X is a series of five consecutive multiple of three whose average is 36 and Y is the series of four consecutive multiple of four, which second smallest number is 13 more than second highest number of X series. Find the sum of smallest number of X series and highest number of Y series?

Correct option is D

Let X is a series consists 3x – 6, 3x – 3, 3x, 3x + 3, 3x + 6
ATQ, 3x – 6 + 3x – 3 + 3x + 3x + 3 + 3x + 6 = 36×5
x = 12
And, X series = 30, 33, 36, 39, 42
Second smallest number of Y = 39 + 13 = 52
Y series = 48, 52, 56, 60
Required sum = 30+60=90