Analytical Reasoning — Syllogism, Puzzles, Venn Diagrams & Statements

Before you tick "neither conclusion follows", run one extra check. If neither conclusion follows on its own, look at whether the two conclusions share the same subject and the same predicate and form a complementary pair — that is, together they cover every possibility, so one of them must be true even though you cannot say which. The three complementary pairs are Some ↔ No, All ↔ Some-not, and Some ↔ Some-not. Remember it as "All fights Some-not; Some fights No." When the pair matches, the answer is "either I or II follows".

Worked example. Statements: Some doctors are teachers. Some teachers are engineers. Conclusions:
I. Some doctors are engineers.
II. No doctor is an engineer. Two particular statements yield nothing, so neither I nor II follows on its own — you can draw the Engineers circle overlapping only the teachers who are not doctors (making II true), or overlapping the doctor-teachers as well (making I true). Both diagrams are legal. But I and II have the same subject (doctors) and the same predicate (engineers) and form a Some ↔ No pair, so the correct answer is either I or II follows.

Possibility conclusions reverse the test completely. A definite conclusion needs every legal diagram; a possibility conclusion needs only one. So possibilities are easy to accept — but they can still be blocked. Statements: All A are B. All B are C. Is "Some C are not A" a possibility? Yes: draw C much larger than B and the outer ring of C contains no A. Is "All C are A" a possibility? Yes: draw all three circles coinciding, which no statement forbids. Now a blocked one — Statements: Some cats are dogs. No dog is a horse. Conclusion: "All cats being horses is a possibility." This is false, because the cats that are dogs can never be horses while the Dog and Horse circles are drawn fully apart, and no legal diagram allows every cat to be a horse.

The paper asks two quite different things under the name "Venn diagram". The first kind gives three words and asks which diagram shows their relationship. The method is to test each pair with the question "is every one of these necessarily one of those?" If yes, nest the circles; if some overlap is possible but not certain, overlap them; if they exclude each other, keep them apart. The second kind gives you numbers and asks how many people fall in a region — that is inclusion–exclusion, handled below.

• Rose, Flower, Plant → three circles nested one inside the other: Rose inside Flower inside Plant.
• Delhi, India, Asia → the same nesting: Delhi inside India inside Asia.
• Pen, Pencil, Stationery → Pen and Pencil are two separate circles, both entirely inside Stationery. They do not overlap, because nothing is both a pen and a pencil.
• Uncle, Brother, Male → both Uncle and Brother lie inside Male, and they overlap each other, because one man can be somebody's uncle and somebody else's brother.
• Doctor, Man, Woman → Man and Woman are fully apart, while Doctor overlaps both, since doctors may be of either sex.

For counting questions the two-set repair is a single subtraction: n(A ∪ B) = n(A) + n(B) − n(A ∩ B), because adding the two group sizes counts the overlap twice. For three sets the signs alternate: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C). But the formula alone will not tell you how many like exactly one item. For that you must fill in all seven regions, and the only safe order is from the centre outward.

Worked three-circle example. In a survey of 200 people: 100 drink tea, 90 drink coffee, 80 drink milk; 40 drink tea and coffee, 30 drink coffee and milk, 25 drink tea and milk, and 15 drink all three. Start at the centre with 15. The pair figures include the centre, so tea-and-coffee only = 40 − 15 = 25, coffee-and-milk only = 30 − 15 = 15, tea-and-milk only = 25 − 15 = 10. Now the singles: tea only = 100 − (25 + 10 + 15) = 50, coffee only = 90 − (25 + 15 + 15) = 35, milk only = 80 − (10 + 15 + 15) = 40.

Now every question is one addition. At least one drink = 50 + 35 + 40 + 25 + 15 + 10 + 15 = 190, which agrees with the formula 100 + 90 + 80 − 40 − 30 − 25 + 15 = 190, so the table is arithmetically sound. Exactly one = 50 + 35 + 40 = 125. Exactly two = 25 + 15 + 10 = 50. At least two = 50 + 15 = 65. None = 200 − 190 = 10. Always run that cross-check: if the seven regions do not add up to the union you obtained from the formula, you have misread a pair value.

These three question types look alike and are graded by three different rules, so the first thing to do is read the instruction line, not the statement. A conclusion is something that must be true given the statement — you may use no outside information at all. An assumption is something that must already have been true for the statement to make sense — it sits behind the statement, unstated. An argument is a reason offered for or against a proposal, and you must judge whether it is strong or weak, not whether you agree with it.

