Let S be the set of integers x such that

I.100 ≤ x ≤200

II. x is odd and

III. x is divisible by 3 but not by 7.

How many elements does S contain?

a. 16

b. 12

c. 11

d. 131. No of integer = 99 (X is greater than 100 but less than 200)

2. No of odd among them = 50 (X is odd)

3. No of multiple of three that are odd = 16 (105 is the first number. Keep adding 6 to it as you want only odd numbers. last number will be 195. It is an AP with number of term = 16)

Now in order to remove the nos which are multiple of 7 just find the multiple of 21 (7*3) between them and not even. First no is 105 and last no is 189. the number of multiple of 21 which are odd are 3 (AP with common diff = 42).

Final answer = 16-3=13

**Write Here**