Number of employees in Department 'X' and 'Y' are 40 and

Solution

ATQ, Sum of weight of all the employees in 'X' initially = 40 × 50 = 2000 kg Sum of weight of all the employees in 'X' now = (40 - e) × 48 = (1920 – 48e) kg Sum of weight of all the employees in 'Y' initially = 54 × 60 = 3240 kg Sum of weight of all the employees in 'Y' now = (54 + e) × 59.375 = (3206.25 + 59.375e) kg So, 2000 + 3240 = 1920 – 48e + 3206.25 + 59.375e Or, 5240 = 5126.25 + 11.375e Or, 113.75 = 11.375e Or, 'e' = 10 Statement I: 10 is the smallest two-digit number. So, statement I is true. Statement II: e + 6 = 10 + 6 = 16 3e - 6 = 3 × 10 - 6 = 24 LCM of 16 and 24 = 48 So, statement II is true. (10 + 1)² = 121 > 120 So, statement III is false. Therefore, only statement III is false.