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Solution to Question A
Hash Function: h(i) = (3 * i + 5) % 11
=> 94, h(94) = (3*94 + 5)%11 = 287%11 = 1
=> 11, h(11) = (3*11 + 5)%11 = 38%11 = 5
=> 39, h(39) = (3*39 + 5)%11 = 1 , So Linear Probing will take it to 2 as 1 is filled
=> 20, h(20) = (3*20 + 5)%11 = 65%11 = 10
=> 16, h(16) = (3*16 + 5)%11 = 53%11 = 9
=> 5, h(5) = (3*5 + 5)%11 = 20%11 = 9, 9 and 10 is full, So it will go to 0
=> 12, h(12) = (3*12 + 5)%11 = 41%11 = 8
=> 44, h(44) = (3*44 + 5)%11 = 5 As 5 is filled it will go to 6
=>13, h(13) = (3*13 + 5)%11 = 44%11 = 0 As 1,2 is also filled it will go to 3
=> 88, h(88) = (3*88 + 5)%11 = 5 As 5,6 is filled it will go to 7
=> 23, h(23) = (3*23 + 5)%11 = 74%11 = 8 As 8,9,10,0,1,2,3 is filled it will go to 4
0:==>5
1:==>94
2:==>39
3:==> 13
4:==>23
5:==>11
6:==>44
7:==>88
8:==>12
9:==>16
10:==>20
Solution to Question B
Hash Function: h'(k) = 7 - (k % 7)
0:==>13
1:==>94