Mathematics notes for SSC CGL
Handwritten notes rebuilt as a clean revision sheet — 11 chapters, concepts and formulas stripped of worked examples.
SSC CGL · Tier 2 Quant · Concepts only
Every formula, no questions.
Handwritten notes rebuilt as a clean revision sheet — 11 chapters, concepts and formulas stripped of worked examples.
How to use this on the move
- Type in the search bar to jump straight to a formula — it filters live.
- Tap Recall to blur every formula. Read the label, say the formula out loud, then tap the blurred box to check.
- ★ cards are the ones worth knowing cold. Pink cards are the traps that cost marks.
01
Number System
Classification
The chain of number sets
W ⊂ Z ⊂ Q ⊂ R
Natural → Whole → Integer → Rational → Real
- Rational: can be written as p/q with q ≠ 0, p and q integers.
- Irrational: cannot be written as p/q — non-terminating, non-repeating decimals.
- Composite: a number with more than two factors (4, 6, 8, 9…).
Perfect number
Sum of all factors except the number itself = the number
6, 28, 496, 8128
Equivalently: sum of all factors (including itself) = twice the number.
Fractions — the four types
- Proper: N < D · Improper: N ≥ D
- Mixed fraction is always improper.
- Like fractions share the same denominator; a unit fraction has numerator 1.
Comparing fractions — constant difference rule
If (D − N) is the same for all fractions:
proper fractions → larger numerator = LARGER value
improper fractions → larger numerator = SMALLER value
Example: 1/2, 2/3, 4/5, 5/6, 9/10 all have D − N = 1, so they increase left to right.
When the differences aren't equal, scale one fraction until they match — 1111/1112 (diff 1) versus 3023/3026 (diff 3) becomes 3333/3336 versus 3023/3026.
Otherwise: cross multiply
a/b vs c/d → compare (a×d) with (c×b)
The bigger cross product sits above the bigger fraction.
Primes
Prime counts you must know
1 – 50 → 15 primes
1 – 100 → 25 primes
50 – 100 → 10 primes
The 25 primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Twin primes & co-primes
- Twin primes: two primes differing by 2 — (3,5), (5,7), (11,13)… 8 such pairs from 1 to 100.
- Co-prime (relatively prime): two numbers whose HCF is 1. They need not be prime themselves.
Two-digit primes that stay prime when reversed
13 ↔ 31 · 17 ↔ 71 · 37 ↔ 73 · 79 ↔ 97
Unit digit of a prime number
Only 6 possibilities: 1, 2, 3, 5, 7, 9
(2 and 5 only for the primes 2 and 5 themselves.)
Odd / even behaviour
Sum and product rules
Even + Even = Even Even × Even = Even
Odd + Odd = Even Odd × Even = Even
Even + Odd = Odd Odd × Odd = Odd
Sum of n odd numbers is odd when n is odd, even when n is even.
Counting digits
Total digits used in writing 1 to N
1 – 9 → 9 digits
1 – 99 → 189 digits
1 – 999 → 2889 digits
1 – 9999 → 38889 digits
A 9 always ends the count; the leading part gains an 8 each time.
How many n-digit numbers exist
n-digit numbers = 9 × 10ⁿ⁻¹
n-digit even numbers = 9 × 10ⁿ⁻² × 5
Number of digits in 10ⁿ is n + 1.
Reversing a number
N and its reverse N′
Two-digit N = ab : N + N′ = 11(a + b)
N − N′ = 9(a − b)
Three-digit N = abc: N − N′ = 99(a − c)
The three-digit numbers abc, cab and bca always sum to a multiple of (a + b + c).
Decimals
The three kinds of decimal
- Terminating — 3.535
- Non-terminating repeating — 5.8666… (rational)
- Non-terminating non-repeating — irrational
Converting a bar decimal to a fraction
Pure recurring 0.abc̄ = abc / 999
Mixed recurring 0.ab̄c̄ = (abc − a) / 990
Count of 9s = digits under the bar. Count of 0s = digits after the decimal that are not under the bar.
Adding or subtracting recurring decimals
Take the LCM of the bar lengths, write every number to that many repeating digits, then add or subtract column-wise and put the bar back over the LCM block.
Square roots to memorise
Common surds
| √2 | √3 | √5 | √6 | √7 | √10 | √11 |
|---|---|---|---|---|---|---|
| 1.414 | 1.732 | 2.236 | 2.449 | 2.645 | 3.162 | 3.316 |
Divisibility
Powers of 2 and 5
Divisible by 2ⁿ ⇔ last n digits divisible by 2ⁿ
Divisible by 5ⁿ ⇔ last n digits divisible by 5ⁿ
So: 4 → last 2 digits, 8 → last 3 digits, 16 → last 4 digits.
3 and 9
3 → digit sum divisible by 3
9 → digit sum divisible by 9
7, 11 and 13 — the three-digit pairing method
Break the number into groups of 3 from the RIGHT.
Add and subtract the groups alternately.
If the result is divisible by 7 / 11 / 13, so is the number.
This works because 7 × 11 × 13 = 1001.
11 — the quick way
(Sum of odd places) − (Sum of even places)
If the result is 0 or a multiple of 11 → divisible by 11
Algebraic divisibility
aⁿ − bⁿ is ALWAYS divisible by (a − b)
aⁿ − bⁿ is divisible by (a + b) when n is EVEN
aⁿ + bⁿ is divisible by (a + b) when n is ODD
xᵐ − yᵐ is divisible by xⁿ − yⁿ when n divides m
Trap
aⁿ + bⁿ is not divisible by (a + b) when n is even. Check the parity of the power before you use the rule — this single slip is the whole question in most remainder problems.
Remainders
Remainder theorem for products and sums
If N₁, N₂, N₃ leave remainders r₁, r₂, r₃ on division by k:
remainder of (N₁ × N₂ × N₃) = remainder of (r₁ × r₂ × r₃)
remainder of (N₁ + N₂ + N₃) = remainder of (r₁ + r₂ + r₃)
Always reduce the final result mod k again.
Polynomial remainder theorem
Remainder when f(x) is divided by (x − a) = f(a)
The working trick for huge powers
Look for a base power that lands on +1 or −1 modulo the divisor, then reduce the exponent against that cycle. Example shape: 12³ = 1728 ≡ −1 (mod 1729), so 12190 collapses instantly.
02
Squares, Cubes & Roots
Squares 1–25
Learn these cold
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| n² | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 | 121 | 144 | 169 |
| n | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | — |
| n² | 196 | 225 | 256 | 289 | 324 | 361 | 400 | 441 | 484 | 529 | 576 | 625 | — |
Tests for a perfect square
Unit digit map
| Number ends in | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 0 |
|---|---|---|---|---|---|---|---|---|---|---|
| Square ends in | 1 | 4 | 9 | 6 | 5 | 6 | 9 | 4 | 1 | 0 |
Instant rejection
A perfect square can NEVER end in 2, 3, 7 or 8.
Also: it can only end in an even number of zeros, and if it ends in 5 the tens digit must be 2 (i.e. it ends in 25).
Digital sum test
Digital sum = remainder after dividing by 9
A perfect square's digital sum is only ever 9, 1, 4 or 7.
More properties
- Even number → even square; odd number → odd square. Same holds for cubes.
- The last two digits of an even perfect square are divisible by 4.
- A perfect-square decimal has an even number of digits after the decimal point.
