The study desk / Compiled from 172 previous-year questions

Every maths formula the last 15 years actually asked.

Not a textbook summary. We read 172 SSC CGL quantitative aptitude questions from 2010–2024 — Tier-I and Tier-II — and wrote down only what those papers rewarded: the formula, the one-line shortcut, and the wrong option the examiner planted next to it.

126

formulas worth knowing cold

72

shortcuts that save a step

44

traps the paper plants

75

question patterns decoded

Why shortcuts, and not just formulas

The SSC CGL 2026 notice puts a 15-minute sectional timer on 25 quantitative aptitude questions in Tier-I. That is 36 seconds per question, with 0.5 marks lost for each wrong answer. Tier-II gives you 60 seconds and takes 1 full mark back.

At that pace, knowing that CI − SI over two years is P(R/100)² is not a convenience — it is the difference between a solved question and a skipped one. That is what this sheet collects.

What the papers leaned on most

Chapters ranked by how many of the 172 analysed questions came from them. Of those questions, 60 are flagged in the source as repeated templates — wordings that come back year after year with the numbers changed.

  1. 10Algebra (identities & equations)14 PYQs
  2. 11Geometry (triangles, circles, tangents)14 PYQs
  3. 13Mensuration 2D & 3D14 PYQs
  4. 01Number System & Simplification12 PYQs
  5. 02HCF / LCM & Surds–Indices12 PYQs

Read this one free

Chapter 01, in full — nothing held back

01

Number System & Simplification

Cyclicity, remainders and factor counting. Almost every question here is a rule you either know in five seconds or cannot do at all.

12 PYQs analysed · 4 repeated templates · 3 at Tier-II · seen in 9 of the 15 years

Formulas (11)

Division identity
Dividend = Divisor × Quotient + Remainder
Sum of first n naturals
Σn=n(n+1)2\Sigma n = \frac{n(n+1)}{2}
Sum of first n squares
Σn2=n(n+1)(2n+1)6\Sigma n^2 = \frac{n(n+1)(2n+1)}{6}
First 10 = 385, first 20 = 2870 — both have been asked directly.
Sum of first n cubes
Σn3=[n(n+1)2]2\Sigma n^3 = \left[\frac{n(n+1)}{2}\right]^2
Odd and even sums
first n odd=n2\text{first } n \text{ odd} = n^2first n even=n(n+1)\text{first } n \text{ even} = n(n+1)
Number of divisors
N=ap bq cr  ⇒  divisors=(p+1)(q+1)(r+1)N = a^p\,b^q\,c^r \;\Rightarrow\; \text{divisors} = (p+1)(q+1)(r+1)
Sum of divisors
σ(N)=ap+1−1a−1×bq+1−1b−1×⋯\sigma(N) = \frac{a^{p+1}-1}{a-1} \times \frac{b^{q+1}-1}{b-1} \times \cdots
Total vs distinct prime factors
total = sum of the powers distinct = count of bases
Highest power of a prime p in n!
⌊np⌋+⌊np2⌋+⌊np3⌋+⋯\left\lfloor\frac{n}{p}\right\rfloor + \left\lfloor\frac{n}{p^2}\right\rfloor + \left\lfloor\frac{n}{p^3}\right\rfloor + \cdots
Trailing zeros of n! = highest power of 5 (2s are always in surplus).
Cube-sum factorisations
a3+b3a2−ab+b2=a+b\frac{a^3+b^3}{a^2-ab+b^2} = a+ba3−b3a2+ab+b2=a−b\frac{a^3-b^3}{a^2+ab+b^2} = a-b
The middle sign of the denominator tells you which one you have.
Order of operations
VBODMAS — Vinculum, Brackets, Of, Division, Multiplication, Addition, Subtraction

Shortcuts (9)

Unit digit by cyclicity
Cycles of length 4: 2→(2,4,8,6), 3→(3,9,7,1), 7→(7,9,3,1), 8→(8,4,2,6). Length 2: 4→(4,6), 9→(9,1). Never change: 0, 1, 5, 6. Take (power mod 4) and read that term.
Largest / smallest n-digit multiple
Largest = (largest n-digit number) − its remainder on ÷D. Smallest = (smallest n-digit number) + (D − remainder).
Add or subtract to make it divisible
"Least to ADD" = D − r. "Least to SUBTRACT" = r. Two different answers from the same division — read which one is wanted.
Second divisor is a factor of the first
Just divide the old remainder by the new divisor. Remainder 29 on ÷56 ⇒ remainder 5 on ÷8.
Remainder of a power
Force the base into (multiple of divisor ± 1). Base ≡ 1 ⇒ remainder always 1. Base ≡ −1 ⇒ remainder 1 for even powers, (D−1) for odd.
Trailing zeros of a factorial
Divide n by 5, 25, 125 … and add the quotients, ignoring remainders. 100! → 20 + 4 = 24. 200! → 40 + 8 + 1 = 49.
Composite divisibility
Split the divisor into co-prime factors and test each. Divisible by 45 ⇔ by 9 and by 5. By 12 ⇔ by 4 and by 3.
Divisibility rules worth holding
4: last two digits. 8: last three. 9: digit sum. 11: alternating digit sum. 7/13: strike off the last digit and subtract twice it (7) or add four times it (13), repeat.
Continued fractions
Always resolve bottom-up. Successive values often run along the Fibonacci ratios (3/2, 5/3, 8/5) — a free sanity check.

Traps the paper plants (3)

  • Power ÷ 4 leaving remainder 0 means the FOURTH term of the cycle, not the first. This single rule decides most unit-digit questions.
  • "Total number of prime factors" is the sum of the powers; "number of factors" is the product of (power + 1). The paper offers both as options.
  • In a sum like 3³³³ + 2²²², work out each unit digit separately and add them, then keep only the last digit of that sum.

If the question says… (5)

Unit digit of aⁿ
n mod 4, read the cyclicity table.
Remainder when aⁿ is divided by p
Rewrite a as (kp ± 1) and use the parity of n.
Number of zeros at the end of n!
Successive division by 5, add quotients.
Least number added to / subtracted from N for divisibility by D
One division: D − r to add, r to subtract.
Value of 1² + 2² + … + n²
n(n+1)(2n+1)/6, cancelling the 6 before multiplying.

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