📚 Study Notes – Average (Quantitative Aptitude)
Target Exam: Odisha Forest Guard (OFG)
Section: Quantitative Aptitude – Average & Ratios
1. Simple Explanation
Average (or mean) is a measure of central tendency that gives a single value representing a set of numbers.
- It is calculated by adding all the numbers and dividing by the count of numbers.
- It is useful when you need a “typical” value that summarizes a data set.
Formula (Unweighted Mean)
[ \text{Average} = \frac{\text{Sum of all values}}{\text{Number of values}} ]
2. Key Points & Variations
| Concept | Formula | Notes |
|---|---|---|
| Simple Mean | ( \bar{x} = \frac{S}{n} ) | (S) = sum, (n) = number of items. |
| Weighted Mean | ( \bar{x}_w = \frac{\sum w_i x_i}{\sum w_i} ) | Use when items have different importance (weights). |
| Increment/Decrease by a Percentage | ( \text{New average} = \bar{x} \pm \frac{p}{100}\bar{x} ) | (p) is percentage change. |
| Average with Known Total | If total after adding a new item is known, use: ( \bar{x}_{new} = \frac{S + new}{n+1} ). | |
| Finding Number of Items | From ( \bar{x} = \frac{S}{n}), rearrange: ( n = \frac{S}{\bar{x}} ). | |
| Average of Averages | Not simply the arithmetic mean of the averages unless the groups have equal sizes. Use weighted formula: | |
| ( \text{Overall Avg} = \frac{\sum (n_i \times \bar{x}_i)}{\sum n_i} ) | (n_i) = size of group i, ( \bar{x}_i) = average of group i. | |
| Median & Mode | Not averages but often asked in aptitude. Median = middle value; Mode = most frequent value. |
Quick Tips for the OFG Exam
- Always check if numbers are integers or decimals – round only at the end (unless specified).
- Use weighted mean when data sets have different sample sizes.
- When a percentage change is mentioned (e.g., “average decreased by 5%”), apply the change directly to the existing average.
- Avoid unnecessary calculations – often you can cancel terms or use proportions.
- Check for “average of averages” problems – apply the weighted formula.
3. Worked Example
Problem (Typical OFG Format)
A batch of 8 forest guard trainees scored the following marks in a written test:
72, 68, 90, 85, 78, 84, 69, 81.
The average score of the batch increased to 80 after 3 more trainees joined and scored 88, 92, and 75.
Find the average score of the new trainees.
Step‑by‑Step Solution
- Compute initial total score
[ S_{\text{old}} = 72 + 68 + 90 + 85 + 78 + 84 + 69 + 81 = 627 ] - Initial average (given as 80 after addition, but we can compute to verify)
[ \bar{x}_{\text{old}} = \frac{627}{8} = 78.375 \quad (\text{not needed for final answer}) ] - Total after adding 3 new trainees
[ S_{\text{new}} = S_{\text{old}} + 88 + 92 + 75 = 627 + 255 = 882 ] - New size
[ n_{\text{new}} = 8 + 3 = 11 ] - Given new average
[ \bar{x}_{\text{new}} = 80 = \frac{882}{11} ] (Check: (882 ÷ 11 = 80) ✅) - Find average of new trainees
[ \bar{x}_{\text{new trainees}} = \frac{88 + 92 + 75}{3} = \frac{255}{3} = 85 ]
Answer: The average score of the new trainees is 85.
4. Quick Revision Summary
| Concept | Formula | What to remember |
|---|---|---|
| Simple Mean | ( \bar{x} = \frac{S}{n} ) | Sum all numbers, divide by count. |
| Weighted Mean | ( \bar{x}_w = \frac{\sum w_i x_i}{\sum w_i} ) | Use when items have different importance. |
| Change by % | ( \bar{x}_{new} = \bar{x} \pm \frac{p}{100}\bar{x} ) | Directly adjust the average. |
| Find Items from Avg | ( n = \frac{S}{\bar{x}} ) | Useful when total and average are known. |
| Average of Averages | ( \frac{\sum n_i \bar{x}_i}{\sum n_i} ) | Weight by group sizes. |
| Check for “average increased/decreased” | Use ( \bar{x}_{new} = \frac{S + \text{new scores}}{n+ \text{new count}} ) |
Final Tip:
When dealing with averages, always keep track of the total sum and the number of items. Most problems become simple algebra once you express everything in terms of these two quantities.