Worked assumption question. Statement: "Download your admit card at least three days before the examination." Assumption I: candidates have access to the internet. Assumption II: no candidate will manage to download the card on the day of the exam. Negate I — if no candidate had internet access, the whole instruction would be pointless, so the statement collapses and I is implicit. Negate II — even if some candidates could download on the day itself, advising an early download still makes perfect sense, so the statement survives and II is not implicit. Notice the giveaway: II contains the absolute word "no", and absolutes are almost never assumptions.

Worked argument question. Should entrance examinations for professional courses be abolished? Argument I: No — when applicants far outnumber seats, an objective written test is the only defensible filter available. Argument II: Yes — students in some other countries do not sit such tests. Argument I is strong: it names the real constraint (more applicants than seats) and explains why the test exists. Argument II is weak: a bare comparison with another country, with no reason why that country's situation resembles ours, is an appeal rather than an argument. Judge relevance and reasoning, never your own opinion on entrance tests.

A cause-and-effect question gives two statements and asks how they are related. There are four standard answers: I is the cause and II the effect; II is the cause and I the effect; both are independent causes; or both are effects of some common cause. The third and fourth options are the ones candidates forget, and they are often correct. Ask two questions in order: which event came first in time? and could a single third event have produced both?

Example 1. Statement I: The government has raised the price of diesel. Statement II: Transport unions have announced a strike. A price rise plainly precedes the reaction, and unions strike because of it, so I is the cause and II the effect. Example 2. Statement I: Many farmers in the district have suffered crop losses. Statement II: Vegetable prices in the district have risen sharply. Here the honest answer is not that one caused the other in the way the options usually mean: a single event — a failed monsoon or a flood — plausibly produced both, so both are effects of a common cause. Learning to spot that shape is worth a mark or two in every paper that uses this type.

Courses of action. A course of action follows only if it is practicable, directly addresses the problem stated, and is not extreme. Statement: a large number of villages in a district have been hit by a sudden flood. Course I: the district administration should immediately set up relief camps on higher ground. Course II: the villagers should be told not to live in low-lying areas in future. Course I follows — it is immediate, feasible and aimed at the stated emergency. Course II does not: it offers no alternative land, cannot be implemented as an instruction, and shifts responsibility to the victims of the flood. Answer: only I follows.

Draw numbered blanks — for seven people write 1 to 7 from left to right — and start with the clue that fixes an exact position, never with a vague one. "X sits at the middle", "X sits third from the left", "X sits at an extreme end" are anchors. Clues like "X sits somewhere to the left of Y" are weak and should be used last, as filters. And note the convention that decides half the questions: when everybody faces north, the reader's left is also their left; when everybody faces south, the two are mirror images and every left becomes a right.

Fully worked single-row puzzle. Seven friends A, B, C, D, E, F and G sit in a row facing north.
(i) A sits exactly in the middle of the row.
(ii) Two persons sit between A and E, and E is to the right of A.
(iii) C is an immediate neighbour of E.
(iv) B sits second to the left of F.
(v) D sits at one of the extreme ends.

Derivation. From (i), A = 4, the middle of seven. From (ii), E lies to the right of A with two people between, so E = 7 — forced, since 4 + 3 = 7 and there is nothing beyond. From (iii), C must neighbour seat 7; seat 8 does not exist, so C = 6, also forced. That leaves seats 1, 2, 3 and 5 for B, D, F and G. From (iv), B = F − 2, and the only pairs available from {1, 2, 3, 5} are (B = 1, F = 3) and (B = 3, F = 5). Take the first: D and G would then occupy 2 and 5, and (v) requires D at seat 1 or 7, both taken — contradiction. So the second holds: B = 3, F = 5, leaving 1 and 2 for D and G, and (v) forces D = 1 and hence G = 2. The arrangement is unique.

Reading off the finished row: Who sits third to the right of G? G is at 2, so count 3, 4, 5 — the answer is F. Who is immediately to the left of A? Seat 3 — B. How many persons sit between B and C? Seats 4 and 5 — two. Every question is now a matter of counting, which is exactly the point of spending forty seconds on the diagram.

Double-row arrangement. One row faces south and the other faces north, so the two rows face each other and each person in one row faces exactly one person in the other. Draw both rows on paper with west on your left and east on your right for both, then remember: a person facing north has west on their left; a person facing south has east on their left. Worked example: A, B, C sit in row 1 facing south while P, Q, R sit in row 2 facing north.
(i) B sits at the middle of row 1.
(ii) A faces R.
(iii) Q sits to the immediate right of R.

