Logical & analytical reasoning

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

The only link to watches is 'some computers', and those computers need not include any mobile or radio. A diagram with the watches placed away from the mobiles fits every statement, so neither 'certainly' conclusion is forced.

1. Some radios are mobiles and all mobiles are computers, so some radios are computers.
2. 'Some computers are watches' does not say which computers.
3. Draw the watches overlapping only the part of 'computers' that has no mobile and no radio — all three statements still hold.
4. In that diagram no radio and no mobile is a watch, so neither conclusion is certain.

• ✗ Conclusion-I Radios reach watches only through two 'some' links; a radio being a watch is possible but not certain.
• ✗ Conclusion-II All mobiles are computers, but the computers that are watches may be the ones that are not mobiles.

Remember · A 'certainly' conclusion must hold in every possible diagram — try to draw one where it fails.

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Answer: (c) Both Conclusion-I and Conclusion-II

The almirahs that are chairs are also tables, so Conclusion-I is certain. Conclusion-II only claims a possibility ('may not'), and since nothing fixes how cats relate to chairs, that possibility is open — so both follow.

1. Some almirahs are chairs and all chairs are tables, so those almirahs are tables — 'some almirahs are tables' is certain.
2. Cats connect to chairs only through 'some almirahs', a particular link, so there is no definite relation between cats and chairs.
3. With no definite relation, a diagram where some cats are not chairs is allowed — so 'some cats may not be chairs' follows.

• ✓ Conclusion-I 'Some almirahs are chairs' + 'All chairs are tables' gives 'some almirahs are tables' with certainty.
• ✓ Conclusion-II It states only a possibility; since the statements do not force every cat to be a chair, the possibility holds.

Remember · 'Some A are B' + 'All B are C' gives 'some A are C'. A 'may / may not' conclusion follows when no definite relation rules it out.

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Answer: (b) 25 km

Tracking the turns, the north–south legs cancel out (10 south, 40 north, 30 south) and the east legs add to 20 + 5 = 25 km. The bank is 25 km due East of the house.

1. Place the house at (0, 0), East as +x and North as +y.
2. 10 km South → (0, −10). Facing South, a left turn faces East: 20 km → (20, −10).
3. Left again faces North: 40 km → (20, 30). Right faces East: 5 km → (25, 30).
4. Right again faces South: 30 km → (25, 0), the bank.
5. The bank is 25 km due East of the house, so the shortest distance is 25 km.

Remember · Add the net East–West and North–South movements separately; use Pythagoras only if both are non-zero.

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Answer: (b) An angle less than 45° North of East

She ends 6 km North and 8 km East of the start. Because the northward part (6) is smaller than the eastward part (8), she is North of East at an angle below 45°.

1. Net North = 12 − 6 = 6 km; East = 8 km.
2. She is 8 km East and 6 km North of the start.
3. tan θ = 6/8 = 0.75, which is less than 1, so the angle from East towards North is less than 45° (about 36.9°).

Remember · Compare the two net components: the direction lies within 45° of the axis with the larger component.

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Answer: (b) Some villagers under 25 years of age are literate.

Fractions of the same village that add up to more than 1 must overlap. Under-25s (4/5) and literates (1/3) together make 17/15, so at least 2/15 of the villagers are both — some under-25s are certainly literate.

1. Under 25: 4/5 of the villagers; literate: 1/3.
2. 4/5 + 1/3 = 12/15 + 5/15 = 17/15, which is more than the whole village.
3. So at least 17/15 − 1 = 2/15 of the villagers are both under 25 and literate — (b) is certain.
4. Check the rest: house-owners (1/2) outnumber literates (1/3), so (a) is impossible; house-owners and under-25s (1/2 + 4/5 = 13/10) must overlap, so (d) is false; (c) cannot be fixed from the data.

• ✗ (a) Half the villagers own houses but only a third are literate, so not all house-owners can be literate.
• ✓ (b) 4/5 + 1/3 exceeds 1, so at least 2/15 of the villagers are both under 25 and literate.
• ✗ (c) Nothing tells us how literacy is spread among paddy growers.
• ✗ (d) 1/2 + 4/5 = 13/10 exceeds 1, so at least 3/10 of the villagers are under 25 and own a house.

Remember · If two fractions of one group add to more than 1, the excess over 1 is the guaranteed minimum overlap.

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Answer: (d) Neither Conclusion 1 nor 2 nor 3 nor 4

Two 'some' statements joined only through blues give no definite link between greens and blacks. None of the four conclusions follows on its own; at most, 1 and 2 form an either-or pair, which no option offers.

1. Both statements are particular ('some') and meet only in blues, so no definite relation between greens and blacks can be drawn.
2. Conclusion-1 and Conclusion-2 are each possible, but neither is certain.
3. Conclusions 3 and 4 ('all') cannot come from 'some' premises.
4. None follows. (1 and 2 together make an either-or pair, but option (a) claims both follow, which is impossible.)

