Analytical Reasoning: Relationships, Deductions and Drawing Valid Conclusions
2. Categorical statements: all, some and none
Four statement forms carry most categorical reasoning: ‘all A are B’, ‘no A is B’, ‘some A are B’, and ‘some A are not B’. ‘All A are B’ lets you move from any A to B, but not back: it does not give ‘all B are A’. ‘No A is B’ works both ways (no B is A). ‘Some A are B’ also works both ways (some B are A), but it does not tell you about the rest, and it does not imply ‘some A are not B’. Reading more into ‘some’ than ‘at least one’ is the commonest categorical error.
| Statement | Valid immediate inference | Common false move |
|---|---|---|
| All A are B | Any particular A is a B | Concluding all B are A |
| No A is B | No B is A (converts) | Concluding some A are B |
| Some A are B | Some B are A (converts) | Concluding some A are not B |
| Some A are not B | At least one A lies outside B | Concluding some B are not A |
3. Ordering, ranking and counting arrangements
Comparative relations build an order through transitivity: if A is taller than B and B is taller than C, then A is taller than C, so the three fall on one line A – B – C. Draw the line, place each new fact on it, and watch for gaps the facts leave open. If the clues connect everyone into a single chain, the order is fully fixed; if two people are never compared, even through a chain, their relative order may stay undetermined.
Counting arrangements uses a few rules. Arranging n distinct items in a row gives n! orders (n factorial: 4! = 24). If two items must sit together, tie them into one block — arrange the block among the rest, then arrange inside it. Worked example: four books in a row with two particular books adjacent gives 3! ways for the three units times 2! ways inside the block, that is 6 × 2 = 12. If one item is fixed at an end, arrange only the remaining items. Count the structure first, then multiply.
4. Grouping, matching and assignment
When each of several people takes one of several options — a drink, a seat, a day — a grid is the clean tool: people down the side, options across the top, a tick or cross in each cell. Each stated clue marks cells. A row with every cell but one crossed out forces the survivor, and marking that tick then crosses the rest of its column, which often cascades. Work the most restrictive clue first, because it removes the most possibilities at once.
A worked grid. Three colleagues — Asha, Binu and Chitra — sit at three desks (window, middle, door). Clues: Asha is not at the door; Binu is not at the window; Chitra is at neither the window nor the middle. Chitra’s row is crossed in two cells, so she is at the door. Then Asha, who cannot be at the door, has window or middle, and Binu, who cannot be at the window, takes the middle by elimination, which leaves the window for Asha. Each clue removed a cell, and the last uncrossed cell was forced. Notice that no clue said where Asha sits; the grid delivered it.
Overlapping groups are counted by inclusion–exclusion. For two groups, the number in at least one is n(A) + n(B) − n(A and B), because adding the two sizes counts the overlap twice. The number in neither is the total minus that figure. Worked example: of 40 members, 22 play chess and 18 play carrom, and 6 play both. At least one game: 22 + 18 − 6 = 34, so 40 − 34 = 6 play neither. Chess only is 22 − 6 = 16 and carrom only is 18 − 6 = 12; check 16 + 12 + 6 + 6 = 40.
5. Conditionals: if-then, necessary and sufficient
A conditional ‘if P then Q’ licenses exactly two moves. Given P, conclude Q (the straight reading). Given not-Q, conclude not-P — this is the contrapositive, and it is always valid. Two tempting moves are invalid: from Q you may not conclude P (affirming the consequent), and from not-P you may not conclude not-Q (denying the antecedent). If the gate locks whenever the alarm is set, then an unlocked gate tells you the alarm was not set; a locked gate tells you nothing about the alarm.
Distinguish necessary from sufficient. If X is required for Y, then X is necessary: without X, no Y, so Y lets you conclude X. But X being necessary does not make it sufficient: having X need not deliver Y, because other requirements may remain. ‘A reserved ticket is needed to board’ means no ticket, no boarding — yet holding a ticket does not by itself prove one may board. Treating a necessary condition as a sufficient one is a classic trap.
6. Relationship puzzles: kinship and direction
For a kinship puzzle, draw a small family tree and read the description from the speaker outward, one link at a time. ‘My mother’s only sister’s son’ is: mother, then her sister (the speaker’s maternal aunt), then that aunt’s son — the speaker’s cousin. Replace each phrase with the exact relative it names before combining, and keep the gender fixed by the words used. The error to avoid is collapsing two links into one, which turns a cousin into a brother or an aunt into a mother.
For a direction puzzle, sketch the path on a mental grid with north up. A right turn rotates you clockwise (north to east to south to west), a left turn anticlockwise. Track the facing direction after each turn, and add up the north–south moves and the east–west moves separately to find the final position. Net north minus south gives the vertical offset, net east minus west the horizontal, and the straight-line distance follows from these two.
Key takeaways
- Sort every candidate conclusion into must-be-true, cannot-be-true and may-be-true; only a forced conclusion answers a ‘what follows’ question, and a ‘could be true’ is a wrong answer.
