Caselets – Quantitative Aptitude (Odisha PSC)
Caselets are short, real‑world problems that require a blend of arithmetic, logic and quick reasoning.
They are a staple of the Odisha PSC exam, often appearing in the Quantitative Aptitude paper (Section A or B).
1. Simple Explanation
| Term | What it Means | Why It Matters |
|---|---|---|
| Caselet | A self‑contained, scenario‑based problem (e.g., “If 30 % of a number is 21, find the number”). | Tests applied mathematics, not just rote formulas. |
| Key Idea | Translate the narrative into equations or simple calculations. | Avoids misreading – the solution is often a single, clean step. |
2. Key Points (What You Must Remember)
| Point | Detail |
|---|---|
| Read Carefully | Identify all numbers and relationships. Highlight them. |
| Convert Words → Numbers | “Three times as many” → 3×, “$20 more” → “+20”. |
| Use Simple Equations | Often a single linear equation or two equations (ratio, proportion). |
| Check Units | E.g., “kg” vs “kg”, “₹” vs “₹”, percentages vs decimals. |
| Look for Quick Tricks | 1. Percentage of a number → multiply by 0.01. 2. Profit/Loss → Selling Price = Cost Price (1 ± %/100). |
| Avoid Over‑analyzing | Many caselets can be solved in 1–2 minutes. |
| Verify with a Quick Substitution | After finding a value, plug it back into the text to ensure consistency. |
3. Types of Caselets Commonly Appeared
| Category | Typical Question | Key Formula / Trick |
|---|---|---|
| Percentages | “30 % of a number is 21.” | x = 21 / 0.30 |
| Ratios & Proportions | “Pencils : Pens = 5 : 3.” | x / y = 5 / 3 |
| Profit / Loss | “Selling price is 120 % of cost.” | SP = CP × 1.20 |
| Speed / Distance | “Two trains meet after 3 h.” | v1 × v2 = (total distance) / 3 |
| Work / Time | “A can finish a job in 5 days, B in 10 days.” | 1/5 + 1/10 = 1/Total |
| Time & Calendar | “If today is Thursday, what day will it be 45 days from now?” | 45 mod 7 = 3 → +3 days |
| Simple Interest | “₹10,000 at 5 % per annum for 3 years.” | SI = P × R × T / 100 |
| Compound Interest | “₹5,000 compounded annually at 8 % for 2 years.” | A = P(1 + r)^n |
| Geometry | “Area of a rectangle is 48 cm², length is 8 cm.” | Area = L × W |
| Miscellaneous | “Price of 3 items equals 2 items + ₹50.” | Set up equation. |
4. Problem‑Solving Approach (Step‑by‑Step)
-
Read the problem twice.
First pass: identify the numbers.
Second pass: identify the relationships. -
Highlight/underline all keywords: total, difference, more than, percentage, time, speed, etc.
-
Translate to algebraic symbols.
E.g. “Number of pencils = Number of pens + 5” →P = Q + 5. -
Write down the equation(s).
Keep it simple; you usually need only one or two equations. -
Solve quickly.
Use mental math where possible (e.g.,30 % of 70 = 21). -
Check consistency.
Plug the solution back into the narrative (does it satisfy all conditions?). -
Answer in the required format.
Usually a single numeric answer, possibly with a unit.
5. Worked Example
Problem (Typical Odisha PSC Caselet)
A shopkeeper sold 300 pens and 500 pencils in a week. The total revenue was ₹16,000. The price of each pen is ₹3 more than the price of each pencil. What is the price of a pencil?
Step‑by‑Step Solution
| Step | Explanation |
|---|---|
| 1. Define variables | Let p = price of a pencil (₹). Then price of a pen = p + 3. |
| 2. Write revenue equation | Revenue = quantity × price. 500p + 300(p + 3) = 16,000 |
| 3. Expand | 500p + 300p + 900 = 16,000 |
| **4 |