Princeton Review Practice Test

I agree that the answer should be E. Two sequences that apply to both conditions are 6-18 and 42-54, in which the former has 5 primes and the latter 4. So that would negate the OA answer C.

Even I think it is E!
6-18 has 4 prime numbers - 7,11,13,17
42-54 has 3 prime numbers - 43,47,53
66-78 has 3 prime numbers - 67,71,73

Given:
S = ( X1,X2,X3..........X13) All < 100.
There are 3 Multiples of 6. there are 13 Numbers so
X1.......X7......X13 has to be multiples of 6 to accomodate at least 3 multiples out of 13 numbers.
Hence S6 = { (6a),......6(a+1)....6(a+2) } , where a=Positive Integer

How many Prime no.in the S6 set ?

1.Both the munltips of 5 are either multiples of 2 or 3
S6-1={6,7,8,9,10,11,12,13,14,15,16,17,18}
Here 10,15 divisible 2,3 respectively
Next if you start from 12.
S6-2=(12,13,14,15,16,17,18,19,20,21,22,23,24)
Next: (25,30), (30,35) : 25/35 Not divisible by either 2 or 3
like this : Only divisible sets are with
36,.. 40,45...48
66,...70,75...78
So we have 4 sets satisfying this condition.

2.Only one of the 2 multiples of 7 is not multipls of 2 or 3
In S6: multiple of 7 are : 7 and 14, 7 is not multiple of 2 or 3
like this 14,21 : hence not valid multiple of 2 or 3

3) 1 and 2
From 1)
S6-1=6...18
S6-2=12...24
S6-3=36...48
S6-4=66....78
From 2)
S6-2 : 14,21 divisble by 2,3 respectively,Hence is not valid
S6-3 : 42 : is only 1 Number as per condition 2) there are 2 numbers of 7 hence not valid.
S6-4:70,77
70 is Multiple of 2, but not 3
77 is not multiple of both 2 and 3
As per condition only one Num is not multiple of 2 or 3 here 70,77 i,e 2 numbers are not multiple of 3. Hence S6-4 is invalid.
S6-1 is the Only set ,No of Prime in this is (7,11,13,17) = 4. Ans

C