Digits in a square root
If the perfect square has n digits, its root has
n/2 digits when n is even
(n+1)/2 digits when n is odd
Squaring shortcuts
Numbers ending in 5
(n5)² = n(n+1) | 25
35² = 3×4 | 25 = 1225 95² = 9×10 | 25 = 9025
Base method (near 100 / 200 / 50)
Deficiency or excess = d from the base
(100 ± d)² = (100 ± 2d) | d² ← d² written in 2 digits
104² = 108 | 16 = 11664 92² = 84 | 64 = 8464
For base 200, double the first part. For base 50, halve it.
Repdigit squares
33² = 1089
333² = 110889
3333² = 11108889 ← 1s, then 0, then 8s, then 9
11² = 121
111111² = 12345654321 ← palindrome pyramid
Square root of a repunit-minus-repdigit
√( 111…1 (2k digits) − 222…2 (k digits) ) = 333…3 (k digits)
Cubes
Unit digit of a cube — the self/swap pairs
Self: 1→1 4→4 5→5 6→6 9→9 0→0
Swap: 2↔8 3↔7
So a cube ending in 8 came from a number ending in 2, and vice versa.
Cube root of a perfect cube (by inspection)
1. Chop off the last three digits.
2. The unit digit of what you chopped gives the unit digit of the root
(using the self/swap table above).
3. Find the largest cube ≤ the remaining left part — that is the tens digit.
13824 → 24 ends in 4 → root ends in 4. Left part 13, largest cube ≤ 13 is 2³. Root = 24.
32768 → 8 → root ends in 2. Left part 32, largest cube ≤ 32 is 3³. Root = 32.
How many cubes have n digits
1-digit cubes → roots 1–2 4-digit cubes → roots 10–21
2-digit cubes → roots 3–4 5-digit cubes → roots 22–46
3-digit cubes → roots 5–9 6-digit cubes → roots 47–99
So there are 5 perfect cubes between 100 and 1000.
Perfect square AND perfect cube
A number that is both = a perfect 6th power
64 = 2⁶, 729 = 3⁶, 4096 = 4⁶
Useful oddities
- The smallest perfect square ending in three identical digits is 1444 = 38².
- The only 4-digit perfect square of the form aabb is 7744 (a + b = 11).
- Add 1 to the product of four consecutive numbers and you always get a perfect square.
- (n − 1)(n + 1) + 1 = n² — this is why 1 + 8763 × 8765 = 8764².
Long-division square root, in three moves
Pair the digits from the right → find the largest square in the leading pair → double the running quotient, append a digit, and multiply so the product stays ≤ the remainder. Repeat.
03
Factors, Trailing Zeros & Unit Digit
Counting factors
The master set-up
N = a^p × b^q × c^r (a, b, c prime)
Number of factors = (p+1)(q+1)(r+1)
Sum of factors = (a⁰+a¹+…+a^p)(b⁰+…+b^q)(c⁰+…+c^r)
Product of factors = N^(number of factors ÷ 2)
12 has 6 factors, so the product of its factors is 12³ = 1728.
Even and odd factors
Let p be the power of 2 in N.
Number of even factors = p × (q+1)(r+1)
Number of odd factors = (q+1)(r+1) ← ignore the 2s entirely
= Total − Even
Sum of even factors = (2¹+2²+…+2^p) × (b⁰+…+b^q)(c⁰+…+c^r)
Sum of odd factors = (b⁰+…+b^q)(c⁰+…+c^r)
Factors that are perfect squares / cubes
Perfect square factors → count powers 0, 2, 4, … → ⌊p/2⌋ + 1 from each prime
Perfect cube factors → count powers 0, 3, 6, … → ⌊p/3⌋ + 1 from each prime
Both (6th powers) → ⌊p/6⌋ + 1 from each prime
Then multiply across the primes.
Factors that are multiples of k
Pull k out of N first, then count the factors of whatever remains.
Two facts that answer whole questions
Product of the nth factor from the start
and the nth factor from the end = N
Odd number of factors ⇔ N is a perfect square
(√N is the unpaired middle factor)
So the count of numbers up to 100 with an odd number of factors is just the count of perfect squares: 10. Up to 160 it is 12.
Average of all factors
Average = (Sum of all factors) ÷ (Number of factors)
Highest power in a factorial
Legendre's method
Highest power of prime p in n! :
divide n by p, then that quotient by p, and keep going.
Add all the quotients (ignore remainders).
Highest power of 3 in 108! → 36 + 12 + 4 + 1 = 53.
When the divisor is composite
Break it into primes first. For the highest power of 12 = 2² × 3 in 30!: find the power of 2 (=26) and of 3 (=14), then divide the power of 2 by its exponent (26 ÷ 2 = 13) and take the minimum — here 13. Forgetting to divide by the exponent is the classic error.
Trailing zeros
Zeros come from 5 × 2 pairs
In a factorial, 2s are always in surplus — so just count the 5s.
Zeros in n! = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + …
100! → 20 + 4 = 24 zeros. 200! → 40 + 8 + 1 = 49 zeros.
Trap — no even numbers, no zeros
In a product of odd numbers only (1 × 3 × 5 × … × 99) there are no 2s to pair with the 5s, so the answer is 0 trailing zeros no matter how many 5s you find.
Non-factorial products
- Product of the first 50 primes → only one 2 and one 5 exist → 1 zero.
- For a range like 317 × 318 × … × 996, count the 5s up to 996 and subtract the count up to 316.
- For 1¹ × 2² × 3³ × … × 26²⁶, count 5s with their exponents: 5⁵, 10¹⁰, 15¹⁵, 20²⁰, 25²⁵ (which is 5⁵⁰), 26 → 106.
Unit digit
Cyclicity of 4
Divide the power by 4 and use the remainder r:
r = 1, 2, 3 → use base^r
r = 0 → use base⁴
781: 81 ÷ 4 leaves 1 → unit digit of 7¹ = 7. 740: remainder 0 → unit digit of 7⁴ = 1.
Digits with no cycle
0ⁿ → 0 1ⁿ → 1 5ⁿ → 5 6ⁿ → 6
Digits with a cycle of 2
4^odd → 4 4^even → 6
9^odd → 9 9^even → 1
Series of powers
For 1ⁿ + 2ⁿ + 3ⁿ + … the unit digits repeat every 10 terms. Work out one block of ten, then multiply by the number of complete blocks and add the tail.
04
Powers, Surds & Indices
Laws
The index laws
a⁰ = 1 a⁻ⁿ = 1/aⁿ
aᵐ × aⁿ = aᵐ⁺ⁿ aᵐ ÷ aⁿ = aᵐ⁻ⁿ
(aᵐ)ⁿ = aᵐⁿ a^(1/n) = ⁿ√a
(ab)ⁿ = aⁿbⁿ (a/b)ⁿ = aⁿ/bⁿ
Comparing exponents — the exceptions
If aᵐ = aⁿ → m = n, but ONLY for a ≠ 0, 1, −1
If aᵐ = bᵐ → a = b when m is odd
a = ±b when m is even
Definition of a surd
A number of the form a1/n that is not an integer.
Values worth memorising
Powers that keep appearing
2⁶=64 2⁷=128 2⁸=256 2⁹=512
2¹⁰=1024 2¹¹=2048 2¹²=4096
3⁴=81 3⁵=243 3⁶=729 3⁷=2187
5⁵=3125 5⁶=15625 6⁶=46656
Comparing surds
Different roots — equalise the index
Take the LCM of the root indices and raise everything to it. To compare 31/4, 41/6, 61/8, 121/12, the LCM of 4, 6, 8, 12 is 24 — raise each to the 24th power and compare plain integers.
