📚 Quantitative Aptitude – Caselets
Target Audience: West Bengal Police Constable (WB‑PC) Examination
Depth: MIT‑level – first principles, rigorous reasoning, edge cases, and a challenging problem
Caselets are short, self‑contained quantitative puzzles that test quick reasoning, mental arithmetic, and a solid grasp of fundamental mathematical concepts.
They often appear in the “Case‑Based Reasoning” or “Direct Application” sections of aptitude tests.
1. What are Caselets?
| Feature | Description |
|---|---|
| Length | 1–3 lines (single paragraph) |
| Data | Minimal; often a single equation or a small set (≤ 4 numbers) |
| Answer | 1‑digit, 2‑digit, or a short phrase (e.g., “odd”, “first”) |
| Skill Tested | Logical deduction, pattern recognition, basic arithmetic, algebra, combinatorics, probability, geometry (rare). |
| Time | 10–20 seconds per case (fast recall, no heavy calculation). |
2. Core Mathematical Tools
| Tool | Why It Matters | Typical Use in Caselets |
|---|---|---|
| Algebraic Manipulation | Expressing relationships, eliminating variables | Solving “X + Y = 30, X – Y = 4” |
| Modular Arithmetic | Working with remainders (e.g., “sum of digits”) | “Find the 10th digit in a repeating decimal” |
| Combinatorics | Counting arrangements, subsets | “How many ways to choose 3 officers from 5?” |
| Probability Basics | Expected value, odds | “Probability of drawing a red card” |
| Number Properties | Parity, divisibility, multiples | “Is the number divisible by 6?” |
| Geometry (if present) | Simple Pythagorean, area, perimeter | “Area of a 3‑side polygon” |
Tip: Whenever a problem presents a “sequence” or “pattern,” first identify the underlying rule (arithmetic progression, geometric progression, pattern of digits, etc.). Once the rule is known, extrapolation is trivial.
3. Systematic Problem‑Solving Framework
-
Read Carefully
- Identify all given numbers and constraints.
- Highlight any “unique” terms like “first,” “last,” “odd,” “even,” “multiple of,” etc.
-
Translate into Mathematics
- Convert words into equations or inequalities.
- For example, “Total of three numbers is 24” → (x + y + z = 24).
-
Reduce the Problem
- Use algebraic elimination or substitution to isolate the unknown.
- Look for common factors, parity checks, or modular reductions.
-
Check Edge Cases
- Verify that the solution satisfies all given conditions.
- Consider “boundary” values (e.g., the smallest/largest possible integer, or a case where a divisor is 1).
-
Answer in Required Format
- If a single digit is requested, give that digit.
- If “odd/even” is asked, state the property.
- If a location (“first,” “third”) is requested, give the position number.
4. Edge Cases & Common Pitfalls
| Edge Case | Scenario | Why It Matters | How to Handle |
|---|---|---|---|
| Zero or One | Divisibility by 0 or 1 | Division by zero is undefined; 1 divides everything | Always check that denominators are ≠ 0; if a divisor is 1, the property holds trivially. |
| Negative Numbers | Sometimes “difference” can be negative | Many test takers overlook sign changes | Convert to absolute value or use inequalities. |
| Parity Swaps | “Odd + Even = Odd” | Misreading “even” as “odd” | Keep track of even/odd status throughout. |
| Duplicate Values | “Three distinct numbers” vs “three numbers” | Distinctness can change counting | Verify if “distinct” is explicitly stated. |
| Boundary Constraints | “Within 1–1000” | A derived value might exceed limits | Check final answer against stated bounds. |
5. Illustrative Caselets (with Full Solutions)
Caselet 1
“Four consecutive even numbers sum to 144. What is the second number?”
Solution
Let the numbers be (x, x+2, x+4, x+6).
[
x + (x+2) + (x+4) + (x+6) = 144 \
4x + 12 = 144 \
4x = 132 \
x = 33
]
Numbers: 33, 35, 37, 39 → but they are odd, contradiction.
Hence no such even numbers exist.
Answer: None.
Caselet 2
“A bag contains 10 cards: 3 red, 4 blue, and 3 green. Two cards are drawn at random without replacement. What is the probability that the first card is red and the second is blue?”
Solution
[
P(\text{Red first}) = \frac{3}{10} \
P(\text{Blue second | Red first}) = \frac{4}{9} \
P = \frac{3}{10}\cdot\frac{4}{9} = \frac{12}{90} = \frac{2}{15}
]
Answer: ( \frac{2}{15} ).
Caselet 3
“A 5‑digit number is such that the sum of its digits equals 27 and it is divisible by 9. The third digit from the left is 5. What is the smallest possible value of this number?”
Solution
Divisibility by 9 → sum of digits ≡ 0 mod 9.
Given sum = 27 → already divisible.
Let digits be (a,b,5,d,e).
(a +