Missing Number
🔒 Log in to trackFigure-based puzzles — paired grids, single grids, circles — where one cell is hidden behind a numerical rule. Every rule connects the cells of a row (or segment), and the same rule is demonstrated by at least two complete instances. The winning habit is two-line verification: fit the rule on one complete row, confirm on the other, then apply. Digit-sum and digit-reversal variants reward students who switch families quickly when plain arithmetic stalls.
One page per subtopic: detailed notes, every question type, formulas, tricks and practice sets.
Every formula on one printable page, grouped by subtopic.
4 exam-level questions worked step by step.
45 questions — untimed practice or a timed test with analysis.
Track record in the exam
Questions per shift in recent SSC CGL papers.
Test difficulty mix (45 questions)
Question patterns exams keep repeating
Taken from previous-year papers. If a pattern is marked "very common", expect to see it in your exam.
Row-rule hunt in a grid
very commonA 3×3 grid with one missing cell; rows (or columns) share a hidden rule.
How to solve: Test bridges in order: a+b, |a−b|, ab, 2ab, ab±k, a²+b², (a+b)² — fit on every complete row before applying.
Example: Rows: (6,3,15), (8,4,28), (9,5,?)
Plain a+b fails (9, 12). Divide out: 15 = 5×3, 28 = 7×4 — the multipliers 5, 7 are one less than the first numbers → c = (a−1)×b → (9−1)×5 = 40.
Two grids, one rule
commonTwo complete grids and an incomplete third.
How to solve: Derive the rule on the first grid, verify it on the second, then apply to the third.
Example: Grid 1: (2,3,7),(3,4,13). Grid 2: (4,5,?),(5,6,31)
c = a×b + 1 → 31 ✓ → 4×5 + 1 = 21.
Square-based row rule
commonOutput cells sit next to perfect squares (26, 48, 50, 63...).
How to solve: Square the first (or second) input and read the leftover: a² ± b, a² ± 1, a² + b².
Example: Rows: (4,6,10), (5,3,22), (7,2,?)
a + b fits row 1 only. a² − b: 16−6 = 10 ✓, 25−3 = 22 ✓ → 49−2 = 47.
Circle / triangle segments
commonNumbers placed in sectors of a circle or corners of a triangle with a centre value.
How to solve: Treat each sector as a row, or pair opposite sectors; test constant sum, then centre = sum/product of corners.
Example: Triangles (corners → centre): (2,3,4 → 24), (1,5,6 → 30), (3,2,5 → ?)
Centre = product of corners: 2×3×4 = 24 ✓, 1×5×6 = 30 ✓ → 3×2×5 = 30.
Reversed-digit rule
commonMirror-image digits across the row (23 beside 32); outputs look like scrambled sums.
How to solve: Test R(a) + R(b) and R(a + b); confirm on both complete rows.
Example: Rows: (23,14,73), (45,12,75), (62,23,?)
R(23)+R(14) = 32+41 = 73 ✓; R(45)+R(12) = 54+21 = 75 ✓ → 26+32 = 58.
Digit-product rule
occasionalOutputs are small and factor neatly into the digits of the inputs.
How to solve: P(n) = product of digits; test P(a), P(a) + P(b), then weighted mixtures.
Example: Rows: (24,35,23), (43,26,24), (52,33,?)
P(24)+P(35) = 8+15 = 23 ✓; P(43)+P(26) = 12+12 = 24 ✓ → P(52)+P(33) = 10+9 = 19.
Digit-sum grid
commonThird numbers tiny compared with the values.
How to solve: S(a)+S(b) first, then S(a+b); fit on all complete rows.
Example: Rows: (29,14,18), (38,25,20), (56,34,?)
S(29)+S(14) = 11+5 = 16, but the answer is 18 — a constant +2 gap. Check row 2: S(38)+S(25) = 11+7 = 18, +2 = 20 ✓ → rule is S(a)+S(b)+2 → 11+7+2 = 20.