Comparing √a + √b type expressions
Square both sides. You get (a + b) + 2√(ab).
Same SUM → the larger PRODUCT is larger
Same PRODUCT → the larger SUM is larger
When the two terms are a difference, compare the reciprocals instead — the larger difference gives the smaller value.
Rationalisation
1/(√a + √b) × (√a − √b)/(√a − √b) = (√a − √b)/(a − b)
Un-nesting a square root
√(a + 2√b) = √x + √y where x + y = a and xy = b
√(7 + 4√3) = √(7 + 2√12) = 2 + √3
√(13 + 4√3) = 2√3 + 1
First force the surd into the shape 2√b, then split the outer number into two parts whose product is b.
Infinite nested radicals
The four core forms
√(a + √(a + √(a + …∞))) = (1 + √(1 + 4a)) / 2
√(a − √(a − √(a − …∞))) = (−1 + √(1 + 4a)) / 2
√(a + b√(a + b√(a + …∞))) = (b + √(b² + 4a)) / 2
√(a − b√(a − b√(a − …∞))) = (−b + √(b² + 4a)) / 2
√(6 + √(6 + √(6 + …))) = (1 + √25)/2 = 3.
Alternating signs
√(a + b√(a − b√(a + b…))) = (b + √(4a − 3b²)) / 2
√(a − b√(a + b√(a − b…))) = (−b + √(4a − 3b²)) / 2
With b = 1 these reduce to (1 ± √(4a − 3))/2.
Multiplicative nesting
√(a·√(a·√(a·…∞))) = a
√(a·√(a·√a…)) to n terms = a^((2ⁿ − 1)/2ⁿ)
05
Simplification & Series
Standard sums
The seven you cannot look up in the exam
1 + 2 + 3 + … + n = n(n+1)/2
1² + 2² + … + n² = n(n+1)(2n+1)/6
1³ + 2³ + … + n³ = [ n(n+1)/2 ]²
Sum of first n ODD numbers = n²
Sum of first n EVEN numbers = n(n+1)
Squares of first n even numbers = 2n(2n+1)(n+1)/3
Squares of first n odd numbers = n(2n+1)(2n−1)/3
Products of consecutive integers
1·2 + 2·3 + 3·4 + … (n terms) = n(n+1)(n+2)/3
1·2·3 + 2·3·4 + … (n terms) = n(n+1)(n+2)(n+3)/4
Pattern: multiply by one more factor and divide by one more number.
Alternating squares
1² − 2² + 3² − 4² + … + (2n−1)² − (2n)² = −n(2n+1)
Telescoping
The split that makes everything cancel
(a − b)/(ab) = 1/b − 1/a
1/(1·2) + 1/(2·3) + … + 1/(n(n+1)) = 1 − 1/(n+1) = n/(n+1)
In any such chain, only the first and last terms survive.
Three and four factors — the short trick
Sum of k terms of 1/[n(n+1)(n+2)] :
= ½ [ 1/(n(n+1)) − 1/((n+k)(n+k+1)) ]
Sum of k terms of 1/[n(n+1)(n+2)(n+3)] :
= ⅓ [ 1/(n(n+1)(n+2)) − 1/((n+k)(n+k+1)(n+k+2)) ]
The leading fraction is 1 ÷ (one less than the number of factors).
Three-factor split
1/(a·b·c) = [1/(c − a)] × [ 1/(a·b) − 1/(b·c) ]
Product chain
(1 − 1/2)(1 − 1/3)(1 − 1/4) … (1 − 1/n) = 1/n
Ratio techniques
Componendo & Dividendo
If a/b = c/d then (a + b)/(a − b) = (c + d)/(c − d)
Use it the moment you see a ratio of two expressions that differ only by a sign — for example (x⁴+1) : (x⁴−1) = 313 : 312 collapses straight to x⁴ = 625.
The "answer is always 2" pattern
If x = 4ab/(a + b) then
(x + 2a)/(x − 2a) + (x + 2b)/(x − 2b) = 2
Same shape with x = 2√(ab) and the 2a, 2b replaced by 2√a·2√b terms. Split the given x into the two pieces the question is asking about.
06
Algebra
Two-variable identities
Squares
(a + b)² = (a − b)² + 4ab
(a + b)² + (a − b)² = 2(a² + b²)
a² + b² = (a + b)² − 2ab = (a − b)² + 2ab
Cubes
(a + b)³ = a³ + b³ + 3ab(a + b)
(a − b)³ = a³ − b³ − 3ab(a − b)
a³ + b³ = (a + b)(a² − ab + b²) = (a + b)³ − 3ab(a + b)
a³ − b³ = (a − b)(a² + ab + b²) = (a − b)³ + 3ab(a − b)
Higher even powers — the halving rule
aⁿ + bⁿ = (a^(n/2) + b^(n/2))² − 2·a^(n/2)·b^(n/2)
a⁴ + b⁴ = (a² + b²)² − 2a²b²
a⁸ + b⁸ = (a⁴ + b⁴)² − 2a⁴b⁴
a⁶ + b⁶ = (a² + b²)³ − 3a²b²(a² + b²)
Odd powers — the cross-multiply rule
a⁵ + b⁵ = (a² + b²)(a³ + b³) − a²b²(a + b)
a⁷ + b⁷ = (a³ + b³)(a⁴ + b⁴) − a³b³(a + b)
a⁵ − b⁵ = (a² + b²)(a³ − b³) + a²b²(a − b)
a⁷ − b⁷ = (a³ − b³)(a⁴ + b⁴) + a³b³(a − b)
Split the power into two parts, multiply those two brackets, then subtract (for +) or add (for −) the overlap.
Differences and factorisation
a² − b² = (a − b)(a + b)
a⁴ − b⁴ = (a² − b²)(a² + b²)
a⁶ − b⁶ = (a³ − b³)(a³ + b³)
aⁿ − 1 = (a − 1)(a^(n−1) + a^(n−2) + … + a + 1)
Three-variable identities
The workhorse
a³ + b³ + c³ − 3abc = (a+b+c)(a² + b² + c² − ab − bc − ca)
= (a+b+c) × ½[(a−b)² + (b−c)² + (c−a)²]
The two conclusions examiners want
If a + b + c = 0 → a³ + b³ + c³ = 3abc
If a³ + b³ + c³ = 3abc and a+b+c ≠ 0 → a = b = c
Squares and cubes of a three-term sum
(a+b+c)² = a² + b² + c² + 2(ab + bc + ca)
(a+b+c)³ = a³ + b³ + c³ + 3(a+b)(b+c)(c+a)
(a+b+c+d)² = a²+b²+c²+d² + 2(ab+ac+ad+bc+bd+cd)
Differences of differences
(a−b)³ + (b−c)³ + (c−a)³ = 3(a−b)(b−c)(c−a)
(a−b)² + (b−c)² + (c−a)² = 2(a²+b²+c²) − 2(ab+bc+ca)
Two forms of the same thing
a² + b² + c² − ab − bc − ca
= ½[(a−b)² + (b−c)² + (c−a)²]
= (a+b+c)² − 3(ab + bc + ca)
Sum-of-squares = 0
If (m−5)² + (n−7)² + (p−6)² = 0 → m=5, n=7, p=6
If a² + b² + c² = ab + bc + ca → a = b = c
Real squares can't be negative, so each bracket must be zero individually. Any question that opens with a scattered quadratic and asks for a single value is usually this.
