Kerala TET – Quantitative Aptitude
Topic: Tables
(A concise, study‑ready guide for the TET exam)
1. Simple Explanation
In quantitative aptitude, a table is a rectangular arrangement of numbers or data with rows (horizontal) and columns (vertical).
- Rows usually represent the first variable (e.g., students, months, items).
- Columns usually represent the second variable (e.g., marks obtained, sales, prices).
- The intersection of a row and a column contains the value of the two variables together.
Tables are used to:
| Purpose | Typical Example |
|---|---|
| Present data clearly | Monthly sales of a shop |
| Show relationships | Factorial: n! for n = 1 … 7 |
| Enable quick comparison | Speed test times of 5 runners |
| Facilitate calculation | Frequency distribution for a test |
2. Key Points
| # | Key Point | Why It Matters |
|---|---|---|
| 1 | Read headers carefully – They tell you what each row/column represents. | Misreading leads to wrong calculations. |
| 2 | Identify patterns – Arithmetic, geometric, factorial, or quadratic patterns often appear. | Recognizing the pattern saves time. |
| 3 | Use the “two‑step” approach – First, find the missing row/column sum or average; second, use it to fill the missing element. | Many TET questions require this technique. |
| 4 | Check for consistency – Sum of a column should equal the sum of its corresponding row if the table is balanced. | Helps spot errors. |
| 5 | Cross‑reference with given formulas – e.g., mean = (sum of values)/(number of observations). | Quick calculation of totals, averages, etc. |
| 6 | Leverage symmetry – In many tables, the value at (i, j) equals the value at (j, i). | Useful for symmetry‑based questions. |
| 7 | Use algebraic substitution – When a row/column is missing, set up an equation using the known totals. | Turns a table question into a simple algebra problem. |
3. Worked Example
Problem (Typical of Kerala TET):
The following table shows the number of students who scored each mark (0–10) in a test.
Score 0 1 2 3 4 5 6 7 8 9 10 Total Students 2 3 5 4 7 8 6 5 4 3 3 ? Find the total number of students and the average score.
Step‑by‑Step Solution
| Step | What to Do | Calculation |
|---|---|---|
| 1 | Sum the column “Total” – Add all numbers in the “Students” row. | 2 + 3 + 5 + 4 + 7 + 8 + 6 + 5 + 4 + 3 + 3 = 52 students. |
| 2 | Compute the average score – Use (\text{Avg} = \frac{\sum (\text{Score} \times \text{Students})}{\text{Total Students}}). | ((0×2)+(1×3)+(2×5)+(3×4)+(4×7)+(5×8)+(6×6)+(7×5)+(8×4)+(9×3)+(10×3) = 0+3+10+12+28+40+36+35+32+27+30 = 261). Average = (261 ÷ 52 ≈ 5.02). |
| 3 | Check for consistency – Verify that the average is reasonable (between |