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    Question

    6 men and 10 boys can finish a task in 18 days. If each

    boy is between 1/3rd to equal the efficiency of a man, find the maximum and minimum number of days 3 men and 27 boys would need to do the same task.
    A 14 days and 9.6 days Correct Answer Incorrect Answer
    B 21 days and 8.5 days Correct Answer Incorrect Answer
    C 14.5 days and 28 days Correct Answer Incorrect Answer
    D 14 days and 8 days Correct Answer Incorrect Answer
    E None of these Correct Answer Incorrect Answer

    Solution

    ATQ,

    Case 1: When each boy is 1/3rd as efficient as a man 6 men + 10 boys = 6 men + (10/3) men = (18 + 10)/3 = 28/3 men 3 men + 27 boys = 3 men + (27/3) men = 3 + 9 = 12 men (28/3) men take 18 days So, 12 men will take = (18 × 28/3) / 12 = (168)/12 = 14 days Case 2: When each boy is equal to a man 6 men + 10 boys = 6 + 10 = 16 men 3 men + 27 boys = 3 + 27 = 30 men 16 men take 18 days So, 30 men will take = (18 × 16)/30 = 288/30 = 9.6 days Therefore,  Maximum days = 14 and  Minimum days = 9.6

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