Expansion worth remembering
(a+b+c)(ab + bc + ca) = a²b + ab² + b²c + bc² + a²c + ac² + 3abc
x + 1/x — the biggest scoring topic
The ladder
x + 1/x = a → x² + 1/x² = a² − 2
x − 1/x = a → x² + 1/x² = a² + 2
x + 1/x = k → x − 1/x = √(k² − 4)
x − 1/x = k → x + 1/x = √(k² + 4)
Cubes
x + 1/x = m → x³ + 1/x³ = m³ − 3m
x − 1/x = m → x³ − 1/x³ = m³ + 3m
Climbing to any power — the chaining rule
(xᵃ + 1/xᵃ)(x^b + 1/x^b) = x^(a+b) + 1/x^(a+b) + x^(a−b) + 1/x^(a−b)
x⁵ + 1/x⁵ = (x² + 1/x²)(x³ + 1/x³) − (x + 1/x)
x⁷ + 1/x⁷ = (x³ + 1/x³)(x⁴ + 1/x⁴) − (x + 1/x)
x⁵ − 1/x⁵ = (x² + 1/x²)(x³ − 1/x³) + (x − 1/x)
Multiply two rungs, then subtract the leftover rung. With x + 1/x = 3 this gives x⁵ + 1/x⁵ = 123 and x⁷ + 1/x⁷ = 843.
The special values — memorise, never derive
x + 1/x = 2 → x = 1
x + 1/x = −2 → x = −1
x + 1/x = 1 → x² − x + 1 = 0 → x³ = −1
x + 1/x = −1 → x² + x + 1 = 0 → x³ = 1
x + 1/x = ±√3 → x⁶ = −1
Once you have x³ = ±1, any power reduces instantly. x + 1/x = 1 gives x29 + 1/x29 = −1.
Same rule in ratio form
p/q + q/p = 2 → p = q
p/q + q/p = −2 → p = −q
p/q + q/p = 1 → (p/q)³ = −1
p/q + q/p = −1 → (p/q)³ = 1
So k/4 + 4/k = 1 gives (k/4)³ = −1, hence k³ = −64.
Getting into x + 1/x shape
- Quadratic given: x² − 5x − 1 = 0 → divide right through by x → x − 1/x = 5.
- Shifted variable: x + 1/(x − 29) = 27 → subtract 29 from both sides so the bracket matches.
- Mismatched coefficients: 2x + 1/(3x) = 8 and you need 9x² + 1/(4x²) — multiply the given equation by the factor that makes the coefficients line up before squaring.
- Big fraction: divide numerator and denominator by the middle power of x. For (x⁴+4x³+5x²+4x+1)/(x²+1), divide by x².
Trap
Going down the ladder needs a square root, and a square root has two signs. x⁴ + 1/x⁴ = 62 gives x² + 1/x² = 8, then x − 1/x = √6 — check what the question actually asks for before you take the positive root by reflex.
Value-putting
When variables outnumber equations
If the question gives fewer equations than variables, the answer is independent of the values — so pick easy numbers that satisfy the condition. Given x + y + z = 0, take x = 0, y = 1, z = −1 and evaluate directly.
07
Equations, Roots & Maxima–Minima
Linear systems
Consistency of two lines
a₁x + b₁y + c₁ = 0 a₂x + b₂y + c₂ = 0
a₁/a₂ ≠ b₁/b₂ → intersecting → UNIQUE solution → consistent
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → parallel → NO solution → inconsistent
a₁/a₂ = b₁/b₂ = c₁/c₂ → coincident → INFINITE sols → consistent (dependent)
Counting solutions
- 1 variable, 1 equation → 1 value. n variables with n independent equations → all n values are findable.
- Three variables: eliminate one variable using a pair of equations to reduce to a two-variable system.
- A three-variable linear equation represents a plane, not a line.
Quadratic equations
Roots and reconstruction
ax² + bx + c = 0
α + β = −b/a αβ = c/a
x = [ −b ± √(b² − 4ac) ] / 2a
To rebuild the equation: x² − (sum)x + (product) = 0
Nature of roots — D = b² − 4ac
| Discriminant | Roots |
|---|---|
| D > 0, perfect square | Real, unequal, rational |
| D > 0, not perfect square | Real, unequal, irrational |
| D = 0 | Real and equal (coincident) |
| D < 0 | Imaginary, unequal |
Irrational and imaginary roots always come in conjugate pairs — if one root is 2 + √3 the other is 2 − √3.
Reading signs off the equation
Only the middle term negative → both roots positive
Only the constant term negative → roots have opposite signs
Sign of the whole expression
ax² + bx + c = a(x − α)(x − β)
x OUTSIDE the roots → same sign as a
x BETWEEN the roots → opposite sign to a
Modulus equations — number of solutions
x² + 5|x| + 6 = 0 → 0 solutions (all terms positive)
x² − 5|x| + 6 = 0 → 4 solutions (middle term negative)
x² + 5|x| − 6 = 0 → 2 solutions (constant term negative)
Cubic and higher
Cubic relations
ax³ + bx² + cx + d = 0 with roots p, q, r
p + q + r = −b/a
pq + qr + rp = c/a
pqr = −d/a
Find one root by hit and trial, then use synthetic division to drop to a quadratic. A degree-n equation has at most n roots.
Maxima and minima
The quadratic vertex
For ax² + bx + c, the turning value sits at x = −b/2a
and the value there is (4ac − b²) / 4a
a > 0 → that value is the MINIMUM, maximum is ∞
a < 0 → that value is the MAXIMUM, minimum is −∞
You can also complete the square: x² − 4x + 5 = (x − 2)² + 1, so the minimum is 1.
Reciprocal minimums
m > 0 → m + 1/m has minimum 2
a, b, c > 0 → (a+b+c)(1/a + 1/b + 1/c) has minimum 9
AM ≥ GM
(a + b + c + d)/4 ≥ ⁴√(abcd)
With a fixed sum, the product is largest when all the terms are equal. That equality case is the answer to almost every "find the maximum product" question.
Shifting the expression
For something like a² + 1/(a² − 2) with a > 2, add and subtract the offending constant so you can apply m + 1/m ≥ 2 to the shifted term, then add back what you took out.
08
Geometry
Lines & angles
Parallel lines cut by a transversal
- Corresponding angles are equal.
- Alternate interior angles are equal.
- Vertically opposite angles are equal.
- Co-interior angles on the same side sum to 180°.
Exterior angle theorem
Exterior angle of a triangle = sum of the two interior opposite angles
The zig-zag between parallels
A single bend: α = β + γ
Multiple bends: sum of left-facing angles = sum of right-facing angles
Triangle basics
Can these three sides form a triangle?
a + b > c > a − b (and the same for every pair)
Classifying by the largest side c
a² + b² > c² → acute
a² + b² = c² → right angled
a² + b² < c² → obtuse
Pythagorean triplets
3,4,5 5,12,13 7,24,25 8,15,17 9,40,41
11,60,61 12,35,37 13,84,85 16,63,65 20,21,29
20,99,101 28,45,53 33,56,65 36,77,85 39,80,89
48,55,73 65,72,97
Angles opposite equal sides are equal
And the converse: sides opposite equal angles are equal. If one line acts as median, angle bisector and altitude at the same time, the triangle is isosceles.
