πŸ“š Chapters

Aptitude & Reasoning

Unit 1 - Number System & Arithmetic

πŸ“– Full Notes
⚑ Quick Tricks
πŸ”’Chapter 1.1 β€” Divisibility Rules
πŸ“Š DIVISIBILITY RULES
2: Last digit even | 3: Sum Γ· 3 | 4: Last 2 digits Γ· 4
7: Special method | 8: Last 3 digits Γ· 8 | 9: Sum Γ· 9
11: (Odd - Even) Γ· 11 | 1001 = 7 Γ— 11 Γ— 13
All Divisibility Rules
2: Last digit even (0,2,4,6,8) 3: Sum of digits divisible by 3 4: Last 2 digits divisible by 4 5: Last digit 0 or 5 6: Divisible by both 2 AND 3 7: Remove last digit, double it, subtract from remaining 8: Last 3 digits divisible by 8 9: Sum of digits divisible by 9 10: Last digit is 0 11: (Sum of odd positions - Sum of even positions) Γ· 11
Examples:
Rule 3: 132 β†’ 1+3+2=6 β†’ 6Γ·3 βœ“
Rule 7: 133 β†’ 13-(3Γ—2)=7 β†’ 7Γ·7 βœ“
Rule 11: 4587 β†’ (7+5)-(8+4)=0 βœ“
Composite Numbers:
6 = 2Γ—3: Check both 2 AND 3
12 = 3Γ—4: Check both 3 AND 4
15 = 3Γ—5: Check both 3 AND 5

πŸ’‘ Key Fact: 1001 = 7 Γ— 11 Γ— 13
βž— Chapter 1.2 β€” Remainders
1. Concept of Positive and Negative Remainders
Remainder:
Dividend = Divisor Γ— Quotient + Remainder
Positive Remainder: 23 Γ· 5 = 4 R 3
Negative Remainder: 23 = 5 Γ— 5 - 2 (R = -2)
πŸ’‘ TRICK: Negative remainder = Divisor - Positive remainder
Use when remainder > divisor/2 for easier calculations
2. Fermat's Little Theorem
FERMAT'S THEOREM:
If p is PRIME and a is not divisible by p:
a^(p-1) ≑ 1 (mod p)
Example 1: 2^10 Γ· 11
p=11 (prime), a=2
2^(11-1) ≑ 1 (mod 11)
Remainder = 1 βœ“

Example 2: 5^40 Γ· 41
5^40 ≑ 1 (mod 41)
Remainder = 1 βœ“

Example 3: 3^40 Γ· 7
3^6 ≑ 1 (mod 7)
3^40 = (3^6)^6 Γ— 3^4
= 1 Γ— 81 mod 7 = 4
Remainder = 4 βœ“
3. Euler's Theorem
EULER'S TOTIENT Ο†(n):
Count of numbers coprime with n

β€’ Prime p: Ο†(p) = p-1
β€’ Two primes: Ο†(pΓ—q) = (p-1)(q-1)

EULER'S THEOREM:
If gcd(a,n)=1: a^Ο†(n) ≑ 1 (mod n)
Example 1: Find Ο†(10)
10 = 2Γ—5
Ο†(10) = (2-1)(5-1) = 4

Example 2: 7^40 Γ· 10
Ο†(10) = 4
7^4 ≑ 1 (mod 10)
7^40 = (7^4)^10 ≑ 1
Remainder = 1 βœ“
Common Ο†(n): Ο†(10) = 4, Ο†(12) = 4, Ο†(100) = 40
4. Wilson's Theorem
WILSON'S THEOREM:
If p is PRIME:
(p-1)! ≑ -1 (mod p)
Or: (p-1)! Γ· p gives remainder (p-1)
Example 1: 6! Γ· 7
p=7 (prime)
6! ≑ -1 ≑ 6 (mod 7)
Remainder = 6 βœ“

Example 2: 10! Γ· 11
Remainder = 10 βœ“

Example 3: 12! Γ· 13
Remainder = 12 βœ“
5. Miscellaneous Problems & Data Sufficiency
Problem 1: Last digit of 3^1000?
Last digit = 3^1000 mod 10
Ο†(10) = 4, so 3^4 ≑ 1 (mod 10)
3^1000 = (3^4)^250 ≑ 1
Answer: 1 βœ“

Problem 2: 2^100 Γ· 7
2^6 ≑ 1 (mod 7)
100 = 6Γ—16 + 4
2^100 = (2^6)^16 Γ— 2^4 ≑ 16 ≑ 2 (mod 7)
Answer: 2 βœ“
Data Sufficiency Tip:
If n = kd + r (where d is divisor), then remainder is r
Example: n = 15k + 3, then n mod 5 = 3
6. Quick Reference
THEOREM SUMMARY: Fermat's: a^(p-1) ≑ 1 (mod p) [p prime] Euler's: a^Ο†(n) ≑ 1 (mod n) [gcd(a,n)=1] Wilson's: (p-1)! ≑ -1 (mod p) [p prime] WHEN TO USE: β€’ Large power + prime divisor β†’ Fermat's β€’ Large power + composite β†’ Euler's β€’ Factorial + prime β†’ Wilson's
πŸ”’ Chapter 1.3 β€” Factors & Factorials
1. Concept of Factors
Factor: A number that divides another number completely (remainder = 0)

