Quantitative Aptitude – Number System
(For Chhattisgarh Police Sub‑Inspector (SI) Examination)
Aim of this section – To test a candidate’s ability to manipulate numbers, apply arithmetic rules, and solve problems that involve properties of integers, fractions, decimals, percentages, ratios, and powers.
1. Simple Explanation
| Term | What it means | Example |
|---|---|---|
| Integer | Whole number, positive or negative, no fractional part | ‑7, 0, 15 |
| Fraction | Part of a whole expressed as a ratio of two integers | ⅔ = 2 ÷ 3 |
| Decimal | Numbers expressed in base‑10 using a decimal point | 0.75 = 75/100 |
| Percentage | A number expressed per hundred | 25 % = 25/100 = 0.25 |
| Ratio | Comparative relationship between two numbers | 3:4 (or 3 ∶ 4) |
| Proportion | Equality of two ratios | a:b = c:d ⇒ ad = bc |
| Index / Power | Repeated multiplication | 5² = 5×5 = 25 |
| Root | The inverse of an index | √25 = 5, ∛27 = 3 |
| Prime Number | Integer > 1 with exactly two distinct divisors (1 and itself) | 2, 3, 5, 7 … |
| Composite Number | Integer > 1 with more than two divisors | 4, 6, 8 … |
| GCD (Greatest Common Divisor) | Largest integer dividing two numbers without remainder | GCD(18, 24) = 6 |
| LCM (Least Common Multiple) | Smallest integer divisible by two numbers | LCM(4, 6) = 12 |
| Divisibility Rules | Quick checks to see if a number is divisible by another | A number is divisible by 3 if the sum of its digits is divisible by 3. |
2. Key Points (Bullet‑style)
| Category | Points |
|---|---|
| Integers & Fractions | • Simplify fractions by dividing numerator & denominator by GCD. • Convert mixed numbers to improper fractions before operations. |
| Decimals & Percentages | • 1 % = 0.01. • To convert 12 % to a decimal: 12/100 = 0.12. • To find a percent of a number: (percent ÷ 100) × number. |
| Ratio & Proportion | • For a:b = c:d, cross‑multiply: ad = bc. • Use to find “part” when total and ratio are known. |
| Powers & Roots | • 10⁰ = 1; 10¹ = 10; 10² = 100. • For square roots, look for perfect squares (1,4,9,…). • For cube roots, look for perfect cubes (1,8,27,…). |
| Prime & Composite | • A prime has no divisors other than 1 and itself. • Use trial division up to √n. |
| GCD & LCM | • GCD: use Euclidean algorithm (a,b) → (b, a mod b). • LCM: (a × b) ÷ GCD(a,b). |
| Divisibility Rules | • 2: last digit even. • 3: sum of digits divisible by 3. • 4: last two digits divisible by 4. • 5: last digit 0 or 5. • 6: divisible by 2 and 3. • 9: sum of digits divisible by 9. • 10: last digit 0. |
| Special Numbers | • Repdigits – numbers like 111, 2222. • Armstrong (Narcissistic) Numbers – sum of digits each raised to the power of the number of digits equals the number itself (e.g. 153 = 1³+5³+3³). |
| Common Pitfalls | • Forgetting to convert percentages to decimals. • Mixing up GCD and LCM while solving simultaneous equations. • Ignoring that 0 is divisible by any non‑zero integer but has no unique prime factorization. |
3. Worked Example
Problem:
A machine can produce 240 units in 6 hours.
(a) How many units will it produce in 18 hours?
(b) If the machine is run at 75 % of its capacity, how many units will it produce in 18 hours?
(c) Find the GCD and LCM of 240 and 360.
(a) Production in 18 hours
-
Rate per hour = Total units ÷ Total hours
( \displaystyle \frac{240}{6}=40) units/hour. -
Units in 18 hours = Rate × 18
( 40 \times 18 = 720) units.
(b) Production at 75 % Capacity
-
Effective rate = 75 % of 40 units/hour
( 0.75 \times 40 = 30) units/hour. -
Units in 18 hours = 30 × 18 = 540 units.
(c) GCD & LCM of 240 & 360
| Step | Calculation |
|---|---|
| GCD | Euclidean algorithm: |
| 360 ÷ 240 = 1 remainder 120 → GCD(240,120) | |
| 240 ÷ 120 = 2 remainder 0 → GCD = 120 | |
| LCM | ( \displaystyle \frac{240 \times 360}{120} = \frac{86,400}{120} = 720) → |