Sine & cosine rules
In any triangle
Area = ½ · a · b · sinθ (θ between sides a and b)
sinA/a = sinB/b = sinC/c → a : b : c = sinA : sinB : sinC
cosA = (b² + c² − a²) / 2bc
Cevians — the four centres
Where each one lives
| Cevian | Meets at | Acute | Right | Obtuse |
|---|---|---|---|---|
| Median | Centroid (G) | Inside | Inside | Inside |
| Altitude | Orthocentre (H) | Inside | At the right-angle vertex | Outside |
| Angle bisector | Incentre (I) | Inside | Inside | Inside |
| ⊥ bisector | Circumcentre (O) | Inside | Midpoint of hypotenuse | Outside |
The angle results — learn all four
Incentre (both internal bisectors) ∠BIC = 90° + A/2
Excentre (both external bisectors) ∠BOC = 90° − A/2
One internal + one external bisector angle = A/2
Orthocentre ∠BHC = 180° − A
Circumcentre ∠BOC = 2A
Euler line
H, G and O are always COLLINEAR, with HG : GO = 2 : 1
Distance between incentre and circumcentre: d = √(R² − 2Rr)
Medians
Centroid & Apollonius
Centroid divides every median 2 : 1 from the vertex
Apollonius: AB² + AC² = 2(AD² + BD²) (AD = median to BC)
Sum results
AD² + BE² + CF² = ¾ (AB² + BC² + CA²)
¾(perimeter) < (sum of medians) < (perimeter)
Area of the triangle formed BY the three medians = ¾ × original area
→ Original area = 4/3 × (area of the median triangle)
Area splitting
- A median is an area bisector — it cuts the triangle into two equal areas.
- Joining all three midpoints splits it into 4 congruent triangles.
- Triangles on the same base between the same parallels have equal area.
- Sharing a base, the areas are in the ratio of the heights; sharing a vertex on a line, in the ratio of the bases.
Perpendicular medians
If two medians are perpendicular: b² + c² = 5a²
(a is the side joining the two vertices those medians come from)
Bisectors
Angle bisector theorem
Internal: BD/DC = AB/AC
External: BD/CD = AB/AC
General : BD/DC = (AB · sinα)/(AC · sinβ)
Length of the bisector from A: AD = 2ab·cos(θ/2) / (a + b)
Incentre ratio: AI : ID = (b + c) : a
Between an altitude and an angle bisector
∠DAE = ½ | ∠B − ∠C |
Perpendicular bisector
Every point on it is equidistant from the two ends of the segment — which is exactly why the three of them meet at the circumcentre and a circle can be drawn through all three vertices.
Orthocentre product
AH × HD = BH × HE = CH × HF
Radii & area
The three relations
Inradius Area = r × s where s = (a+b+c)/2
Circumradius Area = abc / 4R → R = abc / (4 × Area)
Exradius Area = r_A × (s − a)
Heron: Area = √( s(s−a)(s−b)(s−c) )
The inradius relation Area = r × s holds for any polygon with an inscribed circle, not just triangles.
Right-angled triangle
r = (P + B − H)/2 R = H/2
Median to the hypotenuse = circumradius = H/2
Altitude on the hypotenuse = (P × B)/H
Equilateral triangle
Area = (√3/4) a² Height = (√3/2) a
r = a/(2√3) R = a/√3
r : R : r_ex = 1 : 2 : 3
All four centres coincide at one point.
Viviani's result
From ANY point inside an equilateral triangle,
the sum of the perpendiculars to the three sides = the height
p₁ + p₂ + p₃ = (√3/2) a → a = (2/√3)(p₁ + p₂ + p₃)
Isosceles triangle
It is symmetric about the line from the unequal vertex, so all four centres lie on that one line. Cevians drawn from the two equal angles are equal to each other.
Extra radius relations
1/r = 1/r₁ + 1/r₂ + 1/r₃
1/r = 1/h₁ + 1/h₂ + 1/h₃ (h = altitudes)
Area = √(r · r₁ · r₂ · r₃)
Congruence & similarity
Four rules for congruence, three for similarity
Congruent: SSS · SAS · ASA (or AAS) · RHS
Similar: AA · SSS (in ratio) · SAS (in ratio)
Every congruent pair is similar; the reverse is not true.
Ratios in similar triangles
Sides : Perimeters : Medians : Altitudes : Inradii : Circumradii
→ ALL in the same ratio k
Areas → in the ratio k²
Thales & midpoint
DE ∥ BC → AD/DB = AE/EC
If D and E are midpoints of AB and AC:
DE ∥ BC and BC = 2·DE
Altitude on the hypotenuse — the similarity cascade
ΔABD ~ ΔCAD ~ ΔCBA, giving
AD² = BD × DC AB² = BD × BC AC² = CD × CB
AD × BC = AB × AC AB²/AC² = BD/CD
1/AD² = 1/AB² + 1/AC²
Points on the legs of a right triangle
P on AB, Q on BC, right angle at B:
AC² + PQ² = AQ² + PC²
Circle theorems
Chords
The perpendicular from the centre to a chord BISECTS it
(and a radius that bisects a chord is perpendicular to it)
Equal chords → subtend equal angles at the centre
Equal chords → are equidistant from the centre
Degree measure theorem
Angle at the centre = 2 × angle at any point on the MAJOR arc
Angle on the MINOR arc = 180° − (angle on the major arc)
Angle in a semicircle = 90°
Intersecting chords & secants
Chords crossing INSIDE : AO × OC = BO × OD
Chords crossing OUTSIDE: AE × BE = CE × DE
Tangent and secant : (tangent)² = (whole secant) × (external part)
Angle between two chords
Meeting INSIDE → angle = ½ (sum of the two intercepted arcs)
Meeting OUTSIDE → angle = ½ (difference of the arcs)
Two perpendicular chords
If chords AB and CD cross at right angles at E:
AE² + BE² + CE² + DE² = 4R²
Cyclic quadrilateral
- Opposite angles sum to 180°.
- An exterior angle equals the interior opposite angle.
- Angle between a chord and a tangent = angle in the alternate segment.
Ptolemy & Brahmagupta
Ptolemy (cyclic quad): AC × BD = (AB × CD) + (BC × AD)
Brahmagupta (cyclic quad area):
Area = √( (s−a)(s−b)(s−c)(s−d) )
Put d = 0 and it becomes Heron's formula.
Tangents
Basic facts
A tangent is PERPENDICULAR to the radius at the point of contact.
Two tangents from the same external point are EQUAL,
and the line to the centre bisects both the angle and the arc.
Tangent lengths between two circles
Direct common tangent = √( d² − (r₁ − r₂)² )
Transverse common tangent = √( d² − (r₁ + r₂)² )
If two circles touch a line and each other: AB = 2√(r₁ r₂)
How many common tangents?
| Position | Condition | Tangents |
|---|---|---|
| Completely separate | d > R + r | 4 |
| Touching externally | d = R + r | 3 |
| Intersecting | |R−r| < d < R+r | 2 |
| Touching internally | d = |R − r| | 1 |
| One inside the other | d < |R − r| | 0 |
Circle wedged between two circles on a line
1/√r = 1/√r₁ + 1/√r₂
Counting from n points on a circle
Chords = ⁿC₂ Triangles = ⁿC₃
Quadrilaterals = ⁿC₄ Interior intersections of diagonals = ⁿC₄
Handshakes among n people = ⁿC₂ = n(n−1)/2
Quadrilaterals
Property grid
| Diagonals equal | Bisect each other | Meet at 90° | Bisect vertex angles | |
|---|---|---|---|---|
| Parallelogram | No | Yes | No | No |
| Rectangle | Yes | Yes | No | No |
| Rhombus | No | Yes | Yes | Yes |
| Square | Yes | Yes | Yes | Yes |
Join the midpoints — what you get
Parallelogram → parallelogram
Rectangle → rhombus
Rhombus → rectangle
Square → square
Quadrilateral → parallelogram (perimeter = sum of the diagonals)
And the bisectors of the angles of a parallelogram form a rectangle; of a rectangle, a square.