If a divides b completely, then a is a factor of b
Example: Factors of 12
12 = 1Γ—12, 2Γ—6, 3Γ—4
Factors: 1, 2, 3, 4, 6, 12 (6 factors)
Prime Factorization:
Every number can be expressed as product of prime numbers

If N = p₁^a Γ— pβ‚‚^b Γ— p₃^c Γ— ...
(where p₁, pβ‚‚, p₃ are prime numbers)
Example:
60 = 2Β² Γ— 3ΒΉ Γ— 5ΒΉ
360 = 2Β³ Γ— 3Β² Γ— 5ΒΉ
1000 = 2Β³ Γ— 5Β³
2. Number of Factors
FORMULA:

If N = p₁^a Γ— pβ‚‚^b Γ— p₃^c

Number of factors = (a+1) Γ— (b+1) Γ— (c+1)
Example 1: Find number of factors of 60

60 = 2Β² Γ— 3ΒΉ Γ— 5ΒΉ
Number of factors = (2+1) Γ— (1+1) Γ— (1+1)
= 3 Γ— 2 Γ— 2 = 12 factors βœ“

Factors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Example 2: Find number of factors of 360

360 = 2Β³ Γ— 3Β² Γ— 5ΒΉ
Number of factors = (3+1) Γ— (2+1) Γ— (1+1)
= 4 Γ— 3 Γ— 2 = 24 factors βœ“
Special Cases:
πŸ’‘ Number of Even Factors:
If N = 2^a Γ— (odd part), then
Even factors = a Γ— (factors of odd part)

πŸ’‘ Number of Odd Factors:
Find factors of odd part only

Example: 60 = 2Β² Γ— 15
Odd factors = factors of 15 = (1+1)Γ—(1+1) = 4
(Factors: 1, 3, 5, 15)
3. Sum of Factors
FORMULA:

If N = p₁^a Γ— pβ‚‚^b Γ— p₃^c

Sum of factors = [(p₁^(a+1) - 1)/(p₁-1)] Γ— [(pβ‚‚^(b+1) - 1)/(pβ‚‚-1)] Γ— [(p₃^(c+1) - 1)/(p₃-1)]
Example: Find sum of factors of 12

12 = 2Β² Γ— 3ΒΉ

Sum = [(2Β³ - 1)/(2-1)] Γ— [(3Β² - 1)/(3-1)]
= [(8-1)/1] Γ— [(9-1)/2]
= 7 Γ— 4 = 28 βœ“

Verification: 1+2+3+4+6+12 = 28 βœ“
πŸ’‘ Sum of Proper Divisors:
(All factors except number itself)
Sum of proper divisors = Sum of all factors - N
4. Product of Factors
FORMULA:

If N has k factors, then

Product of all factors = N^(k/2)

where k = number of factors
Example: Find product of factors of 12

12 = 2Β² Γ— 3ΒΉ
Number of factors = 3 Γ— 2 = 6

Product = 12^(6/2) = 12Β³ = 1728 βœ“

Verification: 1Γ—2Γ—3Γ—4Γ—6Γ—12 = 1728 βœ“
5. Concept of Factorials
Factorial (n!): Product of all positive integers from 1 to n

n! = n Γ— (n-1) Γ— (n-2) Γ— ... Γ— 3 Γ— 2 Γ— 1
Examples:
5! = 5 Γ— 4 Γ— 3 Γ— 2 Γ— 1 = 120
6! = 6 Γ— 5 Γ— 4 Γ— 3 Γ— 2 Γ— 1 = 720
7! = 7 Γ— 6! = 7 Γ— 720 = 5040

Special: 0! = 1, 1! = 1
6. Number of Zeros in Product/Factorial
Concept: Trailing zeros come from 10 = 2 Γ— 5
Since 2s are abundant, count number of 5s in prime factorization
FORMULA for n!:

Number of zeros = ⌊n/5βŒ‹ + ⌊n/25βŒ‹ + ⌊n/125βŒ‹ + ⌊n/625βŒ‹ + ...

(Keep dividing by powers of 5 until quotient becomes 0)
Example 1: Number of zeros in 25!

⌊25/5βŒ‹ = 5
⌊25/25βŒ‹ = 1
⌊25/125βŒ‹ = 0 (stop)

Total zeros = 5 + 1 = 6 βœ“
Example 2: Number of zeros in 100!

⌊100/5βŒ‹ = 20
⌊100/25βŒ‹ = 4
⌊100/125βŒ‹ = 0

Total zeros = 20 + 4 = 24 βœ“
πŸ’‘ QUICK TRICK:
For number of zeros in product of consecutive numbers (not factorial):
Count pairs of (2,5), (4,5), (6,10), etc.
7. Finding Maximum Power in Factorial
Question Type: Find highest power of a prime p that divides n!
FORMULA:

Highest power of prime p in n! =
⌊n/pβŒ‹ + ⌊n/pΒ²βŒ‹ + ⌊n/pΒ³βŒ‹ + ⌊n/pβ΄βŒ‹ + ...
Example 1: Highest power of 3 in 10!

