Explanation:

Alright, here's how to look at this problem:

- When you measure the deviations of a set of numbers from a number “a,” the sum is \( \sum(x_i - a) \).

- If you have two reference points—say, 100 and 92—you can relate their deviation sums:

Let the sum from 100 be \( S_{100} \) and from 92 be \( S_{92} \).

\( S_{92} = S_{100} + n \times (100 - 92) \).

- You’re told:

- \( S_{100} = -20 \)

- \( S_{92} = 140 \)

- Plugging in:

\( 140 = -20 + n \times 8 \)

\( n \times 8 = 160 \Rightarrow n = 20 \)

- Now, let’s find the sum of deviations from 99:

\( S_{99} = S_{100} + n \times (100 - 99) = -20 + 20 \times 1 = 0 \)

- The options:

- 1: 0

- 2: 10

- 3: 20

- 4: 40

The correct answer is Option 1: 0

Bottom line—if you shift your reference point by k units, you add \( n \times k \) to your previous sum. With n=20 and a shift of 1 from 100 to 99, that's just 20. Add to -20, and you get 0. Simple as that.