There are a total of 55 bats (cricket bats + baseball bats) in two bags, 'P' and 'Q.' Bag 'P' contains 12 cricket bats and 'x' baseball bats, while bag 'Q' contains 'y' cricket bats and 10 baseball bats. What is the difference in the number of bats between the two bags? Statement I: After transferring 3 cricket bats from bag 'Q' to bag 'P,' the probability of picking a cricket bat from bag 'P' becomes (1/2). Statement II: If all Baseball bats from bag 'P' are moved to bag 'Q,' the number of Baseball bats in bag 'Q' will be 7 more than the number of cricket bats in that bag.

ATQ, We have, 12 + x + y + 10 = 55 Or, x + y = 55 - 22 = 33 --------- (I) Statement I: Number of cricket bats in bag 'P' now = 12 + 3 = 15 Probability of drawing a cricket bats from bag 'P' now = 15 ÷ (15 + x) So, 15 ÷ (15 + x) = (1/2) Or, 30 = 15 + x Or, 'x' = 15 On putting value of 'x' in equation (I) , We get, 15 + y = 33 Or, 'y' = 18 So, number of bats in bag 'P' = 12 + x = 12 + 15 = 27 Also, number of bats in bag 'Q' = y + 10 = 18 + 10 = 28 Required difference = 28 - 27 = 1 So, data in statement I alone is sufficient to answer the question. Statement II: Number of Baseball bats in bag 'Q' now = (10 + x) So, 10 + x = b + 7 Or, y - x = 3 ..........(III) On adding equations (I) and (III) , we get, 'y' = 18 And, putting value of 'y' in equation (I) , we get, 'x' = 33 - 18 = 15 So, number of bats in bag 'P' = 12 + a = 12 + 15 = 27 Also, number of bats in bag 'Q' = b + 10 = 18 + 10 = 28 Required difference = 28 - 27 = 1 So, data in statement II alone is sufficient to answer the question. Therefore data either in statement I alone or in statement II alone is sufficient to answer the question