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    Question

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

    54, respectively. Average weight of all the employees in 'X' and 'Y' is 50 kg and 60 kg respectively. When 'e' employees of average weight 56 kg are transferred from 'X' to 'Y', average weight of all the employees in 'X' and 'Y' decreases by 2 kg and 625 gm, respectively. Which of the following statement(s) is/are false regarding the information provided above? I. 'e' is the smallest two-digit number. II. LCM of (e + 6) and (3e - 6) is 48. III. (e + 1)2 < 120
    A Only III Correct Answer Incorrect Answer
    B Only I and II Correct Answer Incorrect Answer
    C Only I Correct Answer Incorrect Answer
    D Only II and III Correct Answer Incorrect Answer
    E All of I, II and III Correct Answer Incorrect Answer

    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.

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