Question

The median of 11 different positive integers is 15 and seven of those 11 integers are 8, 12, 20, 6, 14, 22, and 13.
Statement I: The difference between the averages of four largest integers and four smallest integers is 13.25.
Statement II: The average of all the 11 integers is 16.
Which of the following statements would be sufficient to find the largest possible integer of these numbers?

• A. Statement I only.
• B. Statement II only.
• C. Both Statement I and Statement II are required.
• D. Neither Statement I nor Statement II is sufficient.
• E. Either Statement I or Statement II is sufficient.

Hint:

With a fixed median, count how many values must lie on each side. To maximize the largest value, minimize the other unknowns while respecting distinctness and side constraints.

Answer 1

Answer: (E)

Known set (including the median) so far: \[ \{6,8,12,13,14,15,20,22\}. \] Since the median is \(15\) and five numbers \((6,8,12,13,14)\) are already below \(15\), the remaining three unknown integers \(x < y < z\) must all be \( > 15\).

Using Statement I alone

Smallest four are \(6,8,12,13\) (sum \(=39\)). Largest four are \(y,20,22,z\). Given: \[ \frac{y+20+22+z}{4} - \frac{6+8+12+13}{4} = 13.25 \] \[ (y+20+22+z) - 39 = 53 \quad \Rightarrow \quad y+z=50. \] To maximize \(z\), minimize \(y\) subject to \(x<y\) and \(x,y>15\), distinct. Choose \(x=16,\ y=17 \Rightarrow z=33\). Thus the largest possible integer is fixed at \(z=33\). \[ \Rightarrow \ \textbf{Statement I is sufficient.} \]

Using Statement II alone

Average of all 11 is \(16 \Rightarrow\) total sum \(=176\). Known sum: \[ 6+8+12+13+14+15+20+22=110. \] Hence \[ x+y+z = 176-110 = 66. \] To maximize \(z\), minimize \(x,y\) with \( >15\), distinct: take \(x=16,\ y=17 \Rightarrow z=66-33=33\). Again the largest possible integer is uniquely \(33\). \[ \Rightarrow \ \textbf{Statement II is sufficient.} \]

Final Answer

\[ \boxed{\text{Either statement alone suffices } \Rightarrow \text{Option (E).}} \]