Classification (Odd One Out)

high importance~3 Q in Tier 19 formulas⚡ 9 shortcuts6 subtopics

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Number pairs and sets odd one out

Instead of single numbers, each option is a pair (12 : 144) or a set of three numbers (3, 9, 27). Three options are built by the same operation; one is not. Recent SSC and Railway papers ask this form often, with a note that operations must use the whole numbers (no splitting 13 into 1 and 3).

Method: find the operation from the first option (square, ×3 chain, product of outer numbers...), test it on every option, and mark the one that fails.

📖 Notes 🧩 5 question types ⚡ Formulas 🎯 12 practice Q

Detailed notes

What are number-pair and number-set questions?

Here each option is not one number but a small group of numbers: a pair like 12 : 144 or a set like (3, 9, 27). Three options follow the same operation; one does not. The question is still "find the odd one", but now you look for the operation that links the numbers inside each option.

Everyday example: (2, 4, 8), (3, 6, 12), (5, 10, 20), (4, 8, 15). In the first three each number is double the previous one. In the last set 8 × 2 = 16, not 15. So (4, 8, 15) is odd.

The whole-number note

Many papers add: "Operations should be performed on the whole numbers, without breaking them into their digits." This means 13 must be used as 13. You may not treat it as 1 and 3. Rules like "sum of digits" are not allowed in such questions.

Type 1: pairs (a : b)

The second number is made from the first. Common rules:

$b = a^2$ (12 : 144), $b = a^3$
$b = a^2 \pm k$ (4 : 19 is $4^2 + 3$ )
$b = a^3 \pm 1$ (5 : 126 is $5^3 + 1$ )
$b = ka \pm r$ (7 : 22 is 3 × 7 + 1)

Type 2: chain sets (a, b, c)

The same step is used twice: a → b → c. For (2, 7, 17): 2 × 2 + 3 = 7 and 7 × 2 + 3 = 17. So the rule is "×2 + 3". Check the rule on both steps of every set.

Type 3: middle from the outer numbers (a, m, c)

The middle number is made from the first and last numbers. Common rules: $m = a \times c$ (4, 20, 5), $m = a^2 + c^2$ (5, 34, 3 because 25 + 9 = 34), $m = (a + c) \times k$ .

Type 4: third from the first two (a, b, c)

The last number is made from the first two: $c = 2(a + b)$ as in (12, 8, 40), or $c = a \times b - 1$ as in (7, 3, 20).

Type 5: select the set that belongs

Two sets are given, for example (3, 12, 48) and (2, 8, 32). You must pick the option that follows the same rule (here ×4 twice). Only one option fits.

Method

Look at sizes to guess the family (squares, product, chain).
Write the rule using the first option.
Test it on the second option. If it fails, change the rule.
Test the remaining options. Exactly one should fail (or, in Type 5, exactly one should fit).

Worked example

Odd set: (6, 11, 21), (8, 15, 29), (5, 9, 17), (7, 13, 26).

6 → 11 is 6 × 2 − 1; 11 → 21 is 11 × 2 − 1. Rule: ×2 − 1 twice.
8 → 15 → 29 ✓; 5 → 9 → 17 ✓.
7 → 13 ✓ but 13 × 2 − 1 = 25, not 26. Answer (7, 13, 26).

Quick revision

Type	Rule to try first	Example
Pair	$a^2$ , $a^3$ , $a^2 \pm k$ , $ka \pm r$	12 : 144
Chain set	same step twice	(2, 7, 17): ×2 + 3
Middle from outer	$a \times c$ , $a^2 + c^2$	(4, 20, 5)
Third from first two	$2(a+b)$ , $ab - 1$	(12, 8, 40)
Belongs to group	find rule of given sets, pick the fit	(5, 20, 80)
Note present	never split into digits	13 stays 13

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Number pairs with one wrong relation (a : b)very common3 practice Q

How to spot it:

Options are pairs like 12 : 144, 4 : 19. The second number is a square, cube or simple function of the first.

b = a^2 \pm k \;\text{or}\; b = a^3 \pm k \;\text{or}\; b = ka \pm r

Guess the rule from the first pair (square? cube? square + k?).
Confirm on a second pair.
The pair that fails is odd.

Why it works: one pair is built with the wrong constant or the wrong power.

Example: Find the odd pair: 12 : 144, 15 : 225, 9 : 81, 11 : 111.

144 = 12², 225 = 15², 81 = 9². 11² = 121, not 111. Answer: 11 : 111.

Type 2: Chain sets (a, b, c): same operation twicecommon3 practice Q

How to spot it:

Sets of three numbers that grow steadily (3, 9, 27) or (2, 7, 17).