Derivation. Number the positions 1, 2, 3 from west to east in both rows; facing each other means position i of row 1 faces position i of row 2. From (i), B = 2. From (ii), R sits at the same position number as A. From (iii), Q faces north, so Q's right is east and Q = R + 1. If A were at 3 then R = 3 and Q would need position 4, which does not exist. So A = 1, hence R = 1, Q = 2, P = 3, and C = 3. Unique. Now the trap question: who is to the immediate left of B? B faces south, so B's left is east, which is position 3 — the answer is C, not A. And who faces C? Position 3 of row 2 — P.

Number the seats clockwise 1 to n and fix one person at seat 1 — you may do this freely, because only relative positions matter on a circle. Then everything hinges on one fact you must derive once and never doubt again. Sit at the top of the circle facing the centre: you are facing "downwards", so your left hand points the way the numbers increase, that is clockwise, and your right hand points anticlockwise. Therefore facing the centre: left = clockwise, right = anticlockwise. If instead everyone faces outward, both are reversed: left = anticlockwise, right = clockwise. In a circle of n people, the seat opposite seat p is seat p + n/2, defined only when n is even.

Fully worked circular puzzle. Eight friends A, B, C, D, E, F, G and H sit around a circular table facing the centre.
(i) C sits second to the left of A.
(ii) B sits opposite A.
(iii) E sits between A and C.
(iv) G is an immediate neighbour of both C and B.
(v) D sits second to the right of A. (vi) F is an immediate neighbour of B.

Derivation. Fix A = 1 and number clockwise. From (i), left is clockwise, so C = 1 + 2 = 3. From (ii), opposite means plus four, so B = 1 + 4 = 5. From (iii), the only seat between 1 and 3 is 2, so E = 2. From (iv), the neighbours of C at seat 3 are seats 2 and 4; seat 2 is E, so G = 4 — and seat 4 is indeed also a neighbour of B at seat 5, so the clue is consistent. From (v), right is anticlockwise, so D = 1 − 2 = −1, which wraps round to seat 7. From (vi), the neighbours of B are seats 4 and 6; seat 4 is G, so F = 6. The single remaining seat, 8, must be H. Clockwise the order is A, E, C, G, B, F, D, H — and no step above involved a choice, so the arrangement is unique.

Reading it off: Who sits third to the left of F? F is at seat 6; left is clockwise, so count 7, 8, 1 — the answer is A. Who sits opposite G? G is at 4, and 4 + 4 = 8, so H. Who is immediately to the right of D? Right is anticlockwise from seat 7, which is seat 6 — F. Had you used the wrong convention on this last one you would have said H, and that is precisely the option the paper would have offered you.

When a puzzle attaches two or more attributes to each person — a floor and a subject, a box and a colour, a day and a city — stop trying to hold it in your head and draw an elimination grid: people down the side, positions across the top, a cross in every cell a clue rules out and a tick in every cell a clue confirms. The grid does two things nothing else does. It shows you at a glance when a row has only one blank cell left, which forces a tick; and it shows you when a column has only one blank cell left, which forces the same. Alternate between rows and columns until the grid closes.

Fully worked floor puzzle. Five students — Amit, Bina, Chetan, Divya and Esha — live on the five floors of a building, floor 1 being the lowest, and each studies a different subject among History, Maths, Physics, Botany and Economics.
(1) Chetan lives on the lowest floor.
(2) Bina lives on the topmost floor.
(3) Amit lives immediately above Esha.
(4) Divya does not live on an odd-numbered floor.
(5) Exactly one person lives between Divya and Esha.
(6) The person on the third floor studies Physics.
(7) Chetan studies neither Botany nor Maths.
(8) Bina studies Economics.
(9) The Botany student lives immediately below the Physics student.

Derivation of the floors. Clues (1) and (2) give Chetan = 1 and Bina = 5, leaving floors 2, 3 and 4 for Amit, Divya and Esha. Clue (3) demands Amit = Esha + 1, and within {2, 3, 4} that allows only two possibilities: Esha = 2 with Amit = 3, leaving Divya = 4; or Esha = 3 with Amit = 4, leaving Divya = 2. Clue (4) does not separate them, since both 4 and 2 are even — this is the point at which a careless solver stops and guesses. Clue (5) settles it: with Divya = 4 and Esha = 2 there is exactly one person (on floor 3) between them, but with Divya = 2 and Esha = 3 they are adjacent and nobody is between. So Esha = 2, Amit = 3, Divya = 4, uniquely.

