Logical Reasoning: Puzzles and Arrangements
CSAT analytical reasoning — linear and circular seating, blood relations, direction sense, ranking, coding–decoding, number and letter series, clocks and calendars, with worked examples.
SyllabusCSAT · Logical reasoning and analytical ability
Prerequisites: Logical Reasoning — overview · Syllogisms and Arguments
The Hook: What Day Was 26 January 1950?
India became independent on 15 August 1947, a Friday, and the Constitution came into force on 26 January 1950. Without a calendar, can you work out the weekday of Republic Day from Independence Day alone?
You can, in under a minute, with one idea — odd days. CSAT’s analytical-ability questions (calendars, clocks, seating, relations, directions, series) all work like this: one small rule, applied cleanly on a rough diagram. By the end of this page you will have the answer (it is worked out under Calendars below).
The Core Concept
The universal method for every puzzle in this family:
graph TD
A[Read all conditions once] --> B[Start with the fixed condition: a definite position, an end, opposite seats]
B --> C[Add conditions that link to what is already placed]
C --> D{Branches?}
D -->|Yes| E[Draw each case side by side and test remaining conditions]
D -->|No| F[Fill the rest]
E --> F
F --> G[Check every condition once against the final diagram]1. Linear seating
Facing north, a person’s right is your right as you look at the diagram (east). Facing south, left and right flip.
Worked example. Six friends J, K, L, M, N and O sit in a row facing north.
- K is third from the left end.
- M sits to the immediate right of K.
- N sits at the extreme right end.
- J sits at one of the ends.
- O sits somewhere to the left of L.
Working: Fix K at position 3 and M at 4 (conditions 1, 2). N takes position 6 (condition 3), so J must take position 1 (condition 4). Positions 2 and 5 remain for O and L; O must be to the left of L, so O = 2, L = 5.
| Position (left → right) | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Person | J | O | K | M | L | N |
2. Circular seating
When everyone faces the centre, a person’s left is the clockwise neighbour and their right is the anticlockwise neighbour. (When everyone faces outward, this reverses.)
Worked example. Six friends P, Q, R, S, T and U sit around a circular table facing the centre. Q is to the immediate right of P. S sits opposite P. T is to the immediate left of S. R is not adjacent to S.
Working: Place P. Q is P’s right, i.e. the anticlockwise neighbour. S is opposite P. T is S’s left, i.e. the clockwise neighbour of S. The two seats left are P’s clockwise neighbour and the seat between it and S. R cannot sit next to S, so R sits next to P and U next to S.
Clockwise from P: P → R → U → S → T → Q → (back to P). So R is opposite T, and U is opposite Q.
3. Blood relations
Draw a family tree with four conventions: generations on separate levels, a horizontal line for siblings, a double line for spouses, and +/− for male/female.
Worked example. A is B’s sister. C is B’s mother. D is C’s father. How is A related to D? A and B are C’s children; D is C’s father. So A is D’s grandchild, and A is female (sister) — granddaughter.
4. Direction sense
| Facing | Turn right → | Turn left → |
|---|---|---|
| North | East | West |
| East | South | North |
| South | West | East |
| West | North | South |
Use coordinates: east = +x, north = +y. The straight-line distance is $\sqrt{x^2 + y^2}$.
Worked example. Arun walks 4 km north, turns right and walks 3 km. How far is he from the start? Position $(3, 4)$, distance $= \sqrt{9 + 16} = 5$ km, towards the north-east.
Shadow rule: in the morning the sun is in the east, so shadows fall to the west; in the evening shadows fall to the east.
5. Ranking and ordering
$$\text{Total} = (\text{rank from top}) + (\text{rank from bottom}) - 1$$Worked example. Rohan is 9th from the top and 26th from the bottom of his class. Class size $= 9 + 26 - 1 = 34$.
People between two positions $p < q$ counted from the same end: $q - p - 1$.
6. Coding–decoding
Know the alphabet positions cold. Anchors: EJOTY = 5, 10, 15, 20, 25. A letter and its opposite (A↔Z, B↔Y…) have positions adding to 27.
| Pattern | Example |
|---|---|
| Shift each letter by a fixed number | FIRE → GJSF (each +1) |
| Opposite letters | CAT → XZG (C=3→24=X, A→Z, T=20→7=G) |
| Sum of positions | CAT = 3 + 1 + 20 = 24, so DOG = 4 + 15 + 7 = 26 |
| Reverse the word, then shift | Check both orders when a simple shift fails |
7. Number and letter series
Test in this order: differences → second differences → ratios → squares/cubes → alternating (two interleaved series).
| Series | Rule | Next term |
|---|---|---|
| 2, 6, 12, 20, 30, ? | $n(n+1)$ | 42 |
| 1, 8, 27, 64, ? | $n^3$ | 125 |
| 5, 11, 23, 47, ? | $\times 2 + 1$ | 95 |
| A, C, F, J, O, ? | gaps +2, +3, +4, +5, +6 | U |
8. Clocks
The minute hand moves $6^\circ$ per minute; the hour hand moves $0.5^\circ$ per minute. The angle between the hands at $H$ hours $M$ minutes:
$$\theta = |30H - 5.5M|$$(take $360^\circ - \theta$ if the result exceeds $180^\circ$).
