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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
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