Analytical Reasoning: Relationships, Deductions and Drawing Valid Conclusions

This is the first chapter of the Humanities and Social Sciences (XH) paper for Section XH0.3 of XH0 Reasoning and Comprehension, the part every XH candidate answers. It answers the heading “Section XH0.3: Analytical Reasoning — Ability to understand relationship in statements or short passages and being able to draw reasonable conclusions/inferences from them.” Analytical reasoning starts from a small set of stated relationships — who is taller, which follows which, if-this-then-that — and asks what is forced. The skill is to separate what must be true from what merely may be true, and to read that off cleanly rather than by guesswork. This chapter teaches categorical statements, ordering and ranking, grouping and assignment, conditional deduction, and relationship puzzles such as kinship and direction, with worked numerical examples. A few items are numerical puzzles whose answer is a plain number. Every worked question states its relationships inside the stem.

1. Deduction: reading what must follow

Analytical reasoning is deduction: given a set of relationships, you work out everything they force and nothing they do not. The single most useful habit is to sort every candidate conclusion into three buckets. A conclusion must be true if there is no way the relationships can all hold while it is false. It cannot be true if the relationships rule it out. And it may be true if the relationships neither force it nor forbid it — which, for a question asking what follows, counts as a wrong answer. A ‘could be true’ is not a ‘must be true’.

🧠 Three buckets: must, cannot, may
For every option, try to build a case that satisfies all the given relationships yet makes the option false. If you succeed, the option is only ‘may be true’ and is not the answer to a ‘must follow’ question. If you cannot — if every allowed case makes it true — it must be true. This single test settles most analytical items.

A worked walk-through. Statements: “Every clerk in the records office has passed the typing test. Some of those who passed the typing test are also night-shift staff.” Candidate: “Some clerks are night-shift staff.” Try to break it. Suppose all the night-shift people who passed the typing test are accountants and not clerks: the two statements still hold, yet the candidate is false. So it is only ‘may be true’. Now test “Every clerk has passed the typing test”: it is simply the stated rule itself, so it must be true. The method never asks whether a candidate sounds likely, only whether a consistent world can falsify it.

Before reasoning, translate the words into something compact: a chain for an order (D > A > B), a short list of if-then arrows for conditions, or a small grid for assignments. The translation removes the noise of the language and leaves the structure, and most mistakes happen in the translation, not in the logic that follows it.

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.

What each categorical form does and does not yield
StatementValid immediate inferenceCommon false move
All A are BAny particular A is a BConcluding all B are A
No A is BNo B is A (converts)Concluding some A are B
Some A are BSome B are A (converts)Concluding some A are not B
Some A are not BAt least one A lies outside BConcluding some B are not A
⚠️ The middle term must be shared fully once
From ‘all choir members read music’ and ‘some who read music play the violin’, you cannot conclude that any choir member plays the violin: the violinists among music-readers might all lie outside the choir. Two ‘some’ or a ‘some’ overlapping the wrong part of a shared class produce no valid conclusion.

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.

⚠️ An unconnected pair is not ordered
If the clues tell you A > B and C > B but never compare A with C, then A and C have no forced order, and ‘A is taller than C’ is only ‘may be true’. Before answering ‘who is second’, check that the chain actually reaches that position without a gap; otherwise the honest answer may be that it cannot be determined.

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.

🎯 Elimination does the work
You rarely need to place someone by proving where they go; it is easier to cross out everywhere they cannot go until one cell is left. A constraint like ‘X is not on Tuesday’ removes a cell directly, and ‘if R then S’ removes the case where R is chosen without S. The last cell standing in a row or column is the forced assignment.

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.

⚠️ Necessary is not sufficient, and the converse is not the rule
Keep two separations clear. First, meeting a necessary condition does not guarantee the outcome. Second, ‘if P then Q’ is not the same as ‘if Q then P’; the only reversal you may use is the contrapositive ‘if not-Q then not-P’. Most wrong conditional answers are a converse or a denied antecedent dressed up as a conclusion.

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.

⚠️ Read the link, not the family
“My father’s brother’s daughter” is a female paternal cousin, not a sister: the link passes through the father’s brother, not through the father. “My mother’s brother” is a maternal uncle, while “my mother’s sister’s husband” is the husband of a maternal aunt. When a stem fixes only some of the facts, an answer that needs an unstated fact — such as the sex of a person the stem never labels — cannot be the key.

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.

  1. All the paintings in the east wing are watercolours. The painting titled ‘Dawn’ hangs in the east wing. Which conclusion follows with certainty?

    1. The painting ‘Dawn’ is a watercolour.
    2. Every watercolour in the gallery hangs in the east wing.
    3. Some paintings in the east wing are not watercolours.
    4. The painting ‘Dawn’ is the only watercolour in the east wing.
    Show answer

    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.
  2. Some of the volunteers are also students. All the students are under 25. Which statement must be true?

    1. Every volunteer is under 25.
    2. At least one volunteer is under 25.
    3. No volunteer is older than 25 years of age.
    4. All of the students are volunteers as well.
    Show answer

    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'.
  3. 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?

    1. Bela
    2. Anil
    3. Chetan
    4. It cannot be determined from the information given
    Show answer

    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.
  4. 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.

    Show answer

    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.
  5. 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?

    1. The alarm had definitely been set that night.
    2. The gate’s locking mechanism is permanently broken.
    3. Whenever the gate fails to lock, the alarm has malfunctioned.
    4. The alarm had not been set that night.
    Show answer

    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.
  6. To board the express train, a passenger must hold a reserved ticket. Vijay holds a reserved ticket. Which conclusion is justified?

    1. The rule given does not settle whether Vijay may board.
    2. Vijay may certainly board the express train.
    3. Vijay may certainly not board the express train.
    4. A passenger without a reserved ticket may still board the train.
    Show answer

    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.
  7. 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.

    Show answer

    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.
  8. Given: no reptile is warm-blooded; all snakes are reptiles; some pets are snakes. Which of the following conclusions must follow?

    1. No snake is warm-blooded
    2. Some pets are reptiles
    3. Some pets are not warm-blooded
    4. All reptiles are snakes
    Show answer

    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.
  9. 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?

    1. X
    2. Y
    3. Z
    4. It cannot be determined from the information given
    Show answer

    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.
  10. 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.

    Show answer

    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.
  11. 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?

    1. Her brother (her mother’s son)
    2. Her maternal uncle (her mother’s brother)
    3. Her cousin (her maternal aunt’s son)
    4. Her nephew (her sibling’s son)
    Show answer

    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.
  12. 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?

    1. 4 km to the west of the starting point
    2. 6 km to the north of the starting point
    3. 5 km to the north-east of the starting point
    4. 4 km to the east of the starting point
    Show answer

    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.
  13. 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?

    1. The white box is the heaviest
    2. The green box is heavier than the yellow
    3. The blue box is the lightest
    4. The red box is heavier than the green
    Show answer

    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.
  14. 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?

    1. That file was not urgent.
    2. That file was urgent but was overlooked.
    3. That file had been marked red.
    4. Every file that is not urgent is cleared the next day instead.
    Show answer

    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.
  15. 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?

    1. P and Q
    2. P and R
    3. R and S
    4. Q and S
    Show answer

    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.
  16. 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?

    1. Yes; it follows that some choir members must play the violin.
    2. No; it does not follow that any choir member plays the violin.
    3. Yes; in fact it follows that all choir members play the violin.
    4. No; it follows that no choir member is able to play the violin.
    Show answer

    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.
  17. 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.

    Show answer

    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.
  18. 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?

    1. A
    2. D
    3. B
    4. E
    Show answer

    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.