• ✗ Conclusion-1 Greens and blacks may or may not overlap; nothing forces it.
• ✗ Conclusion-2 A diagram with some greens being blacks fits the statements, so 'no green is black' is not certain.
• ✗ Conclusion-3 An 'all' conclusion cannot follow from two 'some' statements.
• ✗ Conclusion-4 Likewise, nothing says every black is green.

Remember · Two particular ('some') premises never give a definite conclusion; use the either-or rule only when an option offers 'either … or'.

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Answer: (b) BIDNHTSSX

The code moves letters alternately one step back and one step forward (M→L, A→B, T→S, H→I …). Applied to CHEMISTRY, this gives BIDNHTSSX.

1. MATHEMATICS → LBSIDNZUHDR: M−1 = L, A+1 = B, T−1 = S, H+1 = I, E−1 = D, M+1 = N, A−1 = Z … — alternately −1 and +1.
2. CHEMISTRY: C−1 = B, H+1 = I, E−1 = D, M+1 = N, I−1 = H, S+1 = T, T−1 = S, R+1 = S, Y−1 = X.
3. Code: BIDNHTSSX.
4. Check: (c) BIDLHTSSX differs only in the 4th letter — M+1 is N, not L.

Remember · In letter codes, write the shift under each letter; alternating shifts (−1, +1) are common. Compare options only where they differ.

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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: (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: (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: (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) Only III

'Some rats are cats' converts directly into 'Some cats are rats'. The other two conclusions are 'no' statements that would need a definite chain, but the links to cows and to rats run through 'some' statements, which settle nothing.

1. Statement 1 converts: some rats are cats → some cats are rats. Conclusion III follows.
2. Only the cats that are dogs are kept away from cows (no dog is a cow); the other cats may or may not be cows, so Conclusion I is not certain.
3. Rats reach dogs only through cats, by two 'some' links; a rat may or may not be a dog, so Conclusion II is not certain.
4. Only III follows.

• ✗ Conclusion I Cats that are not dogs could still be cows; nothing rules it out.
• ✗ Conclusion II Two 'some' statements give no definite link between rats and dogs, either way.
• ✓ Conclusion III It is the direct converse of 'Some rats are cats'.

Remember · A 'some' statement converts both ways; a 'no' conclusion needs an 'all' or 'no' chain — 'some' links never prove it.

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Answer: (b) Only statement 3 is sufficient

All four colours are used on five floors, so exactly one colour repeats. If floors I and V are both red, red is that repeat, and floors II, III and IV must take the other three colours, one each — no two neighbours can match. Statements 1 and 2 leave the end floors free to copy a neighbour.

1. Five floors, four colours, all four used: one colour appears on two floors, the others once each.
2. Statement 3: I and V are red, so red is the repeated colour; II, III and IV take the other three colours, all different and none red.
3. Hence I–II, II–III, III–IV and IV–V all differ: statement 3 alone is sufficient.
4. Statement 1: II, III, IV differ, but floor I may copy floor II (e.g. blue, blue, green, yellow, red) — not sufficient.
5. Statement 2 (II ≠ IV) is already implied by statement 1, so 1 and 2 together are still not sufficient; 2 alone says nothing about any adjacent pair.

• ✗ 1. Not sufficient: it leaves the pairs I–II and IV–V unchecked.
• ✗ 2. Not sufficient: floors II and IV are not neighbours, so it controls no consecutive pair.
• ✓ 3. Sufficient: it forces II, III and IV to be the three non-red colours, all different.

Remember · 'Painted using four different colours' means all four are used; with five floors exactly one colour repeats — build from that count.

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Answer: (d) 100 km

West and South are at right angles, so Q, P and R form a right triangle with the right angle at P. QR is the hypotenuse of a 60–80–100 triangle.

1. Q is due West and R due South of P, so angle QPR = 90°.
2. QR² = 60² + 80² = 3600 + 6400 = 10000.
3. QR = 100 km.
4. Check: 60 : 80 : 100 = 3 : 4 : 5.

Remember · Perpendicular directions make a right triangle; spot 3–4–5 multiples before squaring.

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Answer: (d) 1, 2 and 3

The data fixes only the 40% who do not play cricket: they all play football and so all own two-wheelers. Nothing is said about whether cricketers own two-wheelers or also play football, so none of the three conclusions can be drawn.

1. Non-cricketers = 40%; all of them play football, so all of them have two-wheelers.
2. The 60% cricketers are not described: some may own two-wheelers, some may also play football.
3. Conclusions 1 and 2 need every cricketer to be without a two-wheeler — not given.
4. Conclusion 3 needs cricketers to avoid football — not given; the rule runs only from 'not cricket' to 'football'.
5. So 1, 2 and 3 all cannot be drawn.

