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


(1) \(10-y < y - \frac{x +10}{2}\)

(2) y is 4 times as large as x

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


Statement 1: \(10−y<y−\frac{x+10}{2}\)
    • 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 \(\frac{1}{2}*\frac{x+10}{2} + \frac{1}{2}*\frac{x+10}{2} < y ⟹ \frac{x+10}{2} < y\)
Hence, statement 1 is sufficient and we can eliminate answer Options B, C and E.
Statement 2: y is 4 times as large as x
    • 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.
Hence, statement 2 is not sufficient.
Thus, the correct answer is Option A.
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