⌊10/3βŒ‹ = 3
⌊10/9βŒ‹ = 1
⌊10/27βŒ‹ = 0

Highest power = 3 + 1 = 4
So 3⁴ divides 10! βœ“
Example 2: Highest power of 2 in 20!

⌊20/2βŒ‹ = 10
⌊20/4βŒ‹ = 5
⌊20/8βŒ‹ = 2
⌊20/16βŒ‹ = 1
⌊20/32βŒ‹ = 0

Highest power = 10 + 5 + 2 + 1 = 18
So 2¹⁸ divides 20! βœ“
For Composite Numbers:
Steps:
1. Find prime factorization of composite number
2. Find highest power of each prime in n!
3. Divide by powers in composite number
4. Take minimum
Example: Highest power of 12 in 15!

12 = 2Β² Γ— 3ΒΉ

Power of 2 in 15!:
⌊15/2βŒ‹ + ⌊15/4βŒ‹ + ⌊15/8βŒ‹ = 7 + 3 + 1 = 11
Available pairs of 2Β² = 11/2 = 5

Power of 3 in 15!:
⌊15/3βŒ‹ + ⌊15/9βŒ‹ = 5 + 1 = 6
Available 3ΒΉ = 6

Minimum(5, 6) = 5
Highest power of 12 = 5 βœ“
8. Practice Problems
Problem 1: How many factors does 144 have?

144 = 2⁴ Γ— 3Β²
Factors = (4+1) Γ— (2+1) = 5 Γ— 3 = 15 βœ“
Problem 2: Number of zeros in 50!

⌊50/5βŒ‹ + ⌊50/25βŒ‹ + ⌊50/125βŒ‹
= 10 + 2 + 0 = 12 zeros βœ“
Problem 3: Highest power of 7 in 50!

⌊50/7βŒ‹ + ⌊50/49βŒ‹
= 7 + 1 = 8
7⁸ divides 50! βœ“
9. Quick Formula Sheet
FACTORS FORMULAS: If N = p₁^a Γ— pβ‚‚^b Γ— p₃^c Number of factors = (a+1)(b+1)(c+1) Sum of factors = [(p₁^(a+1)-1)/(p₁-1)] Γ— [(pβ‚‚^(b+1)-1)/(pβ‚‚-1)] Γ— ... Product of factors = N^(k/2) where k = number of factors FACTORIAL FORMULAS: Zeros in n! = ⌊n/5βŒ‹ + ⌊n/25βŒ‹ + ⌊n/125βŒ‹ + ... Highest power of p in n! = ⌊n/pβŒ‹ + ⌊n/pΒ²βŒ‹ + ⌊n/pΒ³βŒ‹ + ...
πŸ”Ÿ Chapter 1.4 β€” Unit Digit and Last Two Digits
1. Concept of Unit Digit
Unit Digit: The rightmost digit (ones place) of a number

Also called: Last digit or Units place
Examples:
β€’ Unit digit of 2583 = 3
β€’ Unit digit of 45678 = 8
β€’ Unit digit of 1000 = 0
Finding unit digit = Finding remainder when divided by 10
Example: 2583 Γ· 10 β†’ Remainder = 3 (unit digit)
2. Cyclicity of Digits
Cyclicity: Pattern in which unit digits repeat for successive powers of a number
CYCLICITY TABLE: Digit | Powers | Cycle | Cyclicity ----- | ------------------- | ----- | --------- 0 | 0, 0, 0, 0... | 0 | 1 1 | 1, 1, 1, 1... | 1 | 1 2 | 2, 4, 8, 6, 2... | 2,4,8,6 | 4 3 | 3, 9, 7, 1, 3... | 3,9,7,1 | 4 4 | 4, 6, 4, 6... | 4,6 | 2 5 | 5, 5, 5, 5... | 5 | 1 6 | 6, 6, 6, 6... | 6 | 1 7 | 7, 9, 3, 1, 7... | 7,9,3,1 | 4 8 | 8, 4, 2, 6, 8... | 8,4,2,6 | 4 9 | 9, 1, 9, 1... | 9,1 | 2
πŸ’‘ REMEMBER:
β€’ 0, 1, 5, 6 β†’ Always same (cyclicity 1)
β€’ 4, 9 β†’ Alternates (cyclicity 2)
β€’ 2, 3, 7, 8 β†’ Cycle of 4 (cyclicity 4)
3. Finding Unit Digit of Powers
METHOD:
1. Find cyclicity of base's unit digit
2. Divide power by cyclicity
3. Use remainder to find position in cycle
4. If remainder = 0, use last element of cycle
Example 1: Unit digit of 2^37

Unit digit of base = 2
Cycle of 2: 2, 4, 8, 6 (cyclicity = 4)

37 Γ· 4 β†’ Remainder = 1
Position 1 in cycle = 2
Answer: 2 βœ“
Example 2: Unit digit of 3^125

Unit digit of base = 3
Cycle of 3: 3, 9, 7, 1 (cyclicity = 4)

125 Γ· 4 β†’ Remainder = 1
Position 1 in cycle = 3
Answer: 3 βœ“
Example 3: Unit digit of 7^100

Unit digit of base = 7
Cycle of 7: 7, 9, 3, 1 (cyclicity = 4)

100 Γ· 4 β†’ Remainder = 0
When remainder = 0, take last element = 1
Answer: 1 βœ“
Example 4: Unit digit of 234^567

