Understanding the Problem: Sum of Deviations

The question asks us to find the number of items, denoted by 'n', given information about the sum of deviations of these 'n' numbers from two different values, 10 and 20. The sum of deviations from 10 is given as 'p', and the sum of deviations from 20 is given as 'q'. We are also provided with an equation relating p and q: \(\left(p - q\right)^2 = 10000\).

The deviation of a single number \(x_i\) from a constant 'a' is defined as \(\left(x_i - a\right)\). The sum of deviations of n numbers \(x_1, x_2, \dots, x_n\) from a constant 'a' is the sum of the individual deviations:

\(\sum_{i=1}^{n} \left(x_i - a\right)\)

Formulating the Sums of Deviations

Based on the problem description, we can write the expressions for p and q:

• p: The sum of deviations of the n numbers from 10.

  Mathematically, \(p = \sum_{i=1}^{n} \left(x_i - 10\right)\).

  We can expand this sum:

  \(p = \sum_{i=1}^{n} x_i - \sum_{i=1}^{n} 10\)

  The sum of a constant 'c' n times is simply \(n \times c\). So, \(\sum_{i=1}^{n} 10 = 10n\).

  Thus, \(p = \sum_{i=1}^{n} x_i - 10n\).

• q: The sum of deviations of the n numbers from 20.

  Mathematically, \(q = \sum_{i=1}^{n} \left(x_i - 20\right)\).

  We can expand this sum:

  \(q = \sum_{i=1}^{n} x_i - \sum_{i=1}^{n} 20\)

  The sum of a constant 20 n times is \(20n\). So, \(\sum_{i=1}^{n} 20 = 20n\).

  Thus, \(q = \sum_{i=1}^{n} x_i - 20n\).

Using the Given Relationship to Find n

We are given the relationship \(\left(p - q\right)^2 = 10000\). Let's first find the expression for \(\left(p - q\right)\).

Substitute the expressions for p and q:

\(p - q = \left(\sum_{i=1}^{n} x_i - 10n\right) - \left(\sum_{i=1}^{n} x_i - 20n\right)\)

Carefully remove the parentheses. Remember that subtracting a negative term results in adding a positive term:

\(p - q = \sum_{i=1}^{n} x_i - 10n - \sum_{i=1}^{n} x_i + 20n\)

Notice that the term \(\sum_{i=1}^{n} x_i\) cancels out:

\(p - q = -10n + 20n\)

\(p - q = 10n\)

Now, substitute this expression for \(\left(p - q\right)\) into the given equation:

\(\left(10n\right)^2 = 10000\)

Square the term on the left side:

\(100n^2 = 10000\)

Now, solve for \(n^2\) by dividing both sides by 100:

\(n^2 = \frac{10000}{100}\)

\(n^2 = 100\)

To find n, take the square root of both sides:

\(n = \sqrt{100}\)

\(n = \pm 10\)

Since 'n' represents the number of items, it must be a positive value. Therefore, we take the positive root.

\(n = 10\)

Conclusion

The value of n is 10.

Let's check this against the given options:

• Option 1: 10
• Option 2: 20
• Option 3: 50
• Option 4: 100

Our calculated value \(n=10\) matches Option 1.

Revision Table: Key Calculations

Additional Information: Sum of Deviations Property

An important property related to the sum of deviations is that the sum of deviations of a set of numbers from their arithmetic mean is always zero.

Let \(\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i\) be the mean of the n numbers.

The sum of deviations from the mean is \(\sum_{i=1}^{n} (x_i - \bar{x})\).

Expanding this sum: \(\sum_{i=1}^{n} x_i - \sum_{i=1}^{n} \bar{x} = \sum x_i - n\bar{x}\).

Since \(\bar{x} = \frac{\sum x_i}{n}\), we have \(n\bar{x} = \sum x_i\).

So, \(\sum x_i - n\bar{x} = \sum x_i - \sum x_i = 0\).

In this problem, the sum of deviations from 10 (p) and from 20 (q) are generally not zero, because 10 and 20 are not necessarily the mean of the n numbers. The difference \((p-q)\), however, depends only on n and the constants used (10 and 20), not on the numbers \(x_i\) themselves.