• ✗ 1. Cannot be drawn: at most 60% lack two-wheelers; fewer do if any cricketer owns one.
• ✗ 2. Cannot be drawn: cricketers' vehicles are never mentioned.
• ✗ 3. Cannot be drawn: a cricketer may also play football; the data does not forbid it.

Remember · 'Whoever is not A does B' says nothing about the members of A; claims about that group cannot be drawn.

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Answer: (b) 17 m

Plotting the walks from A's house, A's office is at (42, 2) and B's at (34, −13). They are 8 m apart east–west and 15 m apart north–south, so the straight-line distance is 17 m.

1. Take A's house as (0, 0), East as +x and North as +y.
2. 20 m + 10 m East: both reach (30, 0).
3. A turns left (North) 2 m → (30, 2), then right (East) 12 m → A's office at (42, 2).
4. B turns right (South) 8 m → (30, −8), left (East) 4 m → (34, −8), right (South) 5 m → B's office at (34, −13).
5. Separation: 42 − 34 = 8 m East–West and 2 − (−13) = 15 m North–South.
6. Distance = √(8² + 15²) = √289 = 17 m.

Remember · Use coordinates for direction puzzles; left and right depend on the direction currently faced.

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Answer: (d) 1, 2 and 3

F is the father of B and D. G is F's widowed sister-in-law, so her late husband was F's brother, and C, B's cousin, is her child. The one slot left in the family — the mother of B and D — belongs to E, who is therefore F's wife, A's grandmother and C's aunt.

1. F is D's father; B and D are siblings, so F is B's father too, and A (B's daughter) is F's granddaughter.
2. G is a widow and F's sister-in-law, so her husband — F's brother — is no more.
3. C is B's cousin, so C is the child of G and F's late brother.
4. E completes the family as F's wife, the mother of B and D; the three females are then G, A and E.
5. So E is F's wife, A's grandmother and, as the wife of C's uncle, C's aunt.

• ✓ 1. E is the mother of B and D, hence F's wife.
• ✓ 2. A is the daughter of E's son or daughter B, so E is A's grandmother.
• ✓ 3. C is the child of F's brother; F's wife E is C's aunt.

Remember · Place each stated relation on a family tree; the leftover member usually fills the one slot the tree still needs.

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Answer: (d) ‘Whose weight is greatest’ cannot be determined

S1 makes Sohan lighter than both Mohan and Rohan. S2 only says Mohan weighs less than twice Rohan, which allows Mohan to be heavier or lighter than Rohan. So the heaviest person cannot be fixed.

1. S1 (twice Sohan is below each of the others): 2S < M and 2S < R, so Sohan is the lightest.
2. S2: 2R > M and 2R > S — Mohan may be anything below twice Rohan.
3. Example 1: S = 10, R = 30, M = 50 → 20 < 50, 20 < 30, 60 > 50, 60 > 10; Mohan is heaviest.
4. Example 2: S = 10, R = 30, M = 25 → all conditions still hold; Rohan is heaviest.
5. The greatest weight cannot be determined.

Remember · Test inequality puzzles with two sets of numbers; if both fit the conditions but give different answers, it cannot be determined.

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Answer: (b) Weight of Sohan is least

S1 says twice Sohan's weight is less than Mohan's and less than Rohan's. Since weights are positive, Sohan's own weight is even smaller, so Sohan is the lightest.

1. S1: 2S < M and 2S < R.
2. Weights are positive, so S < 2S; hence S < M and S < R.
3. Sohan is lighter than both others: Sohan's weight is the least.
4. S2 is not needed for this item.

Remember · If twice a quantity is below the others, the quantity itself certainly is — for positive values, x < 2x.

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Answer: (a) CCIDD

Each letter of MUMBAI is moved back by 1, 2, 3, 4, 5 and 6 places in turn, wrapping from A round to Z. Moving the letters of DELHI back by 1 to 5 gives CCIDD.

1. M − 1 = L, U − 2 = S, M − 3 = J, B − 4 = X (B → A → Z → Y → X), A − 5 = V, I − 6 = C.
2. DELHI: D − 1 = C, E − 2 = C, L − 3 = I, H − 4 = D, I − 5 = D.
3. Code: CCIDD.
4. Check: (d) CCIFE agrees in the first three letters but fails at H − 4 = D.

Remember · In letter codes, write the shift for each position; a growing shift (−1, −2, −3 …) is common, and letters wrap from A to Z.

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Answer: (b) 92323

Each letter stands for a fixed digit. From the two words, H = 9, A = 2 and M = 3, so HAMAM reads 9-2-3-2-3.