Unit digit of base = 4
Cycle of 4: 4, 6 (cyclicity = 2)

567 Γ· 2 β†’ Remainder = 1
Position 1 in cycle = 4
Answer: 4 βœ“
4. Unit Digit of Products/Sums
RULE:
Unit digit of product = Unit digit of (product of unit digits)
Example 1: Unit digit of (23 Γ— 45 Γ— 67)

Unit digits: 3 Γ— 5 Γ— 7
= 15 Γ— 7
= 105
Unit digit = 5 βœ“
Example 2: Unit digit of (2^5 Γ— 3^7 Γ— 5^3)

Unit digit of 2^5: 5 Γ· 4 β†’ R=1 β†’ 2
Unit digit of 3^7: 7 Γ· 4 β†’ R=3 β†’ 7
Unit digit of 5^3: Always 5

Product: 2 Γ— 7 Γ— 5 = 70
Answer: 0 βœ“
5. Finding Last Two Digits
Last Two Digits: Rightmost two digits (tens and ones place)

Last two digits = Remainder when divided by 100
Method 1: Using Binomial Theorem
For numbers ending in 1, 3, 7, 9:
Express as (multiple of 10 Β± small number)^n
Example: Last two digits of 17^20

17 = (20 - 3)
17^20 = (20 - 3)^20

Using binomial, we need only last two terms:
... + 20C19 Γ— 20ΒΉ Γ— 3ΒΉ + 20C20 Γ— 3^0
= ... + 20 Γ— 20 Γ— 3 + 1
= ... + 1200 + 1
= ... + 1201
Last two digits = 01 βœ“
Method 2: Pattern Recognition
πŸ’‘ SHORTCUT for Powers of 2:

2^10 = 1024 β†’ Last two digits = 24
2^20 = (2^10)^2 β†’ 24^2 = 576 β†’ 76

Pattern repeats every 20 powers
Example 1: Last two digits of 2^50

50 = 2Γ—20 + 10
Last two digits of 2^10 = 24
Answer: 24 βœ“
Example 2: Last two digits of 3^100

Using Euler's theorem:
Ο†(100) = 40
3^40 ≑ 1 (mod 100)
3^100 = (3^40)^2 Γ— 3^20
= 1 Γ— 3^20

Calculate 3^20 mod 100:
3^10 = 59049 β†’ Last two = 49
3^20 = 49^2 = 2401
Answer: 01 βœ“
6. Unit Digit of Higher Powers
Tower/Tetration: Finding unit digit of expressions like 2^(3^4)
METHOD:
Work from top to bottom, simplifying the power using cyclicity
Example 1: Unit digit of 3^(4^5)

Step 1: Find 4^5 mod 4 (cyclicity of 3)
4 ≑ 0 (mod 4)
So 4^5 ≑ 0 (mod 4)

Step 2: When power ≑ 0 (mod 4), use 4th position
Cycle of 3: 3, 9, 7, 1
4th position = 1
Answer: 1 βœ“
Example 2: Unit digit of 7^(6^5)

Step 1: Find 6^5 mod 4 (cyclicity of 7)
6 ≑ 2 (mod 4)
6^5 = (2^5) mod 4 = 32 mod 4 = 0

Step 2: Use 4th position in cycle of 7
Cycle of 7: 7, 9, 3, 1
Answer: 1 βœ“
7. Practice Problems
Problem 1: Unit digit of 8^123

Cycle of 8: 8, 4, 2, 6 (cyclicity 4)
123 Γ· 4 β†’ R = 3
Position 3 = 2
Answer: 2 βœ“
Problem 2: Unit digit of (13^15 Γ— 17^19)

Unit digit of 13^15:
Cycle of 3: 3,9,7,1
15 Γ· 4 β†’ R = 3 β†’ 7

Unit digit of 17^19:
Cycle of 7: 7,9,3,1
19 Γ· 4 β†’ R = 3 β†’ 3

Product: 7 Γ— 3 = 21
Answer: 1 βœ“
Problem 3: Last two digits of 23^45

23 = (20 + 3)
Use pattern or Euler's theorem
Ο†(100) = 40
45 = 40 + 5
23^45 = 23^40 Γ— 23^5 ≑ 1 Γ— 23^5
Calculate 23^5 mod 100
Answer: 07 βœ“
8. Quick Reference
CYCLICITY PATTERNS: Always Same (Cyclicity 1): 0 β†’ 0 1 β†’ 1 5 β†’ 5 6 β†’ 6 Alternates (Cyclicity 2): 4 β†’ 4, 6 9 β†’ 9, 1 Cycle of 4: 2 β†’ 2, 4, 8, 6 3 β†’ 3, 9, 7, 1 7 β†’ 7, 9, 3, 1 8 β†’ 8, 4, 2, 6 QUICK TIPS: β€’ Any number^even ending in 5 β†’ Last digit = 25 β€’ Any odd number^4 β†’ Last digit = 1 β€’ Power of 2: Pattern repeats every 20 β€’ Power of 3: Pattern repeats every 20
βš–οΈ Chapter 1.5 β€” Inequality & Symboperation
1. Concept of Inequalities
Inequality: Mathematical statement showing relationship between two expressions that are not equal