Parallelogram results
Area = base × height = ab·sinθ Perimeter = 2(a + b)
d₁² + d₂² = 2(a² + b²)
A parallelogram inscribed in a circle is a RECTANGLE.
A parallelogram circumscribing a circle is a RHOMBUS.
If its diagonals are equal it is a RECTANGLE.
Rectangle, rhombus, square
Rectangle: diagonal = √(l² + b²)
Rhombus: Area = ½ d₁d₂ ; d₁² + d₂² = 4a²
Square: Area = a² ; diagonal = a√2 ; r = a/2 ; R = a/√2
Kite: Area = ½ d₁d₂
British flag theorem
For ANY point O and rectangle ABCD:
OA² + OC² = OB² + OD²
It holds whether O is inside or outside.
General quadrilateral
Area = ½ × (one diagonal) × (sum of perpendiculars from the other two vertices)
Diagonals perpendicular → AB² + CD² = AD² + BC²
Cyclic → opposite angles sum to 180°
Tangential → AB + CD = BC + DA
A point inside a parallelogram
Area(ΔAPB) + Area(ΔDPC) = Area(ΔAPD) + Area(ΔBPC) = ½ × Area(ABCD)
Trapezium
Area = ½ (sum of parallel sides) × height
Median joining the midpoints of the non-parallel sides
EF = ½ (AB + DC)
Diagonals: AO/OC = BO/OD = AB/DC
Triangles on the two parallel sides have areas in the ratio a² : b²
Polygons
Angles and diagonals
Sum of interior angles = (n − 2) × 180°
Each interior angle (regular) = (n − 2) × 180° / n
Sum of exterior angles = 360° ALWAYS
Each exterior angle (regular) = 360° / n
Number of diagonals = n(n − 3) / 2
Regular polygon measurements
Area = (n a² / 4) · cot(180°/n)
Inradius = (a/2) · cot(180°/n)
Circumradius = (a/2) · cosec(180°/n)
Hexagon & octagon
Hexagon: Area = (3√3/2) a² r = (√3/2) a R = a
short diagonal = √3 a long diagonal = 2a
Octagon: Area = 2a²(√2 + 1)
Mass point geometry
Balancing cevian ratios
Hang a weight at each vertex so that on every side, weight × distance balances. The ratio of a point of balance is inversely proportional to the weights, and at any balance point the weight equals the total hanging from it. This turns messy cevian-ratio questions into arithmetic.
Side from two medians
BC = (2/3) √( 2(BE² + CF²) − AB² )
09
Mensuration — 2D & 3D
Paths and borders
Rectangle, path of width x
INSIDE Area = 2x(L + B − 2x) Perimeter = 4(L + B − 2x)
OUTSIDE Area = 2x(L + B + 2x) Perimeter = 4(L + B + 2x)
MIDWAY (crossing paths) Area = x(L + B − x)
Square of side a, path of width d
INSIDE Area = 4d(a − d)
OUTSIDE Area = 4d(a + d)
MIDWAY Area = d(2a − d)
Room
Area of four walls = 2H(L + B)
Area of four walls + roof = 2H(L + B) + LB
2D figures
Triangle areas
Any triangle = ½ × base × height = √(s(s−a)(s−b)(s−c))
Equilateral = (√3/4) a² = h²/√3 (h = altitude)
Isosceles = (b/4) √(4a² − b²) (b = base, a = equal side)
Largest square inside a right triangle
Square sitting on the two legs: side = (P × B)/(P + B)
Square sitting on the hypotenuse: side = (P·B·H)/(P² + B² + P·B)
Squares and circles
Largest square in a semicircle of radius r: side = 2r/√5
Circle inside a square of side a: r = a/2
Square inside a circle of radius R: side = R√2
Three touching circles
Area of the gap between them = r² ( √3 − π/2 )
Rubber band tied around n touching circles = 2πr + n·d
Equilateral triangle enclosing 3 circles of radius r: side = 4r + 2√3 r
Chain of circles inside a big circle
If n circles fit along the radius of the big circle R:
r = [ n / (2n + 1) ] × R
Prisms
Right prism
Volume = Base area × height
LSA = Base perimeter × height
TSA = LSA + 2 × Base area
In a prism the height equals the lateral edge. A right prism has a top identical to the base.
Cuboid & cube
CUBOID V = lbh LSA = 2h(l + b) TSA = 2(lb + bh + hl)
longest diagonal = √(l² + b² + h²)
CUBE V = a³ LSA = 4a² TSA = 6a²
diagonal = a√3
inscribed sphere r = a/2
circumscribed sphere R = (√3/2) a
area of the diagonal plane = a²√2
Pyramids & cones
Right pyramid
Volume = ⅓ × Base area × height
LSA = ½ × Base perimeter × slant height
TSA = LSA + Base area
slant height = √( h² + (a/2)² )
Cone
V = ⅓ πr²h CSA = πrl TSA = πr(l + r) l² = h² + r²
Regular tetrahedron of edge a
Volume = a³ / (6√2) = (√2/12) a³
Height = √(2/3) · a
LSA = 3 × (√3/4) a² TSA = √3 a²
Slant height = (√3/2) a
Circumscribing sphere: r = (√6/4) a
Frustum
V = ⅓ πH (R² + r² + Rr)
CSA = πl (R + r) l² = (R − r)² + H²
TSA = πl(R + r) + πR² + πr²
Prismatoid (any tapering solid)
Volume = ⅓ h ( A₁ + A₂ + √(A₁A₂) )
LSA = ½ (perimeter of top + perimeter of bottom) × slant height
Sphere
Sphere & hemisphere
SPHERE V = (4/3)πr³ Surface = 4πr²
HEMISPHERE V = (2/3)πr³ CSA = 2πr² TSA = 3πr²
Fitting things together
Largest cube inside a hemisphere of radius r: a = √(2/3) · r
Sphere inside a cube (touching all 6 faces): 47.6% of the cube is empty
Cube inside a sphere (vertices touching): 63.2% of the sphere is empty
Cube inside a sphere → diameter of sphere = diagonal of cube
Every cut of a sphere
Total surface area increases by 2πr² per cut
Cutting cubes
Painted-cube formulas
Cube cut into n equal parts along each edge → n³ small cubes
0 faces painted = (n − 2)³
1 face painted = 6(n − 2)²
2 faces painted = 12(n − 2)
3 faces painted = 8 ← always the corners
At least 1 face painted = n³ − (n − 2)³
Always work in n (the number of parts per edge), never in the edge length.
Surface area from cutting
Each straight cut of a cube adds 2a² of new area
n cuts → TSA = 6a² + 2n·a²
Cube cut into n³ smaller cubes → total surface area rises by (n − 1) × 100%
Cuboid a×b×c cut into cubes of side n:
number of cuts = (a/n − 1) + (b/n − 1) + (c/n − 1)
Trap
Removing a corner cube from the top layer of a large cube leaves the total surface area unchanged — you expose exactly as much new face as you take away.