(a,\ f(a),\ f(f(a)))

Find the operation from a to b.
Check that the same operation turns b into c.
The set where the second step fails is odd.

Why it works: the setter usually spoils only the last number of one set.

Example: Find the odd set: (3, 9, 27), (4, 16, 64), (5, 25, 125), (6, 36, 206).

Each set is (n, n², n³). 6³ = 216, not 206. Answer: (6, 36, 206).

Type 3: Middle number from the two outer numberscommon2 practice Q

How to spot it:

Sets of three where the middle number is much larger than the two outer numbers.

m = a \times c \;\text{or}\; m = a^2 + c^2

Try middle = first × last.
If not, try first² + last², or (first + last) × k.
The set that fails is odd.

Why it works: the outer numbers are the inputs; the middle is the output.

Example: Find the odd set: (4, 20, 5), (3, 21, 7), (6, 48, 8), (9, 45, 6).

4 × 5 = 20, 3 × 7 = 21, 6 × 8 = 48, 9 × 6 = 54 ≠ 45. Answer: (9, 45, 6).

Type 4: Third number from the first twooccasional2 practice Q

How to spot it:

Sets where the last number is the largest and looks like a sum or product of the first two.

c = k(a + b) \;\text{or}\; c = ab \pm r

Try c = a + b, c = k(a + b), c = a × b ± r.
Confirm the rule on two sets.
The set that breaks it is odd.

Why it works: the first two numbers are inputs and the third is the output.

Example: Find the odd set: (12, 8, 40), (15, 5, 40), (9, 7, 32), (10, 6, 30).

c = 2(a + b): 2 × 20 = 40 ✓, 2 × 20 = 40 ✓, 2 × 16 = 32 ✓, 2 × 16 = 32 ≠ 30. Answer: (10, 6, 30).

Type 5: Select the set / pair that belongs to the given groupcommon2 practice Q

How to spot it:

The question gives two sets or pairs as a model and asks which option is related in the same way.

Find the rule that fits BOTH model sets.
Test every option.
Exactly one option fits - choose it.

Why it works: this is classification turned around - three options break the rule, one follows it.

Example: Which set is like (3, 12, 48) and (2, 8, 32)? (5, 20, 80), (4, 12, 36), (6, 24, 72), (7, 28, 84).

Model rule: ×4, ×4. Only (5, 20, 80) fits (20 × 4 = 80). Answer: (5, 20, 80).

Formulas

Pair rule

b = f(a),\; f(a) \in \{a^2,\ a^3,\ a^2 \pm k,\ ka \pm r\}

Find f from one pair, check the other three.

Chain triad

(a,\ f(a),\ f(f(a)))

Same operation applied twice, e.g. ×3: (3, 9, 27).

Outer-to-middle triad

(a,\ g(a, c),\ c),\; g = ac \text{ or } a^2 + c^2

Middle number made from the two outer numbers.

First-two-to-third triad

(a,\ b,\ h(a, b)),\; h = k(a+b) \text{ or } ab \pm r

Third number made from the first two.

Shortcut tricks

⚡ Size tells the operation

Compare sizes: if b is about a², think square; if c ≈ 3 × b and b ≈ 3 × a, think a ×3 chain; if the middle number is large and the outer ones small, think product of the outer numbers.

Example: Q. Odd set: (4, 20, 5), (3, 21, 7), (6, 48, 8), (9, 45, 6).

Middle is big, outer small → try product: 4 × 5 = 20 ✓, 3 × 7 = 21 ✓, 6 × 8 = 48 ✓, 9 × 6 = 54 ≠ 45. Odd set (9, 45, 6).

⚡ Test the rule on two options before trusting it

A rule found from one option may be a coincidence. Confirm it on a second option, then check the remaining two. The one that fails is the answer.

Example: Q. Odd pair: 4 : 19, 6 : 39, 7 : 52, 8 : 66.

4 : 19 could be 4 × 4 + 3 or 4 × 5 − 1. Test on 6 : 39: 6² + 3 = 39 ✓ (6 × 7 − 1 = 41 ✗). Rule a² + 3. 7² + 3 = 52 ✓, 8² + 3 = 67 ≠ 66. Odd pair 8 : 66.

Where students lose marks

Splitting numbers into digits (reading 12 as 1 and 2) when the note says to use whole numbers.
Fixing the rule from only one option - a single pair fits many rules.
Choosing the option whose numbers are simply the largest, instead of the one that breaks the rule.
In 'select the set like the given sets' questions, picking a set that is different instead of the one that fits.

Practice sets — 12 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.