Symbols:
> : Greater than
< : Less than
β‰₯ : Greater than or equal to
≀ : Less than or equal to
β‰  : Not equal to
Examples:
β€’ 5 > 3 (5 is greater than 3)
β€’ x < 10 (x is less than 10)
β€’ y β‰₯ 0 (y is greater than or equal to 0)
β€’ a + b ≀ 100 (sum is at most 100)
2. Properties of Inequalities
Property 1: Adding/Subtracting same number
If a > b, then a + c > b + c
If a > b, then a - c > b - c

Property 2: Multiplying/Dividing by positive number
If a > b and c > 0, then a Γ— c > b Γ— c
If a > b and c > 0, then a Γ· c > b Γ· c

Property 3: Multiplying/Dividing by negative number
If a > b and c < 0, then a Γ— c < b Γ— c (sign reverses)
If a > b and c < 0, then a Γ· c < b Γ· c (sign reverses)
Example:
If x > 5
Add 3: x + 3 > 5 + 3 β†’ x + 3 > 8 βœ“
Multiply by 2: 2x > 10 βœ“
Multiply by -1: -x < -5 (sign reverses) βœ“
πŸ’‘ REMEMBER:
When multiplying/dividing by NEGATIVE number β†’ FLIP the inequality sign!
3. Solving Linear Inequalities
METHOD:
1. Isolate variable on one side
2. Perform same operations on both sides
3. Remember to flip sign if multiplying/dividing by negative
Example 1: Solve 2x + 3 > 11

2x + 3 > 11
2x > 11 - 3
2x > 8
x > 4 βœ“
Example 2: Solve -3x ≀ 12

-3x ≀ 12
Divide by -3 (flip sign!)
x β‰₯ -4 βœ“
Example 3: Solve 5 - 2x < 15

5 - 2x < 15
-2x < 15 - 5
-2x < 10
Divide by -2 (flip sign!)
x > -5 βœ“
4. Coded Inequalities (Passage Based)
Coded Inequality: Inequalities represented by symbols (like @, #, $, %, &) instead of standard symbols
Example Question:

Given Code:
A @ B means A > B
A # B means A < B
A $ B means A β‰₯ B
A % B means A ≀ B
A & B means A = B

Statement: P @ Q, Q $ R, R # S
Question: Which is true?

Solution:
P @ Q β†’ P > Q
Q $ R β†’ Q β‰₯ R
R # S β†’ R < S

Combining: P > Q β‰₯ R < S
Conclusion: P > R βœ“, but cannot compare P and S
πŸ’‘ QUICK APPROACH:
1. Decode all symbols first
2. Write in sequence
3. Check each option systematically
4. Look for direct comparisons
5. Symboperation (Basic Operations)
Symboperation: Problems where mathematical operations (+, -, Γ—, Γ·) are replaced by symbols

Need to decode the pattern and solve
BODMAS Rule:
Order of Operations:

B - Brackets ( )
O - Orders (powers, roots)
D - Division Γ·
M - Multiplication Γ—
A - Addition +
S - Subtraction -
Example: 10 + 5 Γ— 3 - 4 Γ· 2

Step 1: Multiplication β†’ 5 Γ— 3 = 15
Step 2: Division β†’ 4 Γ· 2 = 2
Step 3: Addition β†’ 10 + 15 = 25
Step 4: Subtraction β†’ 25 - 2 = 23 βœ“
6. Symbol Substitution Problems
Type 1: Given Code

Code:
A @ B = A + B
A # B = A - B
A $ B = A Γ— B
A % B = A Γ· B

Question: Find value of (12 @ 8) $ (10 # 4)

Solution:
(12 @ 8) = 12 + 8 = 20
(10 # 4) = 10 - 4 = 6
20 $ 6 = 20 Γ— 6 = 120 βœ“
Type 2: Find Missing Operation

Question: If 5 ? 3 ? 2 = 13, find the operations

Try combinations:
5 + 3 + 2 = 10 βœ—
5 + 3 Γ— 2 = 5 + 6 = 11 βœ—
5 Γ— 3 - 2 = 15 - 2 = 13 βœ“

Answer: 5 Γ— 3 - 2 = 13
Type 3: Make Equation True

Question: Make true: 8 _ 4 _ 2 = 10

Try:
8 + 4 - 2 = 10 βœ“
8 - 4 + 6 βœ— (but 6 not available)
8 Γ— 4 Γ· 2 = 16 βœ—

Answer: 8 + 4 - 2 = 10
7. Mixed Practice Problems
Problem 1 (Inequality):
If 3x - 5 ≀ 10, find maximum integer value of x

3x - 5 ≀ 10
3x ≀ 15
x ≀ 5
Answer: 5 βœ“
Problem 2 (Coded Inequality):

Code: P > Q, Q = R, R < S
Question: Is P > S?

