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Bunuel
4 men and 10 women were assigned to complete a work. Can they complete the work within 15 days? Assume that all men are equally efficient and all women are equally efficient.

(1) 4 men and 10 women employed initially completed 1/3 of the work in less than 6 days.
(2) If 2 additional men and 2 additional women had been employed, they would have completed 1/2 of the work in 5 days (the additional men and the additional women have efficiencies equal to the men and women employed initially, respectively).

Solution:
Pre Analysis:
  • Work by 4 men in 15 days \(=nRT=4\times M\times 15=60M\) where M is the rate of each man
  • Work by 10 women in 15 days \(=nRT=10\times W\times 15=150W\) where W is the rate of each woman
  • We are asked if \(60M+150W=Work \) or not

Statement 1: 4 men and 10 women employed initially completed 1/3 of the work in less than 6 days
  • If they (4 men and 10 women) complete the 1/3 work in 5 days, then they will complete the work in \(5\times 3=15\) days
  • However, if they complete it in less than that, they won't be able to complete the full work in 15 days
  • Thus, statement 1 alone is not sufficient and we can eliminate options A and D

Statement 2: If 2 additional men and 2 additional women had been employed, they would have completed 1/2 of the work in 5 days
  • Work by 6 men in 5 days \(=nRT=6\times M\times 5=30M\)
  • Work by 12 women in 5 days \(=nRT=12\times W\times 5=60W\)
  • According to the statement \(30M+60W=\frac{Work}{2}⇒60M+120W=Work\)
  • This means \(60M+150W=Work \) is not true and we can answer the question

Hence the right answer is Option B
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