Suppose that the mean and median of the non-negative numbers 21, 8, 17, a, 51, 103, b, 13,…

Question

Suppose that the mean and median of the non-negative numbers $21, 8, 17, a, 51, 103, b, 13, 67$ ($a > b$), are 40 and 21, respectively. If the mean deviation about the median is 26, then $2a$ is equal to:

Accepted Answer

Here $n=9$. Mean $= 40$:
$$\frac{21+8+17+a+51+103+b+13+67}{9} = 40 \implies 280+a+b = 360 \implies a+b = 80 \quad \text{(Eq. 1)}$$

Median: for $n=9$, median is the $5^{th}$ term $= 21$. Arranging the known numbers in ascending order: $8, 13, 17, 21, 51, 67, 103$; there are exactly three known numbers less than 21 and three greater. For 21 to remain the 5th term among all 9 values, since $a > b$, we need $a \ge 21$ and $b \le 21$.

Mean deviation about median $= 26$:
$$\frac{|21-21|+|8-21|+|17-21|+|a-21|+|51-21|+|103-21|+|b-21|+|13-21|+|67-21|}{9} = 26$$
$$0+13+4+|a-21|+30+82+|b-21|+8+46 = 234 \implies |a-21|+|b-21| = 51$$

Since $a \ge 21$: $|a-21| = a-21$; since $b \le 21$: $|b-21| = 21-b$.
$$(a-21)+(21-b) = 51 \implies a-b = 51 \quad \text{(Eq. 2)}$$

Adding Eq. 1 and Eq. 2: $2a = 131$.

Solutions · 1