Set A contains five consecutive multiples of 4 and Set B

Solution

ATQ,

Set A:

Five consecutive multiples of 4 starting from 4x,4x+4,4x+8,4x+12,4x+16. Highest = 4x+16. Set B: Four consecutive odd numbers starting from 2y+1: 2y+1,2y+3,2y+5,2y+7. Highest = 2y+7. Given Conditions: Highest values of A and B sum to 53: (4x+16)+(2y+7)=53⟹4x+2y=30(1) Smallest of B is 21 less than second smallest of A: 2y+1=(4x+4)−21⟹2y=4x−18(2) Solve equations: From (1) and (2): x=6, y=3. Values: Set A: 24,28,32,36,40. Set B: 7,9,11,13. 2nd highest of A = 36, 2nd smallest of B = 9. Difference:36 − 9 = 27 Answer: 27