If x is an integer between 0 and 10 is y less than the average of x an

Solution

Step 1: Analyse Question Stem

• x is an integer.
• \(0 < x < 10 ⟹ 1 ≤ x ≤9\)
• We need to find if y is less than the average of x and 10

o Now, average of x and 10 \(= \frac{x+10}{2}\)

 Therefore, \(\frac{1 + 10}{2 }≤ \frac{x + 10}{2} ≤ \frac{9+ 10}{2} ⟹ 5.5 ≤ \frac{x + 10}{2} ≤ 9.5\)

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE

• According to this statement: \(2 (5 -y) < -\frac{x +10}{2 }⟹ \frac{1}{2}*\frac{x + 10}{2 }< y – 5\)

o \(⟹ 5 + \frac{1}{2}*\frac{x+10}{2} < y ……….(i)\)
• Since, \(5 > \frac{1}{2}*maximum\space of\space \frac{x + 10}{2} ……….(ii)\)
• So, from (i) and (ii), we can write,

o \(\frac{1}{2}*\frac{x+10}{2} + \frac{1}{2}*\frac{x+10}{2} < y ⟹ \frac{x+10}{2} < y\)

• We know, \(1 ≤ x ≤9 ⟹ 4 ≤ 4x= y ≤36\)
• Thus, when \(x = 1, y = 4 \) and \(\frac{x+10}{2} = 5.5 > 4\)

o Here y is greater than the average of x and 10
• However, when \(x = 9, y = 36 \) and \(\frac{x+10}{2} = 9.5 < 36\)

o Here y is less than the average of x and 10.
• Results of the above two cases are contradictory.