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.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 à 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 à 16)/30 = 288/30 = 9.6 days Therefore, Maximum days = 14 and Minimum days = 9.6
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