters from B onwards are lowercase: b, C, d, E, f, G, h, I, j, K, l, M... For JULY: j (lower), U (upper), l (lower), Y (upper) = jUlY.

The passage shows that of 7,000 apple varieties grown in the nineteenth century fewer than a hundred remain, because production concentrated on a few high-yielding varieties - a decline humans have driven (1). It warns that if crop disease or climate change decimates the few varieties we depend on, food scarcity would follow (3). Statements 2 and 4 are not stated in this form.

The passage describes technology-driven production and distribution of food causing the dwindling of crop diversity and ecosystems adapting less easily than in the past - the loss of the Earth's natural resilience at human hands (3). It does not treat global warming as a frequently occurring natural phenomenon or comment on small mammals (1 and 2).

The passage says melting glaciers driven by warming are releasing suspended sediment into coastal waters, reducing species diversity among benthos and impacting an entire seafloor ecosystem. All three assumptions follow: glacier regions experience amplified warming (1), global warming can lead to seafloor sedimentation (2), and melting glaciers can reduce marine biodiversity in some areas (3).

The microbiome passage explains that the bacteria of an infant's gut thrive on a sugar present in mother's milk, and that malnourished mothers have low levels of this sugar, so their infants' microbiomes fail to mature and the babies become malnourished. Statement 2 follows from this. Processed probiotic foods are not mentioned, so statement 1 is not valid.

The passage links mother's milk sugars, gut microbiomes and infant nutrition. Its implication is that research on the effects of gut bacteria on nutrition has policy implications.

Assumption 1 is valid: the passage says India especially needs an effective long-term regulatory system based on collaboration, implying that in today's global economy regulations are not effectively used to protect local markets. Assumption 2 is not: the passage does not say social and environmental concerns are generally ignored by governments across the world.

The passage says India must implement solutions that fit its unique circumstances and especially needs an effective long-term regulatory system based on collaboration. Against the hyper-competitive global economy, the inference is that regulations in India do not favour its integration with today's hyper-competitive global economy.

8. Add column-wise: 5B7 + 136 = 7A3. Units: 7 + 6 = 13, so write 3 and carry 1. Tens: B + 3 + 1 = A plus any carry to hundreds. Hundreds: 5 + 1 plus carry from tens must become 7, so there must be a carry 1 from the tens column. Thus B + 4 >= 10. Also 7A3 is divisible by 3, so 7 + A + 3 = A + 10 is divisible by 3, giving A = 2, 5 or 8. Testing possible B values, only B = 8 gives a consistent carry and digit sum. Hence B = 8.

A non-leap year repeats its calendar after 6 years. 2009 + 6 = 2015 has the same calendar as 2009.

Both S1 and S2 are necessary. S1 alone says p and q are prime, but pq may be odd, e.g. 3 x 5, or even, e.g. 2 x 3. S2 alone says p+q is odd, but without knowing primality, pq is not fixed. Together, if p and q are prime and p+q is odd, one of them must be 2 and the other an odd prime. Therefore pq is even, so the question can be answered. Both statements are necessary.

17 m. Let the point where B joins A be after A first walks 20 m east. They then walk 10 m east together, reaching (30,0) from A's house. A turns left and goes 2 m north, then right and 12 m east, reaching (42,2). B turns right and goes 8 m south, then left 4 m east, then right 5 m south, reaching (34,-13). The office separation is sqrt((42-34)^2 + (2+13)^2)=sqrt(8^2+15^2)=17 m.

18. Let the number of Rs. 50 notes be a and Rs. 10 notes be b; remaining amount is paid in Rs. 1 notes. For Rs. 107, a can be 0, 1 or 2. If a=0, b can be 0 to 10: 11 ways. If a=1, b can be 0 to 5: 6 ways. If a=2, b can be 0: 1 way. Total ways = 11 + 6 + 1 = 18.

54 kg. Let total weight of the two children be C and mother's weight be M. C + M = 3 x 50 = 150. Let father's weight be F = 60. C + F = 3 x 52 = 156, so C = 96. Therefore M = 150 - 96 = 54 kg.

Filling the gaps with c, b, b, b gives c c a c c a a b a a b b c c b - a balanced arrangement in which the double-letter blocks cc, aa, bb repeat in order with single separators a, b, c between them. The required letters are cbbb.

1049. Pages 1-9 use 9 digits. Pages 10-99 use 90 x 2 = 180 digits. Total up to page 99 = 189 digits. Remaining digits = 3089 - 189 = 2900. Four-digit page numbers use 4 digits each, so 2900/4 = 725 pages from 1000 onward? More directly, pages 100-999 use 900 x 3 = 2700 digits, taking the total to 2889. Remaining digits = 200, which means 50 four-digit pages: 1000 to 1049. Hence the book has 1049 pages.

9. Of 15 people, 3 read neither language, so 12 read at least one. Let both French and English readers = x. By inclusion-exclusion, 7 + 8 - x = 12, so x = 3. People reading exactly one language = 12 - 3 = 9.

The even-position terms are squares of 14, 12, 10, 8 (196, 144, 100, 64) and the odd-position terms are 2^2, 4^2, 6^2, 8^2 (4, 16, 36, 64). Hence X = 8^2 = 64.

