Quantitative Aptitude – Caselets
(Chandigarh Police Sub‑Inspector – SI – Exam)
What are Caselets?
Caselets are short, story‑based problems (often word problems) that test your ability to extract relevant quantitative information, set up the correct mathematical relationships, and solve for the unknowns. They usually involve everyday situations (traffic, salaries, distances, rates, etc.) and require a mix of reasoning, arithmetic, and algebra.
1. Simple Explanation
| Aspect | What to Do |
|---|---|
| Read Carefully | Identify every piece of numerical data given. |
| Define Variables | Assign symbols (x, y, …) to unknown quantities. |
| Set Up Equations | Translate the story into algebraic relations (ratios, percentages, time‑speed, etc.). |
| Solve | Use algebraic manipulation, mental math, or quick estimation. |
| Check | Verify that the answer fits all conditions of the problem. |
2. Key Points to Remember
| Key Point | Why It Matters | Quick Tip |
|---|---|---|
| Identify the “Question” | The question often hides in a sentence that doesn’t appear to ask for a number. | Highlight the interrogative clause. |
| List All Given Numbers | Helps prevent missing a critical piece of data. | Write them in a separate “Numbers” box. |
| Look for Ratios & Proportions | Many caselets involve relative comparisons. | Use “x : y = a : b” → cross‑multiply. |
| Use the “Unit” Approach | Converting units early avoids mistakes (e.g., km → m). | Keep a mini‑conversion table handy. |
| Check for “Common Factors” | Simplify before multiplying to reduce mental load. | Reduce fractions first. |
| Beware of “Trick” Words | Words like “more than”, “less than”, “difference” signal subtraction or addition. | Translate into algebraic operators. |
| Estimate Before Calculating | A quick sanity check ensures you’re on the right track. | Round numbers to nearest 10–100. |
| Use Tables & Diagrams | Visual representation clarifies relationships. | Sketch a quick diagram if the problem involves distances or timelines. |
| Watch for Unit Consistency | Mixing units (e.g., km and m) can lead to wrong answers. | Convert all units to a common base. |
| Time & Speed Problems | Use ( \text{Speed} = \frac{\text{Distance}}{\text{Time}} ). | Convert time to hours or minutes consistently. |
| Percentage Problems | For “increase/decrease by x%”, use ( \text{New} = \text{Old} \times (1 ± \frac{x}{100})). | Memorize 10% = 0.10, 25% = 0.25 etc. |
| Compound Interest | Use ( A = P(1 + \frac{r}{100})^n ) for simple cases. | For quick SI test, use ( \text{SI} = \frac{P \times R \times T}{100} ). |
| Geometry (Area, Volume) | Many caselets involve shapes; memorize formulas. | Use Pythagoras for right triangles, ( V = πr^2h ) for cylinders. |
| Prime & Factorisation | Useful for “least common multiple” or “greatest common divisor” problems. | Remember 2, 3, 5, 7 are the first primes. |
3. Worked Example
Caselet
“A police officer was tasked with patrolling a 120‑km stretch of highway. He started at 8:00 AM and covered the first part of the highway at a constant speed of 60 km/h. After a 30‑minute break, he resumed his patrol at a higher speed that was 20% greater than his initial speed. How long, in minutes, will it take him to complete the remaining distance?”
Step‑by‑Step Solution
-
Read & Extract Data
- Total distance = 120 km
- First speed ( v_1 = 60,\text{km/h} )
- Break = 30 min = 0.5 h
- Second speed ( v_2 = 1.20 \times v_1 = 72,\text{km/h} )
-
Set up equations
- Distance covered in first part:
[ d_1 = v_1 \times t_1 ] Where ( t_1 ) is time in hours. - Remaining distance:
[ d_{\text{rem}} = 120 - d_1 ] - Time to cover remaining distance at ( v_2 ):
[ t_{\text{rem}} = \frac{d_{\text{rem}}}{v_2} ]
- Distance covered in first part:
-
Solve for ( t_1 )
- The officer started at 8:00 AM and finished the first part before the break.
- To find ( t_1 ), we need the distance he covered before the break.
- Since he didn’t specify when the break occurred, we can use the fact that the break is 30 min after finishing the first part.
- Assume he started the break immediately after finishing the first part.
- Therefore, the time from start to end of first part ( = t_1 ).
- Without extra info, we can compute the maximum possible first part distance if the break happened right after 8:30 AM? No.
- Alternate approach: Recognise that the break duration is independent of distance. We can solve directly for ( t