The average of five positive numbers is 56. If the first number is three-fourth of the sum of the last four numbers, then the average of the last four numbers is:
40
The problem involves the concept of average, specifically the average of five positive numbers. The average is calculated by dividing the sum of the numbers by the count of the numbers.
We are given that the average of five positive numbers is 56. Let the five positive numbers be \(n_1, n_2, n_3, n_4,\) and \(n_5\).
The formula for the average is:
$$\text{Average} = \frac{\text{Sum of numbers}}{\text{Count of numbers}}$$
In this case, the sum of the five numbers is \(n_1 + n_2 + n_3 + n_4 + n_5\), and the count is 5. The average is given as 56.
So, we can write the equation:
$$\frac{n_1 + n_2 + n_3 + n_4 + n_5}{5} = 56$$
From the average formula, we can find the sum of the five numbers:
$$\text{Sum of numbers} = \text{Average} \times \text{Count of numbers}$$
Sum of the five numbers \( = 56 \times 5 \)
Sum of the five numbers \( = 280 \)
So, \(n_1 + n_2 + n_3 + n_4 + n_5 = 280\).
The question gives a relationship between the first number and the sum of the last four numbers. It states that the first number (\(n_1\)) is three-fourth of the sum of the last four numbers (\(n_2 + n_3 + n_4 + n_5\)).
Let \(S_4\) be the sum of the last four numbers. So, \(S_4 = n_2 + n_3 + n_4 + n_5\).
According to the problem statement:
$$n_1 = \frac{3}{4} \times (n_2 + n_3 + n_4 + n_5)$$
Substituting \(S_4\):
$$n_1 = \frac{3}{4} S_4$$
We know that the total sum of the five numbers is 280:
$$n_1 + (n_2 + n_3 + n_4 + n_5) = 280$$
Substituting \(S_4 = n_2 + n_3 + n_4 + n_5\), we get:
$$n_1 + S_4 = 280$$
Now, substitute the relationship \(n_1 = \frac{3}{4} S_4\) into this equation:
$$\frac{3}{4} S_4 + S_4 = 280$$
Combine the terms involving \(S_4\):
$$\left(\frac{3}{4} + 1\right) S_4 = 280$$
$$\left(\frac{3}{4} + \frac{4}{4}\right) S_4 = 280$$
$$\frac{7}{4} S_4 = 280$$
To find \(S_4\), multiply both sides by \(\frac{4}{7}\):
$$S_4 = 280 \times \frac{4}{7}$$
$$S_4 = \frac{280}{7} \times 4$$
$$S_4 = 40 \times 4$$
$$S_4 = 160$$
So, the sum of the last four numbers is 160.
The question asks for the average of the last four numbers. The last four numbers are \(n_2, n_3, n_4,\) and \(n_5\), and their sum is \(S_4 = 160\). The count of these numbers is 4.
The average of the last four numbers is:
$$\text{Average of last four numbers} = \frac{\text{Sum of last four numbers}}{\text{Count of last four numbers}}$$
$$\text{Average of last four numbers} = \frac{S_4}{4}$$
$$\text{Average of last four numbers} = \frac{160}{4}$$
$$\text{Average of last four numbers} = 40$$
Therefore, the average of the last four numbers is 40.
Let's check the options:
Our calculated average matches Option 4.
| Concept | Formula / Relationship | Value |
|---|---|---|
| Average of 5 numbers | \(\frac{\text{Sum of 5 numbers}}{5}\) | 56 |
| Sum of 5 numbers | Average × Count | \(56 \times 5 = 280\) |
| Relationship \(n_1\) and \(S_4\) | \(n_1 = \frac{3}{4} S_4\) | - |
| Total Sum | \(n_1 + S_4\) | 280 |
| Equation for \(S_4\) | \(\frac{3}{4} S_4 + S_4 = 280\) | - |
| Sum of last 4 numbers (\(S_4\)) | Solved from equation | 160 |
| Average of last 4 numbers | \(\frac{S_4}{4}\) | \(\frac{160}{4} = 40\) |
| Term | Definition | How it's used here |
|---|---|---|
| Average (Arithmetic Mean) | Sum of values divided by the count of values. | Given for 5 numbers, calculated for 4 numbers. |
| Sum | The total result of adding numbers together. | Calculated for 5 numbers and for the last 4 numbers. |
| Positive Numbers | Numbers greater than zero. | The context of the problem involves positive numbers. |
Solving average word problems often involves translating the given information into algebraic equations. Key steps usually include:
Understanding how the sum, average, and count relate to each other is fundamental to solving these types of quantitative aptitude problems.
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