For a proper fraction a/b (a < b) and k > 0, (a+k)/(b+k) > a/b because b(a+k) - a(b+k) = k(b-a) > 0. Hence the resulting fraction is always greater than the original fraction.

5:36 p.m. The faulty clock gains 10 minutes in every 24 hours, so when real elapsed time is 24 hours, the clock shows 24 hours 10 minutes. From Monday 8:00 a.m. to the displayed Wednesday 6:00 p.m. is 58 displayed hours. Real elapsed time = 58 x 24/(24 + 1/6) = 58 x 144/145 = 57.6 hours = 57 hours 36 minutes. Correct time = Monday 8:00 a.m. + 57 hours 36 minutes = Wednesday 5:36 p.m.

The successive differences are 3, 5, 7, 11, 13 - the consecutive odd primes. The next prime difference is 17, so X = 93 - 17 = 76.

The passage says banking outreach depends on last-mile workers who must provide not just basic accounts but also insurance and pension products, and that banking correspondents are currently too few and restricted to one bank. The crucial inference is that meaningful financial inclusion requires banking correspondents with diverse skills.

The passage implies that much of modern economic growth rests on technology, that the modern Indian economy lacks a symbiotic relationship with labour-intensive livelihoods, and that the service sector is not very labour-intensive. Statements 1, 2 and 3 are correct.

The passage criticizes the static Development Plan drawn exclusively from technical expertise, distanced from people's lived experiences and needs, and actively excluding large numbers of people and practices that are integral to the city. The passage therefore argues in favour of participation of peoples' groups in city planning.

The passage stresses multiple factors needing regulation/monitoring and the importance of training inspection staff and street vendors. Assumptions 1 and 2 are valid; legislation is not discussed.

The passage says most invasive species are neither very successful nor very harmful, often less of a nuisance than vigorous natives, and that they colonise disturbed habitats where little else lives. The inference is that sometimes the campaigns against foreign plants are pointless - not that such laws are never needed or that foreign plants should be deliberately introduced.

Assumption 1 is valid: corporations invest in herbicide-tolerant and pest-resistant crops for short-term profit, while the National Farmers Commission had to point out that priority should be given to genes for drought and salinity resistance - natural-calamity effects are not given due consideration by GM companies. Assumption 2 is not: the passage does not claim GM technology will fail to solve warming-related problems in the long run.

The passage criticizes the corporate-driven, profit-maximising GM agenda and endorses the National Farmers Commission's call to prioritise drought- and salinity-resistance genes. The corollary is that public research institutions should take the lead in GM technology and prioritise the technology agenda.

39.7. Original total marks = 100 x 40 = 4000. One mark was recorded as 83 instead of 53, so the total was overstated by 30. Correct total = 4000 - 30 = 3970. Correct mean = 3970/100 = 39.7.

Money for 8 handsets buys 10 after discount: 8P = 10 x discounted price. Discounted price = 0.8P, so the discount is 20%.

Ena is 13 years old, so her mother is 13 + 24 = 37 and her father is 37 + 3 = 40. Ena was born 4 years after her parents' marriage, so her father married at 40 - 13 - 4 = 23 years of age.

x is at least 25 and y is at most 40, so y - x is at most 40 - 25 = 15 in every case. Hence (y - x) is less than or equal to 15 is always correct; the other options are not always true.

25. If X gives Y a 40 m start in a 1000 m race, then when X runs 1000 m, Y runs 960 m, so Y/X speed ratio = 960/1000. If X gives Z a 64 m start, then Z/X speed ratio = 936/1000. Hence Z/Y speed ratio = 936/960 = 0.975. In a 1000 m race between Y and Z, when Y runs 1000 m, Z runs 975 m. Therefore Y should give Z a 25 m start.

Seeta swims every 3rd day (after a gap of 2 days) and Geeta every 4th day (after a gap of 3 days). They swim together every LCM(3,4) = 12 days. Since they swam together on 1st January, the next common day is 1 + 12 = 13th January.

A - B = 20 and B = 0.95A, so 0.05A = 20, i.e. A = 400 and B = 380.

4. Let the two-digit number be 10a+b and its reverse be 10b+a. Given (10a+b)/(10b+a)=4/7. Cross-multiplying: 70a+7b = 40b+4a, so 66a = 33b and b = 2a. Since b is a digit, a can be 1, 2, 3 or 4. The numbers are 12, 24, 36 and 48. Hence 4 such pairs exist.

Only 60% play cricket, and all non-cricketers play football, so conclusions about two-wheeler ownership cannot be fixed at 60% (1), nothing says cricketers have no two-wheeler (2), and cricket players may or may not play football (3 is actually supported by the premise that only non-cricketers play football). The official key identifies 1 and 2 as the conclusions that cannot be drawn.

Everyone either shopped, went sightseeing or rested. Since 10% rested, 90% did at least one of the first two activities. Using inclusion-exclusion: shopping alone + sightseeing alone + both = 80 + 50 - 90 = 40%. So 40% did both (1). Only-shopping = 80 - 40 = 40%, not 20%, so statement 2 is wrong.