P > Q = R < S
P > R and R < S
Cannot determine relation between P and S
Answer: Cannot say
Problem 3 (Symboperation):

If @ means +, # means -, $ means Γ—
Find: (8 @ 4) $ (6 # 2)

(8 + 4) Γ— (6 - 2)
= 12 Γ— 4
= 48 βœ“
Problem 4 (BODMAS):
Solve: 20 - 5 + 3 Γ— 4 Γ· 2

Step 1: 3 Γ— 4 = 12
Step 2: 12 Γ· 2 = 6
Step 3: 20 - 5 = 15
Step 4: 15 + 6 = 21 βœ“
8. Common Mistakes to Avoid
⚠️ WATCH OUT:

Mistake 1: Forgetting to flip inequality sign
-2x > 6 β†’ x > -3 βœ— (Wrong!)
-2x > 6 β†’ x < -3 βœ“ (Correct - flip sign)

Mistake 2: Wrong BODMAS order
5 + 3 Γ— 2 = 16 βœ— (Wrong! Did addition first)
5 + 3 Γ— 2 = 5 + 6 = 11 βœ“ (Correct - multiply first)

Mistake 3: Mixing up symbols
Always decode first, then solve

Mistake 4: In coded inequalities
Not checking if comparison is possible
A > B, C < D doesn't mean A > C
9. Quick Reference Guide
INEQUALITY RULES: Add/Subtract: No change in sign Multiply/Divide by positive: No change Multiply/Divide by negative: FLIP SIGN BODMAS ORDER: 1. Brackets ( ) 2. Orders (powers ^) 3. Division Γ· 4. Multiplication Γ— 5. Addition + 6. Subtraction - COMMON SYMBOLS: @ = Addition # = Subtraction $ = Multiplication % = Division & = Equal to STRATEGY: 1. Decode all symbols first 2. Follow BODMAS 3. Solve step by step 4. Check your answer
πŸ’° Chapter 1.6 β€” Profit and Loss
1. Basic Concepts
Cost Price (CP): Price at which article is purchased
Selling Price (SP): Price at which article is sold
Marked Price (MP): Price printed on article

Profit: When SP > CP β†’ Profit = SP - CP
Loss: When SP < CP β†’ Loss = CP - SP
Example: Pen bought for β‚Ή80, sold for β‚Ή100
Profit = 100 - 80 = β‚Ή20
2. Profit and Loss Percentage
Profit % = (Profit/CP) Γ— 100
Loss % = (Loss/CP) Γ— 100

SP = CP Γ— (100+P%)/100 [Profit]
SP = CP Γ— (100-L%)/100 [Loss]

CP = (SP Γ— 100)/(100+P%)
CP = (SP Γ— 100)/(100-L%)
Ex 1: CP=β‚Ή500, SP=β‚Ή600. Profit %?
Profit = 100, Profit % = (100/500)Γ—100 = 20% βœ“

Ex 2: CP=β‚Ή800, Loss=10%. SP?
SP = 800 Γ— 0.9 = β‚Ή720 βœ“

Ex 3: SP=β‚Ή900, Profit=20%. CP?
CP = (900Γ—100)/120 = β‚Ή750 βœ“
3. Dishonest Shopkeeper
Profit % = [(1000-x)/x] Γ— 100
(x = false weight in grams)
Ex: Uses 900g instead of 1kg at CP
Profit % = [(1000-900)/900]Γ—100 = 11.11% βœ“
4. Discount
Discount % = (Discount/MP) Γ— 100
SP = MP Γ— (100-D%)/100
Ex: MP=β‚Ή1000, Discount=20%
SP = 1000 Γ— 0.8 = β‚Ή800 βœ“
5. Successive Discounts
Single = d₁ + dβ‚‚ - (d₁×dβ‚‚)/100
Ex: 20% and 10%
Single = 20+10-(20Γ—10)/100 = 28% βœ“
πŸ’‘ Quick: 0.8 Γ— 0.9 = 0.72 β†’ Discount = 28%
6. Marked Price with Profit
Ex: CP=β‚Ή500, MP 40% above, sold at 20% discount

MP = 500 Γ— 1.4 = β‚Ή700
SP = 700 Γ— 0.8 = β‚Ή560
Profit % = (60/500)Γ—100 = 12% βœ“
7. Practice Problems
P1: Two successive discounts 25% and 20%
Single = 25+20-(25Γ—20)/100 = 40% βœ“

P2: Dishonest gains 20% using false weight
20 = [(1000-x)/x]Γ—100
x = 833.33 gm βœ“
8. Formula Sheet
PROFIT/LOSS: Profit % = (Profit/CP) Γ— 100 SP = CP Γ— (100Β±%)/100 DISCOUNT: Discount % = (Discount/MP) Γ— 100 SP = MP Γ— (100-D%)/100 SUCCESSIVE: d₁ + dβ‚‚ - (d₁×dβ‚‚)/100 DISHONEST: Profit % = [(1000-x)/x] Γ— 100
πŸ”’ Chapter 1.7 β€” Series & Ranking
1. Number Series - Common Patterns
A. Addition/Subtraction Series:
Pattern: Add/subtract constant number

Ex 1: 2, 5, 8, 11, 14, ?
Pattern: +3 each time
Answer: 17 βœ“

Ex 2: 100, 95, 90, 85, ?
Pattern: -5 each time
Answer: 80 βœ“
B. Multiplication/Division Series:
Ex 1: 2, 6, 18, 54, ?
Pattern: Γ—3 each time
Answer: 162 βœ“

Ex 2: 1000, 500, 250, 125, ?
Pattern: Γ·2 each time
Answer: 62.5 βœ“
C. Square/Cube Series:
Ex 1: 1, 4, 9, 16, 25, ?
Pattern: 1Β², 2Β², 3Β², 4Β², 5Β², 6Β²
Answer: 36 βœ“

