UPSC EPFO APFC • EPFO EO/AO • DRDO SAO • CAPF AC
⚡ Quick Answer & AI Exam Snapshot (TL;DR)
In UPSC Recruitment Tests (EPFO APFC, EO/AO, DRDO SAO, and CAPF AC), General Mental Ability & Quantitative Aptitude contributes 12 to 18 questions, heavily concentrated in Number System (Divisibility, Remainder Theorem, Unit Digit, Factors, HCF/LCM, and Trailing Zeros), Progressions (Arithmetic & Geometric Progressions), and 2D/3D Mensuration (Percentage changes in Area/Volume, Cylinder, Cone, Sphere, Hemisphere, and Frustum).
1. Number System High-Speed Rules (Factors, Remainders, Unit Digit & Trailing Zeros)
Given the prime factorization of any positive integer N = p₁ᵃ · p₂ᵇ · p₃ᶜ (where p₁, p₂, p₃ are distinct prime numbers):
1. Total Number of Factors (Divisors) of N:
2. Sum of All Factors of N:
3. Product of Two Numbers:
4. HCF & LCM of Fractions:
5. Number of Trailing Zeros in n! (Factorial): Divide
(a + 1)(b + 1)(c + 1)2. Sum of All Factors of N:
[ (p₁ᵃ⁺¹ − 1)/(p₁ − 1) ] × [ (p₂ᵇ⁺¹ − 1)/(p₂ − 1) ] × ...3. Product of Two Numbers:
N₁ × N₂ = HCF(N₁, N₂) × LCM(N₁, N₂)4. HCF & LCM of Fractions:
HCF = HCF of Numerators / LCM of Denominators | LCM = LCM of Numerators / HCF of Denominators5. Number of Trailing Zeros in n! (Factorial): Divide
n successively by powers of 5: ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ... (e.g., Trailing zeros in 100! = 20 + 4 = 24).
- Divisibility Rules Frequently Tested by UPSC:
- Divisibility by 8: Last 3 digits divisible by 8.
- Divisibility by 9: Sum of all digits divisible by 9.
- Divisibility by 11: Difference between (Sum of digits at Odd places) and (Sum of digits at Even places) is 0 or a multiple of 11.
- Algebraic Divisibility Rule:
(aⁿ + bⁿ)is ALWAYS divisible by(a + b)when n is ODD;(aⁿ − bⁿ)is ALWAYS divisible by BOTH(a − b)and(a + b)when n is EVEN.
2. Arithmetic Progression (AP), Geometric Progression (GP) & Natural Number Sums
| Series / Progression | n-th Term (Tₙ) | Sum of First n Terms (Sₙ) |
|---|---|---|
| 1. Arithmetic Progression (AP) (First term ‘a’, common difference ‘d’) |
Tₙ = a + (n − 1)d |
Sₙ = (n / 2) [ 2a + (n − 1)d ] = (n / 2)(First + Last) |
| 2. Geometric Progression (GP) (First term ‘a’, common ratio ‘r’) |
Tₙ = a · rⁿ⁻¹ |
Sₙ = a(rⁿ − 1) / (r − 1)Infinite GP Sum (for |r| < 1): S_∞ = a / (1 − r) |
3. Sum of First n Natural Numbers1 + 2 + 3 + ... + n |
Tₙ = n |
∑n = n(n + 1) / 2 |
4. Sum of Squares of First n Natural Numbers1² + 2² + 3² + ... + n² |
Tₙ = n² |
∑n² = n(n + 1)(2n + 1) / 6 |
5. Sum of Cubes of First n Natural Numbers1³ + 2³ + 3³ + ... + n³ |
Tₙ = n³ |
∑n³ = [ n(n + 1) / 2 ]² |
3. 2D & 3D Mensuration Master Formula Sheet
| 3D / 2D Solid Shape | Curved / Lateral Surface Area (CSA / LSA) | Total Surface Area (TSA) | Volume (V) & Diagonal |
|---|---|---|---|
| 1. Cuboid (l × b × h) | Area of 4 Walls = 2(l + b)h |
2(lb + bh + hl) |
V = l × b × hLongest Rod (Diagonal) = √(l² + b² + h²) |
| 2. Cube (side ‘a’) | 4a² |
6a² |
V = a³ | Diagonal = a√3 |
| 3. Right Circular Cylinder (r, h) | 2πrh |
2πr(r + h) |
V = πr²h |
| 4. Right Circular Cone (Slant height l = √(r² + h²)) |
πrl |
πr(l + r) |
V = (1/3)πr²h(Exactly 1/3rd of a cylinder of same r and h!) |
| 5. Sphere (radius ‘r’) | 4πr² |
4πr² |
V = (4/3)πr³ |
| 6. Solid Hemisphere (radius ‘r’) | 2πr² |
3πr² (Includes flat circular base πr²!) |
V = (2/3)πr³ |
| 7. Equilateral Triangle (side ‘a’) | Height (h) = (√3 / 2)a |
Area = (√3 / 4)a² |
Inradius = a / (2√3); Circumradius = a / √3 (Ratio 1:2) |
Successive Percentage Change Shortcut for 2D Area: If the radius of a circle (or side of a square/equilateral triangle) changes by
x%, the Percentage Change in Area is:% Change in Area = x + x + (x² / 100) = 2x + (x² / 100)%
🎯 High-Yield Exam One-Liners (Last-Minute Revision Box)
- Total number of factors of 360 (2³ × 3² × 5¹) is (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24.