- ‘All A are B’ does not give ‘all B are A’; ‘no A is B’ and ‘some A are B’ both convert; ‘some’ means at least one and never implies ‘some are not’.
- Transitivity builds an order; a pair never connected by the clues has no forced order. Count arrangements with n!, tie adjacent items into a block, and fix any end-constraint first.
- For assignments use a grid and let elimination do the work: the last uncrossed cell in a row or column is forced; apply the most restrictive clue first.
- ‘If P then Q’ yields Q from P and not-P from not-Q (contrapositive); it never yields P from Q or not-Q from not-P. A necessary condition is not a sufficient one.
- Solve kinship by a family tree read link by link, and direction by a grid where a right turn goes clockwise; keep facing direction and north–south versus east–west moves separate.
Practice questions (18)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
All the paintings in the east wing are watercolours. The painting titled ‘Dawn’ hangs in the east wing. Which conclusion follows with certainty?
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Answer: A — The painting ‘Dawn’ is a watercolour.
Since all east-wing paintings are watercolours and ‘Dawn’ is in the east wing, ‘Dawn’ is a watercolour. Concluding that every watercolour hangs in the east wing reverses 'all A are B' into 'all B are A'. Saying some east-wing paintings are not watercolours contradicts the given rule that all of them are watercolours, and calling 'Dawn' the only watercolour adds information not given.Some of the volunteers are also students. All the students are under 25. Which statement must be true?
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Answer: B — At least one volunteer is under 25.
The volunteers who are students are under 25, so at least one volunteer is under 25. The claim that every volunteer is under 25, or that none is over 25, goes beyond the student-volunteers to all volunteers, which is not forced. 'All students are volunteers' reverses 'some volunteers are students'.In a race with no ties, Anil finished ahead of Bela, Bela finished ahead of Chetan, and Deepa finished ahead of Anil. Among these four, who finished last?
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Answer: C — Chetan
The chain is Deepa ahead of Anil ahead of Bela ahead of Chetan, so the full order is Deepa, Anil, Bela, Chetan and Chetan is last. Bela and Anil both have someone behind them, and the clues connect all four into one chain, so the result is fully determined, not undetermined.Five friends have different heights. Ravi is taller than Sita; Sita is taller than Meena; Tara is shorter than Meena; and Uday is the tallest of all five. Counting from the tallest as position 1, what is Sita’s position? Give the number only.
Numerical answer — type the value.
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Answer: 3
Uday is the tallest, so he is position 1. The chain Ravi > Sita > Meena > Tara, with only Uday above Ravi, gives the full order Uday, Ravi, Sita, Meena, Tara. Sita is therefore in position 3. The usual slip is to forget that Uday sits above Ravi and place Sita second.At a certain house, whenever the alarm is set at night, the gate locks itself. On one particular night the gate did not lock itself. Which conclusion follows?
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Answer: D — The alarm had not been set that night.
By the contrapositive of ‘alarm set implies gate locks’, a gate that did not lock means the alarm was not set that night. Concluding that the alarm was set contradicts this: had it been set, the gate would have locked. Nothing in the information says the mechanism is broken, since the alarm not being set already accounts for the open gate, and the general claim about every failure to lock is an unsupported generalisation.To board the express train, a passenger must hold a reserved ticket. Vijay holds a reserved ticket. Which conclusion is justified?
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Answer: A — The rule given does not settle whether Vijay may board.
A reserved ticket is stated only as a necessary condition. Holding one removes one barrier, but the rule never says it is the only requirement, so it neither guarantees that Vijay may board nor rules it out. Concluding that he may certainly board treats the necessary condition as sufficient. Concluding that he may certainly not board invents an obstacle the rule never mentions. Saying a ticketless passenger may board contradicts the necessary condition.In a class of 50 students, 30 study French and 25 study German. If 10 students study both French and German, how many study neither language? Give the number only.
Numerical answer — type the value.
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Answer: 5
The number studying at least one language is 30 + 25 − 10 = 45 (adding the two groups double-counts the 10 who do both, so subtract them once). Neither = 50 − 45 = 5. Forgetting to subtract the overlap gives 55, which already exceeds the class size and flags the slip.Given: no reptile is warm-blooded; all snakes are reptiles; some pets are snakes. Which of the following conclusions must follow?
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Answer: A — No snake is warm-blooded; B — Some pets are reptiles; C — Some pets are not warm-blooded
All snakes are reptiles and no reptile is warm-blooded, so no snake is warm-blooded. Some pets are snakes and every snake is a reptile, so some pets are reptiles, and those pets are not warm-blooded. ‘All reptiles are snakes’ reverses ‘all snakes are reptiles’, which the premises do not allow.Three friends — X, Y and Z — each ordered a different drink from among tea, coffee and juice. X did not order tea, and Z ordered juice. Who ordered tea?
