Quantitative Aptitude – Average
Exam Focus: Himachal Pradesh Forest Guard (HPFG)
Topic: Mean / Average
(All content is in plain‑English, with clear examples and quick‑review notes.)
1. Simple Explanation
- Average (Mean) is a single value that summarizes a set of numbers by representing their typical or central value.
- It is calculated as total sum ÷ number of items.
- The average gives you an idea of how all the values behave together (not just extremes).
Formula
[ \text{Average} = \frac{\text{Sum of all numbers}}{\text{Count of numbers}} ]
2. Key Points & Useful Rules
| Concept | What it Means | Formula / Quick Tip |
|---|---|---|
| Arithmetic Mean | Standard average of a list of numbers | ( \displaystyle \bar{x} = \frac{S}{n} ) |
| Weighted Mean | Numbers have different “importance” (weights) | ( \displaystyle \bar{x}_w = \frac{\sum w_i x_i}{\sum w_i} ) |
| Average of Two Numbers | ( \frac{a+b}{2} ) | Shortcut: ( \frac{a+b}{2} = \frac{a}{2} + \frac{b}{2} ) |
| Average of Three Numbers | ( \frac{a+b+c}{3} ) | Shortcut: ( \frac{a+b+c}{3} = \frac{a}{3} + \frac{b}{3} + \frac{c}{3} ) |
| Average of Two Groups | Combine totals & counts | (\displaystyle \frac{S_1+S_2}{n_1+n_2}) |
| Finding Sum from Average | ( S = \text{Average} \times n ) | |
| Finding Count from Sum & Average | ( n = \frac{S}{\text{Average}} ) | |
| Removing or Adding Numbers | Adjust total sum & count | New sum = Old sum ± new numbers; New average = New sum ÷ New count |
| Percentages & Averages | Average of percentages sometimes equals overall percentage | Only true if percentages represent proportions of equal totals |
2.1. Common Averages Problems
- Average of whole vs parts – e.g., average of 10 numbers is 20; after removing 3 numbers, new average is 18 → find sum of removed numbers.
- Combining averages – e.g., two groups of students, group A average 75, group B average 68, find overall average.
- Weighted average – e.g., exam marks: 30% in quiz, 70% in final; quiz average 80, final average 70 → overall average.
- Average of percentages – e.g., 40% of 20 students passed, 60% of 30 students passed → overall pass percentage.
3. Worked Example
Problem (typical HPFG style)
The average score of 12 students in a test is 78. After adding 3 new students whose scores are 85, 90, and 80, the new average becomes 80. What was the total score of the original 12 students?
Step‑by‑Step Solution
-
Write what we know
- Original average (A_1 = 78)
- Number of original students (n_1 = 12)
- New students’ scores: (85, 90, 80) → sum (S_{\text{new}} = 85+90+80 = 255)
- New average (A_2 = 80)
- New count (n_2 = 12 + 3 = 15)
-
Find total score of the original group
[ S_1 = A_1 \times n_1 = 78 \times 12 = 936 ] -
Find total score after adding the new students
[ S_2 = A_2 \times n_2 = 80 \times 15 = 1200 ] -
Check consistency
[ S_2 = S_1 + S_{\text{new}} ;;\Rightarrow;; 1200 = 936 + 255 ] Indeed, (936 + 255 = 1191) – oh! we made a mistake, so adjust. -
Correct the calculation
Compute (S_1) again:
[ S_1 = 78 \times 12 = 936 ] Compute (S_2) again:
[ S_2 = 80 \times 15 = 1200 ] Now compute sum of new scores (already 255).
Check: (936 + 255 = 1191), not 1200.
The discrepancy shows that the new average cannot be 80 with those added scores.
→ Take a pause – maybe the problem statement had a misprint; but let’s proceed with the correct approach. -
Assuming the new average is indeed 80, we find the required sum of the new scores to satisfy the condition:
[ S_{\text{new}} = S_2 - S_1 = 1200 - 936 = 264 ]
The sum of