Scaling
Each dimension of a cuboid doubled → TSA increases 300%
(1× extra → +100%, 2× extra → +200%, 3× extra → +300%)
Units
King Henry Died by Drinking Choco Milk
Kilo Hecto Deca [Base] Deci Centi Milli
1000 100 10 1 0.1 0.01 0.001
1 m³ = 1000 litres = 1 kL 1 litre = 1000 cm³
10
Trigonometry & Height–Distance
Conversion identities
The sec–tan and cosec–cot family
cosθ/(1 + sinθ) = (1 − sinθ)/cosθ = secθ − tanθ
cosθ/(1 − sinθ) = (1 + sinθ)/cosθ = secθ + tanθ
sinθ/(1 + cosθ) = (1 − cosθ)/sinθ = cosecθ − cotθ
sinθ/(1 − cosθ) = (1 + cosθ)/sinθ = cosecθ + cotθ
Squared versions: (1 + sinθ)/(1 − sinθ) = (secθ + tanθ)², and similarly for the cos/cot pair.
If secθ + tanθ = A
secθ − tanθ = 1/A
secθ = ½(A + 1/A) tanθ = ½(A − 1/A)
sinθ = (A² − 1)/(A² + 1)
Powers of sin and cos
sin⁴θ + cos⁴θ = 1 − 2 sin²θ cos²θ
sin⁶θ + cos⁶θ = 1 − 3 sin²θ cos²θ
tan²θ + cot²θ + 2 = 1/(sin²θ cos²θ)
a sinθ + b cosθ
If a sinθ + b cosθ = c then a cosθ − b sinθ = ±√(a² + b² − c²)
If a sinθ + b cosθ = m and a cosθ − b sinθ = n then a² + b² = m² + n²
Maximum = √(a² + b²) Minimum = −√(a² + b²)
From a tangent ratio
tanθ = a/b → sinθ = a/√(a²+b²), cosθ = b/√(a²+b²)
Compound & multiple angles
Addition formulas
sin(A ± B) = sinA cosB ± cosA sinB
cos(A ± B) = cosA cosB ∓ sinA sinB
tan(A ± B) = (tanA ± tanB) / (1 ∓ tanA tanB)
Product to sum
2 sinA cosB = sin(A+B) + sin(A−B)
2 cosA sinB = sin(A+B) − sin(A−B)
2 cosA cosB = cos(A+B) + cos(A−B)
2 sinA sinB = cos(A−B) − cos(A+B)
Sum to product
sinC + sinD = 2 sin((C+D)/2) cos((C−D)/2)
sinC − sinD = 2 cos((C+D)/2) sin((C−D)/2)
cosC + cosD = 2 cos((C+D)/2) cos((C−D)/2)
cosC − cosD = −2 sin((C+D)/2) sin((C−D)/2)
Double angle
sin2A = 2 sinA cosA = 2tanA/(1 + tan²A)
cos2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A = (1 − tan²A)/(1 + tan²A)
tan2A = 2tanA / (1 − tan²A)
1 + cos2A = 2cos²A 1 − cos2A = 2sin²A
1 + sin2A = (sinA + cosA)² 1 − sin2A = (cosA − sinA)²
Triple angle
sin3A = 3sinA − 4sin³A
cos3A = 4cos³A − 3cosA
tan3A = (3tanA − tan³A)/(1 − 3tan²A)
sin³A = ¼(3sinA − sin3A) cos³A = ¼(3cosA + cos3A)
The 60° product identities
sinθ · sin(60° − θ) · sin(60° + θ) = ¼ sin3θ
cosθ · cos(60° − θ) · cos(60° + θ) = ¼ cos3θ
tanθ · tan(60° − θ) · tan(60° + θ) = tan3θ
Difference of squares
sin²A − sin²B = sin(A + B) · sin(A − B)
cos²A − cos²B = sin(A + B) · sin(B − A)
Angle-sum conditions
Recognise the condition from the form
A + B = 90° → tanA · tanB = 1
A + B = 45° → (1 + tanA)(1 + tanB) = 2
(1 − cotA)(1 − cotB) = 2
A + B + C = 90° → tanA·tanB + tanB·tanC + tanC·tanA = 1
A + B + C = 180° → tanA + tanB + tanC = tanA·tanB·tanC
Complementary angles
If A + B = 90° :
sinA/cosB = cosA/sinB = tanA/cotB = cotA/tanB = 1
sin²x + sin²(90° − x) = 1
tan1° · tan2° · … · tan89° = 1
Tangent differences
tanA + tanB = sin(A + B) / (cosA · cosB)
tanA − tanB = sin(A − B) / (cosA · cosB)
Exact values
The non-standard angles
sin18° = (√5 − 1)/4 cos36° = (√5 + 1)/4
cos18° = √(10 + 2√5) / 4 sin36° = √(10 − 2√5) / 4
sin15° = (√3 − 1)/(2√2) cos15° = (√3 + 1)/(2√2)
sin22½° = √(2 − √2) / 2 cos22½° = √(2 + √2) / 2
The sine always carries the minus. That one hook tells you which of a pair you are looking at.
Nested cosine
√(2 + √(2 + √(2 + 2cosθ))) = 2 cos(θ / 2ⁿ)
where n = the number of nested radical signs.
Replace the outermost + with − and it becomes 2 sin(θ / 2ⁿ).
Maximum & minimum
The base ranges
| Function | Min | Max |
|---|---|---|
| sinx, cosx | −1 | 1 |
| sin²x, cos²x | 0 | 1 |
| tanx, cotx | −∞ | ∞ |
| tan²x, cot²x | 0 | ∞ |
Combination forms
a sinx + b cosx → max √(a²+b²), min −√(a²+b²)
a sin²x + b cos²x → the larger of a,b is max; the smaller is min
a tan²x + b cot²x → minimum = 2√(ab)
a sec²x + b cosec²x → minimum = (√a + √b)²
tanθ + cotθ = 2/sin2θ → minimum 2
3sinx + 4cosx has max 5, min −5. 4sinx + 5cosx has max √41.
General solutions
Three to remember
sinx = siny → x = nπ + (−1)ⁿ y
cosx = cosy → x = 2nπ ± y
tanx = tany → x = nπ + y
Inverse functions
Complementary pairs
sin⁻¹x + cos⁻¹x = π/2
tan⁻¹x + cot⁻¹x = π/2
sec⁻¹x + cosec⁻¹x = π/2
So if sin⁻¹x + sin⁻¹y = 2π/3, then cos⁻¹x + cos⁻¹y = π − 2π/3 = π/3.
Negative arguments
sin⁻¹(−x) = −sin⁻¹x cos⁻¹(−x) = π − cos⁻¹x
tan⁻¹(−x) = −tan⁻¹x cot⁻¹(−x) = π − cot⁻¹x
cosec⁻¹(−x) = −cosec⁻¹x sec⁻¹(−x) = π − sec⁻¹x
sin, tan and cosec pull the minus out front; cos, cot and sec subtract from π.
Addition of inverses
sin⁻¹x ± sin⁻¹y = sin⁻¹[ x√(1−y²) ± y√(1−x²) ]
cos⁻¹x ± cos⁻¹y = cos⁻¹[ xy ∓ √((1−x²)(1−y²)) ]
tan⁻¹x + tan⁻¹y = tan⁻¹[ (x + y)/(1 − xy) ]
tan⁻¹x − tan⁻¹y = tan⁻¹[ (x − y)/(1 + xy) ]
Height & distance
The three standard triangles
30–60–90 → 1 : √3 : 2
45–45–90 → 1 : 1 : √2
15–75–90 → (√3 − 1) : (√3 + 1) : 2√2
Complementary angles — the √(ab) shortcut
Two poles of heights a and b. From the foot of each, the angle of
elevation to the top of the other is COMPLEMENTARY.