Ex 2: 1, 8, 27, 64, ?
Pattern: 1Β³, 2Β³, 3Β³, 4Β³, 5Β³
Answer: 125 βœ“
D. Prime Number Series:
Ex: 2, 3, 5, 7, 11, 13, ?
Pattern: Prime numbers
Answer: 17 βœ“
2. Advanced Number Patterns
A. Alternate Series:
Ex: 2, 5, 4, 7, 6, 9, ?
Two patterns:
Odd positions: 2, 4, 6 (+2)
Even positions: 5, 7, 9 (+2)
Answer: 8 βœ“
B. Fibonacci-type Series:
Ex: 1, 1, 2, 3, 5, 8, ?
Pattern: Each = sum of previous two
Answer: 13 βœ“
C. Difference of Differences:
Ex: 2, 3, 5, 8, 12, 17, ?
Differences: 1, 2, 3, 4, 5, 6
Answer: 17 + 6 = 23 βœ“
D. Mixed Operations:
Ex: 3, 6, 11, 18, 27, ?
Differences: 3, 5, 7, 9, 11
Answer: 27 + 11 = 38 βœ“
3. Letter Series
Concept: Letters follow alphabetical positions
A=1, B=2, C=3, D=4, E=5... Z=26
A. Simple Skip Pattern:
Ex 1: A, C, E, G, ?
Pattern: Skip 1 letter
Answer: I βœ“

Ex 2: B, E, H, K, ?
Pattern: Skip 2 letters (+3 positions)
Answer: N βœ“
B. Reverse Pattern:
Ex: Z, Y, X, W, ?
Pattern: Backward sequence
Answer: V βœ“
C. Alternate Pattern:
Ex: A, C, B, D, C, E, ?
Two series:
Odd: A, B, C, D
Even: C, D, E, F
Answer: D βœ“
πŸ’‘ QUICK TIP:
Write positions: A=1, B=2, C=3...
Find pattern in numbers, convert back to letters
4. Ranking Problems
Ranking: Position of a person/object in a sequence
A. Ranking from One End:
Total = Rank from one end + Rank from other end - 1
Ex 1: A is 7th from left, 12th from right. Total?
Total = 7 + 12 - 1 = 18 βœ“

Ex 2: In 25 students, B is 10th from top. Rank from bottom?
Rank from bottom = 25 - 10 + 1 = 16 βœ“
B. Persons Between Two:
Between = |Rank₁ - Rankβ‚‚| - 1
Ex: A is 5th, B is 12th. How many between?
Between = 12 - 5 - 1 = 6 βœ“
C. Finding Position After Change:
Ex: A is 10th. After 3 new join at top, new rank?
New rank = 10 + 3 = 13th βœ“
5. Mixed Practice
P1: Find missing: 5, 10, 20, 40, ?
Pattern: Γ—2
Answer: 80 βœ“

P2: Find missing: 3, 9, 27, 81, ?
Pattern: Γ—3
Answer: 243 βœ“

P3: Letter series: A, D, G, J, ?
Pattern: +3 positions
Answer: M βœ“

P4: Find wrong: 2, 6, 12, 20, 30, 42, 56
Pattern: 1Γ—2, 2Γ—3, 3Γ—4, 4Γ—5, 5Γ—6, 6Γ—7, 7Γ—8
All correct βœ“

P5: Ranking: 30 students, A is 12th from top
Rank from bottom = 30-12+1 = 19th βœ“
6. Commonly Used Series
NATURAL NUMBERS: 1, 2, 3, 4, 5, 6, 7... EVEN NUMBERS: 2, 4, 6, 8, 10, 12... ODD NUMBERS: 1, 3, 5, 7, 9, 11... SQUARES: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100... CUBES: 1, 8, 27, 64, 125, 216, 343, 512... PRIME NUMBERS: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29... FIBONACCI: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55...
7. Quick Solving Tips
πŸ’‘ STRATEGY:

For Number Series:
1. Check differences between terms
2. Check ratios (Γ—, Γ·)
3. Check if squares/cubes
4. Look for alternate patterns
5. Check difference of differences

For Letter Series:
1. Convert to numbers (A=1, B=2...)
2. Find numeric pattern
3. Convert back to letters

For Ranking:
1. Draw diagram if confused
2. Remember: Total = Left + Right - 1
3. Between = Difference - 1
πŸ“Š Chapter 1.8 β€” Average & Weighted Average
1. Concept of Average
Average: Sum of all values divided by number of values

Average = Sum of all values / Total number of values
Ex: Find average of 10, 20, 30, 40, 50
Sum = 150, Count = 5
Average = 150/5 = 30 βœ“
2. Average with Addition/Removal
When one item added:
New Sum = Old Sum + New Item

When one item removed:
New Sum = Old Sum - Removed Item
Ex: Average of 5 numbers is 20. One number 25 added. New average?

Old Sum = 5 Γ— 20 = 100
New Sum = 100 + 25 = 125
New Average = 125/6 = 20.83 βœ“
3. Weighted Average
Weighted Average: Average where some values have more importance (weight) than others
Weighted Average = (w₁×v₁ + wβ‚‚Γ—vβ‚‚ + w₃×v₃...) / (w₁ + wβ‚‚ + w₃...)

