There are a total of 55 bats

Solution

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