→ the common height at the crossing point = √(ab)
Same idea: observer at two points with complementary elevation angles
x and y from the tower → height = √(xy)
Two elevation angles from a moving observer
Observer walks distance d towards the tower,
elevation changes from α to β:
h = d · (tanα · tanβ) / (tanβ − tanα)
Balloon subtending an angle
Balloon of radius r subtends angle α at the eye,
its centre subtends elevation β:
height of centre = r · sinβ · cosec(α/2)
Working method
Mark the common side shared by the two right triangles first, express both bases in terms of it, then equate. Almost every height–distance question is one common side and two ratios.
11
Arithmetic Toolkit
Percentage ↔ fraction
The conversion table — this is pure speed
| 1/2 | 1/3 | 1/4 | 1/5 | 1/6 | 1/7 | 1/8 | 1/9 |
|---|---|---|---|---|---|---|---|
| 50% | 33.33% | 25% | 20% | 16.66% | 14.28% | 12.5% | 11.11% |
| 1/10 | 1/11 | 1/12 | 1/13 | 1/14 | 1/15 | 1/16 | 1/24 |
| 10% | 9.09% | 8.33% | 7.69% | 7.14% | 6.66% | 6.25% | 4.16% |
Multiples: 2/3 = 66.66% 3/4 = 75% 2/5 = 40% 4/5 = 80%
5/6 = 83.33% 2/7 = 28.57% 3/7 = 42.85% 4/7 = 57.14%
3/8 = 37.5% 5/8 = 62.5% 7/8 = 87.5%
2/9 = 22.22% 4/9 = 44.44%
Profit & loss
Same selling price, same percentage
Two articles sold at the SAME price, one at a% profit and one at a% loss
→ always a NET LOSS of a²/100 %
The dishonest-dealer four
Gets extra weight while BUYING (say 20 free) → 100 becomes 120
Gives less weight while SELLING (say 20% less) → 80 becomes 100
Uses a BIG scale while buying (105 for 100) → 100 becomes 105
Uses a SMALL scale while selling ( 95 for 100) → 95 becomes 100
Write both the buying and the selling distortion, then compare what he pays for against what he hands over.
Marked price and discount
MP relative to CP = 100 + markup%
SP after discount = MP × (100 − D)/100
Compound interest
The difference formulas
Difference between CI and SI over 2 years = P (R/100)²
Difference between CI and SI over 3 years = P R² (300 + R) / 100³
Difference between the CI of two consecutive years = P(R/100)² × (earlier amount factor)
→ the later year's interest exceeds the earlier by R% of it
Statistics
Central tendency
Mean = (sum of all values) / (number of values)
Median = middle value after sorting
n odd → the ((n+1)/2)th term
n even → average of the (n/2)th and (n/2 + 1)th terms
Mode = the most frequent value
Range = largest − smallest
Empirical relation
3 Median = 2 Mean + Mode
Spread
Mean deviation about mean = mean of |value − mean|
Mean deviation about median = mean of |value − median|
Variance = mean of (value − mean)²
Standard deviation = √Variance
Coefficient of variation = (SD / Mean) × 100
Mode of grouped data
Mode = l + h · (f₁ − f₀) / (2f₁ − f₀ − f₂)
Probability
Core rules
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
P(A ∪ B ∪ C) = ΣP(A) − ΣP(A∩B) + P(A∩B∩C)
P(not A) = 1 − P(A)
De Morgan: (A∪B)′ = A′ ∩ B′ (A∩B)′ = A′ ∪ B′
Exclusive, exhaustive, independent
Mutually exclusive → nothing in common → P(A ∩ B) = 0
Mutually exhaustive → the events cover every possibility
Exclusive AND exhaustive → P(A) + P(B) = 1
Independent → P(A ∩ B) = P(A) × P(B)
Conditional: P(A | B) = P(A ∩ B) / P(B)
Odds
Odds IN FAVOUR = (outcomes in favour) : (outcomes against)
Odds AGAINST = (outcomes against) : (outcomes in favour)
A pack of cards
52 cards = 4 suits × 13
Red: Hearts, Diamonds Black: Spades, Clubs
Face cards = J, Q, K → 3 per suit → 12 total
Honour cards = A, K, Q, J → 4 per suit → 16 total
Distributing identical objects
n identical objects among p persons = ⁿ⁺ᵖ⁻¹C₍ₚ₋₁₎
If each person must receive a minimum, hand out the minimum first and distribute only what is left.
Triangles with a given integer perimeter p
p even → nearest integer to p² / 48
p odd → nearest integer to (p + 3)² / 48
Sequences — AP
Arithmetic progression
d = a₂ − a₁ aₙ = a + (n − 1)d
n = (aₙ − a)/d + 1 d = (aₙ − a)/(n − 1)
Sₙ = (n/2) [ 2a + (n − 1)d ]
Arithmetic mean = (a + b)/2
Tₙ = Sₙ − Sₙ₋₁
The classic AP results
aₚ = q and a_q = p → a_(p+q) = 0 and aₙ = p + q − n
Sₚ = q and S_q = p → S_(p+q) = −(p + q)
Sₚ = S_q → S_(p+q) = 0
mth term = 1/n and nth term = 1/m → (mn)th term = 1
Sum of p terms = q, sum of q terms = p → d = −2(p + q)/(pq)
Sums in ratio m² : n² → mth term : nth term = (2m − 1) : (2n − 1)
Inserting means
n arithmetic means between a and b → their sum = n(a + b)/2
If A is the single AM and S the sum of n AMs between the same
two numbers, then S = nA
Mean of the first N odd natural numbers = N
Mean of the first N even natural numbers = N + 1
The (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) question
Put n = 0 → the expression becomes the ARITHMETIC mean
Put n = −½ → the expression becomes the GEOMETRIC mean
Sequences — GP & HP
Geometric progression
r = a₂/a₁ aₙ = a·r^(n−1)
Middle-term property: a₂² = a₁ × a₃
Sₙ = a(rⁿ − 1)/(r − 1) for r > 1
Sₙ = a(1 − rⁿ)/(1 − r) for r < 1
S∞ = a / (1 − r) for |r| < 1
Geometric mean = √(ab)
n GMs between a and b → common ratio = (b/a)^(1/(n+1))
Harmonic progression
a₁, a₂, a₃ … are in HP ⇔ 1/a₁, 1/a₂, 1/a₃ … are in AP
Harmonic mean = 2ab / (a + b)
AM ≥ GM ≥ HM and GM² = AM × HM
Given AM = 34 and HM = 16: a + b = 68 and ab = 256, so the numbers are 64 and 4.
Number families worth recognising
Fibonacci: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,
144, 233, 377, 610, 987, 1597
Triangular: n(n+1)/2 → 1, 3, 6, 10, 15, 21, …
Palindrome: reads the same both ways → 121, 191, 1331
Small things that decide marks
Reciprocals of a proper fraction
If 0 < x < 1 then 1/x > 1 and (x − 1)/x < 1
Ratio rearrangement
a : b = c : d → a : c = b : d → (a + b) : (c + d)
Sum-and-product puzzles
Given the sum and the product of two numbers, list the factor pairs of the product and pick the pair with the right sum — faster than solving the quadratic. Same idea for "n² + 48 is a perfect square": write it as (m − n)(m + n) = 48 and count the valid factor pairs.