The sum of deviations of n numbers from 10 and 20 are p and q respectively. If (p-q)^{2}=10000, then what is the value of n?

Explanation

Let the numbers be x_i. We are given p = \sum_{i=1}^n (x_i - 10) = \sum x_i - 10n and q = \sum_{i=1}^n (x_i - 20) = \sum x_i - 20n. Subtracting the two gives p - q = 10n. Squaring both sides yields (p-q)^2 = 100n^2. Since (p-q)^2 = 10000, we have 100n^2 = 10000 \implies n^2 = 100 \implies n = 10.

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