Worked example. At 3:40, $\theta = |90 - 220| = 130^\circ$.
| Event | In 12 hours | In 24 hours |
|---|---|---|
| Hands coincide | 11 times | 22 times |
| Hands at right angles | 22 times | 44 times |
| Hands opposite (180°) | 11 times | 22 times |
9. Calendars — odd days
An odd day is the remainder of days after removing complete weeks.
| Period | Odd days |
|---|---|
| Ordinary year (365 days) | 1 |
| Leap year (366 days) | 2 |
| 31-day month / 30-day month | 3 / 2 |
| February (28 / 29 days) | 0 / 1 |
| 100 / 200 / 300 / 400 years | 5 / 3 / 1 / 0 |
Leap year rule: divisible by 4; a century year only if divisible by 400 (2000 was a leap year, 1900 was not).
Hook solved. From 15 August 1947 (Friday) to 15 August 1949 there are two years, one of which contains 29 February 1948 → $2 + 1 = 3$ odd days → Monday, 15 August 1949. From 15 August 1949 to 26 January 1950: $16 + 30 + 31 + 30 + 31 + 26 = 164$ days $= 23 \times 7 + 3$ → 3 odd days → Monday + 3 = Thursday.
Memory Anchors
- EJOTY — E5, J10, O15, T20, Y25.
- Opposite letters sum to 27 — pairs AZ, BY, CX, DW, EV, FU, GT, HS, IR, JQ, KP, LO, MN ("BY" and “LO” are easy hooks).
- Facing centre: Left = cLockwise (both have an L).
- Clock formula: “30 H minus 11/2 M”.
- Odd days: “1-2-3-2” — ordinary year 1, leap year 2, 31-day month 3, 30-day month 2.
For Prelims
- Facing the centre in a circle: left = clockwise, right = anticlockwise.
- Total in a ranking = top rank + bottom rank − 1.
- Hands of a clock coincide 22 times and are at right angles 44 times in a day.
- A century year is a leap year only if divisible by 400.
- An ordinary year moves a date forward by one weekday; a leap year (when 29 February falls in the span) by two.
Speed Strategy
- Draw first, read second. Draw the row or circle before reading the conditions in detail; fill in fixed positions as you read.
- Start with the most restrictive condition (a definite position, an end, “opposite”), not with the first condition in the list.
- Branch on paper, not in your head. If two placements are possible, draw both; the next condition usually kills one.
- Series: if differences and ratios fail within 30 seconds, check for two alternating series, then move on.
- What to skip: long puzzles with eight or more people and four sub-questions are worth attempting only if you clear the first two conditions quickly — otherwise come back at the end.
Common Traps
Practice
Prelims (CSAT)
Pointing to a man, a woman named Meera said, “His mother is the only daughter of my mother.” How is Meera related to the man?
Solution
“The only daughter of my mother” — said by a woman — is Meera herself. So Meera is the man’s mother.
Answer: C
A person walks 5 km north, turns right and walks 3 km, then turns right again and walks 9 km. How far and in which direction is he from his starting point?
Solution
North 5 → $(0, 5)$. Right turn from north = east, 3 km → $(3, 5)$. Right turn from east = south, 9 km → $(3, -4)$.
Distance $= \sqrt{3^2 + 4^2} = 5$ km. He is east and south of the start — south-east.
Answer: A
In a row of 40 students, Anil is 12th from the left end and Sunil is 18th from the right end. How many students sit between Anil and Sunil?
Solution
Sunil’s position from the left $= 40 - 18 + 1 = 23$. Students between positions 12 and 23 $= 23 - 12 - 1 = 10$.
Check: positions 13 to 22 inclusive are 10 seats.
Answer: D
What comes next in the series: 3, 7, 15, 31, 63, ?
Solution
Each term $= 2 \times$ previous $+ 1$: $3 \to 7 \to 15 \to 31 \to 63 \to 127$. (Equivalently, the terms are $2^n - 1$.)
Answer: C
In a certain code, COLD is written as DPME. How is WARM written in that code?
Solution
Each letter moves one place forward: C→D, O→P, L→M, D→E. So W→X, A→B, R→S, M→N, giving XBSN.
Answer: B
Quick Revision
| Point | One-liner |
|---|---|
| Facing centre | Left = clockwise, right = anticlockwise |
| Facing north | Right = east on the diagram |
| Distance | $\sqrt{x^2 + y^2}$ using east/north coordinates |
| Ranking | Total = top + bottom − 1; between = $q - p - 1$ |
| Coding | EJOTY; opposite letters sum to 27 |
| Series | Differences → ratios → powers → alternating |
| Clock angle | $\lvert 30H - 5.5M \rvert$ |
| Coincide / right angle per day | 22 / 44 |
| Odd days | Year 1, leap year 2, 31-day month 3, 30-day month 2 |
| Leap century | Divisible by 400 |
Key Takeaways
- Every arrangement puzzle starts from the most restrictive condition.
- Draw facing directions explicitly — most errors are left–right errors.
- Odd days solve any calendar question; the clock formula solves any angle question.
- Series questions yield to a fixed checking order.
- Verify the final diagram against every condition before answering.
Related Topics
- Logical Reasoning: Syllogisms and Arguments
- Logical Reasoning — quick revision
- Logical Reasoning — practice questions
- Number System — remainders behind calendar and clock questions
- Data Interpretation