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Answer: B — Y
Z took juice, so tea is left for X or Y. X did not take tea, so X took coffee, which leaves tea for Y. The assignment is fully forced: Z juice, X coffee, Y tea, so it is determined and Z did not order tea.Four books — a novel, a dictionary, an atlas and a diary — are to be placed in a row on a shelf. The dictionary and the atlas must stand next to each other. In how many different orders can the four books be arranged? Give the number only.
Numerical answer — type the value.
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Answer: 12
Tie the dictionary and the atlas into one block, giving three units to arrange: the block, the novel and the diary. These three arrange in 3! = 6 ways, and within the block the two books can be in 2! = 2 orders, so the total is 6 × 2 = 12. Forgetting the internal 2 gives 6; treating all four as free gives 24.Pointing to a man in a photograph, Neha said, “He is the son of my mother’s only sister.” How is the man related to Neha?
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Answer: C — Her cousin (her maternal aunt’s son)
Neha’s mother’s only sister is Neha’s maternal aunt, and that aunt’s son is Neha’s cousin. He would be her brother only if he were her own mother’s son; her maternal uncle would be her mother’s brother, not her aunt’s son; and her nephew would be a child of her own sibling.A person starts at a point and walks 3 km due north. She then turns right and walks 4 km, then turns right again and walks 3 km. Where is she now, relative to her starting point?
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Answer: D — 4 km to the east of the starting point
North 3 km reaches (0, 3). A right turn faces east; 4 km reaches (4, 3). A second right turn faces south; 3 km returns to (4, 0). The north and south moves cancel, leaving her 4 km due east of the start. The 6 km north and 5 km north-east answers ignore that the southward leg undoes the northward one.Five boxes have different weights. The red box is heavier than the blue; the blue is heavier than the green; the yellow is lighter than the green; and the white is heavier than the red. Which of the following must be true?
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Answer: A — The white box is the heaviest; B — The green box is heavier than the yellow; D — The red box is heavier than the green
The relations chain into white > red > blue > green > yellow. White is therefore heaviest, green is heavier than yellow, and red is heavier than green. Blue is not the lightest, since green and yellow are both lighter than it.In an office, every urgent file is marked red, and every red-marked file is cleared the same day. One file was not cleared the same day. Which conclusion follows with certainty?
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Answer: A — That file was not urgent.
Urgent implies red implies same-day; by the contrapositive, not-same-day implies not-red implies not-urgent, so the file was not urgent. Saying it was urgent but overlooked contradicts the chain, and saying it was red contradicts 'red implies same-day'. The claim about all non-urgent files being cleared the next day adds a rule that was never stated.A committee of exactly two is to be chosen from four people P, Q, R and S, under two rules: P and Q cannot both be on it; and if R is on it, then S must also be on it. Which of the following committees are allowed?
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Answer: C — R and S; D — Q and S
The committee of P and Q breaks the rule that the two cannot both serve. The committee of P and R has R without S, breaking ‘if R then S’. The committee of R and S has R with S and no P–Q clash, and the committee of Q and S has no R to trigger the ‘if R then S’ rule and no P–Q clash, so those two are allowed.All members of the choir can read music, and some people who can read music play the violin. Does it follow that some choir members play the violin?
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Answer: B — No; it does not follow that any choir member plays the violin.
Choir members are among the music-readers, but the music-readers who play the violin could all be people outside the choir, so no overlap is forced; the conclusion does not follow. 'Some choir members must play' and 'all play' both assert a forced overlap that the premises do not supply, and 'no choir member can play' wrongly rules the overlap out entirely.Five students P, Q, R, S and T are to sit in a row of five chairs. P and Q must not sit next to each other. In how many different ways can the five students be seated? Give the number only.
Numerical answer — type the value.
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Answer: 72
All seatings: 5! = 120. Seatings with P and Q adjacent: tie them into one block, arrange the 4 units in 4! = 24 ways and the pair inside the block in 2 ways, giving 24 × 2 = 48. Not adjacent = 120 − 48 = 72. Forgetting the 2 orders inside the block gives 120 − 24 = 96, and forgetting to subtract gives 120.Five colleagues A, B, C, D and E sit in a row of five seats numbered 1 to 5 from left to right, so a higher seat number means further to the right. B sits exactly two seats to the right of A. D sits immediately to the left of E. C sits in neither seat 1 nor seat 5. A does not sit in seat 1. Who sits immediately to the right of C?
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Answer: C — B
B and A are two seats apart, so A is in seat 1, 2 or 3, and not in seat 1. If A is in seat 2 and B in seat 4, C must take seat 3 (not an end), leaving seats 1 and 5 for D and E, which are not adjacent, so this fails. So A is in seat 3 and B in seat 5. C cannot be in seat 1, and if C is in seat 2 then D and E are left with seats 1 and 4, which are not adjacent. So C is in seat 4, and D and E fill seats 1 and 2 as D then E. The row is D, E, A, C, B, and B sits immediately to the right of C.