Derivation of the subjects. Clue (6) puts Physics on floor 3, which is Amit. Clue (9) then puts Botany immediately below, on floor 2 — that is Esha. Clue (8) gives Bina Economics. Clue (7) bars Chetan from Botany and Maths; Botany is already Esha's and Economics is already Bina's, so Chetan must take History, and the only subject left, Maths, goes to Divya. Final answer: floor 1 Chetan–History, floor 2 Esha–Botany, floor 3 Amit–Physics, floor 4 Divya–Maths, floor 5 Bina–Economics. Every clue is satisfied and nothing was chosen arbitrarily, so this is the only solution.

Day and month scheduling works identically, with days as the columns. Five employees P, Q, R, S and T take leave on five different days from Monday to Friday.
(a) S takes leave on Friday.
(b) P takes leave on the day immediately after R.
(c) Q takes leave two days after P. Try each start for R: R = Monday gives P = Tuesday and Q = Thursday, which leaves Wednesday for T — consistent. R = Tuesday gives P = Wednesday and Q = Friday, but Friday is S's — rejected. R = Wednesday gives P = Thursday and Q = Saturday, which does not exist — rejected. So the unique schedule is R Monday, P Tuesday, T Wednesday, Q Thursday, S Friday.

In a matrix coding question two grids are printed. Matrix I has its rows and columns numbered 0 to 4 and Matrix II has them numbered 5 to 9, so a two-digit code never belongs to both. A letter is coded by row digit first, then column digit — and because each letter appears five times in its matrix, every letter has five valid codes. Your job is to find the option in which all the codes are valid for the letters of the given word, in order.

Read the codes for a letter by walking down its diagonal. A appears at 00, 14, 23, 32 and 41; E at 01, 10, 24, 33, 42; I at 02, 11, 20, 34, 43; O at 03, 12, 21, 30, 44; U at 04, 13, 22, 31, 40. In Matrix II, B appears at 55, 69, 78, 87, 96; C at 56, 65, 79, 88, 97; D at 57, 66, 75, 89, 98; F at 58, 67, 76, 85, 99; G at 59, 68, 77, 86, 95. So BAG can legitimately be written 55-00-59, or 78-23-86, or any other mixture of one valid code per letter — and CUE can be written 56-04-33. The efficient exam technique is to check the last code of each option first, because that is where the paper hides the single wrong digit.

Mathematical operations questions replace the arithmetic symbols with other symbols and ask you to evaluate. The only two mistakes possible are copying a substitution wrongly and forgetting BODMAS. The order is Brackets, Of or Order (powers), Division, Multiplication, Addition, Subtraction — and it applies to the real operations after substitution, never to the printed symbols. That is exactly where marks are lost: candidates substitute correctly and then work left to right.

Worked substitution. Using the table above, evaluate 12 + 6 − 3 × 4 ÷ 2. Rewrite it in real operations: 12 × 6 ÷ 3 + 4 − 2. Now BODMAS: division and multiplication first, left to right, so 12 × 6 = 72 and 72 ÷ 3 = 24; then addition and subtraction, giving 24 + 4 − 2 = 26. A letter-coded version works identically: if P means +, Q means −, R means × and S means ÷, then 18 S 6 R 4 Q 3 P 5 becomes 18 ÷ 6 × 4 − 3 + 5 = 3 × 4 − 3 + 5 = 12 − 3 + 5 = 14.

Interchange of signs. "Which equation becomes correct when ÷ and + are interchanged?" Test the option 16 + 8 ÷ 6 = 8: after interchanging, it reads 16 ÷ 8 + 6, which is 2 + 6 = 8 — correct. Interchange of numbers. "Which two numbers must be interchanged to make 2 + 5 × 3 − 4 = 17 correct?" As printed the left side is 2 + 15 − 4 = 13, so something must change. Swapping 2 and 4 gives 4 + 5 × 3 − 2 = 4 + 15 − 2 = 17 — correct. All five other swaps fail (5 and 3 give 13 again, 2 and 5 give 7, 2 and 3 give 9, 5 and 4 give 9, 3 and 4 give 19), so the answer is unique. Balancing an equation. "Fill 20 * 4 * 2 = 12": try the pair (−, ×) to get 20 − 4 × 2 = 20 − 8 = 12 — correct, because BODMAS does the multiplication first.