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ATQ, Quantity I: Total coins in the bag = 5 Total number of coins which are withdrawn from the bag = ‘p’ Number of ways in which ‘p’ coins can be drawn from the bag = 5Cp = 10 5!/p!(5-p)!=10 120/p!(5-p)!=10 p! (5 – p)! = 12 Maximum value ‘p’ can take is 5. When p = 1, then p! (5 – p)! = 1! (5 – 1)! = 24 When p = 2, then p! (5 – p)! = 2! (5 – 2)! = 12 When p = 3, then p! (5 – p)! = 3! (5 – 3)! = 12 When p = 4, then p! (5 – p)! = 4! (5 – 4)! = 24 When p = 5, then p! (5 – p)! = 5! (5 – 5)! = 1 Hence, possible values of ‘p’ = 2 and 3. Now, q = p + 2p + 3 = 11 (when p = 2) and 18 (when p = 3) Hence, possible values of y are 11 and 18. Quantity II: Case 1: When Ajay is more efficient than Vijay. Ratio of efficiency of time taken by Ajay to Vijay is 5: 3. Let time taken Ajay and Vijay alone to finish the work is ‘3p’ days and ‘5p’ days respectively. According to the question: 1/3p+1/5p=1/11.25=4/45 (5+3)/15p=4/45 p = 6 Time taken by Ajay alone to finish the task = 3p = 18 days Case 2: When Ajay is less efficient than Vijay. Ratio of efficiency of time taken by Ajay to Vijay is 3: 5. Let time taken Ajay and Vijay alone to finish the work is ‘5p’ days and ‘3p’ days respectively. According to the question: 1/5p+1/3p=1/11.25=4/45 (3+5)/15p=4/45 p = 6 Time taken by Ajay alone to finish the work = 5p = 30 days Hence, Quantity I ≤ Quantity II
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