📢 Too many exams? Don’t know which one suits you best? Book Your Free Expert 👉 call Now!
  • google app store apple app store
    • ✖

      Question

      Tanu can complete a task in 'A' days on her own, while

      together with Manu, they can complete the same task in (A - 6) days. Manu, working alone, can finish the task in 72 days. What would be the time taken by Tanu to finish the task if she works with 33(1/3)% more efficiency than usual?
      A 32 days Correct Answer Incorrect Answer
      B 20 days Correct Answer Incorrect Answer
      C 18 days Correct Answer Incorrect Answer
      D 16 days Correct Answer Incorrect Answer
      E 8 days Correct Answer Incorrect Answer

      Solution

      Work done by Tanu in One day = (1/A) units Work done by Tanu and Manu together in one day = [1/(A - 6) ] units Work done by Manu in one day = (1/72) units ATQ, (1/A) + (1/72) = [1/(A - 6) ] Or, (1/72) = {1/(A - 6) } - (1/A) Or, A X (A - 6) = 72 X 6 Or, A2 - 6A = 432 Or, A2 - 6A = 432 Or, A2 - 6A - 432 = 0 Or, A2 - 24A + 18A - 432 = 0 Or, A(A - 24) + 18(A - 24) = 0 Or, (A - 24) (A + 18) = 0 So, 'A' = 24 or -18 But since time taken cannot be negative, so 'A' = 24 So, time taken by Tanu to complete the work = 'A' days = 24 days Therefore, required time by Tanu with increased efficiency = 24 x (3/4) = 18 days 

      Practice Next
      More Time and work Questions

      Relevant for Exams:

      ask-question