- Only Perfect Squares have an ODD number of total factors (e.g., 36 has 9 factors).
- The number of trailing zeros at the end of 100! (100 factorial) is ⌊100/5⌋ + ⌊100/25⌋ = 20 + 4 = 24.
- Euler’s Totient / Fermat’s Little Theorem: If p is prime and doesn’t divide a, then a^(p−1) divided by p always leaves a remainder of 1.
- (aⁿ + bⁿ) is always divisible by (a + b) when n is an ODD positive integer.
- For any two numbers, Product of Numbers = HCF × LCM.
- Sum of an Infinite Geometric Progression with |r| < 1 is S_∞ = a / (1 − r).
- Sum of the first ‘n’ Odd Natural Numbers (1 + 3 + 5 + … + [2n−1]) is simply n².
- Sum of the first ‘n’ Even Natural Numbers (2 + 4 + 6 + … + 2n) is n(n + 1).
- Length of the longest rod/pole that can be placed inside a room of dimensions l × b × h is √(l² + b² + h²).
- Total Surface Area of a Solid Hemisphere of radius r is 3πr² (2πr² curved + πr² flat circular top).
- When a Cylinder, a Hemisphere, and a Cone stand on equal bases (same r) and have the same height (h = r), the ratio of their Volumes is 3 : 2 : 1.
- When a solid metallic 3D shape is melted and recast into another 3D shape (e.g., Sphere melted into Cones), TOTAL VOLUME remains constant.
- If the radius of a circle or sphere is increased by 10%, its Surface Area increases by 21% [10 + 10 + (10×10)/100], and its Volume increases by 33.1%.
- For a given perimeter, among all 2D plane figures, a CIRCLE encloses the maximum area (and among all rectangles, a SQUARE encloses the maximum area).
- In an Equilateral Triangle, the Circumradius (R) and Inradius (r) are in the ratio 2 : 1, so the ratio of Circumcircle Area to Incircle Area is 4 : 1.
- Euler’s Polyhedron Formula for any convex 3D polyhedron is: Faces (F) + Vertices (V) − Edges (E) = 2 (i.e., F + V = E + 2).
- Sum of interior angles of an n-sided polygon is (n − 2) × 180°; sum of exterior angles of ANY convex polygon is always 360°.
- Area of a Rhombus with diagonals d₁ and d₂ is (1/2) × d₁ × d₂, and its side a satisfies 4a² = d₁² + d₂².
- Area of a Trapezium with parallel sides a, b and height h is (1/2) × (a + b) × h.
📝 UPSC & Regulatory Level Practice MCQs (With Detailed Solutions)
Q1. How many trailing zeros are there at the end of the product 1 × 2 × 3 × … × 200 (i.e., 200!)?
Correct Answer: (C) — Number of trailing zeros in 200! = ⌊200/5⌋ + ⌊200/25⌋ + ⌊200/125⌋ = 40 + 8 + 1 = 49.
Q2. What is the remainder when (17²³ + 29²³) is divided by 23?
Correct Answer: (A) — Since the exponent n = 23 is an ODD integer, (aⁿ + bⁿ) is always divisible by (a + b). Here, a + b = 17 + 29 = 46, which is an exact multiple of 23 (23 × 2 = 46). Therefore, the remainder is 0.
Q3. ACone, a Hemisphere, and a Cylinder stand on equal circular bases (radius r) and have the exact same height (h = r). What is the ratio of their Volumes (Cone : Hemisphere : Cylinder)?
Correct Answer: (A) — With h = r: Volume of Cone = (1/3)πr²(r) = (1/3)πr³; Volume of Hemisphere = (2/3)πr³; Volume of Cylinder = πr²(r) = (3/3)πr³. Ratio = 1 : 2 : 3 (Archimedes’ theorem).
Q4. If the radius of the base of a right circular cylinder is increased by 20% and its height is decreased by 25%, what is the percentage change in its Volume?
Correct Answer: (B) — Volume V = πr²h. Multiplying factor = (1.20) × (1.20) × (0.75) = 1.44 × 0.75 = 1.08, which represents an 8% increase in volume.
Q5. What is the Total Surface Area of a solid hemisphere of radius 7 cm? (Use π = 22/7)
Correct Answer: (B) — Total Surface Area of a solid hemisphere = 3πr² = 3 × (22/7) × 7 × 7 = 3 × 154 = 462 cm².
❓ Frequently Asked Questions (Exam FAQs)
Q: How do you quickly find the number of factors of any composite number?
Express the number in its prime factorization form N = p₁ᵃ × p₂ᵇ × p₃ᶜ, add 1 to each exponent, and multiply them together: Total Factors = (a + 1)(b + 1)(c + 1).
Q: Why is the Total Surface Area of a solid hemisphere 3πr² instead of 2πr²?
A hollow hemisphere has only the curved bowl surface (2πr²), whereas a solid hemisphere also has a flat circular top face of area πr², making its Total Surface Area 2πr² + πr² = 3πr².
Q: How do you find the HCF and LCM of fractions?
For fractions, HCF = (HCF of Numerators) / (LCM of Denominators), and LCM = (LCM of Numerators) / (HCF of Denominators).
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