Plug in numbers.

Pick C

Let N = 60,

N is a multiple of 20 -- YES

N+6 is a multiple of 3 - YES


Is the integer N a multiple of 15? --YES

Another example N = 120


N is a multiple of 20 -- YES

N+6 is a multiple of 3 - YES

Is the integer N a multiple of 15? --YES

St 1 & St2 alone is insufficient

from 1 N=20k (k=1,2,3....)
when N=60 then multiple of 15 but when N=80 the not a multiple of 15. So Not Suff.

from 2 N+6 = 3k (k=2,3,4....)
6 is a multiple of 3 so N will be also a multiple of 3 (Say 9=3+6 so N=3)
but we cannot say as some cases it will be multiple of 15 and some cases it will not be. So NS.

Combine both
if N=20k when k=1 then N=20 at the same time from 2 N+6 = 26 which will not be a multiple of 3
if N=20k when k=2 then N=40 at the same time from 2 N+6 = 46 which will not be a multiple of 3
if N=20k when k=3 then N=60 at the same time from 2 N+6 = 66 which will be a multiple of 3
if N=20k when k=4 then N=80 at the same time from 2 N+6 = 86 which will not be a multiple of 3
if N=20k when k=5 then N=100 at the same time from 2 N+6 = 106 which will not be a multiple of 3
if N=20k when k=6 then N=120 at the same time from 2 N+6 = 126 which will be a multiple of 3...

So when both (1) and (2) are satisfied we are getting a proper result.

IMOC