In everyday life, we often want to find a single value that represents a group of numbers. This value is called the average. The average gives us an idea of the "central" or "typical" value among a set of numbers. For example, if you want to know the average marks scored by students in a class or the average price of fruits bought, the concept of average helps us summarize the data into one meaningful number.
Understanding averages is crucial not only in daily life but also in competitive exams, where questions often test your ability to quickly and accurately calculate averages in various contexts.
The basic idea behind average is simple: it is the total of all values divided by how many values there are.
The average of a set of numbers is defined as the sum of all the numbers divided by the total count of those numbers.
Mathematically, if you have numbers \(x_1, x_2, x_3, \ldots, x_n\), their average \(A\) is given by:
This formula can be rearranged to find the sum if the average and number of terms are known:
Or to find the number of observations if the sum and average are known:
This diagram shows how the sum and number of terms relate to the average: Average = Sum / Number of Terms.
Sometimes, not all observations are equally important. For example, if you buy 3 kg of apples at one price and 2 kg at another price, the average price per kg is not just the simple average of the two prices. Instead, you must consider the quantity (weight) of each purchase. This is where weighted average comes in.
The weighted average gives different importance (weights) to different observations.
If \(x_1, x_2, \ldots, x_n\) are the values and \(w_1, w_2, \ldots, w_n\) are their respective weights, then the weighted average \(A_w\) is:
| Quantity (kg) | Price per kg (Rs.) | Total Cost (Rs.) |
|---|---|---|
| 3 | 120 | 360 |
| 2 | 150 | 300 |
| 5 (Total) | 660 (Total) |
The weighted average price per kg is:
Step 1: Add all the marks:
75 + 80 + 85 + 90 + 95 = 425
Step 2: Count the number of subjects:
Number of subjects = 5
Step 3: Use the average formula:
\( \text{Average} = \frac{\text{Sum of marks}}{\text{Number of subjects}} = \frac{425}{5} = 85 \)
Answer: The average marks are 85.
Step 1: Calculate the total sum of all 5 tests using the average:
\( \text{Total sum} = \text{Average} \times \text{Number of tests} = 72 \times 5 = 360 \)
Step 2: Add the known scores:
70 + 75 + 68 + 74 = 287
Step 3: Find the missing score:
\( \text{Missing score} = \text{Total sum} - \text{Sum of known scores} = 360 - 287 = 73 \)
Answer: The score of the fifth test is 73.
Step 1: Calculate total cost for each purchase:
3 kg x Rs.120 = Rs.360
2 kg x Rs.150 = Rs.300
Step 2: Calculate total quantity and total cost:
Total quantity = 3 + 2 = 5 kg
Total cost = Rs.360 + Rs.300 = Rs.660
Step 3: Calculate weighted average price:
\( \text{Average price} = \frac{\text{Total cost}}{\text{Total quantity}} = \frac{660}{5} = 132 \)
Answer: The average price per kg is Rs.132.
Step 1: Calculate total marks for each class:
Class A total = 40 x 70 = 2800
Class B total = 60 x 80 = 4800
Step 2: Calculate combined total marks and total number of students:
Total marks = 2800 + 4800 = 7600
Total students = 40 + 60 = 100
Step 3: Calculate combined average:
\( \text{Combined average} = \frac{7600}{100} = 76 \)
Answer: The combined average marks are 76.
Step 1: Calculate time taken for each part:
Time for first 60 km = \( \frac{60}{40} = 1.5 \) hours
Time for next 60 km = \( \frac{60}{60} = 1 \) hour
Step 2: Calculate total distance and total time:
Total distance = 60 + 60 = 120 km
Total time = 1.5 + 1 = 2.5 hours
Step 3: Calculate average speed:
\( \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{120}{2.5} = 48 \) km/h
Answer: The average speed for the entire journey is 48 km/h.
When to use: When solving multi-step average problems.
When to use: When combining groups with different averages.
When to use: When dealing with items bought in different quantities or frequencies.
When to use: When solving average speed problems with different speeds over different distances or times.
When to use: When a single value is unknown in a set.
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