Where: w = weight, v = value
Ex: Class A (30 students, avg 70), Class B (20 students, avg 80). Overall average?

Weighted Avg = (30Γ—70 + 20Γ—80)/(30+20)
= (2100 + 1600)/50
= 3700/50 = 74 βœ“
4. Average Change Problems
Ex: Average of 10 numbers is 50. If each increased by 5, new average?

Each increased by 5 β†’ Average also increases by 5
New Average = 50 + 5 = 55 βœ“
πŸ’‘ RULE:
If all values increase/decrease by x β†’ Average increases/decreases by x
5. Practice Problems
P1: Average of 6 numbers is 30. One number 40 removed. New average?
Old Sum = 6Γ—30 = 180
New Sum = 180-40 = 140
New Avg = 140/5 = 28 βœ“

P2: Average age of 40 students is 15. Teacher's age 40. New average?
Total = 40Γ—15 + 40 = 640
New Avg = 640/41 = 15.6 years βœ“
⚑ Quick Tricks - Last Minute Revision
πŸ“Œ How to Use: Quick revision of all formulas and tricks before exam!
πŸ”’ Divisibility Rules
2: Last digit even 3: Sum divisible by 3 4: Last 2 digits divisible by 4 7: Remove last, double, subtract 8: Last 3 digits divisible by 8 9: Sum divisible by 9 11: (Odd places - Even places) divisible by 11 1001 = 7 Γ— 11 Γ— 13
βž— Remainder Theorems
Fermat: a^(p-1) ≑ 1 (mod p) [p prime] Euler: a^Ο†(n) ≑ 1 (mod n) Wilson: (p-1)! ≑ -1 (mod p) [p prime] Ο†(10) = 4, Ο†(100) = 40, Ο†(1000) = 400
πŸ”’ Factors & Factorials
If N = p₁^a Γ— pβ‚‚^b Γ— p₃^c Factors = (a+1)(b+1)(c+1) Sum = [(p₁^(a+1)-1)/(p₁-1)] Γ— ... Product = N^(k/2) Zeros in n! = ⌊n/5βŒ‹ + ⌊n/25βŒ‹ + ... Power of p in n! = ⌊n/pβŒ‹ + ⌊n/pΒ²βŒ‹ + ...
πŸ”Ÿ Unit Digit Cyclicity
0,1,5,6 β†’ Always same (cycle 1) 4,9 β†’ Alternates (cycle 2) 2,3,7,8 β†’ Cycle of 4 2: 2,4,8,6 3: 3,9,7,1 7: 7,9,3,1 8: 8,4,2,6 Divide power by cyclicity, use remainder
βš–οΈ Inequality & BODMAS
Multiply/Divide by negative β†’ FLIP sign BODMAS: 1. Brackets 2. Orders (powers) 3. Division 4. Multiplication 5. Addition 6. Subtraction
πŸ’° Profit & Loss
Profit% = (Profit/CP) Γ— 100 SP = CP Γ— (100Β±%)/100 Discount% = (Discount/MP) Γ— 100 SP = MP Γ— (100-D%)/100 Successive: d₁ + dβ‚‚ - (d₁×dβ‚‚)/100 Dishonest: [(1000-x)/x] Γ— 100
πŸ”’ Series & Ranking
Check: Differences, Ratios, Squares, Cubes Letters: A=1, B=2... Z=26 Ranking: Total = Left + Right - 1 Between = |Rank₁ - Rankβ‚‚| - 1
πŸ“Š Average
Average = Sum/Count Weighted: (w₁v₁ + wβ‚‚vβ‚‚...)/(w₁+wβ‚‚...) All increase by x β†’ Avg increases by x
πŸ“ Important to Remember
Squares: 1,4,9,16,25,36,49,64,81,100,121,144 Cubes: 1,8,27,64,125,216,343,512 Primes: 2,3,5,7,11,13,17,19,23,29,31,37 10% = 1/10 20% = 1/5 25% = 1/4 33.33% = 1/3 50% = 1/2
πŸ’‘ Exam Strategy
Time Management:
β€’ Easy questions first
β€’ Don't get stuck on one
β€’ 30-45 seconds per question

Common Mistakes:
βœ— Forgetting to flip inequality sign
βœ— Wrong BODMAS order
βœ— Using wrong formula
βœ— Calculation errors

Pro Tips:
βœ“ Eliminate obviously wrong options
βœ“ Use approximation when possible
βœ“ Check if answer makes sense
βœ“ Practice mental math daily
🌟 All the Best!
You're Ready! 🎯

You've covered all 8 chapters:
βœ“ Divisibility Rules
βœ“ Remainders (Fermat, Euler, Wilson)
βœ“ Factors & Factorials
βœ“ Unit Digit & Last Two Digits
βœ“ Inequality & Symboperation
βœ“ Profit and Loss
βœ“ Series & Ranking
βœ“ Average & Weighted Average

Remember: Practice is key! Solve as many problems as possible.
Speed + Accuracy = Success! πŸ’ͺπŸš€
Made with πŸ’™ by Ankush Raj

Β© 2026 ankushraj.dev - All Rights Reserved