1. RAMON = 12345: R = 1, A = 2, M = 3, O = 4, N = 5.
2. DINESH = 675849: D = 6, I = 7, N = 5, E = 8, S = 4, H = 9.
3. HAMAM = H, A, M, A, M = 9, 2, 3, 2, 3 = 92323.

Remember · For letter-to-digit codes, map each letter once and read the new word letter by letter; repeated letters repeat their digit.

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Answer: (b) Pen < Book

Pen < Pencil < Book is an unbroken chain, so Pen < Book always. Cap is tied only through Book > Cap, which says nothing about how Cap compares with Pen or Pencil.

1. Join the first two relations: Pen < Pencil < Book, so Pen < Book.
2. Book > Cap points the other way; Cap could be above or below Pen and Pencil.
3. So (a), (c) and (d) are not always true, while (b) is.

Remember · A comparison is certain only if an unbroken chain of signs points the same way between the two items.

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Answer: (c) 6

Every boy wears at least a shirt or pants, so adding shirts and pants counts the boys in full uniform twice. 11 + 14 − 19 = 6 boys wear both.

1. Every one of the 19 boys wears a shirt, pants or both.
2. Shirts + pants = 11 + 14 = 25, which counts the boys wearing both twice.
3. Both = 25 − 19 = 6.
4. Check: 5 with shirts only + 8 with pants only + 6 with both = 19.

Remember · For two overlapping groups that together cover everyone: both = first group + second group − total.

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Answer: (b) There is at least one supporter of ‘party Y’ who knew some supporters of ‘party X’ as a friend.

The information is about a group of party X supporters who knew Z and backed his campaign strategy, and it says some of them had friends in party Y. So at least one person in party Y has a party X supporter as a friend.

1. Call the group described G: supporters of party X who knew Z and supported his campaign strategy.
2. 'Some of them had friends in party Y' tells us that G is not empty and that at least one member of G has a friend in party Y.
3. Friendship works both ways, so that person in party Y knows a supporter of party X as a friend — option (b).
4. Check (c) and (d): G itself consists of party X supporters who knew Z and backed his strategy, so both 'No …' statements are false.

• ✗ (a) The information says nothing about how party Y's supporters viewed the alliance; it speaks only of party X supporters.
• ✓ (b) Some members of the group had friends in party Y, so at least one person in party Y counts a party X supporter as a friend.
• ✗ (c) The group described consists of party X supporters who did support Z's strategy, so this is false.
• ✗ (d) The same group knew Z, so it is false that no party X supporter knew Z.

Remember · 'Some of them' confirms the group exists. A 'No …' statement fails as soon as one member of that group is shown to exist.

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Answer: (c) 3 only

The information confirms that some party X supporters knew Z and supported his strategy. That makes statement 1 true and statement 3 false. Statement 2, about supporters who opposed the strategy, is neither confirmed nor contradicted, so only statement 3 can be called not correct.

1. From 'some of them had friends in party Y', there exist party X supporters who knew Z and supported his strategy.
2. Statement 1: some party X supporters knew Z — true.
3. Statement 3: it says no party X supporter backed Z's strategy — this contradicts the information, so it is not correct.
4. Statement 2: the information is silent about supporters who opposed the strategy, so it cannot be marked incorrect; it is simply not established.
5. Only statement 3 is not correct.

• ✓ 1. The supporters described knew Z, so 'some supporters of party X knew Z' must hold.
• ✓ 2. Not contradicted: the information says nothing about supporters who opposed the strategy, so this may well be true and cannot be called incorrect.
• ✗ 3. The information describes party X supporters who did support Z's campaign strategy, so this statement is false.

Remember · 'Not correct' means contradicted by the information. A statement the information is merely silent about is not proved wrong.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) Only conclusion I follows from the statement.

The statement says that even a natural gift like a good voice must be practised to improve and excel — that is, a gift needs nurturing. It says nothing about people whose voice is not good, so only Conclusion I follows.

1. The statement: a good voice is a gift, yet it must be practised to improve and excel.
2. Conclusion I generalises this correctly — a natural gift still needs nurturing — so it follows.
3. Conclusion II is about people without a good voice, whom the statement never mentions, so it does not follow.

• ✓ I Practising 'to improve and excel' is exactly the nurturing and care the conclusion speaks of.
• ✗ II The statement concerns only people who have a good voice; whether others should practise lies outside it, however sensible it sounds.

Remember · A conclusion must come from the statement itself; one about a group the statement never covers does not follow.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) Q is the father of P.

P + R means P is the son of R, and R − Q means R is the wife of Q. So Q is the husband of P's mother R, which makes Q the father of P.

1. P + R → P is the son of R.
2. R − Q → R is the wife of Q, so Q is R's husband.
3. P is the son of R, and R's husband is Q, so Q is the father of P.

Remember · Decode coded relations one link at a time, left to right, noting each person's gender as you go.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error