Study Notes – Tables
(Quantitative Aptitude – Ladakh Police Constable Exam)
1. Simple Explanation
In quantitative aptitude, “tables” are compact data sets used to solve problems quickly.
Typical table types you’ll encounter:
| Table Type | What it shows | Common use in questions |
|---|---|---|
| Speed–Time–Distance | Speeds, times, distances of a moving object | Find missing distance, time or speed |
| Interest / Rate of Return | Principal, rate, time, interest | Compute simple or compound interest |
| Percentage / Ratio | Percentages, ratios | Convert percentages, find parts of a whole |
| Exchange / Currency | Conversion rates | Convert between currencies |
| Weight / Volume | Weights, volumes, densities | Find missing quantity |
| Temperature / Fahrenheit‑Celsius | Temperature conversions | Convert between units |
Tables are built to let you read off values without lengthy calculations. The key is understanding the relationships between the columns and using the appropriate formula.
2. Key Points
| # | Point | Why It Matters |
|---|---|---|
| 1. Identify the variables | Recognize which column is missing (speed, time, distance, etc.) | You’ll know which formula to use |
| 2. Use the right formula | - (d = s \times t) - (t = \frac{d}{s}) - (s = \frac{d}{t}) |
Direct calculation from the table |
| 3. Check units | Speed in km/h, distance in km, time in h | Avoids errors when multiplying/dividing |
| 4. Look for patterns | Incremental changes, percentages, ratios | Helps predict missing values |
| 5. Cross‑check with two methods | e.g., compute time from distance & speed, then verify | Reduces mistakes |
| 6. Use “common sense” limits | Speed > 0, time > 0, distance > 0 | Unphysical values indicate a mistake |
| 7. Memorise common tables | Speed–time tables for everyday vehicles, standard interest tables | Saves time in exam |
| 8. Practice “derivation” | Given a table of speeds, derive the distance table | Strengthens understanding of relationships |
3. Worked Example
Problem
A cyclist travels at a constant speed. The table below gives the distance covered at 2‑hour intervals. Find the time needed to cover a distance of 120 km.
| Time (h) | Distance (km) |
|---|---|
| 0 | 0 |
| 2 | 30 |
| 4 | 60 |
| 6 | 90 |
| 8 | 120 |
Solution
-
Identify the pattern
The distance increases by 30 km every 2 h → Speed = ( \frac{30\text{ km}}{2\text{ h}} = 15\text{ km/h}). -
Confirm with the table
( 15 \text{ km/h} \times 4 \text{ h} = 60 \text{ km}) ✔️
( 15 \text{ km/h} \times 8 \text{ h} = 120 \text{ km}) ✔️ -
Find unknown time
Use ( t = \frac{d}{s} ).
( t = \frac{120\text{ km}}{15\text{ km/h}} = 8\text{ h}).
Answer: The cyclist needs 8 hours to cover 120 km.
Quick Check:
Distance table already contains 120 km at 8 h, which matches the computed time.
4. Quick Revision Summary
| Topic | Key Formula | Tip |
|---|---|---|
| Speed–Time–Distance | (d = s \times t) | Check units; if two values are given, compute the third. |
| Time | (t = \frac{d}{s}) | Useful when distance & speed known. |
| Speed | (s = \frac{d}{t}) | Useful when distance & time known. |
| Interest (Simple) | (I = P \times r \times t) | (r) in decimal form. |
| Interest (Compound) | (A = P (1 + \frac{r}{n})^{nt}) | Use when compounding frequency given. |
| Percentages | (x% \text{ of } N = \frac{x}{100}\times N) | Convert percent to decimal before multiplying. |
| Exchange Rates | ( \text{Amount in Currency B} = \frac{\text{Amount in Currency A}}{\text{Rate}}) | Rate usually “units of A per 1 unit of B”. |
| Temperature Conversion | (C = \frac{5}{9}(F-32)) | (F = \frac{9}{5}C + 32 |